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November 2004 Electromagnetic Theory and Special Relativity Robert de Mello Koch1,3 and Neil Turok2,3 Department of Physics and Centre for Theoretical Physics1 , University of the Witwatersrand, Wits, 2050, South Africa robert@aims.ac.za Centre for Mathematical Sciences2 , University of Cambridge, Wilberforce Road, Cambridge, CB3 0WA, UK neil@aims.ac.za African Institute for Mathematical Sciences3 , 6 Melrose Road, Muizenberg 7945, South Africa. Notes from the course presented by the authors at the African Institute for Mathematical Sciences, November 2004. 1. Introduction to ﬁelds What are we going to do in this section? In this section we will start by studying the grav- itational force exerted by the earth on a particle. This allows us to introduce the concept of a ﬁeld, which will play an important role in our study of electricity and magnetism. Before we study electricity and magnetism, we are going to spend some time discussing gravity. We do this because we are much more familiar with gravity from our everyday lives and because a lot of what we learn from gravity is directly applicable to electromagnetism. One of the questions which we are often interested in, is determining the trajectory that a particle will follow through space. At any instant in time, we specify the position of our particle by three numbers (x(t), y(t), z(t)). These three numbers taken together are the position vector of the particle. If we introduce three unit vectors ˆ (directed along the i ˆ positive x axis), ˆ (directed along the positive y axis) and k (directed along the positive z j axis), we can write ˆ x = x(t)ˆ + y(t)ˆ + z(t)k. i j The notation x is much more eﬃcient compared to always writing the three coordinates (x(t), y(t), z(t)). As an example, lets start by thinking about the gravitational forces acting on a massive particle. Why are we concentrating on the forces? Well, the motivation for this comes from classical mechanics. Newton has taught us that if a force F acts on our particle and if our particle has a mass m, then it will experience an acceleration a according to the formula d2 x F = ma = m . dt2 where ˆ F = Fxˆ + Fy ˆ + Fz k. i j I can write this one vector equation as three ordinary diﬀerential equations for the three coordinates of our particle, d2 x d2 y d2 z m = Fx , m = Fy , m = Fz . dt2 dt2 dt2 The initial conditions for solving these three ordinary diﬀerential equations are given by the initial position and initial velocity of our particles. So, what forces act on our particle? Well, close to the surface of the earth, we would measure that our particle accelerates at about 9.8m·s−2 . If we moved from the surface of the earth, say on a rocket destined for the moon, and repeated our measurements, we’d ﬁnd that the magnitude of the acceleration of our particle would become smaller and smaller the further we were from the earth. At each diﬀerent position in space, depending on how far we are from the earth, we measure a diﬀerent acceleration for our particle. This implies that at each diﬀerent position in space, the magnitude of the force acting on our particle changes. In fact, the time at which we make our measurements also plays a role. Indeed, other massive objects (like for example the moon) will also exert a force on our particle. The magnitude and direction of the force exerted by the moon on our particle will depend on the position of the moon with respect to our particle. The position of the moon with respect to our particle is time dependent. The new idea here is that of a ﬁeld. When we talk about a ﬁeld in mathematical physics, we are talking about a quantity that has a value at each location in space and at each instant in time. It may have a vector value (like the force exerted by the earth in the above example) or it may be a number (like the temperature T (x, t) at each point in this room at each instant in time). This discussion suggests that instead of saying that the earth exerts a force on a falling object, it is more useful to say that the earth sets up a gravitational force ﬁeld. Any object near the earth is acted upon by the gravitational force ﬁeld at that location. Did you get it? If you know what a ﬁeld is, you got the point of this section. 2. More on the Gravitational Field What are we going to do in this section? In this section we summarise the usual description of the gravitational force ﬁeld, in terms of the gravitational ﬁeld. This idea is important - by copying it, we will later be lead to describe the force between charged particles in terms of an electric ﬁeld. You might remember the formula F (x, t) = mg(x, t), from your previous studies, where F is the force acting on a particle of mass m and you probably called g the acceleration due to gravity. In this formula, we have explicitly indicated that both F and g are ﬁelds; the mass of the particle m is not a ﬁeld. This formula is a very useful way to think about nature. To see why it is so useful, imagine that we ﬁrst solve the problem of a ball moving in the gravitational ﬁeld. To compute what force is acting on the ball, we have to (i) specify the mass of the ball and (ii) specify the gravitational ﬁeld. To do (i) is easy - we just need to determine a single number. To do (ii) is awful - we need to specify a vector for every location in space at every single instant in time! That is a lot of work. Now imagine that we want to solve a second problem. The problem of a pen (say) moving in the earth’s gravitational ﬁeld. To compute what force is acting on the pen, we have to (i) specify the mass of the pen and (ii) specify the gravitational ﬁeld. To do (i) is easy - we again just need to specify a single number. Here comes the important part - we have already done (ii) when we studied the problem of the ball! This saves us a huge amount of work! Implicit in this way of thinking is a split of the system into an object which moves in the ﬁeld set up by some source, and further, we treat both pieces separately. This not quite exact, but it is an excellent approximation. First, lets understand why it is not exact. Then we will ﬁgure out how good the approximation is. We have assumed that we can calculate the gravitational ﬁeld g for the source alone and that we’ll get the same answer if we calculate things in the presence of the object moving in the ﬁeld. When we calculate the gravitational ﬁeld for the earth on its own, it is at rest and remains at rest. When we compute the gravitational ﬁeld in the presence of our object, the earth exerts a force on the object and the object exerts a force on the earth. This implies that the earth will accelerate towards our object - the ﬁeld g produced by the earth when it is at rest will not be the same as the ﬁeld produced by the earth when it is accelerating. This is the reason why our answer is only approximate. How good is the approximation? It will be a good approximation as long as the acceleration of the earth is small. Is this the case? Imagine that our ball has a mass of 0.2kg. Then the force the earth exerts on our ball is approximately 2N. By Newton’s third law, this is equal to the force that our ball will exert on the earth. The earth has a mass of 6×1024 kg. Thus, the acceleration of the earth is |F | 2 1 |a| = m = 6×1024 = 3 × 10−24 m·s−2 . This is a small acceleration and hence our approximation will be very accurate. Did you get it? If you understand that F = mg (i) assumes that we can split the system into a source which produces the ﬁeld and an object which reacts to the ﬁeld, (ii) that this split saves a huge amount of work and that (iii) this is not exact, but it is an extremely good approximation (for the example we studied), then you have got the point of this section. 3. The Electric Field What are we going to do in this section? In this section we are interested in studying the force between charged particles. Our approach will copy the description of the gravitational force ﬁeld, in terms of the gravitational ﬁeld. To describe the force between charged particles we will introduce an electric ﬁeld. We know that charged particles exert forces on each other. You saw that in class with the charged tapes and with your electro scopes. We already have a very good way to describe the gravitational force ﬁeld. By copying what we did for the gravitational ﬁeld, we write F = q E. This is a very natural guess and it is what we do in physics - we try to copy previous successes. We are copying what we did before by splitting the electric force experienced by a particle into two factors. The charge q of our particle replaces the mass m of our particle that we had in our previous formula. Again, q is a single number associated with the object that experiences the ﬁeld. Further, the electric ﬁeld E is what replaces the gravitational ﬁeld g in our gravitational formulas. So, we have replaced a number with a number (m → q) and a ﬁeld with a ﬁeld (g → E). Further, we are still splitting things up into a source that produces a ﬁeld and an object that experiences the ﬁeld. Just like in the gravitational case, we expect that this split is going to provide a very useful way of thinking about the problem. Also, just like in the gravity case, we expect that this will not be exact - we may be making a (hopefully!) small approximation. We measure the charge q in coulombs. In SI units the charge of a single proton is 1.6×10−19 C. Did you get it? If you understand that our experience with gravity has suggested that we should introduce the electric ﬁeld E, you have understood the point of this section. 4. Point charges: The Coulomb Force Law What are we going to do in this section? In this section we will study the force between two point charges. A simple situation we could think about involves the interaction between two point charges. By a point charge we mean an object whose radius is very small compared to the distances between it and all other objects of interest. From careful experiments we know that 1. like charges repel and unlike charges attract; 2. the force acts along the line joining the two point charges; 3. if our point charges have charge q1 , q2 the force has a magnitude 1 |q1 q2 | |F | = . 4π 0 r2 This is a complete speciﬁcation of the force, since we have given both its magnitude and 1 its direction. The number 4π 0 is a constant equal to 9×109 N·m2 ·C−2 . Make sure that 1 you could get the units of 4π 0 yourself. We can give a graphical representation of this force ﬁeld. We could, for example, draw the force ﬁeld felt by q1 due to interaction with q2 and also the electric ﬁeld set up by q2 . The rules for our graphical representation are 1. the tail of the arrow is at the location where the force is felt; 2. the direction of the arrow shows the direction of F (or E); 3. the length of the arrow shows the magnitude |F | (or |E|). As an example, we plot the electric ﬁeld for a positive point charge and a negative point charge below. Fig. 1: The electric ﬁeld from a positive point charge. Fig. 2: The electric ﬁeld from a negative point charge. Make sure that you understand why the electric ﬁeld is directed radially out of a positive point charge and radially into a negative point charge. To understand this, all you need to know is F = q E and that unlike charges attract and like charges repel. You should also understand why the arrows get smaller as we move away from the point charge. An important point to note here is that a point particle is not aﬀected by its own electric ﬁeld, i.e. a point charge does not exert a force on itself. Physically, this makes sense because the charge can’t get itself moving. There would not even be a way to decide in what direction the charge should move! Did you get it? If you know how to compute the force between two point charges and you know how to compute the electric ﬁeld set up be a point charge, you have understood the point of this section. 5. Diﬀerences Between Gravity and Electromagnetism What are we going to do in this section? We have studied gravitational forces to get some insight into forces acting between charged objects. It is important to realize however, that despite the similarities there are also important diﬀerences between the two types of forces. By experimenting with tapes in class, we have seen that the Coulomb force can be both attractive and repulsive. This is diﬀerent to gravity, where the force is always attractive. Another comparison that we can make is the ratio of the strength of the gravitational attraction between two electrons to the electric repulsion between the two. Since both 1 forces have a r2 dependence, it does not matter how far apart the electrons are when we make this comparison. The result is |Fgrav | Gm1 m2 = 1 ∼ 10−36 . |Felec | 4π 0 q1 q2 Thus, gravity is a much much weaker force that the electric force. For this reason, we are going to make another approximation - we will assume that we can ignore the gravitational forces acting between our particles. This is an excellent approximation. 6. Many Charges: The Superposition principle What are we going to do in this section? In this section we study the electric ﬁeld set up by a collection of charges. We have already obtained the formula for the electric ﬁeld set up by a point charge. In nearly all of the applications we are interested in, we will have more than one point charge. How do we compute the electric ﬁeld set up by a collection of point charges? There is a very powerful tool for these problems called the superposition principle. The superposition principle states that the net electric ﬁeld at a location in space is equal to the vector sum of individual electric ﬁelds contributed by all charged particles located elsewhere. Thus, the electric ﬁeld contributed by a charged particle is unaﬀected by the presence of other charged particles. To understand the superposition principle it is useful to do an example. The next section gives an example of how to use the superposition principle. Did you get it? If you can work out the electric ﬁeld due to a collection of point charges using only the formula for the electric ﬁeld set up by a single point charge and the super- position principle, you have understood this section. 7. Electric Dipoles What are we going to do in this section? In this section the electric ﬁeld of an electric dipole is studied. Two equal but opposite charges separated by a distance s are called an electric dipole. The ﬁgure below shows an example of an electric dipole. We are interested in computing the value of the electric ﬁeld at point A, which lies on the axis the dipole. 00 00 11 11 −q +q 000 111 00 00 11 11 0 1 A 1111111111111111111 0000000000000000000 s r Fig. 3: The two charges in the electric dipole are separated by a distance s. The point A is a distance r from the centre of the dipole. Just by looking at the system we can get a good idea about the electric ﬁeld at A. Since the positive charge is closer to A it produces a larger electric ﬁeld than the negative charge. The electric ﬁeld due to the positive charge at A points away from the dipole; the electric ﬁeld from the negative charge at A points towards the dipole. When we sum the two we’ll get a much smaller electric ﬁeld at A, pointing away from the dipole. You may be surprised that the dipole which has total charge zero is able to produce an electric ﬁeld at point A. There is a non-zero electric ﬁeld only because the two charges don’t sit at the same point. The electric ﬁeld due to q is (ˆ is a unit vector pointing from the negative charge to the i positive charge) 1 q ˆ Eq = 2 i. 4π 0 r − s 2 The electric ﬁeld due to −q is 1 q ˆ Eq = − 2 i. 4π 0 r + s 2 The superposition principle tells us that EA is given by the sum Eq + E−q . Summing the two contributions we ﬁnd 1 2qsr ˆ EA = i. 4π 0 r + s 2 s 2 2 r− 2 Does this result make sense? Two things which we can check immediately are the direction of the ﬁeld and the dimensions of the answer. The electric ﬁeld is in the ˆ direction, that i is, away from the dipole. This agrees with our discussion above. Are the dimensions of the 1 result correct? Previously to get the electric ﬁeld we multiplied 4π 0 by a charge divided 1 by a length squared. It is obvious that the factor multiplying 4π 0 in our above formula does have the dimensions of charge divided by length squared. We are often studying a situation for which r >> s. In this case we can get a much simpler formula for the ﬁeld from a dipole. In this limit we approximate s 2 r± ≈ r2 . 2 With this approximation 1 2qs ˆ EA = i. 4π 0 r3 Note that the ﬁeld of a dipole has a diﬀerent r dependence to the ﬁeld for a point charge. We can write this electric ﬁeld as 1 2qs ˆ 1 2p EA = 3 i= , 4π 0 r 4π 0 r3 where p = qsˆ We call p the dipole moment. Along the axis of the dipole, the ﬁeld of i. the dipole is parallel to the dipole moment. The dipole moment and electric ﬁeld are not parallel in general. Far from the dipole, the dipole moment is the only quantity you need to determine to completely specify the ﬁeld of the dipole. p is the product of the charge q with the vector sˆ which stretches from the position of the negative point charge to the i position of the positive point charge. Here is another example that you can work through: imagine that we want to compute the electric ﬁeld of the dipole at the position B as shown in the ﬁgure. You need to work out the ﬁeld due to the positive and negative charges at point B and sum them to get the 0 1 B r 11 11 00 00+q 11 11 00 00 −q s Fig. 4: The two charges in the electric dipole are separated by a distance s. The point B is a distance r from the centre of the dipole. total electric ﬁeld. With some work you should be able to convince yourself that the ﬁeld of the positive charge at B is 1 q − 2 ˆ + rˆ s i j Eq = , 4π 0 r2 + s 2 s 2 2 r2 + 2 the ﬁeld of the negative charge at B is (ˆ points from the centre of the dipole towards B) j 1 q − 2 ˆ − rˆ s i j E−q = , 4π 0 r2 + s 2 s 2 2 r2 + 2 and hence that the electric ﬁeld at B is 1 sqˆ i EB = Eq + E−q = − 3 . 4π 0 s 2 2 r2 + 2 Notice that this ﬁeld has no ˆ component. Can you understand this in a simple way? You j should again convince yourself that when r >> s, this ﬁeld is completely speciﬁed by the dipole moment p. Did you get it? If you can compute the ﬁeld of the dipole at any point in space and understand the behaviour of the dipole ﬁeld in the limit that you are very far from the dipole (compared to the distance between the two charges in the dipole), then you have understood this section. 8. Matter What are we going to do in this section? Matter is made from building blocks which are charged. In this section we will review what these building blocks are. we are doing this because we’d like to use what we have learnt about the interaction between charges to understand the properties of matter. All matter is made up of atoms. An atom consists of a positively charged nucleus surrounded by a negatively charged electron cloud. The nucleus itself consists of positively charged protons and electrically neutral neutrons. A single atom has a radius of about 10−10 m. This implies that it has a volume of about ∼ r3 = 10−30 m3 . Consequently in a 1cm3 =10−6 m3 cube of a metal lattice, there are about 1024 atoms. The nucleus of the atom is about 10−15 m, that is, about 100 000 times smaller than the atom itself. Most of the mass of the atom is concentrated in the nucleus; indeed, the proton and neutron have roughly the same mass and the proton is about 2000 times more massive than the electron. The charge of an electron is −1.6 × 10−19 C; the charge of a proton is 1.6 × 10−19 C. Atoms have the same number of electrons and protons and hence are electrically neutral. However, it is possible for an atom to become an ion when it accepts or donates electrons. A sodium ion for example has 11 protons and 10 electrons. As a result, it has a net charge of 1.6 × 10−19 C. We can often model ions as point charges. It is also possible that atoms bind to form molecules. For example a hydrogen atom can bind with a chlorine atom to form a HCl molecule. The hydrogen atom in the molecule is slightly positive; the chlorine atom is slightly negative. Thus, the hydrogen chloride molecule can be modelled as a dipole. Did you get it? What you should get out of this section is an appreciation that matter is built out of components which interact electrically. 9. Interaction of a Neutral Atom with a Monopole What are we going to do in this section? In this section we will study the force exerted by a neutral atom on a point charge, when the neutral atom is placed in the electric ﬁeld of the point charge. The electron cloud of an atom can move relative to the nucleus. As a result, when a neutral atom is placed in the electric ﬁeld of, for example, a positive point charge, the electron cloud moves slightly towards the point charge whilst the nucleus moves slightly away from the point charge. As a result, one side of the atom is slightly negative and the other side is slightly positive. We say that the atom is polarised. We have already seen that far from a dipole, the electric ﬁeld of the dipole is completely speciﬁed by the dipole moment of the dipole. Since the atom is so tiny, it is often an excellent approximation to assume that we are far from the dipole. In this case, to specify the electric ﬁeld of a polarised atom, we only need to give its dipole moment. It has been found experimentally, that for almost all materials, the dipole moment of the polarised atoms in the material is directly proportional to the applied electric ﬁeld. We write this as p = αE, with p the dipole moment of the polarised atoms, E the applied electric ﬁeld and α an experimentally measured material dependent constant called the polarisability of the ma- terial. You should convince yourself that α has units C·m/(N/C). When a neutral atom is polarised, by how much does the electron cloud shift relative to the nucleus? A typical atomic polarisability is 10−40 C·m/(N/C). We will imagine that our atom is placed in an electric ﬁeld with magnitude |E| = 3 × 106 N/C. This is a rather large electric ﬁeld - it is the electric ﬁeld at which air ionises - the strength of electric ﬁeld you’d need to see sparks travel through the air. Putting the numbers together we have |p| = 10−40 × 3 × 106 = 3 × 10−34 C·m. The electrons closest to the nucleus feel the electric ﬁeld of the nucleus the most. The outermost electrons feel the weakest nuclear electric ﬁeld. Thus, when we apply the electric ﬁeld, it will have the largest eﬀect on the outermost electron. This outermost electron sees the rest of the atom as something with the charge of a single proton. If we assume that our dipole is built from charges with a charge of 1.6 × 10−19 C, we ﬁnd that the charges are separated by a distance 3×10−34 s= 1.6×10−19 ∼ 2 × 10−15 m. This is about the size of an atomic nucleus - a very tiny shift of the electron cloud. To compute the force that a neutral atom exerts on a point charge, we need to compute the strength of the electric ﬁeld of the dipole at the location of the point charge. q + _ + r Fig. 5: An atom polarised by the ﬁeld of a point charge. We consider the case that the atom is far from the point charge so that we can approximate the dipole ﬁeld that the point charge feels as 1 2|p| . |Edipole | = (9.1) 4π 0 r3 We need to compute |p|. The atom is polarised by the electric ﬁeld of the point charge. The strength of this ﬁeld, at the location of the atom is 1 q |Eq | = . 4π 0 r2 Thus, in terms of the polarisability of the atom we have 1 αq |p| = α|Eq | = . 4π 0 r2 Plugging this into (9.1) we ﬁnd 2 1 2αq |Edipole | = . 4π 0 r5 Thus, the force exerted by the neutral atom on the point charge is 2 1 2αq 2 q|Edipole | = . 4π 0 r5 1 1 The r5 dependence in our formula comes about from the r2 fall oﬀ of the ﬁeld of the point 1 charge which polarises the atom times the r3 fall oﬀ of the ﬁeld from the dipole. Did you get it? If you have an understanding of why the force the neutral atom exerts on the point charge depends on r as r−5 , you have understood this section. 10. Insulators What are we going to do in this section? In this section we will study materials which are made up of molecules with tightly bound electrons. We are interested in understanding how these materials respond to an electric ﬁeld. We will now study materials which are made up of molecules with tightly bound electrons. Since all electrons are ﬁrmly bound to the atoms/molecules making up the ma- terial (most charged particles move much less that 10−10 m) these materials can electrically “insulate” one charged object from another. If we expose our insulator to an external electric ﬁeld, the electric ﬁeld will cause all of the (normally unpolarised) atoms in the insulator to polarise. In any given molecule, the electrons didn’t move very far at all. In addition, the molecules within the material are also not free to move. Thus, no single charged particle moved very far at all. The net eﬀect however can be big because there are so many molecules in the insulator to be aﬀected. The polarised molecules align with the E ﬁeld that is polarising them. Thus, even at equilibrium, there can be a non-zero electric ﬁeld inside the insulator. + − + − + − + − + + − + − Fig. 6: Polarisation of an insulating material by the electric ﬁeld of a point charge Another question we could ask is where any excess charge is located. Since there are no mobile charge carriers, excess charges will stay where they are. They could be located in the interior of the insulator or bound to a particular point on the surface without spreading along the surface. Did you get it? If you know what an insulator is and how it responds to an electric ﬁeld, then you understood this section. This includes knowing what the electric ﬁeld inside an insulator at equilibrium is, understanding how the insulator polarises and where excess charge in the insulator is located. 11. Conductors What are we going to do in this section? In this section we will study materials which do have mobile charge carriers. We are interested in understanding how these materials respond to an electric ﬁeld. An example of a conductor is a metal. Metals are good conductors. In this case, the mobile charge carriers are electrons. The atoms in a solid piece of metal are arranged in a regular three dimensional geometry, a lattice. The inner electrons of each atom are tightly bound to the nucleus; some of the outer electrons of each atom participate in chemical bonds and about one electron per atom join a sea of mobile electrons which are free to move throughout the entire macroscopic piece of metal. These mobile electrons look a lot like an ideal gas. In fact, some simple models treat the mobile electron sea as an ideal gas. How does a conductor respond to an external electric ﬁeld? The external electric ﬁeld exerts a force on the mobile electrons. Electrons move in response to these forces and a net charge starts to build up on the edges of the conductor. A good way to think about this is that the electron sea shifts with the result that we have excess electrons building up on one side of the conductor making it negatively charged and a deﬁciency of electrons on the other side of the conductor leaving it with a net positive charge. Denoting only the excesses of charge, we have the situation shown below. + + − − − + − + − + − + − − + + + − + − − + − + − + + − − Fig. 7: A conductor polarises in response to an external electric ﬁeld. The ﬁeld due to the excess charge is opposite to the ﬁeld due to the point charge. Thus, the net electric ﬁeld in the conductor starts to decrease as excess charge builds up. How much excess charge will build up? A natural guess is that enough excess charge will build up so that the ﬁeld due to the excess charge exactly cancels the applied external ﬁeld. In fact, we can prove that this is the case. When we say that the conductor has reached static equilibrium, we mean the mobile electrons inside the conductor are no longer moving. When we reach this point, no further excess charge will build up. Actually, there will always be some electron motion. When we say that the electrons have stopped moving, we are really talking about the average electron motion. We will prove that once the system reaches static equilibrium, the excess charge build up exactly cancels the applied external ﬁeld. We’ll do the proof by contradiction - that is we’ll make an assumption and show that the consequences of this assumption are incorrect. This shows that our assumption must have been incorrect. Imagine that the system has reached static equilibrium and assume that the net electric ﬁeld inside the conductor is non-zero. If the net electric ﬁeld inside the conductor is non-zero, the electrons inside the conductor will feel a non-zero force. Since they are free to move, this implies that they will accelerate. But this contradicts our starting point - the system has reached static equilibrium so that electrons are not moving. Thus, our assumption that the net electric ﬁeld inside the conductor is non-zero when the conductor has reached static equilibrium must have been false. The net electric ﬁeld inside a conductor which has reached static equilibrium is zero. Another question we could ask is where any excess charge is located. Imagine for example that we have an excess of electrons in the conductor. Since the charge in a conductor is free to move, the excess electrons will repel each other and move away from each other. In the ﬁnal equilibrium conﬁguration the electrons will be as far as possible from each other. This occurs when the excess charge distributes itself uniformly on the surface of the conductor. Did you get it? If you know what a conductor is and how it responds to an electric ﬁeld, then you understood this section. This includes knowing what the electric ﬁeld inside a conductor at equilibrium is, understanding how the conductor polarises and where excess charge in the conductor is located. 12. Charged Tapes What are we going to do in this section? In class we did an experiment involving charged tapes. By ripping one tape oﬀ of a second tape, the tapes became charged. In this section we will use what we already know about charges to discuss the results of this experiment in detail. Recall the experiment we did in class: we stuck a ﬁrst piece of sticky tape to the table. We then stuck a second piece of tape onto the ﬁrst piece and ripped the second piece oﬀ. We then did exactly the same thing with a third tape. It turned out that the second and third tapes repelled each other. From this we can infer that they are charged. The charge on the tape is a result of one tape pulling electrons oﬀ of the other tape. One thing we can try to calculate, is the fraction of atoms that lost one of their electrons. Our calculation is set up like this: (i) ﬁrst, estimate the force of repulsion between the two tapes; (ii) from this force of repulsion we can determine the charge on each tape; (iii) from this charge we know how many electrons are on each tape and hence what fraction of atoms lost an electron. Something else that you should keep in mind is that we are not trying to do an exact calculation. We are just trying to get some sort of an idea about how large the charge on each tape is. We’ll be happy if our answer is good to a factor of 10! Consider the case that the two tapes that are repelling each other have come to rest. There are three forces acting on each tape: an electrical repulsion (we’ll denote the magnitude of this force by Fe ), the force due to gravity (with magnitude given by mg) and the force that your hand exerts on the tape (with magnitude given by T ). Consult the ﬁgure below. T θ Fe mg Fig. 8: One of the two tapes which were repelling each other. Since the tape is not accelerating, the forces acting on it must sum to zero. This gives us two equations T cos θ = mg, T sin θ = Fe . We can estimate mg and θ; we want to know Fe . Dividing the two equations eliminates T - the quantity that we are not interested in (in the next equation we use the fact that θ is small) Fe = tan θmg ≈ mgθ. Combining this with Coulomb’s Law we have (this is an approximation - all of the point charges on the two tapes are not the same distance r from each other) 1 q2 Fe = = mgθ. 4π 0 r2 Thus, the charge on each tape is q= 4π 0 r2 mgθ. The quantity we could compute was q 2 . We can’t tell if the tape has extra electrons (giving it a net negative charge) or if it is missing electrons (giving it a net positive charge). 1 Now, lets estimate q. First, 4π 0 = 9 × 109 ≈ 1010 N·m2 ·C−2 . The distance between −2 our tapes was about r =5cm=5× 10 m. The mass of our tape was about 0.1g=1×10−4 kg. The deﬂection of the tape was roughly 0.1 radians. Finally, g is about 10 m·s−2 . Thus, q≈ 10−10 × (5 × 10−2 )2 × 10−4 × 10 × 0.1 = 5 × 10−9 C. One electron has a charge of 1.6 × 10−19 C, so that this charge translates into about 5 × 10−9 n= ≈ 3 × 1010 1.6 × 10−19 electrons. Next we need to estimate how many atoms there are in our tape. Lets imagine that the portion of our tape that is charged has a width of about 1 cm and a length of about 3 cm. This translate into an area of about 3 × 10−4 m2 . Each atom has a radius of about 10−10 m and hence an area of about r2 = 10−20 m2 . Thus, on the charged portion of the tape there are about 3 × 10−4 N= = 3 × 1016 10−20 atoms. The fraction of atoms f that lost (or gained, we don’t know) an electron is n f= = 10−6 , N so that roughly one atom in a million will lose an electron. Did you get it? In this section we applied the physics we have developed so far to study a real problem. If you understand how the quantities in the real problem are related to the quantities appearing in the formulas we have developed, you have understood the point of this section. 13. Charged Electroscope What are we going to do in this section? In class you built electroscopes. You were able to show that a charged cup attracts the needle of your electroscope. In this section we will study the physics behind this experiment. In class, you charged a polystyrene cup by rubbing it with a woolen scarf. When the cup was brought close to the needle of your electroscope, it attracted the needle. Your needle has a net charge of zero, so why is there a force of attraction? To understand this eﬀect, note that the needle of the electroscope is a conductor. When you bring your cup close to the needle, you are polarising the conductor. This eﬀectively turns it into one big dipole. It is the interaction between this dipole and your charged cup that is responsible for attracting the needle. Although we don’t know whether the cup is positively or negatively charged, we’ll assume it is positively charged in this section. This is an arbitrary choice and it certainly won’t aﬀect any of the answers we will obtain. In this section we will use a result that we will prove later. Consider two inﬁnite parallel uniformly charged plates as shown below. The plates need not really be inﬁnite. All that we need is that the gap between the plates is much much smaller than the length of the plates. The plates are characterised by a charge density Q σ= . A If our plates really were inﬁnite, both q and A would be inﬁnite, but the ratio σ would be ﬁnite. The units of σ is coulombs per meter squared. The result that we will use now and prove later says that the electric ﬁelds between the plates takes a constant value for all points between the two plates. This value is equal to σ |E| = . 0 You should check that the dimensions on both sides of this formula match. − + − + − + − + − + − + − + − + − + − + − + − + − + − + − Fig. 9: Two inﬁnite uniformly charged plates. As we have already stated, when the cup is brought close to the needle of the electro- scope, the needle polarises. We already know that (because the needle is a conductor) the build up of excess charge on the surface of the needle will halt when the ﬁeld from this excess charge exactly cancels the applied ﬁeld of the cup. If we approximated the needle of the electroscope by an inﬁnitely long needle, we know that the magnitude of the ﬁeld |Q| inside the needle is given by A 0, where A is the area of the face of the needle and Q is the excess charge that builds up on the surface of the needle. This ﬁeld must exactly cancel the electric ﬁeld of the cup, so that (again this is an approximation - we are assuming that all charges on the cup are the same distance from the needle of the electroscope) 1 |q| |Q| |E| = = . (13.1) 4π 0 r2 A 0 In this formula, q is the net charge on the cup and r is the distance from the cup to the needle. We now want to calculate the force that the cup exerts on the needle. We can equally compute the force that the needle exerts on the cup, because according to Newton’s third law, these forces are equal and opposite. To compute the force that the cup exerts on the needle, we need an expression for the ﬁeld set up by the dipole at the location of the cup. Since the distance from the cup to the dipole is much larger than the thickness of the dipole, we can use the following approximation for the ﬁeld of the dipole 1 2|p| 1 2|Q|t |E| = 3 = . 4π 0 r 4π 0 r3 needle + − + − Q + − −Q + + − + + − + + − + + − + − + − + − + − + t − + − r + − Fig. 10: The polarised needle of the electroscope and the charged cup are shown. In this ﬁgure the thickness of the needle (t) and the distance from the needle to the dipole (r) are given. The ﬁgure is exaggerated; in reality, t is much much smaller than r. The face of the needle (which has an area A) is not visible because we have a side view of the needle. In this last formula, t is the thickness of the needle, Q is the excess charge on the surface of the needle and r is the distance from the needle to the cup. Multiplying this by the charge q of the cup and using (13.1) to obtain an expression for Q, we ﬁnd 1 2Aq 2 t F = , 4π 0 4πr5 for the magnitude of force exerted by the needle on the cup. This is of course equal to the force that the cup exerts on the needle. Following the argument we gave for the tapes we obtain F = mgθ. Consequently 1 2Aq 2 t F = mgθ = . 4π 0 4πr5 Rearranging a little, we ﬁnd mgθ(4π 0 )4πr5 q= . 2At We can now estimate q. m is about 1g=10−3 kg, g is about 10m·s−2 and θ is about 0.1 radians so that mgθ ≈ 10−3 . r is about centimetres so that r5 ≈ 10−10 m5 . The area of the needle is of the order of cm2 =10−4 m2 and the thickness of the needle is of the order of mm=10−3 m. Also, 4π ≈ 10. Consequently 10−3 10−10 1010−10 q≈ 10−4 10−3 ≈ 3 × 10−8 C. This corresponds to 3 × 10−8 n= ≈ 2 × 1011 1.6 × 10−19 electrons. We again want to calculate the fraction of atoms that gained or lost an electron. The area of our cup is roughly 4cm×5cm=20cm2 = 2 × 10−3 m2 . Again, the radius of an atom is roughly 10−10 m so that the area of a single atom is about 10−20 m2 . Thus, the number of electrons in the surface of the cup is roughly 2 × 10−3 N≈ = 2 × 1017 . 10−20 Thus, the fraction of atoms that lost an electron is 2 × 1011 f= 17 = 10−6 , 2 × 10 which agrees with the answer we obtained in the previous section. (Because our calculations are so rough, even if we had obtained f = 10−5 or f = 10−7 , we would still claim to have agreement.) When the needle of the electroscope polarises, the electron sea shifts relative to the regular lattice of metal ions in the needle. By how much does the sea shift? This is the last number we will calculate. If the electron sea shifts by a distance d, the volume of excess charge will be Ad where A is the area of the face of the needle. Since we have roughly n = 1030 atoms per meter cubed, we will have a total of nAd = 1030 Ad excess electrons corresponding to an excess charge of Q = nAde = 1030 Ad × 1.6 × 10−19 C, where e is the charge of one electron. Thus, the charge density σ is nAde σ= = nde = 1, 6 × 1011 d. A Inserting this into (13.1) we obtain Q 1 q 1 3 × 10−8 σ= = 1, 6 × 1011 d = ≈ = 3 × 10−5 . A 4π r2 10 10−4 Thus, d ≈ 0.2 × 10−15 m - that is the electron sea shifts by a fraction of the radius of the nucleus! Did you get it? In this section we again studied a real problem. If you can connect the symbols in the various formulas with the quantities in the real problem and understand why (for example) the needle is attracted by the charged cup, you have understood this section. 14. Continuous Distributions of Charge What are we going to do in this section? Up till now we have considered only sources involving a ﬁnite number of charges. In this section we will consider a problem involving millions and millions of charges. Imagine that we want to compute the electric ﬁeld set up by a conﬁguration involving millions of individual particles. In this situation we can describe the source as a continuous charge distribution. To compute the electric ﬁeld, we will again use the superposition principle. The basic plan we follow is: 1. Divide the charge distribution up into pieces dq. 2. Compute the electric ﬁeld dE due to one piece. 3. Add up all the contributions to get the total electric ﬁeld. To illustrate this method, we will compute the electric ﬁeld set up by a uniformly charged thin rod. By a thin rod, we mean that we can ignore the width of the rod when we do our calculations. The length of the rod is L. Before we start, lets sketch the ﬁeld that we expect from the rod. We will assume that the rod has a total positive charge of Q. The point charge was spherically symmetric, so that the ﬁeld set up by a point charge also has spherical symmetry. The rod has a cylindrical symmetry, so the ﬁeld set up by the rod will also have this cylindrical symmetry. At the centre of the rod there is an additional symmetry - we can rotate the rod about its midpoint. Away from the centre of the rod, rotating the rod about its midpoint is not a symmetry. Using these arguments and what we already know about the electric ﬁeld set up by positive charges, we might expect the ﬁeld pattern shown in the ﬁgure below. Fig. 11: The expected electric ﬁeld close to a positively charged rod. Now that we have some idea of what we expect, lets try to compute things more explicitly. Consider a slice of the rod as shown in the ﬁgure below. We will be computing the electric ﬁeld at the observation point shown, set up by this slice. To compute the magnitude of the electric ﬁeld, we will need the distance from dq to the observation point. From the sketch below, we see that this distance is r= x2 + (y − yo )2 . o dq dy r y−y o y xo yo dE Fig. 12: The electric ﬁeld due to a small slice of the charged rod is shown. The midpoint of the rod lies at x = y = 0. Thus, the magnitude of the electric ﬁeld is given by 1 dq |dE| = 2 + (y − y )2 . 4π 0 xo o ˆ To get the direction of the electric ﬁeld, we need the unit vector r in the direction of the vector r shown below. First, we compute r (look at the ﬁgure to get this) r = (xo , yo − y, 0). The unit vector is thus (xo , yo − y, 0) ˆ r= . x2 + (y − yo )2 o The electric ﬁeld now becomes 1 dq dE = 3 (xo , yo − y, 0). 4π 0 (xo 2 + (y − y )2 ) 2 o Finally, we want to express the charge dq in terms of the length of the slice dy. We can do this by multiplying the length of our slice by the charge density Q dq = dy. L Thus, our ﬁnal expression is Q 1 1 dE = (xo , yo − y, 0)dy. 2 + (y − y )2 ) 3 L 4π 0 (xo 2 o The total electric ﬁeld is now obtained by integrating over the length of the rod L 2 Q 1 1 E= (xo , yo − y, 0)dy. 2 + (y − y )2 ) 3 L 4π 0 (xo −L 2 0 2 This is a tough integral to do analytically. You could of course do it numerically and hence obtain an expression for the electric ﬁeld at any location you wanted. As we mentioned above, when yo = 0, we expect that the y component of the ﬁeld vanishes. For this special point, we can perform the above integral analytically. The only non-zero component of the electric ﬁeld is L 2 Q xo dy Ex = −L 2 4π 0 L (x2 + y 2 ) 3 o 2 L 1 Qxo y 2 = 4π 0 L x2 o x2 + y 2 o −L 2 1 Qxo y = . 4π 0 L L 2 x2 o x2 + o 2 Did you get it? We studied the electric ﬁeld set up by a continuous charge distribution. The basic idea in this section was to break the continuous charge distribution into inﬁnitesimal segments, compute the electric ﬁeld from each segment and then sum these electric ﬁelds to obtain the electric ﬁeld from the original charge distribution. Of course, this is another application of the superposition principle. 15. Gauss’s Law What are we going to do in this section? In the last section we saw that we could use the superposition principle to determine the electric ﬁeld of a continuous charge distribution. In this section we will obtain a very useful result that will often provide a simpler way to obtain the electric ﬁeld of a continuous charge distribution. To obtain Gauss’s law, we start by considering a distribution of charges with total charge Q enclosed by a surface S. We can compute the value of the electric ﬁeld (of the distribution of charges) at any given point on the surface. Gauss’s law is stated in terms of an integral of this electric ﬁeld over the surface S. To do this integral, imagine breaking the surface S up into many tiny pieces, each of area dA. The pieces are so small that they each look like a tiny, ﬂat, two dimensional area. Consequently, to each of these little area elements we can associate a unique unit vector which is normal to the small ﬂat area and points outwards. These small area elements times these outwardly directed unit vectors are denoted by dS. We can now state Gauss’s law Q E · dS = . S 0 To prove Gauss’s law we will use the superposition principle. Any charge distribution can be built as a sum over point charges. The superposition principle tells us that we can consider each of these point charges separately. Thus, if we prove the result for a single point charge, the result for an arbitrary distribution of charge is also true. As a warm up exercise, lets prove the result for a point charge and the special case that S is a sphere of radius R. In this case the electric ﬁeld is e r ˆ E= , 4π 0 R2 ˆ with r a unit vector pointing in the radial direction. The outward normal to our tiny area elements are also directed in the radial direction dS = rdA = rR2 dΩ, ˆ ˆ with dΩ an inﬁnitesimal solid angle. Thus, e r ˆ e e E · dA = · rR2 dΩ = ˆ dΩ = . 4π 0 R2 4π 0 0 In the above we used the fact that the integral dΩ = 4π. After that simple warm up, we now consider an arbitrary surface S. If we consider an arbitrary surface S, the inﬁnitesimal area elements dS will no longer point in the radial ˆ direction r. In this case, the area of the element will be larger than an area dA⊥ with ˆ the same solid angle but oriented so that it points in the direction of r. We would like an expression for |dAperp | in terms of |dA|. Let the angle between the unit normal of dA and π ˆ r be θ. When the angle θ = 0, |dA| = |dA⊥ |. When θ = 2 ˆ radians, the vector r lies in the surface and |dA⊥ | = 0. 1111111 0000000 0000000 1111111 1111111 0000000 1111111 0000000 1111111 0000000 0000000 1111111 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 0000000 1111111 1111111 0000000 1111111 0000000 Fig. 13: The perpendicular area is shaded. The angle between dA and dA⊥ ˆ is θ. The angle between the dA and the unit vector r is also θ. It is now not hard to see that |dA⊥ | |dA| = . cos θ The reason why we like to express things in terms of |dA⊥ | is that we known |dA⊥ | = R2 dΩ where the small area element is a distance R from the origin and it traces out a solid angle ˆ dΩ. Since the electric ﬁeld is in the r direction, 1 e 2 e E · dA = |E||dA| cos θ = |E||dA⊥ | = 2 R dΩ = dΩ. 4π 0 R 4π 0 Thus, e e E · dA = dΩ = . 4π 0 0 This proves the formula for a point charge and an arbitrary surface S. Thus, as explained above, by the superposition principle this proves the theorem in general. Did you get it? The basic result is that if we integrate the outward normal component of the electric ﬁeld over a closed surface S, the integral is equal to the total charge enclosed by S divided by 0. If you understand this, then you got the point of this section. 16. Continuous Distributions of Charge Again What are we going to do in this section? In our treatment of the electroscope we used the ﬁeld between two inﬁnite charged plates. In this section we will derive this result. Consider a uniformly charged inﬁnite sheet. By applying Gauss’s law with a clever choice for the surface S we will be able to derive an expression for the electric ﬁeld of the sheet. You could also break the sheet into small charges dq and use the superposition principle. It turn out that using Gauss’s law is a much simpler calculation. Before we do the computation, lets try to get some idea of what the electric ﬁeld set up by this charged plate looks like. Since our charged plate is inﬁnite, there is nothing in the geometry of the charge distribution that distinguishes between parallel to the plate or anti-parallel to the plate. Thus, by symmetry of the charge distribution alone we don’t + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + Fig. 14: The ﬁeld from an inﬁnite uniformly charged plate. The surface S is the cylinder shown. The two ends of the cylinder are the same distance from the charged plate. expect that the electric ﬁeld set up by the plate will have a component parallel to the plate. Thus, we might expect the ﬁeld shown in the ﬁgure. The next step in the calculation involves choosing the surface S. A convenient choice is the cylinder shown. We can split the integral over S into two contributions - the contribution coming from integrating along the length of the cylinder and the contribution coming from integrating over the ends of the cylinder. First consider the contribution from the integral over the length of the cylinder. The direction of the electric ﬁeld is parallel to the sides of the cylinder. The direction of dS is normal to the sides of the cylinder. Consequently E · dS is zero. We see that the entire contribution to the integral comes from the ends of the cylinder. Since the ends of the cylinder are at a ﬁxed distance from the charged plate, the strength of the electric ﬁeld will be constant over the end of the cylinder. The cylinder is placed so that the two ends of the cylinder are the same distance from the charged plate. At both ends, the electric ﬁeld is parallel to the area element dS. Thus, E · dA = E · dA = 2AE, S ends where E is the magnitude of the electric ﬁeld at the ends of the cylinder and A is the area of the end of the cylinder. Gauss’s law now tells us that Q 2AE = . 0 Note that the area A of the end of the cylinder is also equal to the area of the plate that Q is enclosed by S. Thus, A will be the surface charge density of the plates, measured in C·m−2 . Solving for E we obtain Q σ E= = . 2A 0 2 0 You should check that E has the correct units. This formula tells you that the magnitude of the electric ﬁeld is independent of your distance from the plates. The result we used for the needle of our electroscope involved two inﬁnite plates with opposite charge. Using the superposition principle, we can get the total ﬁeld by adding the ﬁeld from the positively charged plate to the ﬁeld from the negatively charged plate. The ﬁeld from the positive plate is directed outwards from the plate. The ﬁeld for the negative plate is directed into the plate. Thus, in the region in between the two plates the ﬁeld of the positive plate and the negative plate sum to give twice the original ﬁeld σ E= . 0 In the region outside the two plates the electric ﬁelds are opposite in direction and sum to zero. + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − Fig. 15: Two inﬁnite plates with opposite charge. The ﬁeld in between the two plates is constant = 0. The ﬁeld outside of the two plates vanishes. Did you get it? In this section we have obtained a concrete result - the electric ﬁeld between two uniformly charged inﬁnite plates. We have also demonstrated that Gauss’s law can be a powerful tool for obtaining the electric ﬁeld from a continuous distribution of charge. 17. Potential Energy: Gravity What are we going to do in this section? In this section we will introduce the concepts of work and energy. This often provides a method for analysing the dynamics which is simpler than an analysis using forces. The dynamics of point particles is described by Newton’s laws. In this section, we are going to use Newton’s laws to introduce the ideas of energy and work done. These concepts provide an alternative method for analysing the dynamics of particles. Of course, the “work-energy” approach is derived from Newton’s laws, so that we are not saying something that was not already contained in Newton’s laws. The “work-energy” repackaging however, often provides a simpler solution. To illustrate this last point, consider the following problem: we throw a ball upwards. It starts from some initial height hi with some initial speed vi . Since gravity is acting on the ball, it starts to slow down until it comes to rest. The question we ask is, at what height does the ball come to rest? We could answer this question using Newtons laws. Indeed, we know the initial velocity v(0) and initial position x(0). We also know the force F = mg acting on our F particle. Thus, we know the acceleration a = m = g of our particle, for all times. For a small enough we can write (we dropped terms of O( 2 )) dx x( ) = x(0) + (0) = x(0) + v(0), dt dv v( ) = v(0) + (0) = v(0) + a(0) = v(0) + g. dt Thus, we now have the values of the position and velocity of our particle at time . Given these values, we can compute (again we dropped terms of O( 2 )) dx x(2 ) = x( ) + ( ) = x( ) + v( ), dt dv v(2 ) = v( ) + ( ) = v( ) + a( ) = v( ) + g. dt Thus, we now have the values of the position and velocity of our particle at time 2 . Given these values, we can compute the position and velocity of our particle at time 3 , and after that at 4 , and on and on, until we eventually get to the point where the particle has come to rest. Knowing the position of the particle at this point allows us to answer our initial question. What an ugly solution! We just want to know how high the particle goes - one number. To get this we have computed the position and velocity of the ball at all points on the balls trajectory. If we want to be accurate, we need to take to be small, so that we may be computing thousands and thousands of numbers, and only use one of them. Surely we can do better than this? It is the work energy approach that will provide a better solution. The work energy approach integrates Newton’s second law from the initial point on the particles trajectory to the ﬁnal point. Since the result of the integral only depends on the endpoints of the particles trajectory, we expect to get an equation which will directly relate the initial and ﬁnal velocities and positions of our particle. Start from Newtons second law d2 x F =m 2. dt We are going to integrate this equation of motion along the trajectory of the particle. We deﬁne the inﬁnitesimal displacement dx to be a small segment along the trajectory of the particle. It has the same direction as the particles velocity; it has an inﬁnitesimal magnitude dx. Integrating we obtain xf xf tf d2 x d2 x dx F · dx = m · dx = m · dt. xi xi dt2 ti dt2 dt xf We call xi F · dx the work done on the particle. If you take a look at the integrand of the term on the right hand side of this last equation, it is of the form f df = dt 1 d 2 2 dt f . This term is thus easily integrated xf m m F · dx = v · v(tf ) − v · v(ti ). xi 2 2 We see that the right hand side is nothing but the change in kinetic energy of our particle. What is the left hand side? To get some idea, lets plug in the known expression for our force and explicitly evaluate the integral. The result is xf xf F · dx = mg · dx = mg · (xi − xf ). xi xi Lets imagine that the ball starts at xi = (0, 0, hi ) and lands up at xf = (0, 0, hf ). Since the acceleration due to gravity is g = (0, 0, g) we have mg · (xi − xf ) = mg(hi − hf ). This looks like the initial gravitational potential energy minus the ﬁnal gravitational po- tential energy. Thus, we obtain xf m m F · dx = mg(hi − hf ) = v · v(tf ) − v · v(ti ). xi 2 2 Since v(tf ) = 0 we have m mghf = mghi + v · v(ti ). 2 We easily solve this equation for hf - this answers our initial question. This time we have obtained our answer by computing a single number - this is a much better approach to the problem. From this example we can extract the following general result xf m m F · dx = v · v(tf ) − v · v(ti ) = ∆K, xi 2 2 with ∆K the change in kinetic energy of the particle. If in addition we deal with a conservative force we also have xf F · dx = U (xi ) − U (xf ) = −∆U, xi with U (x) the potential energy and ∆U the change in potential energy. Since it is only ∆U that appears in this formula, it is only changes in potential energy that are deﬁned. This is not surprising because the potential is only deﬁned up to a constant. We call the sum of the kinetic energy and the potential energy the mechanical energy of the particle. Notice that when only conservative forces act, the mechanical energy is conserved ∆K + ∆U = 0. Did you get it? The most important result from this section is the relation between potential energy and work done, and it is what we will use in later sections. It is also good if you have some idea as to why the energy and work concepts can be more useful than just using Newton’s laws directly. 18. Potential Energy: Electrostatics What are we going to do in this section? In this section we will obtain a formula for the electrical potential energy. In this section we would like to obtain an expression for the electrical potential energy. The Coulomb force is a conservative force and so we will be able to use the results of the last section. Imagine that we have a pair of point charges, with charge q1 and q2 respectively, separated by a distance r as shown below. We imagine that the charge q1 is ﬁxed at the origin. The charge q2 is free to move. Since the force between the two charges is repulsive, q2 will move oﬀ to inﬁnity. It starts from rest so that it will pick up kinetic energy. Consequently it will lose potential energy. As we have already mentioned, the potential energy is only deﬁned up to a constant. To deﬁne it we set U (r = ∞) = 0. y q1 q2 r x Fig. 16: Two point charges separated by a distance r. The integral we need to compute is ∞ 1 q1 q2 1 q1 q2 ∞ 1 q1 q2 F · dx = dx 2 =− = . r 4π 0 x 4π 0 x r 4π 0 r From the last section we know that ∞ 1 q1 q2 F · dx = U (r) − U (∞) = U (r) = . r 4π 0 r This formula for the potential energy between two charges is the main result of this section. Lets explore this result a little. To get some insight into what potential energy means, note that from the formula xf F · dx = U (xi ) − U (xf ), xi it follows that ∂U ∂U ∂U Fx = − , Fy = − , Fz = − . ∂x ∂y ∂z This formula has a simple geometrical interpretation - the particle will feel a force that pushes it into a region of lower potential energy. As an example, lets consider the potential energy for a pair of point charges which both have positive charge. We know that these particles will repel each other and thus, the particles will move in such a way that their separation increases. The plot of potential energy is shown below. Fig. 17: Potential energy of two positive charges plotted as a function of the separation between the two charges. If the force acting is to decrease the potential energy, we see immediately from the above plot that it must act in such a way that r is increased - i.e. we see that the two point charges do indeed repel each other! We call this a repulsive potential. For the case that q1 and q2 have opposite charges we can again compute the potential energy. In this case, the potential energy is shown below. Fig. 18: Potential energy of two opposite charges plotted as a function of the separation between the two charges. We see that the potential energy is decreased if r decreases. Thus the force that acts will try to decrease the separation between our point charges. This is again exactly what we expect - opposite charges attract. We call this an attractive potential. Having studied the potential energy for two point charges, it is natural to ask what the potential energy for a collection of point charges is. The answer is simple - it is the sum of contributions for each pair of particles. For example, for the three particle system shown below the potential energy is 1 q1 q2 1 q1 q3 1 q2 q3 U= + + . 4π 0 r12 4π 0 r13 4π 0 r23 This formula for the potential energy of a collection of particles is not a new fact - it has the same content as the superposition principle. To demonstrate this point, consider the q3 r13 r23 q1 q2 r12 Fig. 19: Three charged particles. q1 q2 q3 x1 x2 x3 Fig. 20: The potential of these three charged particles is given as a sum over contributions for each pair of particles. case when all three particles lie on the x-axis. The particles have charges q1 , q2 and q3 , and positions x1 , x2 and x3 . Assume for simplicity that all three charges are positive. Using the superposition principle, the force that particle three experiences is given by 1 q1 q3 1 q2 q3 F = 2 + . 4π 0 (x1 − x3 ) 4π 0 (x2 − x3 )2 The potential energy for this collection of particles is 1 q1 q2 1 q1 q3 1 q2 q3 U= + + . 4π 0 x2 − x1 4π 0 x3 − x1 4π 0 x3 − x2 It is easy to check that ∂U F =− , ∂x3 so that the formula for the electric ﬁeld and the formula for the potential energy have the same content. Did you get it? You should be able to write down the electrical potential energy for a pair of point charges and for a collection of point charges. You should also understand the relationship between the potential energy of a particle and the force that acts on the particle. 19. Electric Potential What are we going to do in this section? We have already seen that the electric ﬁeld is a useful concept. The electric ﬁeld is a force per charge. It is natural to study the electrical potential energy per charge, which is the electric potential. We have already seen that it is useful to factorise the force into a charge times an electric ﬁeld F = q E. The potential energy was derived by integrating the force along the particles trajectory. This suggests a natural factorisation of the electrical potential energy into a charge (q) times the electrical potential (φ) U (x) = qφ(x). The electrical potential is measured in J·C−1 =volts. By splitting the force into a product q E we could easily compute the force acting on any charged particle once we had computed the ﬁeld E set up by our source. By splitting our potential energy into a product qφ we can easily compute the potential energy of any charged particle interacting with our source. The formulas relating forces and potential energies generalise to give us formulas relating electric ﬁelds and electric potentials xf E · dx = φ(xi ) − φ(xf ), xi ∂φ ∂φ ∂φ Ex = − , Ey = − , Ez = − . ∂x ∂y ∂z As a simple example, the electrical potential at point B a distance r from a point charge q is given by 1 q φB = 4π 0 r volts. If we place a particle of charge q at B, it has an electrical potential energy of 1 q q U = q φB = 4π 0 r joules. This is the correct result. Potentials add - just like potential energies. Thus, the potential at point C in the ﬁgure is given by 1 q1 1 q2 φC = + . 4π 0 r1 4π 0 r2 q2 r2 C q1 r1 Fig. 21: The potential at point C is a sum of the potential from charge q1 and the potential from charge q2 . To get some idea of how large potentials are, we could compute the potential at a distance 10−10 m from the proton. This will give us an idea of the potential of the electron in a hydrogen atom. The result is −19 1 e φ= 4π 0 r ≈ 1010 1.6×10 10−10 = 16 Volts. This is a reasonable number to work with. The corresponding potential energy of the electron U = −eφ ≈ −1.6 × 10−19 ∗ 16 ≈ −2 × 10−18 Joules is so small that it is a lot less convenient. For this reason, it is often more useful to work with a diﬀerent unit of energy - the electron volt (eV). The electron volt is deﬁned by saying that if an electron moves through a potential diﬀerence of one volt, there is a change in its electrical potential energy of ∆U = (e) × 1 = 1.6 × 10−19 Joules = 1 electron volt. In the remainder of this section we will explore some of the consequences of the equation xf E · dx = φ(xi ) − φ(xf ). xi From this formula, it follows that since the electric ﬁeld inside a conductor at static equi- librium is zero, all points inside the conductor are at the same potential. Consider a pair of uniformly charged inﬁnite plates as shown below. For these plates, the electric ﬁeld within the plates is a constant. Thus, we can write xf xf E · dx = E · dx = E · (xi − xf ) = φ(xi ) − φ(xf ). xi xi + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − Fig. 22: Two inﬁnite plates with opposite charge. The ﬁeld in between the two plates is constant = 0. The ﬁeld outside of the two plates vanishes. As an example, if we are told that the potential diﬀerence (∆φ) between two plates of a capacitor is 6volts and the distance (s) between the plates of the capacitor is 3mm, the electric ﬁeld between the plates is ∆φ 6 E= s = 3×10−3 = 2000volts per meter. We mentioned previously that the zero of potential energy is arbitrary. The zero of the electric potential is also arbitrary. To illustrate this point, consider the two capacitors shown below. 35V 135V 75000V 75100V + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − + + − − 0.2mm 0.2mm Fig. 23: Two capacitors with exactly the same electric ﬁeld between their plates. These two capacitors have the same potential diﬀerence between their plates. Conse- quently, they both have an electric ﬁeld of magnitude ∆φ 100 E= s = 0.2×10−3 = 5 × 105 V·m−1 , between their plates. Clearly, what counts in computing the electric ﬁeld is the potential diﬀerence and not the value of the potential. You know a similar fact from your experience with gravity: it is no harder to walk up from the 75th ﬂoor to the 76th ﬂoor than it is to walk up from the ﬁrst ﬂoor to the second ﬂoor. Did you get it? If you can calculate the electrical potential for a point charge and a collection of point charges, and you understand the relationship between electrical potential and electric ﬁeld, then you got the point of this section. 20. Currents What are we going to do in this section? Up to now we have studied static charges and we have seen that they produce electric ﬁelds. We will now start to study moving charges, that is, currents. The reason we do this is because currents set up a magnetic ﬁeld. We now want to consider moving charges. Consider an electrical wire (which is a conductor) of cross sectional area A. The charges moving in the wire are moving with a speed v. 111 000 + v 000 111 111 000 000 111 + v + v 111 000 111 000 + v + v 111 000 000 111 Cross sectional area A Fig. 24: Charges moving in a wire of cross sectional area A. The charges move in the direction shown with a speed v. When we have moving charges, the quantity we are interested in computing is the amount of dQ charge dQ ﬂowing through the wire in a time dt. We call the ratio dt the current. Current is measured in coulombs per second. We also call a coulomb per second an ampere. In a time dt all of the charges in a volume Avdt will pass a given point on the wire. Thus, if the charges that are moving are positive with charge q and have a density of n+ per m3 , the current is given by dQ n+ vdtAq = = n+ vAq. dt dt A good way to picture what is happening is to think of the electrons moving through wire as water moving in a pipe. Calculating the current is the same as calculating how much water passes through the pipe in a given time. There is an important point to note here. We have described things as if the mobile charge carriers are positively charged. Of course, the mobile charge carriers are electrons which are negatively charged. Consequently, the charged particles (electrons) are actually ﬂowing opposite to the direction shown for the current ﬂow. A simple situation that involves currents, involves connecting a bulb to a battery. + − Fig. 25: A simple circuit involving a bulb. When the battery is connected across the light bulb, it sets up a potential diﬀerence across the light bulb. This implies that there will be a non-zero electric ﬁeld both in the wires connected to the bulb and in the bulb ﬁlament itself. This electric ﬁeld will cause the mobile charge carriers (that is electrons) to accelerate. They lose electrical potential energy but gain kinetic energy. When the electrons move through the bulb’s ﬁlament, they collide with the positive cores (nuclei plus inner electrons). These collisions can be thought of as a “friction” experienced by the electrons. It is this friction that makes the metal get hot and it glows. In this section, we will denote the potential diﬀerence across the battery by V . This is so that we are consistent with most of the literature discussing simple electric circuits. Of course, we have been denoting potentials by φ up to now. So, in this section we will denote potentials by V and not by φ which is the notation we were using. The quantity we would associate to the bulb is a resistance R0 . Then, Ohm’s law tells us that if a current I0 ﬂows through the bulb, the potential diﬀerence V0 across the bulb is given by V0 = I0 R0 . We measure resistance in ohms, usually denoted by the symbol Ω. From Ohm’s law we see that an ohm is a volt per ampere. When we start to consider two (or more) bulbs our circuits become quite interesting. Indeed, for the case of two bulbs we have two diﬀerent ways in which we can make our connections. We say that we can connect the bulbs in series or in parallel. The two possibilities are shown below. + − Fig. 26: Two bulbs connected in parallel. Note that the potential across both bulbs is V0 . + − Fig. 27: Two bulbs connected in series. Note that the potential across both bulbs is V0 . 2 When you built these two diﬀerent circuits, you checked to see how brightly the bulbs glowed. The bulbs connected with the parallel connection glowed more brightly than the bulbs connected in series. Why is this the case? The energy lost by the charge carriers when they traverses the circuit is given by Elost = q∆V with ∆V the potential diﬀerence across the battery and q is the amount of charge that traverses the circuit. We can compute the rate at which this energy is lost, i.e. the power supplied by the charge carriers as dElost dq = ∆V = I∆V. dt dt Power is measured in joules per second = watts. To ﬁgure out which bulb glows brightest, we can compare the power delivered to each bulb. For the case of the bulbs in parallel, each bulb has a potential diﬀerence of V0 across it, so that the current ﬂowing in each bulb is V0 I0 = . R0 Consequently, the power for each bulb is dE V2 = 0. dt R0 This is the same power as we’d have if we just connected a single bulb across the battery. (This is not quite true - the battery also has some internal resistance - this is a small eﬀect which we are neglecting. If we took this into account, each of the bulbs which were connected in parallel would glow slightly less brightly than if we just connected a single bulb across the battery.) Now lets consider the case of bulbs connected in series. In this situation, the total resistance across the battery is R0 + R0 = 2R0 so that the current ﬂowing through the bulbs is V0 I0 = . 2R0 Further, since the potential diﬀerence across both bulbs is V0 , the potential diﬀerence V0 across each bulb must be 2 . Thus, the power delivered to each bulb is dE V0 V0 1 V02 = = . dt 2R0 2 4 R0 Thus, for bulbs connected in series, each bulb will glow with roughly one quarter of the brightness that a single bulb connected across the battery would glow with. Did you get it? You should understand what a current is. You should also understand how current, potential and resistance are related. You should be able to use this knowledge to look at electric circuits and predict how brightly bulbs would glow. 21. Magnetic Fields What are we going to do in this section? In this section we will describe exactly how currents set up magnetic ﬁelds. In class you experimented with a piece of wire and a compass. You were able to determine that when a current ﬂowed in the wire, it deﬂected the compass needle. This implies that when current ﬂowed in the wire, it produced a magnetic ﬁeld. You were also able to determine the direction of the magnetic ﬁeld which was produced. The result for the direction is summarised in the picture below. B B + − Fig. 28: The magnetic ﬁeld set up by a current carrying conductor. The units of the magnetic ﬁeld are tesla (T ). A segment of wire of length dl, carrying a current I sets up a magnetic ﬁeld given by ˆ µ0 dl × r dB = I 2 , 4π r ˆ where dl has magnitude dl and the same direction as the direction of current ﬂow, r points from the short segment dl to the observation point, r is the distance from the short segment µ0 to the observation point and 4π is a constant equal to 10−7 T·m·s·C−1 . This is called the Biot-Savart law for currents. In the above formula, we have used the vector cross product a × b. a × b is the vector with magnitude equal to |a||b| sin θ, and with direction normal to both a and b, given by the right hand rule. Fig. 29: Magnetic ﬁeld of a short thin current carrying wire. Did you get it? If you can compute the direction and magnitude of the magnetic ﬁeld set up by a current carrying wire, you have understood this section. 22. Electric motors: The Lorentz law for currents What are we going to do in this section? In class, you built an electric motor. In this section we will use your experimental results to write down a force law for a current carrying wire in a background magnetic ﬁeld. In class, you built an electric motor. Recall that you took a piece of copper wite and made a loop with many turns as shown below. You then passed a current through your loop, producing a magnetic ﬁeld running through the loop. 1 0 I 1 0 1 0 1 0 1 0 1 0 I Fig. 30: The B ﬁeld in the loop shown is out of the page. Your motor turned because the loop felt a force which tried to align the ﬁeld produced by your loop with the ﬁeld of your magnet. The force law which will accomplish this is dF = Idl × B, where dF is the force felt be a small segment dl of the loop, which is carrying a current I in the presence of the background magnetic ﬁeld B. This is a lovely and simple formula - there are no awkward factors that you need to remember. The reason for this is simply that this is the ﬁrst equation that people wrote down - it was this equation that was used to deﬁne the tesla. 23. Biot-Savart Law and Lorentz Law for Charges What are we going to do in this section? In this section, we will obtain the Biot-Savart Law and Lorentz Law for point charges, by manipulating the corresponding laws for currents. Recall that the current I is given by I = n+ Avq. In this formula n+ is the density of charge carriers (it is a number per unit volume), A is the cross sectional area of the wire, v is the speed of the charge carriers and q is the charge of the charge carriers. What is the magnetic ﬁeld from a single charge carrier? Start from the Biot-Savart law for currents ˆ µ0 dl × r µ0 ˆ dl × r dB = I 2 = n+ Aqv 2 . 4π r 4π r This is the magnetic ﬁeld from a collection of charge carriers. To get the magnetic ﬁeld from a single charge (we’ll denote this by B), we need to divide dB by the number of charge carriers (=n+ A|dl|). Thus, µ0 n+ Aqv dl × rˆ ˆ µ0 dl × r B= 2 = vq 2 , 4π n+ A|dl| r 4π r dl where dl = dl is a unit vector. This unit vector points in the direction of the current, i.e. in the direction of the velocity of the charges so that the velocity v is given by v = dlv. This is the result we wanted: a point charge of charge q, moving with a velocity v sets up a magnetic ﬁeld B given by µ0 v × rˆ B= q 2 , 4π r ˆ where r is the distance from the point charge to the observation point and r points from the charge to the observation point. From your study on motors, you know that the Lorentz law for currents says a segment of wire dl which is carrying a current I in the presence of a background magnetic ﬁeld B, will experience a force dF = Idl × B. Following exactly the same procedure as we did above, we can also turn this law for currents into a law for point charges. The result is F = qv × B, which is called the Lorentz force law for point charges. You should make sure that you can derive this result. This result is quite unusual. You are used to forces depending on positions, not on velocities. You are also used to being able to derive the force as the gradient of a potential - the Lorentz force can not be obtained as the gradient of some potential. Notice also that the force acts perpendicular to both the external magnetic ﬁeld and the velocity of the particle. This has many observable consequences. One of these consequences is that particles moving in a magnetic ﬁeld move in circles. B Fig. 31: A charged particle moving in a magnetic ﬁeld will move in a circle. Convince yourself that the force is always directed towards of the centre of the circle. Does this make sense? 24. Newton is Dead! What are we going to do in this section? In this section, we will show that when magnetic ﬁelds are present, action is not always equal to reaction. Thus, Newton’s third law is not correct. It is however a good approximation. In this section we will consider the interaction between the two point charges shown below. We will be focusing on the magnetic interaction between the two - you already know how to treat the electrical interactions. First, lets compute the magnetic ﬁeld due to particle 1 at the location of particle 2. We are not really interested in magnitudes - just ˆ ˆ directions. We know that the magnetic ﬁeld that 1 creates at 2 is ∝ v1 × r12 where r12 is the vector stretching from particle 1 to particle 2. Using the right hand rule convince yourself that this produces a magnetic ﬁeld going into the page at particle 2’s location. 00 11 00 11 00 11 q1 v1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 v2 0 1 0 1 0 1 q2 Fig. 32: The two point charges shown will interact both electrically and magnetically. What force does particle 2 feel as a result of this magnetic ﬁeld? From the Lorentz force law, we know that the force ∝ v × B. Convince yourself (again using the right hand rule) that this force acts to the left of the page. This is the force exerted on particle 2 by particle 1. If we use Newton’s second law, we would predict that particle 2 would exert a force on particle 1 which would have the same magnitude as the force exerted by 1 on 2, but this force would act to the right of the page. This can’t possibly be true. Indeed, if we compute the magnetic ﬁeld set up by particle 2 at the location of particle 1, we need to compute ˆ ˆ the cross product v2 × r21 where r21 stretches from particle 2 to particle 1. Clearly, v2 and ˆ ˆ r21 are parallel so that v2 × r21 = 0. Thus, particle 2 does not set up a magnetic ﬁeld at the location of particle 1 and consequently, particle 2 does not exert a force on particle 1! Newton’s third law is not valid! This is worrying - one of the consequences of Newton’s third law is the conservation of momentum. You may worry that if Newtons third law no longer holds, perhaps momentum is no longer conserved. In fact, it turns out that momentum is still conserved, but the analysis is subtle. You need to take into account the fact that some momentum is carried away by the ﬁeld itself! Another thing which may be worrying you, is that if Newton’s third law is wrong, how come we thought it was true for so long? The only reasonable explanation is that the terms which spoil Newton’s third law (the magnetic forces) must be much smaller than the terms which do satisfy the third law. This is indeed true. The magnitude of the electric forces (which do satisfy Newton’s third law) divided by the magnetic forces (which don’t) is q2 v2 × B µ0 ˆ q2 q1 4πr2 v2 × (v1 × r) |v1 ||v2 | = = |v1 ||v2 |µ0 0 = . 1 q1 q2 1 q1 q2 9 × 1016 4π 0 r2 4π 0 r2 Thus, the magnetic forces are much much smaller (in magnitude) than the electric forces. You might notice that 9 × 1016 is equal to the value of the speed of light squared - this is not a coincidence as we will see later. 25. Maxwell’s Equations What are we going to do in this section? In this section we will explain the set of equations ﬁrst put together by the Scottish physicist James Clerk Maxwell in the nineteenth century. Maxwell’s equations are the basis for our understanding of all electromagnetic phenomena. They describe space- and time-dependent electric and magnetic ﬁelds, and how they are generated by charges and currents. Maxwell’s equations also describe light and other forms of electromagnetic radiation. Historically, they were the foundation for Einstein’s theory of relativity. 25.1. The ﬁrst two equations: Gauss’s law for electric and magnetic ﬁelds The ﬁrst of Maxwell’s equations is one we have seen before, Gauss’s Law for E: q E · dS = , S,closed 0 where q is the charge contained within the closed surface S. As a simple example, we calculated the electric ﬁeld due to a point charge. The electric ﬁeld is radial, and so the integral taken on a sphere of radius r centred on the charge is just |E|.4πr2 . Equating this to the right hand side, one ﬁnds the usual Coulomb law. As a second example, we used Gauss’s law to calculate the electric ﬁeld of a charged sheet: it was given by |E| = σ/(2 0 ), where σ was the charge per unit area. (You should review that calculation now, if you have forgotten it). The second of Maxwell’s equations is Gauss’s Law for B: B · dS = 0. S,closed What this equation says is that the net magnetic charge inside a closed surface S is always zero. There are no magnetic ’point charges’ in nature. Instead, there are only magnetic dipoles, made of a positive magnetic charge and a negative magnetic charge tied together so they can never be separated. One example is a magnet, which always has a North pole and a South pole. Even if you break a magnet in two, each piece will again have a North pole and a South pole. The earth itself is a giant magnet, and its ﬁeld is illustrated in the diagram. The ﬁeld is used by migrating birds moving from one location to another when the seasons change. The fact that we know the earth’s magnetic ﬁeld allows us to determine the strength of the magnetic ﬁeld from other sources, by comparing with the earth’s magnetic ﬁeld. For example, a bar magnet behaves like a magnetic dipole. By analogy with an electric dipole, we expect the magnetic ﬁeld strength, at distances large compared to the size of the magnet, to be proportional to 1/r3 where r is the distance from the magnet. In class, we found using a compass that the magnetic ﬁeld from a bar magnet became stronger than the magnetic ﬁeld from the earth at a distance of about 20 cm from the bar magnet. Therefore the magnetic ﬁeld of the magnet has a strength 2 × 10−5 Tesla at a distance of 20 cm. Scaling the ﬁeld with r−3 , we can ﬁnd the ﬁeld at about 1 cm from the bar magnet, the location of the nearest side of the coil in our electric motor. The magnetic ﬁeld strength in this location must be (203 ) × 2 × 10−5 or approximately 0.16 Tesla. S N Fig. 33: Earth’s magnetic ﬁeld. Note that the earth’s magnetic South pole is at the geographical north pole: it attracts the North pole of any magnet. The strength of the earth’s magnetic ﬁeld is about 2 × 10−5 Tesla. In Cape Town, it points towards the North, slightly upwards (because we are near the geographical south pole). 25.2. The third Maxwell equation: time-varying magnetic ﬁelds The third of Maxwell’s equations is Faraday’s Law, which tells you how a changing magnetic ﬁeld causes an electric ﬁeld. We won’t derive this law rigorously, but will just give an argument for how it works in a particular situation. Imagine you have a uniform magnetic ﬁeld, caused for example by a large ﬂat magnet, as shown in the diagram. v B Fig. 34: A wire is moved across a uniform magnetic ﬁeld, like cutting grass. As the wire segment moves across the magnetic ﬁeld, cutting across the ﬁeld lines, the charge carriers in the wire feel a Lorentz force qv × B, where q is their charge and v is the velocity of the wire across the ﬁeld. Because of this force, we will get a current I ﬂowing into the paper. (Check that you understand this direction). What we argue next is that we should see the same eﬀect, i.e. a current ﬂowing along the wire in the direction pointing into the paper, when we hold the wire ﬁxed, but move the magnetic ﬁeld across it, for example by moving a magnet causing the ﬁeld: v B Fig. 35: A magnetic ﬁeld is moved across a wire. In this situation, the charge carriers in the wire are not moving initially, as they encounter the magnetic ﬁeld: only the magnetic ﬁeld is moving. Therefore it cannot be the magnetic Lorentz force which is causing the current: there must be an electric ﬁeld, pointing along the wire and into the paper, produced by the changing magnetic ﬁeld. From equating the magnitude of the Lorentz force found in the ﬁrst picture above, with the wire moving, |F | = q|v||B|, to the electric force we infer must be present in the second picture, |F | = q|E|, we see that the motion of a magnetic ﬁeld induces an electric ﬁeld of strength |v||B|. This is an example of Faraday’s law. The general form of Faraday’s law is: d E · dl = − B · dS , C dt S,open where the closed curve C bounds the open surface S, as shown below. C dS S dl Fig. 36: A closed curve C bounds a surface S. The normal direction to the surface is conventionally deﬁned as shown, using the right hand rule. In our example, where the B ﬁeld is moving across the wire, the left hand side of Faraday’s law is |E|l where l is the length of the section of wire cutting across the magnetic ﬁeld, and the right hand side is the rate of change of the magnetic ﬂux passing inside the wire, which is |B| times |v|l. Equating the two, we ﬁnd the result |E| = |B||v|, just the result we argued for, above. We can actually do an experiment to test Faraday’s law using our electric motor coil and paper clips, attached to a magnet but with no battery. We attach a voltmeter to the two paper clips, and ﬂick the coil to make it spin. The wire in each turn of the coil moves rapidly across the face of the magnet. It moves approximately 2 cm in 0.2 of a second ( about 5 revolutions per second), so it is moving at a speed v of about 0.1 meters per second across the magnetic ﬁeld. The length l of wire close to the magnet is about 1 cm. and the strength of the magnetic ﬁeld in this location is about 0.1 Tesla. Therefore the voltage induced across each turn of the wire coil is about l|E| (remember: voltage is potential diﬀerence, which is the integral of the distance). The voltage across each turn of the coil adds up, so we get a factor of N being the number of turns in the coil. We ﬁnd φ = lBvN ≈ 0.01 × 0.1 × 0.1 × 10 = 10−3 Volts, about a millivolt, which is roughly what we found with the voltmeter. The experiment showed us how a coil of wire spinning in a magnetic ﬁeld can be used to generate a voltage. Electric dynamos work this way, in cars (where they re-charge the battery), in electric power stations, in wind energy machines etc. 25.3. The fourth Maxwell equation: time-varying electric ﬁelds Faraday’s law describes time-varying magnetic ﬁelds, and how they produce electric ﬁelds. Maxwell’s fourth equation describes how time-varying electric ﬁelds produce mag- netic ﬁelds. Maxwell’s fourth equation was based on a law previously discovered by Ampere, which was believed to be correct until Maxwell discovered it was incomplete. When Maxwell ﬁxed the problem, he found that the new set of equations described light, as we shall see later. According to Ampere’s law, a wire carrying a current I generates a magnetic ﬁeld, according to the equation B · dl = µ0 I. C The setup is shown in the ﬁgure below: I C Fig. 37: A current I passes through a closed curve C. Ampere’s law states that the integral of the magnetic ﬁeld around the contour C equals µ0 I. An example of the use of this equation is to calculate the magnetic ﬁeld due to a straight wire, carrying a current I. The magnetic ﬁeld goes around the wire, in the direction of the ﬁngers of your right hand if your right thumb is pointing in the direction of the current. If we take the contour C to be a circle of radius r centred on the wire, then the left hand side of Ampere’s law is 2πr|B|. So Ampere’s law tells us that the strength of the magnetic ﬁeld at a distance r from a wire carrying a current I is: |B| = µ0 I/(2πr). Maxwell made a very clever observation about this example. He argued that if we inserted a pair of parallel plates (called a ‘capacitor’) into the wire, as shown in the ﬁgure below, then on a contour C taken around the wire at a large distance from it, one should not notice the presence of the capacitor. Therefore the same result for the magnetic ﬁeld should be obtained for this case. +− +− I +− +− I +− +− +− +− +− +− C Fig. 38: A pair of parallel plates is inserted into the wire carrying a current I. The magnetic ﬁeld is still integrated along the contour C. Now, Maxwell realised there was a problem with Ampere’s law. Current is really the amount of charge passing through a certain area per unit time. In Ampere’s law, this area must be the area S bounded by the contour C. So far, so good. However, when you insert the parallel plates, there are two ways of drawing the surface S. In the ﬁrst, we take the surface S to cut across the wire, for example on the right hand side of the parallel plates. S +− +− I +− +− I +− +− +− +− +− +− C Fig. 39: The surface S is drawn passing through the wire. Ampere’s law gives the correct magnetic ﬁeld for this case. In the second case, we take the surface S to pass between the plates of the capacitor and hence avoid the wire completely. S +− +− I +− +− I +− +− +− +− +− +− C Fig. 40: The surface S passes between the plates of the capacitor, so that no current ﬂows through it. For this second case, the current passing through S is zero, because the electric charge is simply building up on the two plates. So, the right hand side of Ampere’s law would give zero, which is the wrong result. Maxwell realised how to ﬁx Ampere’s law. The point is that the electric ﬁeld inside the two plates is directed across the gap between them, and its magnitude is given by |E| = Q/( 0 A) where A is the area of the plates and Q is the magnitude of the charge on each of them. The current I is the amount of charge passing along the wire, per unit time, so I = dQ/dt. Using the relation between Q and E, we can write this as I= 0 A(d|E|/dt). Maxwell realised that we can add a term involving d|E|/dt to the right hand side of Ampere’s law. If our surface S passes outside the parallel plates, the electric ﬁeld is zero there, so this new term is zero and the equation gives the correct magnetic ﬁeld. If our surface S passes between the plates, I will be zero but our new term will be nonzero. If we choose the coeﬃcient of the new term correctly, it will exactly compensate for the missing current I. By analogy with Faraday’s law, it is natural to add a term involving the integral of dE/dt through a surface S: d B · dl = µ0 I + κ E · dS , C dt S,open d with some constant κ we have to determine. In our example, dt S,open E · dS is A(d|E|/dt), which we want to be equal to I/ 0 . Therefore if we choose the constant κ to equal µ0 0 , the second term in our new, corrected Ampere’s law will give µ0 I, even though there is no current passing through the surface S. In eﬀect, the time-varying electric ﬁeld across S is compensating for the absence of the current. The corrected equation is known as the Ampere-Maxwell law: d B · dl = µ0 I + µ0 0 E · dS . C dt S,open This derivation has been long and subtle, but the result is well worth it because the last term involving the time derivative of the electric ﬂux through S is crucial to understanding light. 25.4. Summary: Maxwell’s equations Maxwell’s equations may now be written down together: 1. Gauss’s Law for E: q E · dS = . S,closed 0 2. Gauss’s Law for B: B · dS = 0. S,closed 3. Faraday’s Law for time-varying magnetic ﬁelds: d E · dl = − B · dS . C dt S,open 4. Ampere-Maxwell Law for time-varying electric ﬁelds: d B · dl = µ0 I + µ0 0 E · dS . C dt S,open These four equations are very, very powerful. They are one of the high points of theoretical physics. This is a good moment to summarise what we have learned so far in this course. We started from electric charges, and learned how they produce electric ﬁelds. Then we turned to electric currents, and how they produce magnetic ﬁelds. Then we started to connect the two - arguing that moving (or time-varying) magnetic ﬁelds should produce electric ﬁelds (Faraday’s law), and ﬁnally we have just argued that time-varying electric ﬁelds should produce magnetic ﬁelds (Ampere-Maxwell law). What is the next step? Having built up the great ediﬁce of Maxwell’s equations, we now apply it to a new and simpler situation. Imagine there are no electric charges, and no currents, at all. Maxwell’s ﬁrst and second equations tell us that the total electric and magnetic ﬂux through any surface S must be zero, but they don’t force the electric and magnetic ﬁelds to be zero. Faraday’s law associates an electric ﬁeld to a time-varying magnetic ﬁeld, and the Ampere-Maxwell law associates a magnetic ﬁeld to a time-varying electric ﬁeld. Thus we are tempted to imagine a situation where the electric and magnetic ﬁelds sustain each other, even though there are no charges and currents. As we shall see in the next lecture, that is how light works. 26. The Diﬀerential form of Maxwell’s Equations What are we going to do in this section? In this section, we will obtain the diﬀerential form of Maxwell’s equations. In many textbooks Maxwell’s equations are written as diﬀerential equations. There are two results we want to use in this section. The ﬁrst result is the divergence theorem. The divergence theorem states that E · dS = · EdV. S V In the above formula, V is a volume bounded by the closed surface S. Thus, the right hand side of the above equation involves a volume integral; the left hand side involves a surface integral. The operator ∂ ∂ ∂ = , , , ∂x ∂y ∂z is manipulated as if it is a vector. Thus, for example ∂Ex ∂Ey ∂Ez ·E = + + . ∂x ∂y ∂z The second result we want to use is Stokes theorem, which states that E · dl = ( × E) · dS. C S On the right hand side of this equation we are integrating over the surface S which is bounded by the contour C; on the left hand side we have a line integral over C. We now turn to Maxwell’s equations. We can write (S is a closed surface) q E · dS = , S 0 as (use the divergence theorem) q · EdV = . V 0 Further, we can write the charge q as the integral of the charge density (ρ) over the volume ρ · EdV = dV. V V 0 This equation has to be true when integrated over any volume V. This can only be true if the integrands on the two side of this equation are equal. Thus, ρ ·E = . 0 This is the diﬀerential form of the ﬁrst of Maxwell’s equations. Following an exactly analogous procedure, you should show that the second of Maxwell’s equations become · B = 0. Now, consider the third of Maxwell’s equations d E · dl = − B · dS . C dt S Recall that the curve C is the boundary of the open surface S. Using Stokes theorem to rewrite the left hand side, we ﬁnd d ( × E) · dS = − B · dS . S dt S Once again, because the above equation is true if we integrate over any surface S, we can write this as the diﬀerential equation ∂B ×E =− . ∂t This is the diﬀerential form of Maxwell’s third equation. Following exactly the same procedure, you should be able to convince yourself that the fourth of Maxwell’s equations can be written as ∂E × B = µ0 J + µ0 0 . ∂t To get this you will need to deﬁne the density of current ﬂowing through the surface S deﬁned by I= J · dS. S What are the units of the current density J? Did you get it? If you can pass from the integral versions of Maxwell’s equations to the diﬀerential versions of Maxwell’s equations, you got the point of this section. 27. Let There be Light! What are we going to do in this section? In this section we would like to see if Maxwell’s equations have any solutions when q = 0 = I. We will ﬁnd a beautiful result - namely that light itself is indeed a solution of Maxwell’s equations in this case. In this section we are going to study Maxwell’s equations in the case that there are no charges and no currents to set up the electric and magnetic ﬁelds. We are interested in seeing if there are any non-zero solutions to Maxwell’s equations in this case. When q = 0 = I Maxwell’s equations become (in the next equation S is any closed surface) E · dS = 0 = B · dS, S S and (S is any open surface whose boundary is C) d d E · dl = − B · dS , B · dl = +µ0 0 E · dS . C dt S C dt S These are the equations we want to solve. How are we going to solve them? This is a rather non-trivial task. We won’t follow a systematic procedure, but rather, we’ll try to guess a decent solution. We could look for solutions with just a non-zero magnetic ﬁeld, but a zero electric ﬁeld. After some work we’d ﬁnd that there is no solution. We might then look for a solution with a non-zero electric ﬁeld and zero magnetic ﬁeld. With more work we’d ﬁnd there is still no solution. The next simplest thing is to have at least one component of the electric and magnetic ﬁeld to be non-zero.There are a huge number of possibilities - we can choose which components we keep non-zero and also how these non-zero components depend on t, x, y and z. It turns out that a good guess is to take E = (0, Ey (t, x), 0) B = (0, 0, Bz (t, x)). With this guess, the electric ﬁeld is perpendicular to the magnetic ﬁeld. z y B E x Fig. 41: In our guess for the solution of Maxwell’s equations, the electric and magnetic ﬁelds are perpendicular. How do we check if this guess is correct and further, determine exactly how Ey (t, x) and Bz (t, x) depend on x and t? Well, any surface S can be built as the surface of a collection of small cubes. Thus, if we could verify our equations for a single small cube, we’d be done. We will consider the following small cube dz dy (x,y,z) dx Fig. 42: A small cube. The small area elements dS are pointing out of the cube. If the ﬁrst of Maxwell’s equations is true, the integral of E·dS integrated over the surface of the cube vanishes. The y direction is into the page. Since the only non-zero component of E is pointing into the page, E · dS is only non-zero for the front and back faces of the cube. For the front face, dS points out of the page and E points into the page so that E · dS is negative. Since |dS| = dxdz, the contribution from the front face is −Ey (t, x)dxdz. For the back face, dS points into of the page and E points into the page so that E · dS is positive. Since |dS| = dxdz, and since the back and front faces are at the same x value, the contribution from the back face is Ey (t, x)dxdz. Thus, E · dS = −Ey (t, x)dxdz + Ey (t, x)dxdz = 0. Thus, Maxwell’s ﬁrst equation is satisﬁed. If the second of Maxwell’s equations is true, the integral of B · dS integrated over the surface of the cube vanishes. Since the only non-zero component of B is pointing in the z direction, B · dS is only non-zero for the top and bottom faces of the cube. For the top face, dS points in the positive z direction i.e. parallel to B so that B · dS is positive. Since |dS| = dxdy, the contribution from the top face is Bz (t, x)dxdy. For the bottom face, dS points in the negative z direction, so that B · dS is negative. Since |dS| = dxdy, and since the top and bottom faces are at the same x value, the contribution from the bottom face is −Bz (t, x)dxdy. Thus, B · dS = Bz (t, x)dxdy − Bz (t, x)dxdy = 0. Thus, Maxwell’s second equation is also satisﬁed. Next, consider Maxwell’s third equation. We will integrate the third equation over the top face of the cube. Ey(x,t) 3 4 2 dy 1 dx Ey(x+dx,t) Fig. 43: Shown is the top face of the cube. The face itself is the surface S; the boundary of the face is the curve C. The direction of C is chosen so that when we use the right hand rule the unit normal dS points out of the cube. First, we evaluate E · dl. Since E points in the y direction, the electric ﬁeld is orthogonal to dl for the segment of the path labelled 1 and hence the contribution from this segment of the path vanishes. The direction of the path segment labelled 2 is parallel to the electric ﬁeld so that the contribution from the segment of the path labelled 2 is Ey (x + dx, t)dy. Once again, along the segment of the path labelled 3 dl is orthogonal to the electric ﬁeld so that again the contribution from this segment of the path vanishes. Finally, for the segment of the path labelled 4, the ﬁeld and dl are pointing in opposite directions so that the contribution from this segment of the paths is −Ey (x, t)dy. Thus, ∂Ey (x, t) E · dl = 0 + Ey (x + dx, t)dy + 0 − Ey (x, t)dy ≈ dxdy. ∂x It is easy to obtain B · dS = Bz dxdy. S This follows immediately upon noting that (i) |dS| = dxdy and that (ii) dS is parallel to B. Using these results, Faraday’s law becomes ∂Ey ∂Bz =− . (27.1) ∂x ∂t Finally, consider the last of Maxwell’s equations. This time, we will choose S to be the front face of the cube. Proceeding exactly as we did for Maxwell’s third equation, we obtain ∂Bz (x, t) B · dl = Bz (x + dx, t)dz − Bz (x, t)dz ≈ dxdz, C ∂x and d ∂Ey µ0 0 E · dS = −µ0 0 . dt ∂t Thus, Maxwell’s fourth equation becomes ∂Ey ∂Bz −µ0 0 = . (27.2) ∂t ∂x Taking the derivative of (27.1) with respect to x we obtain ∂ 2 Ey ∂ 2 Bz =− . (27.3) ∂x2 ∂x∂t Taking the derivative of (27.2) with respect to t we obtain ∂ 2 Bz ∂ 2 Ey = −µ0 0 . ∂t∂x ∂t2 Inserting this result into (27.3), we obtain ∂ 2 Ey ∂ 2 Ey = µ0 0 . ∂x2 ∂t2 The product µ0 0 has the dimensions of an inverse speed squared. In fact, √1 = µ0 0 3 × 108 m·s−1 is equal to the speed of light. This is how Maxwell ﬁrst understood that this equation is describing light waves. We will rewrite this equation as 1 ∂2 ∂2 − Ey (t, x) = 0. (27.4) c2 ∂t2 ∂x2 To solve this equation, move into light cone coordinates 1 + 1 + x± = x ± ct, x= (x + x− ), ct = (x − x− ). 2 2 It is now easy to obtain ∂ 1 ∂ 1 ∂ = ± . ∂x± 2 ∂x c ∂t In terms of light cone coordinates, (27.4) becomes ∂ ∂ −4 E = 0. + ∂x− y ∂x This implies that ∂ ˜ E = f (x− ). − y ∂x Integrating this equation with respect to x− and realizing that the arbitrary constant of integration is now any function of x+ , we obtain Ey = f (x− ) + g(x+ ) = f (x − ct) + g(x + ct), as the general solution to (27.4). What is the physical interpretation of these solutions? Well, if f (u) describes a pulse centred at u = 0, then f (x − ct) describes a pulse centred at x = ct. Thus, f (x − ct) corresponds to a pulse moving along the positive x axis. Similarly, g(x + ct) describes a pulse moving along the negative x axis. There is one loose end remaining - we have obtained the electric ﬁeld; how do we obtain the magnetic ﬁeld. This could be obtained by integrating ∂Ey ∂Bz =− . ∂x ∂t with respect to t. For the electric ﬁeld that we have considered above, it is easy to see that 1 1 Bz = f (x − ct) − g(x + ct). c c You would also need to add the arbitrary constant of integration, which could be any function of x. This arbitrary function is ﬁxed by considering our second equation ∂Bz ∂Ey = −µ0 0 . ∂x ∂t If you were given a solution of Maxwell’s equations, there is a nice method which you can use to determine the direction of propagation of the wave - you can compute the Poynting vector E × B, which points in the direction of propagation. Indeed, it is easy to check that for a right moving wave 1 Ey = f (x − ct), Bz = f (x − ct), c the Poynting vector is given by 1 2 E×B = f (x − ct), 0, 0) , c which does indeed point along the positive x-axis. For a left moving wave 1 Ey = g(x + ct), Bz = − g(x + ct), c the Poynting vector is given by 1 E×B = − g 2 (x + ct), 0, 0) , c which now points along the negative x-axis. We will now study a speciﬁc example, that is, we’ll study a particular choice for the functions f and g. The example we will study is Ey (x, t) = A cos (k(x − ct)) . We call A the amplitude of the wave. You should check that 1 ∂2 ∂2 − Ey (x, t) = 0. c2 ∂t2 ∂x2 A sketch of this wave is shown below. Fig. 44: A sketch of a light wave showing the electric and magnetic ﬁelds which are perpendicular. The wave is periodic is space. The length that it takes to complete a full wave is called the wavelength of the wave. The argument of the cosine in our solution is k(x − ct); if we have two points x1 and x2 which are separated by a wavelength λ (i.e. x1 = x2 + λ), then because the value of the wave is the same at x1 and x2 we must have k(x1 − ct) − k(x2 − ct) = 2π = kλ. We can solve this to obtain 2π 2π k= , λ= . λ k The solution is also periodic in time. If you sit at a speciﬁc location x you will see the wave repeats itself after a time T called the period of the wave. Arguing exactly as above, we must have (t2 = t1 + T ) k(x − ct1 ) − k(x − ct2 ) = 2π = kcT. λ Thus, T = c. Rather than talking of the period of the wave, we like to speak about the 1 frequency ν = T of the wave. The frequency is measured in hertz; 1 hertz = 1 cycle per second. We can now write our above equation rather neatly as νλ = c. When we talk about the polarisation of the wave, we are really talking about the direction of the electric ﬁeld of the wave. In our example, we would say that the polarisation of the wave is ˆ since the electric ﬁeld points along the y axis. You already saw some j examples of polarisers. How do polarisers work? A polariser is a collection of vertical conducting wires. If the polarisation is aligned with the wires, when the light passes through the polariser its electric ﬁeld will move the electrons along the wire and the light loses its energy. It thus is not able to pass through the polariser. If the polarisation is across the wires, light can’t move the electrons and it consequently doesn’t lose any energy and just passes straight through the polariser. It turns out that although Maxwell’s equations give us a beautiful description of the classical properties of light, these equations are only an approximation to nature. The energy carried by light of frequency ν is quantised in integer multiples of hν where h is Planck’s constant =6.6 × 10−34 J·s. Maxwell’s equations do not incorporate this quantisa- tion. 28. Energy carried by a light wave What are we going to do in this section? In this section we will compute the amount of energy carried by a light wave as it propagates. We will do this by obtaining a formula for the energy in an electric ﬁeld. We are going to start with a simple question: how much energy is there in an electric ﬁeld? To answer this question, imagine the following situation: we are going to start with two uniformly charged plates of equal area A and opposite charge ±Q. We imagine that the two plates are initially coincident, and that we pull the positive plate (say) a distance d from the negative plate. − + − + pull here − + − + this plate is fixed Fig. 45: We have pulled the positive plate a distance d away from the negative plate. To pull the positive plate, we need to apply a force to overcome the force of attraction from the negative plate. We don’t want to give the positive plate any kinetic energy, so we pull with a force that exactly cancels the force exerted by the negative plate. The electric ﬁeld of the negative plate has a magnitude Q |E| = , 2 0A so that the force exerted on the positive plate by the negative plate is Q2 |F | = Q|E| = . 2 0A Since we exert this force for a distance d, the net energy we expend in moving the plate is Q2 1 E = |F |d = d = 0 E · E × Ad. 2 0A 2 Where did the energy that we expend go to? Well, before we started pulling, the plates were at rest. After we have ﬁnished pulling the plates are at rest and there is an electric ﬁeld between the plates. It is natural to think that the energy we expended is now stored in the electric ﬁeld we have just created. The electric ﬁeld occupies a volume Ad, so that the energy of the electric ﬁeld per unit volume (that is the energy density of the electric ﬁeld) is 1 ρ= 0 E · E. 2 Of course, if the electric ﬁeld has an energy, then so will the magnetic ﬁeld. It turns out that the energy in the magnetic ﬁeld is equal to 1 ρ= B · B. 2µ0 Given these results, we can now compute the energy carried by light. Imagine we have some area A, perpendicular to the direction in which our light wave is travelling. Our light wave has an energy density ρ = ρe + ρm where ρe is the energy density of the electric ﬁeld and ρm is the energy density of the magnetic ﬁeld. Using our result above, you should be able to check that for light ρe = ρm . We would like to compute the amount of energy passing the area A every second. In one second, light can travel c m·s−1 × 1s=3 × 108 m. Thus, all of the light inside a box with dimensions Am2 × c m·s−1 ×1s will pass the screen. This implies that an energy Area A ο | Fig. 46: The light wave has an energy density ρ. The area A is perpendicular to the direction in which the light travels. E = ρAc = (ρe + ρm )Ac = 2ρe Ac = 0E · EAc, passes the screen every second. Thus, the power per unit area from our light wave is P = 0E · Ec. You can also write this power per unit area as 0 1 P = E·E = E · E. µ0 377Ohms You can use this to estimate the size of the electric ﬁeld in a light wave. For example, imagine that a 100W bulb is glowing. At a distance of 1m from the bulb, the power per unit area is 100W P = , 4π(1m)2 so that 100W |E| = 377Ohms ≈ 60V · m−1 . 4π(1m)2 29. Relativity What are we going to do in this section? In this section we will study the symmetries of Maxwell’s equations and show that they are not the same as the symmetries of Newton’s equations. Special relativity is a natural consequence of Maxwell’s equations. The basic idea here is to explore the group of symmetries of an equation. As a concrete example, start with Newton’s equations. Newton described nature in terms of particles. Each particle is labelled by an integer (say i) and has a position xi . The motion of these particles is summarised in Newton’s second law, which states mi ai = Fij (xi − xj ), j,j=i where Fij is the force exerted exerted on particle i by particle j. Notice that the force depends only on the diﬀerence xi − xj . 00 11 00 11 00 11 00 11 00 11 xi 00 11 00 11 00 11 00 11 i 00 11 00 11 00 11 00 11 00 11 00 11 00 00 11 11 00 11 00 11 00 11 00 11 Fig. 47: A collection of particles is shown. Each particle has a position. We have shown the position and label for the ith particle. In formulating his laws, Newton was inﬂuenced by Galileo. Galileo spent a lot of time thinking about the following problem: imagine there is a ship at sea, travelling with some speed v. Imagine further that there are some ﬂies trapped inside the ship. The question which fascinated Galileo is this: Can we tell the speed v with which the ship is moving by studying the ﬂies? Galileo suggested that, as long as the ship was moving with a constant speed, it would not be possible to tell if the ship was moving or not by just studying the ﬂies. So, if you only study the ﬂies, you could not determine v. flies a ship 1111 11111111 11111111111111111 00000000000000000 00000000 0000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 00000000000000000 00000000 0000 1111 11111111 11111111111111111 00000000000000000 00000000 0000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 00000000000000000 00000000 0000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 00000000000000000 00000000 0000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11111111 11111111111111111 0000 00000000000000000 00000000 1111 11 11111111 0000 00 00000000 00 11 11111111111111111 00000000000000000 00 11 00 11 00 11 00 11 00 11 00 11 00 11 the deep blue sea Fig. 48: Galileo asked if it is possible to tell the speed of the ship by studying the ﬂies in the ship. Newton has this built this into his equations. Since you can’t tell that you are moving, Newton’s laws on the ship must be the same as Newton’s laws for someone who is not on the ship and hence not moving. The person who is not moving assigns a coordinate xi to the ith particle; the person who is on the ship and hence is moving assigns a coordinate xi to the particle. The relationship between the two coordinates is xi (t) = xi (t) + vt, where v is the (constant but otherwise arbitrary) velocity of the ship. We say that this is a symmetry of Newton’s laws because if the xi (t) are solutions of Newton’s laws then so are the xi (t). To see this, note that d2 d2 d2 xi = 2 (xi (t) + vt) = 2 xi (t), dt2 dt dt and xi − xj = xi − xj , i.e. Fij (xi − xj ) = Fij (xi − xj ). You can summarise this by saying something like “all motion is relative.” We call the transformations between the two coordinate system S (that is the coordinates x, t of the person who is not moving) and S (that is the coordinates x , t of the person who is moving) Galilean transformations. In fact, it turns out that Galileo was wrong - the Galilean transformations are not quite a symmetry of nature. To see this, lets return to Maxwell’s equations. If we have one solution of Maxwell’s equations and we perform a Galilean transformation, do we get another solution of Maxwell’s equations? Recall that when we had no charges and no currents, the solution of Maxwell’s equations we found is given by Ey (x, t) = f (x − ct) + g(x + ct). Consider the following Galilean transformation t = t, x = x + vt, y = y, z = z. This transformation is for a velocity parallel to the x-axis. The inverse transformation is t=t, x = x − vt , y=y, z=z, which you can easily check. In terms of the new coordinates, our solution is Ey (x, t) = f (x − (v + c)t ) + g(x + (c − v)t ). This is not a solution of Maxwell’s equations, so that Maxwell’s equations are not invari- ant under the Galilean transformations. We could now ask what replaces the Galilean transformations - that is, what are the symmetry transformations that take one solution of Maxwell’s equations into another solution of Maxwell’s equations? We want to perform a liner transformation on x± , that is, we want a transformation of the form x+ = ax + + bx − , x− = cx + + dx − . To get another solution of Maxwell’s equations, we need to transform our original solution into a new solution. This means that our function which depends only on x− should, after the transformation, only depend on x − and our function which depends only on x+ should, after the transformation, only depend on x + . For this to happen we need b = 0 and c = 0, i.e. our transformations are x+ = ax + , x− = dx − . (29.1) In fact, this transformation is too general. It includes trivial transformations (called scal- ing) which essentially amount to a change of units. To see this, note that x± are measured in meters. If we change to a new set of coordinates, deﬁned by the fact that we now measure x± in centimetres, we would have x ± = 100x± . This looks like a transformation of the form (29.1) with a = d. We are not interested in including these trivial transformations. To exclude these we will require that dx+ dx− = dx + dx − . Since dx+ dx− has the dimensions of length squared, holding this quantity ﬁxed will indeed make sure that we are keeping our units ﬁxed. This will also ensure that dxdt = dx dt . This extra condition forces a = d−1 . Thus, lets set a = eθ , d = e−θ . This implies that x ± ct = e±θ (x ± ct). This can be rearranged to give x = cosh(θ)x + sinh(θ)ct ct = cosh(θ)ct + sinh(θ)x. This is called a Lorentz transformation, and it is the answer to the question we already asked: We asked what replaces the Galilean transformations; the answer is: the Lorentz transformations. In the next plot, we show space time using both ct, x as coordinates and using x± as coordinates. The dashed lines are lines of constant ct and constant x. The solid lines are lines of constant x . 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 ct 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 1 0 0 1 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0 1 1 0 00000000000000000000 11111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 1 0 0 1 00000000000000000000 11111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0 1 1 0 00000000000000000000 11111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 0 1 00000000000000000000 11111111111111111111 0000000000000000000 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00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 1 0 00000000000000000000 11111111111111111111 x− 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 1 0 1 0 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 1 0 1 0 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 1 0 1 0 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 1 0 1 0 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 1 0 1 0 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 Fig. 49: The coordinate system before the Lorentz transformation. The Lorentz transformations “squash” x+ and “stretch” x− as shown below. ct 0000000000000000000 1111111111111111111 00000000000000000000 11111111111111111111 0000000000000000000 1111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 0000000000000000000 1111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 0000000000000000000 1111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 0000000000000000000 1111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 11111111111111111111 00000000000000000000 00000000000000000000 11111111111111111111 11111111111111111111 0000000000000000000 1111111111111111111 00000000000000000000 11111111111111111111 x+ 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0000000000000000000 1111111111111111111 0000000000000000000 1111111111111111111 Fig. 50: The coordinate system after the Lorentz transformation. If we now consider the action on ct and x, we ﬁnd the ﬁgure below. ct 0000000 ct’ 000000 0 0 1111111 0000000 111111 000000 1111111 0000000 1111111 111111 1 111111 000000 1111111 0000000 111111 000000 1111111 0000000 111111 000000 1 111111 1 000000 0 1111111 0000000 1111111 0000000 111111 000000 1 0 111111 1 000000 0 1111111 0000000 1111111 0000000 111111 000000 1 0 111111 1 000000 0 1111111 0000000 1111111 0000000 111111 000000 1 0 1111111 0000000 1111111 0000000 1 0 1 0 111111 1 000000 0 111111 000000 1111111 0000000 1111111 0000000 111111 000000 1 1 0 111111 000000 000000 0 1111111 0000000 1111111 0000000 111111 0 111111 1 0 000000 1 1111111 0000000 1111111 0000000 111111 1 000000 0 1 0 111111 000000 1111111 0000000 1111111 0000000 111111 1 000000 0 111111 000000 1 0 1111111 0000000 1111111 0000000 111111 000000 111111 000000 1 0 111111 1 000000 0 1111111 0000000 1111111 0000000 111111 000000 1 0 1111111 0000000 1111111 0000000 1 0 1 0 111111 1 000000 0 111111 000000 1111111 0000000 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1111111 0000000 111111 1 000000 0 111111 000000 1 0 0000000000000000000000000000000000 1111111111111111111111111111111111 1111111 0000000 1111111 0000000 111111 1 000000 0 111111 000000 1 0 0000000000000000000000000000000000 1111111111111111111111111111111111 1111111 0000000 1111111 0000000 111111 000000 111111 000000 x’ 111111 1 000000 0 1111111111111111111111111111111111 0000000000000000000000000000000000 1111111 0000000 1111111 0000000 111111 000000 1 0 1111111 0000000 1111111 0000000 111111 1 000000 0 1 0 1 0 111111 000000 1111111 0000000 1111111 0000000 111111 000000 111111 1 0 000000 111111 1 000000 0 1111111 0000000 1111111 00000001 0 111111 000000 1 0 1111111 0000000 1111111 0000000 111111 1 000000 0 111111 000000 1 0 1111111 0000000 1111111 0000000 111111 1 000000 0 111111 000000 1 0 1 0 1111111 0000000 1111111 0000000 111111 000000 111111 000000 111111 1 000000 0 1111111 0000000 1111111 0000000 111111 000000 1 111111 0 1111111 0000000 1111111 0000000 000000 0 1 0 1 0 111111 000000 1 1111111 0000000 1111111 0000000 111111 000000 111111 000000 1 0 111111 1 000000 0 1111111 0000000 1111111 0000000 111111 000000 1 111111 0 1111111 0000000 1111111 0000000 1 0 1 0 000000 0 111111 000000 1 1111111 0000000 1111111 0000000 111111 1 1 0 000000 0 111111 000000 1111111 0000000 1111111 0000000 111111 000000 1 1 0 111111 000000 000000 0 1111111 0000000 111111 0 111111 1 0 000000 111111 000000 1 1 0 1 0 1 0 Fig. 51: The coordinate system after the Lorentz transformation. The solid lines are lines of constant ct and x . To get some insight into the above transformations, we would now like to understand the physical meaning of θ. To do this, consider a ﬂy sleeping on the ship. The ﬂy is sleeping at x = 0. In this case, the Lorentz transformations are x = sinh(θ)ct ct = cosh(θ)ct. Thus, the position of the ﬂy in the S coordinate system is x = c tanh(θ)t . Thus, the velocity of the ﬂy in the S coordinate system in v = c tanh(θ). From this and a little algebra we obtain 1 γ = cosh(θ) = . v2 1− c2 This then gives us the following form of the Lorentz transformations vx ct = γ(ct + ), x = γ(x + vt). c v Notice that if v << c, we have γ ≈ 1 and c ≈ 0. In this case our Lorentz transformations reduce to the Galilean transformations. Since our Lorentz transformations take us from a solution in the S coordinate system, which is an arbitrary function of x+ plus a second arbitrary function of x− to a solution in S which is an arbitrary function of x + plus an arbitrary function of x − , we learn that light travels at the same speed in the S coordinate system as in the S coordinate system. Thus, the speed of light is a constant (independent of whether you are in S or S ) in special relativity. Warning to students: There may (with a probability≈ 1) be errors in these notes. If you spot an error, conﬁrm this with a tutor or send us an email, and we’ll correct the notes. Happy learning and be excellent!