Physics, asked by rifle6314, 1 year ago

Derive the einstein velocity-addition formula by performing a lorentz transformation with velocity v on the four-velocity of a particle whose speed in the original frame was w.

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Answered by GENIUS1223
0

The Lorentz transformation

In The Wonderful World and appendix 1, the reasoning is kept as direct as possible.

Much use is made of graphical arguments to back up the mathematical results. Now

we will introduce a more algebraic approach. This is needed in order to go further. In

particular, it will save a lot of trouble in calculations involving a change of reference

frame, and we will learn how to formulate laws of physics so that they obey the Main

Postulates of the theory.

The Lorentz transformation, for which this chapter is named, is the coordinate transformation which replaces the Galilean transformation presented in eq. (2.1).

Let S and S0 be reference frames allowing coordinate systems (t, x, y, z) and (t

0

, x0

, y0

, z0

)

to be defined. Let their corresponding axes be aligned, with the x and x

0 axes along

the line of relative motion, so that S0 has velocity v in the x direction in reference frame

S. Also, let the origins of coordinates and time be chosen so that the origins of the two

reference frames coincide at t = t

0 = 0. Hereafter we refer to this arrangement as the

‘standard configuration’ of a pair of reference frames. In such a standard configuration,

if an event has coordinates (t, x, y, z) in S, then its coordinates in S0 are given by

t

0 = γ(t − vx/c2

) (3.1)

x

0 = γ(−vt + x) (3.2)

y

0 = y (3.3)

z

0 = z (3.4)

29

30 Copyright A. Steane, Oxford University 2010, 2011; not for redistribution.

where γ = γ(v) = 1/(1 − v

2/c2

)

1/2

. This set of simultaneous equations is called the

Lorentz transformation; we will derive it from the Main Postulates of Special Relativity

in section 3.2.

By solving for (t, x, y, z) in terms of (t

0

, x0

, y0

, z0

) you can easily derive the inverse Lorentz

transformation:

t = γ(t

0 + vx0

/c2

) (3.5)

x = γ(vt0 + x

0

) (3.6)

y = y

0

(3.7)

z = z

0

(3.8)

This can also be obtained by replacing v by −v and swapping primed and unprimed

symbols in the first set of equations. This is how it must turn out, since if S0 has velocity

v in S, then S has velocity −v in S0 and both are equally valid inertial frames.

Let us immediately extract from the Lorentz transformation the phenomena of time

dilation and Lorentz contraction. For the former, simply pick two events at the same

spatial location in S, separated by time τ . We may as well pick the origin, x = y = z = 0,

and times t = 0 and t = τ in frame S. Now apply eq. (3.1) to the two events: we find

the first event occurs at time t

0 = 0, and the second at time t

0 = γτ , so the time interval

between them in frame S

0

is γτ , i.e. longer than in the first frame by the factor γ. This

is time dilation.

For Lorentz contraction, one must consider not two events but two worldlines. These are

the worldlines of the two ends, in the x direction, of some object fixed in S. Place the

origin on one of these worldlines, and then the other end lies at x = L0 for all t, where

L0 is the rest length. Now consider these worldlines in the frame S0 and pick the time

t

0 = 0. At this moment, the worldline passing through the origin of S is also at the origin

of S0

, i.e. at x

0 = 0. Using the Lorentz transformation, the other worldline is found at

t

0 = γ(t − vL0/c2

), x0 = γ(−vt + L0). (3.9)

Since we are considering the situation at t

0 = 0 we deduce from the first equation that

t = vL0/c2

. Substituting this into the second equation we obtain x

0 = γL0(1 − v

2/c2

) =

L0/γ. Thus in the primed frame at a given instant the two ends of the object are at

x

0 = 0 and x

0 = L0/γ. Therefore the length of the object is reduced from L0 by a factor

γ. This is Lorentz contraction.

For relativistic addition of velocities, eq. (21.8), consider a particle moving along the x

0

axis with speed u in frame S0

. Its worldline is given by x

0 = ut0

. Substituting in (3.6)

we obtain x = γ(vt0 + ut0

) = γ

2

(v + u)(t − vx/c2

). Solve for x as a function of t and one

obtains x = wt with w as given by (21.8).

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