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The lorentz transformation equations

The Galilean transformation nevertheless violates Einstein’s postulates, because the velocity equations state that a pulse of light moving with speed c along the x -axis would travel at speed c v in the other inertial frame. Specifically, the spherical pulse has radius r = c t at time t in the unprimed frame, and also has radius r = c t at time t in the primed frame. Expressing these relations in Cartesian coordinates gives

x 2 + y 2 + z 2 c 2 t 2 = 0 x 2 + y 2 + z 2 c 2 t 2 = 0 .

The left-hand sides of the two expressions can be set equal because both are zero. Because y = y and z = z , we obtain

x 2 c 2 t 2 = x 2 c 2 t 2 .

This cannot be satisfied for nonzero relative velocity v of the two frames if we assume the Galilean transformation results in t = t with x = x + v t .

To find the correct set of transformation equations, assume the two coordinate systems S and S in [link] . First suppose that an event occurs at ( x , 0 , 0 , t ) in S and at ( x , 0 , 0 , t ) in S , as depicted in the figure.

The axes of frames S and S prime are shown. S has axes x, y, and z. S prime is moving to the right with velocity v and has axes x prime, y prime and z prime. S and S prime are aligned along the horizontal x and x prime axes and are separated by a distance v t. An event on the horizontal x and x prime axes is indicated by a point which is a distance x from the y z plane of the S frame and a distance x prime from the y prime, z prime plane of the S prime frame.
An event occurs at ( x , 0, 0, t ) in S and at ( x , 0 , 0 , t ) in S . The Lorentz transformation equations relate events in the two systems.

Suppose that at the instant that the origins of the coordinate systems in S and S coincide, a flash bulb emits a spherically spreading pulse of light starting from the origin. At time t , an observer in S finds the origin of S to be at x = v t . With the help of a friend in S , the S observer also measures the distance from the event to the origin of S and finds it to be x 1 v 2 / c 2 . This follows because we have already shown the postulates of relativity to imply length contraction. Thus the position of the event in S is

x = v t + x 1 v 2 / c 2

and

x = x v t 1 v 2 / c 2 .

The postulates of relativity imply that the equation relating distance and time of the spherical wave front:

x 2 + y 2 + z 2 c 2 t 2 = 0

must apply both in terms of primed and unprimed coordinates, which was shown above to lead to [link] :

x 2 c 2 t 2 = x 2 c 2 t 2 .

We combine this with the equation relating x and x to obtain the relation between t and t :

t = t v x / c 2 1 v 2 / c 2 .

The equations relating the time and position of the events as seen in S are then

t = t + v x / c 2 1 v 2 / c 2 x = x + v t 1 v 2 / c 2 y = y z = z .

This set of equations, relating the position and time in the two inertial frames, is known as the Lorentz transformation    . They are named in honor of H.A. Lorentz (1853–1928), who first proposed them. Interestingly, he justified the transformation on what was eventually discovered to be a fallacious hypothesis. The correct theoretical basis is Einstein’s special theory of relativity.

The reverse transformation expresses the variables in S in terms of those in S . Simply interchanging the primed and unprimed variables and substituting gives:

t = t v x / c 2 1 v 2 / c 2 x = x v t 1 v 2 / c 2 y = y z = z .

Using the lorentz transformation for time

Spacecraft S is on its way to Alpha Centauri when Spacecraft S passes it at relative speed c /2. The captain of S sends a radio signal that lasts 1.2 s according to that ship’s clock. Use the Lorentz transformation to find the time interval of the signal measured by the communications officer of spaceship S .

Solution

  1. Identify the known: Δ t = t 2 t 1 = 1.2 s ; Δ x = x 2 x 1 = 0 .
  2. Identify the unknown: Δ t = t 2 t 1 .
  3. Express the answer as an equation. The time signal starts as ( x , t 1 ) and stops at ( x , t 2 ) . Note that the x coordinate of both events is the same because the clock is at rest in S . Write the first Lorentz transformation equation in terms of Δ t = t 2 t 1 , Δ x = x 2 x 1 , and similarly for the primed coordinates, as:
    Δ t = Δ t + v Δ x / c 2 1 v 2 c 2 .

    Because the position of the clock in S is fixed, Δ x = 0 , and the time interval Δ t becomes:
    Δ t = Δ t 1 v 2 c 2 .
  4. Do the calculation.
    With Δ t = 1.2 s this gives:
    Δ t = 1.2 s 1 ( 1 2 ) 2 = 1.6 s.

    Note that the Lorentz transformation reproduces the time dilation equation.
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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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