# 5.5 The lorentz transformation  (Page 8/12)

 Page 8 / 12

## Summary

• The Galilean transformation equations describe how, in classical nonrelativistic mechanics, the position, velocity, and accelerations measured in one frame appear in another. Lengths remain unchanged and a single universal time scale is assumed to apply to all inertial frames.
• Newton’s laws of mechanics obey the principle of having the same form in all inertial frames under a Galilean transformation, given by
$x=x\prime +vt,\phantom{\rule{0.5em}{0ex}}y=y\prime ,\phantom{\rule{0.5em}{0ex}}z=z\prime ,\phantom{\rule{0.5em}{0ex}}t=t\prime .$

The concept that times and distances are the same in all inertial frames in the Galilean transformation, however, is inconsistent with the postulates of special relativity.
• The relativistically correct Lorentz transformation equations are
$\begin{array}{ccc}\text{Lorentz transformation}\hfill & & \text{Inverse Lorentz transformation}\hfill \\ t=\frac{t\prime +vx\prime \text{/}{c}^{2}}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill & & t\prime =\frac{t-vx\text{/}{c}^{2}}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill \\ x=\frac{x\prime +vt\prime }{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill & & x\prime =\frac{x-vt}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\hfill \\ y=y\prime \hfill & & y\prime =y\hfill \\ z=z\prime \hfill & & z\prime =z\hfill \end{array}$

We can obtain these equations by requiring an expanding spherical light signal to have the same shape and speed of growth, c , in both reference frames.
• Relativistic phenomena can be explained in terms of the geometrical properties of four-dimensional space-time, in which Lorentz transformations correspond to rotations of axes.
• The Lorentz transformation corresponds to a space-time axis rotation, similar in some ways to a rotation of space axes, but in which the invariant spatial separation is given by $\text{Δ}s$ rather than distances $\text{Δ}r,$ and that the Lorentz transformation involving the time axis does not preserve perpendicularity of axes or the scales along the axes.
• The analysis of relativistic phenomena in terms of space-time diagrams supports the conclusion that these phenomena result from properties of space and time itself, rather than from the laws of electromagnetism.

## Problems

Describe the following physical occurrences as events, that is, in the form ( x , y , z , t ): (a) A postman rings a doorbell of a house precisely at noon. (b) At the same time as the doorbell is rung, a slice of bread pops out of a toaster that is located 10 m from the door in the east direction from the door. (c) Ten seconds later, an airplane arrives at the airport, which is 10 km from the door in the east direction and 2 km to the south.

Describe what happens to the angle $\alpha =\text{tan}\left(v\text{/}c\right),$ and therefore to the transformed axes in [link] , as the relative velocity v of the S and $\text{S}\prime$ frames of reference approaches c .

The angle α approaches $45\text{°},$ and the $t\prime \text{-}$ and $x\prime \text{-axes}$ rotate toward the edge of the light cone.

Describe the shape of the world line on a space-time diagram of (a) an object that remains at rest at a specific position along the x- axis; (b) an object that moves at constant velocity u in the x- direction; (c) an object that begins at rest and accelerates at a constant rate of in the positive x- direction.

A man standing still at a train station watches two boys throwing a baseball in a moving train. Suppose the train is moving east with a constant speed of 20 m/s and one of the boys throws the ball with a speed of 5 m/s with respect to himself toward the other boy, who is 5 m west from him. What is the velocity of the ball as observed by the man on the station?

15 m/s east

When observed from the sun at a particular instant, Earth and Mars appear to move in opposite directions with speeds 108,000 km/h and 86,871 km/h, respectively. What is the speed of Mars at this instant when observed from Earth?

A man is running on a straight road perpendicular to a train track and away from the track at a speed of 12 m/s. The train is moving with a speed of 30 m/s with respect to the track. What is the speed of the man with respect to a passenger sitting at rest in the train?

32 m/s

A man is running on a straight road that makes $30\text{°}$ with the train track. The man is running in the direction on the road that is away from the track at a speed of 12 m/s. The train is moving with a speed of 30 m/s with respect to the track. What is the speed of the man with respect to a passenger sitting at rest in the train?

In a frame at rest with respect to the billiard table, a billiard ball of mass m moving with speed v strikes another billiard ball of mass m at rest. The first ball comes to rest after the collision while the second ball takes off with speed v in the original direction of the motion of the first ball. This shows that momentum is conserved in this frame. (a) Now, describe the same collision from the perspective of a frame that is moving with speed v in the direction of the motion of the first ball. (b) Is the momentum conserved in this frame?

a. The second ball approaches with velocity − v and comes to rest while the other ball continues with velocity − v ; b. This conserves momentum.

In a frame at rest with respect to the billiard table, two billiard balls of same mass m are moving toward each other with the same speed v . After the collision, the two balls come to rest. (a) Show that momentum is conserved in this frame. (b) Now, describe the same collision from the perspective of a frame that is moving with speed v in the direction of the motion of the first ball. (c) Is the momentum conserved in this frame?

In a frame S, two events are observed: event 1: a pion is created at rest at the origin and event 2: the pion disintegrates after time $\tau$ . Another observer in a frame $\text{S}\prime$ is moving in the positive direction along the positive x -axis with a constant speed v and observes the same two events in his frame. The origins of the two frames coincide at $t=t\prime =0.$ (a) Find the positions and timings of these two events in the frame $\text{S}\prime$ (a) according to the Galilean transformation, and (b) according to the Lorentz transformation.

a. $\begin{array}{c}{t}_{1}\prime =0;\phantom{\rule{0.5em}{0ex}}{x}_{1}\prime =0;\\ {t}_{2}\prime =\tau ;\phantom{\rule{0.5em}{0ex}}{x}_{2}\prime =0\end{array};$ b. $\begin{array}{c}{t}_{1}\prime =0;\phantom{\rule{0.5em}{0ex}}{x}_{1}\prime =0;\\ {t}_{2}\prime =\frac{\tau }{\sqrt{1-{v}^{2}\text{/}{c}^{2}}};\phantom{\rule{0.5em}{0ex}}{x}_{2}\prime =\frac{-v\tau }{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}\end{array}$

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