1. Lorentz Transformation
Notes:
This assumes that the two frames are both at origin at t = t’ = 0. (If not, must change t  t - t0, where t0 is the time when the two origins coincided.
v > 0 when S’ is moving in +x direction; v < 0 when S’ moving in –x direction. [Sign should be obvious from x’ equation.]
Interval between two events: x  x2-x1, y  y2-y1, z  z2-z1, t  t2 –t1 :
x’ =  (x – v  t)
y’ =  y
z’ = z
t’ =  (t – v  x/c2)
Similarly for differentials:   d.

2. and uz’ = (1/) [ uz/(1-vux/c2)]
length contraction
Lorentz Trans.
dx’ =  (dx – v dt)
dy’ = dy
dz’ = dz
dt’ =  (dt – v dx/c2)
Galilean Trans.
dx’ = dx – v dt
dy’ = dy
dz’ = dz
dt’ = dt
time dilation
failure of simultaneity at a distance
Suppose an object is moving in S frame: dr/dt = u
What is its speed (u’ = dr’/dt’) in S’ frame?
Galileo: dx’/dt’ = dx/dt –vdt/dt  ux’ = ux – v, (and uy’ = uy, uz’ = uz)
Lorentz: dx’/dt’ = [(dx – v dt)] / [(dt-vdx/c2] ( dt/dt)
 ux’ = (dx/dt - v dt/dt) / (dt/dt – v dx/dt / c2)]
ux’ = (ux – v)/ (1 – vux/c2)
uy’ = dy’/dt’ = dy/[(dt –vdx/c2)] ( dt/dt)
uy’ = (1/) [ uy/(1-vux/c2)]
and uz’ = (1/) [ uz/(1-vux/c2)]
Velocity Transformations

3. Galilean Velocity Trans.
ux’ = ux – v
uy’ = uy
uz’ = uz
Lorentz Velocity Trans.
ux’ = (ux – v)/ (1 – vux/c2)
uy’ = (1/) [ uy/(1 - vux/c2)]
uz’ = (1/) [ uz/(1 - vux/c2)]
Notes:
Lorentz velocity transformation reduces to Galilean if |v| << c (so  1) and |ux| << c.
For ux’: Sign should be easy to learn:
a) Sign of denominator is same as sign of numerator.
b) Numerator for ux’ is the classical expression, so sign of ux’ is the same as classical:
i) If S’ is moving to the right, v > 0  ux’ < ux
ii) If S’ is moving to the left, v < 0  ux’ > ux
For y and z cases, velocities change due to both time dilation (1/) and simultaneity difference of two frames.
For x direction, time dilation and length contraction effects cancel, so only effect is from simultaneity difference.

4. Galilean Velocity Trans.
ux’ = ux – v
Lorentz Velocity Trans.
ux’ = (ux – v)/ (1 – vux/c2)
Suppose S’ is moving toward left with respect to S, so v < 0 and ux’ > ux.
Is it possible that ux’ > c when ux < c?
Example: Let v = c and ux = 0.9c.
Galileo: ux’ = 1.8 c
Lorentz: ux’ = 1.8c/(1+.81) = c ( < c !)

5. Notes:
If ux < c, then ux’ < c; i.e. there is no reference frame in which the speed of the object is > c.
If ux = c
(i.e. the object is “light”)
ux’ = (c-v)/(1-v/c) = c !
This is Einstein’s “invariance of speed of light postulate” – all observers will measure the same speed of light.

6. Lorentz Velocity Trans. ux’ = (ux – v)/ (1 – vux/c2)
uy’ = (1/) [ uy/(1 - vux/c2)] c 
What about a ray of light (speed c) making an angle  with respect of x axis in frame S?
What is its speed and direction in S’?
ux = c cos, uy = c sin
ux’ = [c cos -v] / [1 – v cos/c], uy’ = [1 – v2/c2]1/2 [c sin] / [1 – v cos/c],
Therefore u’2 = ux’2 + uy’2 = {c2 cos2 -2cv cos + v2 + (1 – v2/c2)c2sin2} /[1 – v cos/c]2
u’2 = {c2 -2cv cos + v2(1-sin2)} / [1 – v cos/c]2, but 1 – sin2 = cos2, so
u’2 = c2 [1 – v cos/c]2 / [1 – v cos/c]2 and u’ = c, as expected.
Angle: cos’ = ux’/c = [cos - v/c] / [1 – v/c cos].
Example:  = 45o (cos = 2/2) and v = c/2:
cos’ = (2/2 – ½) / [1 - 2/4] = 0.320
 ’ = 71o

7. Angle: cos’ = ux’/c = [cos - v/c] / [1 – v/c cos].
Example: v = 0.9c and  = 80o
cos = 0.174
cos’ = ( )/[1-(0.9)(0.174)] = /0.843 =
’ = 149.5o
i.e. ux’ can have sign opposite to ux if v is large.

8. Lorentz Velocity Trans.
ux’ = (ux – v)/ (1 – vux/c2)
uy’ = (1/) [ uy/(1 - vux/c2)]
uz’ = (1/) [ uz/(1 - vux/c2)]
Problem: Consider a collision between an
electron and positron in a collider. In the lab frame, it is a head on collision with each particle approaching the other at 0.995c. How fast is the positron approaching the electron as measured in the electron’s rest frame? e- e+
lab frame
electron’s frame

9. Lorentz Velocity Trans. ux’ = (ux – v)/ (1 – vux/c2)
Problem: Consider a collision between an electron and positron in a collider.
In the lab frame, it is a head on collision with each particle approaching the other at 0.995c. How fast is the positron approaching the electron as measured in the electron’s rest frame?
Lorentz Velocity Trans.
ux’ = (ux – v)/ (1 – vux/c2)
S = lab frame
S’ = electron frame
v = c, ux = c
ux’ = - ( ) c / [1+(0.995)2]
ux’ = c
Numerator = Galilean result = c

10. Problem: A spacecraft (C) leaves earth (E) with a constant velocity = 0.6c, such that the origins of E and C coincide at tE = tC = 0. At t1C = 333 s, the spacecraft fires a rocket (R) at a target in front of it, and the rocket hits the target at t2C = 383 s and x2C = 1 x 1010 m.
At what time does the rocket hit the target as measured on earth?
What is the speed of the rocket as measured on earth?
When the rocket hits the target, it explodes. How much time does it take (as measured on earth) for light from the explosion to reach earth?

11. Problem: A spacecraft (S) leaves earth (E) with a constant velocity = 0.6c, such that the origins of E and S coincide at tE = tS = 0. At t1C = 333 s, the spacecraft fires a rocket (R) at a target in front of it, and the rocket hits the target at t2C = 383 s and x2C = 1 x 1010 m.
At what time does the rocket hit the target as measured on earth?
What is the speed of the rocket as measured on earth?
When the rocket hits the target, it explodes. How much time does it take (as measured on earth) for light from the explosion to reach earth?
Note: Neither S nor E measures the proper time or proper length for the rocket’s trip, so to transform from S to E must use Lorentz transformation.
Note that E is traveling at velocity – v with respect to S:
v = 0.6c   = 1 / (1 – v2/c2)1/2
 = 1/(1-0.36)1/2 = 1/0.641/2 = 1/0.8
 = 1.25
a) t2E =  (t2S + vx2C/c2)
t2E = 1.25 [383 + (1 x 1010 x 0.6)/3x108] s = 1.25 x 403 s
t2E = 504 s E S

12. Problem: A spacecraft (S) leaves earth (E) with a constant velocity = 0.6c, such that the origins of E and S coincide at tE = tS = 0. At t1S = 333 s, the spacecraft fires a rocket (R) at a target in front of it, and the rocket hits the target at t2S = 383 s and x2S = 1 x 1010 m.
At what time does the rocket hit the target as measured on earth?
What is the speed of the rocket as measured on earth?
When the rocket hits the target, it explodes. How much time does it take (as measured on earth) for light from the explosion to reach earth?
v = 0.6 c,  = 1.25
b) uE = (uS + v) / (1 + uCv/c2)
uS = (x2S-x1S)/(t2S-t1S), where x1S = 0 (i.e. at t1S, the rocket is on the spacecraft)
uS = (1 x 1010 – 0) m /( )s = 2 x 108 m/s = 2/3 c
uE = ( ) c / [1 + (2/3)(3/5)]
uE = c = 2.71 x 108 m/s
c) t = x2E/c =  [x2S + vt2S]/c
t = 1.25 [(1 x 1010 m / 3 x 108m/s) + (0.6 x 383 m)]
t = 1.25 [ ) s = 1.25 x s
t = 329 s
relativistic correction
nonrelativistic value

13. We saw that if u < c in one frame, u’ < c in all frames.
Is it possible for u > c in a frame?
Newtonian Mechanics: F = dp/dt = d(mu)/dt = m du/dt, so if a constant force continues to act on a mass, its speed will eventually > c. [In Newtonian mechanics, a particle’s inertia (its resistance to changes in its velocity) = m, which is constant.]
Einstein showed that in relativity, one still has F = dp/dt, but now
p  mu/[(1 – (u/c)2]1/2 = mu.
[That is, the particle’s inertia = m, which increases as its speed u increases. Einstein deduced this from considering momentum conservation in collisions, as viewed in different inertial reference frames.]
   as u  c. Therefore, as u  c, its inertia  . It becomes impossible to increase its speed  c, although it can approach c.
Therefore, c is a limiting speed – no mass can go faster than c (in any frame)!

14. F = dp/dt; p  mu / [1 – (u/c)2] ½ =  m u
Consider a constant force and a particle that starts from rest:
u/[(1 – (u/c)2]1/2 = (F/m) t  u2 = (Ft/m)2 (1-u2/c2)
u2 [1+(FT/mc)2] = (Ft/m)2
u/c = (Ft/mc) / [1 + (Ft/mc)2]1/2
“classical regime”

15. “classical regime”

16. p  mu/[(1 – (u/c)2]1/2 = mu
Problem: What is the magnitude of the momentum of a particle with mass m = 3 kg traveling at u = 0.4 c? 0.8c?
u = 0.4c
 = 1/[(1 – (u/c)2]1/2 = 1/ [1 – 0.42]1/2 = 1/(0.84)1/2 = 1/0.917 = 1.09
p = (1.09) (3 kg) (0.4 x 3 x 108 m/s)
p = 3.93 x 108 kgm/s
u = 0.8c
 = 1/[(1 – (u/c)2]1/2 = 1/ [1 – 0.82]1/2 = 1/(0.36)1/2 = 1/0.6 = 1.66
p = (1.66) (3 kg) (0.8 x 3 x 108 m/s)
p = x 108 kgm/s
[Because of , doubling u more than doubles p.]

17. p  mu/[(1 – (u/c)2]1/2 = mu
Problem: Consider a collision between an electron and positron in a collider. In the lab frame, it is a head on collision with each particle approaching the other at 0.995c. What are the magnitudes of the momenta of the electron and positron in the lab frame and in the electron’s frame?
(me = mp = 9.11 x kg)

18. p  mu/[(1 – (u/c)2]1/2 = mu
Problem: Consider a collision between an electron and positron in a collider. In the lab frame, it is a head on collision with each particle approaching the other at 0.995c. What are the magnitudes of the momenta of the electron and positron in the lab frame and in the electron’s frame?
(me = mp = 9.11 x kg)
In the lab frame, each particle has the same m and u =|u|, so each has same (magnitude of) momentum.
 = [1 – (0.995)2]-1/2 = 10.01
pe = pp = (10.01) (9.11 x kg) (0.995 x 3 x 108 m/s)
pe = pp = 2.72 x kgm/s

19. p  mu/[(1 – (u/c)2]1/2 = mu
Problem: Consider a collision between an electron and positron in a collider. In the lab frame, it is a head on collision with each particle approaching the other at 0.995c. What are the magnitudes of the momenta of the electron and positron in the lab frame and in the electron’s frame?
(me = mp = 9.11 x kg)
In the electrons frame, ue = 0 so pe = 0.
The positron’s speed up = c (earlier problem)
p = [1 – ( ]-1/2 = 196
pp = (196) (9.11 x kg) ( x 3 x 108 m/s)
pp = 5.4 x kgm/s
[Note: pp(electron’s frame) / pp(lab frame) = 5.4 x / 2.7 x 10-21) = 20 !
(In classical physics, it would only be double.)]