Relativity by erin.natividad

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									PHY2061 Enriched Physics 2 Lecture Notes                                        Relativity 6



                                                                Relativity 6
Disclaimer: These lecture notes are not meant to replace the course textbook. The
content may be incomplete. Some topics may be unclear. These notes are only meant to
be a study aid and a supplement to your own notes. Please report any inaccuracies to the
professor.


Transformation of the Electric Field
        Electric charge is invariant under motion. In other words, electric charge, like
mass, is a property of a particle and is invariant under transformations. It is a Lorentz
Invariant quantity. If this were not so, physics would look different in different reference
frames. The light emitted by hydrogen would have different colors beyond just the usual
Doppler shift. In fact, hydrogen might not even bind if the positive and electric charges
did not transform the same way.

Thus, Gauss’ Law holds in all reference frames:

   S (t )
             E  da       S '( t )
                                       E  da

Two reference frames with a relative velocity between them would agree on the charge
enclosed by a surface, even if the surface integral is computed using the coordinates of S
or S’.

Consider 2 parallel sheets of charge infinite in extent with opposite charge density:
       . By Gauss’ Law in the rest frame of the two sheets, we can solve for the
electric field arising from each sheet. Take as a Gaussian surface a box enclosing just one
sheet or the other, with area A  xz on the top and bottom faces parallel to the sheet:

                        A                                         A
E A  E A                                      E A  E  A 
                         0                                         0
                                                       
E                                               E 
              2 0                                       2 0

You can neglect the field arising from one sheet affecting the flux through the box
enclosing the other sheet because the net flux will be zero through the surface (no charge
enclosed from the other sheet). By the superposition principle, the two fields add, and we
see that outside the two sheets, E = 0. Between the 2 sheets,:
     
E
     0



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PHY2061 Enriched Physics 2 Lecture Notes                                         Relativity 6


Now consider a Lorentz Transformation to a frame where the 2 sheets are moving at a
velocity v with respect to an observer. How do the previously considered Gaussian
surfaces change? First of all, if we do not change the definition of the surface, the charge
enclosed must remain invariant as stated before. However, the geometry of the box has
changed. The dimension parallel to the direction of motion has contracted, so the area of
the top and bottom surfaces has contracted.

In the S’ frame where the sheets are in motion, Gauss’ Law stated in that frame becomes:

              A
             qenc
E A              
        0    0
But since the dimension parallel to the motion is length contracted, the area in the S’
frame is

         A                          1
A                 where  
                               1  v2 / c2
So

     A       A
E       
            0

 E 
               
               0

In other words, the charge density has increased:    

And the electric field perpendicular to the plates (and to the direction of motion) has
increased:

 
E   E

If we had instead considered a Lorentz Transformation in a direction perpendicular to the
plates (i.e. parallel to the electric field direction), there would have been no length
contraction of the plate dimensions since they are perpendicular to the direction of
motion. So the charge density would not change in the S’ frame, nor would the electric
field:

E  E




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PHY2061 Enriched Physics 2 Lecture Notes                                           Relativity 6



Derivation of the Magnetic Force
         Now consider two lines of charge (1-dimensional) moving in opposite directions
with opposite charge density. In other words, a wire carries a current of positive charges
to the right and negative charges to the left. Call this the lab frame S, and denote it with
unprimed coordinates. In this frame, the velocities of the charges are equal and opposite:
v  v  v0

                                    i                            v=v0
                           ++++++++++++++++
                           
                           +
                       v= -v0
                              Test charge q v

The charge densities are also taken to be equal and opposite:
    

The net effect is a current moving to the right (positive x-axis direction):
i  i  i    v0       v0   2  v0

The charge densities are measured in the lab frame. In the rest frame of the charges, the
charge density would be different because of length contraction. We know from the
previous section that for directions parallel to the direction of motion
                                1
    00 where  0 
                            1  v0 2 / c 2
So,
                       
0         0       
        0           0    0

Now consider placing a test charge, q, next to these two lines of charge. In the lab
frame, the charge densities are equal and opposite; so if the test charge is at rest in the lab
frame, no electric force is felt since there is no net charge on the wire.

But consider the test charge moving with a velocity v parallel to the line charges in the
lab frame (in the same direction as the positive charges). Let’s calculate the electric field
from the wire acting on the charge in its rest frame. Thus, we want to consider a Lorentz
Transformation to the rest frame of the particle.

To find the velocity of the line charges in this new frame where the test charge is at rest
(call this frame S’), we must use the relativistic addition of velocity formulae:




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PHY2061 Enriched Physics 2 Lecture Notes                                         Relativity 6


       v0  v                       v
v            0       c where  
      1  02 1   0 
          vv                           c
           c
       v v         
v  0         0       c
          v0 v 1   0 
     1 2
           c
The linear charge densities in this frame are increased relative to the rest frame of the
moving charges by -factors depending on the velocity of the charges in the S’ frame:
                                       1                 v
   0        where              and  +  
                     0                1               c

                                            1                  v
               where                 and    
           0
                        0                  1                 c

Recall that   is the charge density measured in the lab frame S, whereas  is the
charge density in the S’ frame moving with the test charge.

Now since | v |  | v | , the charge densities will not be equal and opposite (because of
differing amounts of length contraction of the original charge densities). Thus, there now
will be a net charge on the wire, and an electric force on the particle!

The net charge density can be calculated:




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PHY2061 Enriched Physics 2 Lecture Notes                                          Relativity 6


                    
                                      
                             
net                                  
                           0
                                                
        1          1      
                
   0  1   2           2 
                   1    
            

                                          
                                          
           1                   1        
                                        
  0        
                       2
                                  
                                         2

      1  0
                         1  0        
          1  0           1  0  
             1  0              1  0                                 
                                                                          
     0  1  0    0 
                2    2   2 2
                               1  02   2  02  2                       
                                                                            
              2  0                         
                                                  2 0    0
    0       1   1   
                  0
                      2             2           0
                                                                1
 2  0               where as usual  
                                                              1  2

which is indeed not zero.

We can use this net charge density to calculate the electric field at the position of the test
charge (which is at rest in this frame):

Appling Gauss’ Law to an infinitely long line charge, we consider a cylindrical surface:

                          net L
  Er 2 r L 
                           0
          net       2  0 
Er               
         2 0 r      2 0 r 

The force on a test charge of charge q is toward the wire (since the wire has a net
negative charge in this frame) if q>0:

             2q 0 
Fr  qEr 
               2 0 r 
Thus, the particle which was originally at rest, will accelerate toward the wire (in the
direction perpendicular to the Lorentz Transformation).




D. Acosta                                                           Page 5          4/20/2013
PHY2061 Enriched Physics 2 Lecture Notes                                               Relativity 6


Now, let’s transform this force back to the original frame where the test charge was in
motion and the wire had no net charge density. In this frame, the force perpendicular to
the direction of the Lorentz Transformation direction is:

       1
F    F  , so
      
     2q  0 
Fr                    since r   r
      2 0 r
                                                                                1
Now 2  v0  i is just the current carried in the wire. Also, note that K            . So,
                                                                               4 0
       2q  v0v       2Ki      2ki 
Fr               qv  2   qv      
       2 0c r
              2
                      cr        r 
                                                       K
where we have defined a new constant: k 
                                                       c2

The term in parentheses can be defined to be the “magnetic field” arising from a current
in a wire:
                         2ki
 Fr  qvB where B            for this example
                          r
We have just derived the force acting on a charged particle when it moves at a velocity v
with respect to a current i. Note that the force is in a direction perpendicular to the
velocity direction. This force is called the magnetic force, with an associated magnetic
field. However, it really is nothing new. In the rest frame of the test charge, only an
electric field existed. But when transformed to the lab frame, the electric field changed
into a magnetic field that only operates on charges with non-zero velocities.

The magnetic force equation can be written more generally in vector form as:
F  qv × B

The force is the cross-product of the velocity and magnetic field vectors, multiplied by
the charge of the particle:

F  q  vy Bz  vz By  x  q  vx Bz  vz Bx  y  q  vx By  vy Bx  z
                        ˆ                       ˆ                       ˆ

When combined with the electric force law, we have the full Lorentz Force equation:

F  q  E  v×B 




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