# Special Theory of Relativity - PowerPoint - PowerPoint by theoryman

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```									 Special Theory of Relativity

• In ~1895, used simple Galilean Transformations
x’ = x - vt t’ = t
• But observed that the speed of light, c, is always
measured to travel at the same speed even if seen
from different, moving frames
• c = 3 x 108 m/s is finite and is the fastest speed at
which information/energy/particles can travel
• Einstein postulated that the laws of physics are the
same in all inertial frames. With c=constant he
“derived” Lorentz Transformations

light                        u

sees v_l =c        moving frame also sees
v_l =c.NOT v_l = c-u
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“Derive” Lorentz Transform
Bounce light off a mirror. Observe in 2 frames:
A vel=0 respect to light source
A’ velocity = v
observe speed of light = c in both frames ->
c = distance/time = 2L / t       A
c2 = 4(L2 + (x’/2)2) / t’2       A’
Assume linear transform (guess)
x’ = G(x + vt)         let x = 0
t’ = G(t + Bx)          so x’=Gvt t’=Gt
some algebra
(ct’)2 = 4L2 + (Gvt)2 and L = ct /2 and t’=Gt
gives
(Gt)2 = t2(1 + (Gv/c)2) or G2 = 1/(1 - v2/c2)

mirror
A                                     L
A’ x’
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Lorentz Transformations
Define b = u/c and g = 1/sqrt(1 - b2)

x’ = g (x + ut)    u = velocity of transform
y’ = y            between frames is in
z’ = z            x-direction. If do x’ ->x then
t’ = g (t + bx/c) + -> -. Use common sense

can differentiate these to get velocity transforms
vx’ = (vx - u) / (1 -u vx/ c2)
vy’ = vy / g / (1 - u vx / c2)
vz’ = vz / g / (1 - u vx / c2)

usually for v < 0.1c non-relativistic (non-Newtonian)
expressions are OK. Note that 3D space point is
now 4D space-time point (x,y,z,t)

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Time Dilation
• Saw that t’ = g t. The “clock” runs
slower for an observer not in the
“rest” frame
• muons in atmosphere. Lifetime = t
=2.2 x 10-6 sec ct = 0.66 km
decay path = bgtc
b        g            average in lab
.1      1.005         2.2 ms          0.07 km
.5      1.15          2.5 ms          0.4 km
.9       2.29         5.0 ms          1.4 km
.99      7.09          16 ms          4.6 km
.999    22.4           49 ms          15 km
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Time Dilation
• Short-lived particles like tau and B.
Lifetime = 10-12 sec ct = 0.03 mm
• time dilation gives longer path lengths
• measure “second” vertex, determine
“proper time” in rest frame

If measure L=1.25 mm
and v = .995c
L               t(proper)=L/vg = .4 ps

Twin Paradox. If travel to distant planet
at v~c then age less on spaceship then in
“lab” frame
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• Rocket A has v = 0.8c with respect to
DS9. Rocket B have u = 0.9c with
respect to Rocket A. What is velocity
of B with respect to DS9?
DS9       A                  B

V’ =(v-u)/(1-vu/c2)
v’ = (.9+.8)/(1+.9*.8)= .988c
Notes: use common sense on +/-
if v = c and u = c v’ = (c+c)/2= c
P460 - Relativity   6
• Rocket A has v = 0.826c with respect
to DS9. Rocket B have u = 0.635c
with respect to DS9. What is velocity
of A as observed from B?

DS9       A              Think of O as DSP
and O’ as rocket B

B
V’ =(v-u)/(1-vu/c2)
v’ = (.826-.635)/(1-.826*.635)= .4c
If did B from A get -.4c
(.4+.635)/(1+.4*.635)=1.04/1.25 = .826

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Relativistic Kinematics
• E2 = (pc)2 + (mc2)2 E = total energy
m= mass and p=momentum

• natural units E in eV, p in eV/c, m in
eV/c2 -> c = 1 effectively. E2=p2+m2
• kinetic energy K = T = E - m
@ 1/2 mv2 if v << c
• Can show: b = p/E and g = E/m
--> p = bgm if m.ne.0 or p=E if m=0
(many massless particles, photon, gluon and
(almost massless) neutrinos)
• relativistic mass m = gm0 a BAD concept
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“Derive” Kinematics
• dE = Fdx = dp/dt*dx = vdp = vd(gmv)
• assume p=gmv (need relativistic)
• d(gmv) = mgdv + mvdg = mdv/g

v = final
 dE =  mvdv / g = v =0
mv 1  v 2 / c 2 . dv

= gmc2 - mc2
=Total Energy - “rest” energy
= Kinetic Energy

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• What are the momentum, kinetic, and
total energies of a proton with v=.86c?

· V = .86c g = 1/(1.86*.86)0.5 = 1.96

• E = g m = 1.96*938 MeV/c2 = 1840 MeV
• T = E - mc2 = 1840 - 938 = 900 MeV

•   p = bE = gbm = .86*1840 MeV = 1580 MeV/c
or p = (E2 - m2)0.5

Note units: MeV, MeV/c and MeV/c2. Usually never
have to use c = 300,000 km/s

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• Accelerate electron to 0.99c and then
to 0.999c. How much energy is added
at each step?

· V = .99c g = 7.1    v = .999c          g = 22.4

• E = g m = 7.1*0.511 MeV = 3.6 Mev
•        = 22.4*0.511 MeV = 11.4 MeV

• step 1 adds 3.1 MeV and step 2 adds 7.8 MeV even
though velocity change in step 2 is only 0.9%

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Lorentz Transformations
(px,py,px,E) are components of a 4-vector which has
same Lorentz transformation

px’ = g (px + uE/c2)      u = velocity of transform
py’ = py             between frames is in
pz’ = pz            x-direction. If do px’ -> px
E’ = g (E + upx) + -> - Use common sense
also let c = 1
Frame 1                        Frame 2 (cm)

Before and after scatter
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center-of-momentum frame
Sp = 0. Some quantities are invariant when going
from one frame to another:
py and pz are “transverse” momentum
Mtotal = Invariant mass of system dervived from
Etotal and Ptotal as if just one particle
How to get to CM system? Think as if 1 particle
Etotal = E1 + E2 Pxtotal = px1 + px2 (etc)
Mtotal2 = Etotal2 - Ptotal2
gcm = Etotal/Mtotal and bcm = Ptotal/Etotal

1    2 at rest               p1 = p 2

1
2
Lab                          CM
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Particle production
convert kinetic energy into mass - new particles
assume 2 particles 1 and 2 both mass = m
Lab or fixed target Etotal = E1 + E2 = E1 + m2
Ptotal = p1 --> Mtotal2 = Etotal2 - Ptotal2
Mtotal = (E12+2E1*m + m*m - p12).5
Mtotal= (2m*m + 2E1*m)0.5 ~ (2E1*m)0.5
CM: E1 = E2 Etotal = E1+E2 and Ptotal = 0
Mtotal = 2E1

1    2 at rest                 p1 = p 2

1
2
Lab                          CM
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p + p --> p + p + p + pbar
what is the minimum energy to make a proton-
antiproton pair?
• In all frames Mtotal (invariant mass) at threshold is
equal to 4*mp (think of cm frame, all at rest)
Lab Mtotal = (E12+2E1*m + m*m - p12).5
Mtotal= (2m*m + 2E1*m)0.5 = 4m
E1 = (16*mp*mp - 2mp*mp)/2mp = 7mp
CM: Mtotal = 2E1 = 4mp or E1 and E2 each =2mp

1     2 at rest                 p1 = p 2

1
2
Lab                           CM
P460 - Relativity          15
Transform examples
• Trivial: at rest E = m p=0. “boost”
velocity = v
E’ = g(E + bp) = gm
p’ = g(p + bE) = gbm

• moving with velocity v = p/E and then
boost velocity = u (letting c=1)
E’ = g(E + pu)
p’ = g(p + Eu)
calculate v’ = p’/E’ = (p +Eu)/(E+pu)
= (p/E + u)/(1+up/E) = (v+u)/(1+vu)
P460 - Relativity     16
• A p=1 GeV proton hits an electron at rest.
What is the maximum pt and E of the
electron after the reaction?
• Elastic collision. In cm frame, the energy
and momentum before/after collision are the
same. Direction changes. 90 deg = max pt
180 deg = max energy
 bcm = Ptot/Etot = Pp/(Ep + me)
· Pcm = gcmbcm*me         (transform electron to cm)
· Ecme = gcm*me           (“easy” as at rest in lab)
·
• pt max = Pcm as elastic scatter same pt in lab

•   Emax = gcm(Ecm + bcmPcm)              180 deg scatter
= gcm(gcm*me + gbb*me)
= g*g*me(1 + b*b)
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• p=1 GeV proton (or electron) hits a
stationary electron (or proton)
mp = .94 GeV me = .5 MeV
incoming target bcm gcm                Ptmax   Emax
p       e       .7       1.5 .4 MeV 1.7 MeV
p       p       .4        1.2 .4 GeV 1.4 GeV
e       p       .5       1.2 .5 GeV 1.7 GeV
e      e       .9995 30 15 MeV 1 GeV
 Emax is maximum energy transferred to stationary
particle. Ptmax is maximum momentum of (either)
outgoing particle transverse to beam. Ptmax gives
you the maximum scattering angle
•    a proton can’t transfer much energy to the electron
as need to conserve E and P. An electron scattering
off another electron can’t have much Pt as need to
conserve E and P.
P460 - Relativity              18

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