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LORENTZ TRANSFORMATIONS



In this lecture, we assume that the speed of light is independent
 of the speed of the source.

We will use the the factor g as:

1. Time Dilation


Let's consider the figure 1. The reference frame xoy is at rest. The reference frame x'o'y' is moving at the speed v in the direction o towards o'. The frame x'oy' carries a mirror at the distance L from the observer o'. If the observer in the frame x'oy' send a flash in the direction o'y', this flash will reach the mirror and then reflect back to arrive to this observer in the point o'. The time for this travel is :
t' = L/c + L/c = 2L/c (1)

For the observer in the farme at rest xoy, the situation is differente. It's related to the figure 2. When the flash returned back to the observer o', the frame x'o'y' (observer o' along with the mirror) were moved with the speed v. The observer o' moved from o to o'. At this precise position o', this observer o' receives the emitted flash sent the time t before.
The flash has travelled ct/2 ( from o to Mirror) + ct/2 (from Mirror to the o') =ct. We can write: (ct/2)2 = (o'm)2 + L2 (2) o'm = vt/2, L = ct'/2, from the equation (1) Thus: (ct/2)2 = (vt/2)2 + (ct'/2)2 Resolving for t', we get: (ct'/2)2 = (ct/2)2 - (vt/2)2 t' 2 = (2/c) 2 (t/2)2 ( c2 -(v)2) t' 2 = (t)2 ( 1 - (v)2/(c)2) t' 2 = (t)2 ( 1/ g 2) t' = t/g or t = gt' (3)

The factor g is greater than 1. Then t is greater that t'. For the observer at rest, the time is longer than the time in the moving frame is.

2. Length Contraction:

Now, let's consider the mirror attached in o'x' axis as chown in figure 3. For the the moving observer in its moving frame x'o'x', if a flash is sent from o', it will reach Mirror and reflect back with the time t' = 2L'/c. (4)

For the stationary observer, the situation is differente: When the flash leaves from o' and reached Mirror, at this precise time, the observer o' has moved to o' . (supposed o=o' at first).
We have oo' = vt1 and oMirror = ct1 t1 is the time taken by the flash to go from o to Mirror at the position M'. We can write: ct1 = oo' + L ( L is the distance between o' and Mirror for the observer in the stationary xoy frame) Thus: ct 1= vt1 + L (5) The flash reflects back and reaches the observer o' at the position o'' while the observer o' has moved from o' to o'' (The Mirror from M' to M'') the related time to reflect back to the observer to o' is t2. We can write ct2 = L - v t2 (6) The total time taken by the flash for its round trip is t= t1 + t2 from (5) and from (6): t = (L/ (c-v) + L/(c+v) = 2Lc/ c2 - v2 t = g2. 2L/c (7) Solving for L', we have: from (4) and from (3): L' = ct'/2 = (c/2)(t/g) And from (7) L' = g L (8)

The factor g is greater than 1. Then L is smaller that L'. For the observer at rest, the length is shorter than the length in the moving frame is.

3. Lorentz Transformations

Now, for the observer in its farme xo'y', o'M' = o"M'". Let's write it equal to x'. At t'0, the flash leaves o' and at t' it reaches the point x'. Thus, for the moving observer, c( t' - t'0) = L' = x' (9)
For the observer at rest, we have: oM' = x and c(t - t0) = oo' + L = v. (t - t0) + L Then : (t - t0) = L/(c-v) (10) Let's consider : At x = 0, t0= 0. Then ct = x , that is: x = vt +L (11) Solving for L : L = x-vt (12) Using the relation (8) : L' = gL (12) becomes: L' = g(x-vt) or x' = g(x-vt) (13)

(9) gives: t' = x'/c + t'0 Using (13) : t' = g(x-vt)/c + t'0 Using the dilatation formula ( relation(3)): t 0= gt'0 We have then: t' = g(x-vt)/c + t0/g Using the contraction formula ( relation(8)): t 0 = t - L/(c-v) = t - (x-vt)/(c-v) Thus: t' = g(x-vt)/c + t/g - (x-vt)/(c-v)/g = g(x-vt)/c + (ct -x)/(c-v)g = g( (x-vt)/c) + (ct-x)/(c-v).g2 ) = g( (x-vt)/c) +(ct-x)(c+v)/c2) = g/c2(c2t -xv) = g(t -xv/c2)
t' = = g(t -xv/c2) (14)

Since there is no change with y ans z axis, we have y = y' and z= z'
When a reference frame (x'y'z't') moves at a speed v relating to the reference frame at rest (xyzt), the relationships between these coordinates are known as the Lorentz transformations and grouped as fellow:

Remark:

In the case of very low velocities, g is near to 1, furthermore, the Lorentz transformations become Galileo transformations.

4. Lorentz Inverse Transformations

The transformations above are set by the stationary observer for whom the farme x'o'y' is moving at the speed +v. The inverse transormation are obtained when the observers change the roles. That is, for the observer in x'o'y', we have the same transformations related to the frame xoy moving at the speed -v:(changing prime in no prime and vis versa , - in +). We have:

5. Lorentz Velocities Transformations

We are interested now to set the transformations of the components of a certain vector velocity V ( Vx, Vy, Vz) related to V' ( V'x, V'y, V'z) . From the relations above, we have the following differentiations: dx = g(dx' + vdt'`) dt = g(dt' + dx'v/c2) dx/dt = Vx, the component of the velocity over ox. Vx = (V'x + v)/(1 + V'x.v/c2) Where V'x = dx'/dt' is the component of the velocity V' over o'x', and v is the relative velocity of the frame Vy = dy/dt = dy'/dt' = dy'/g(dt' + dx'v/c2) = V'y/g(1 + dx'v/c2/dt') = V'y/g(1 + V'x v/c2) Vz = V'z/g(1 + V'x v/c2)

Remark that, it the moving observer in (x',y',z') referencial frame meseares the velocity of light in this frame and finds V'x = c , the first relation gives also Vx = c. The conclusion is the speed of light is the same in these two referential frames.







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