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PHY3221 Detweiler Lecture #4Relativity

Lorentz transformation

A coordinate rotation:

t′ = t

x′ = x cos θ + y sin θ

y′ = −x sin θ + y cos θ

z′ = z′

Notice that distance is conserved under a coordinate rotation rotation:

x′2 + y′2 = (x cos θ + y sin θ)2 + (−x sin θ + y cos θ)2

= x2 cos2 θ + 2xy cos θ sin θ + y2 sin2 θ + x2 sin2 θ − 2xy cos θ sin θ + y cos2 θ

= x2(cos2 θ + sin2 θ) + y2(sin2 θ + cos θ)

= x2 + y2.

This might seem obvious because the mathematics just confirms your everyday experiences.The Galilean transformation:

t′ = t

x′ = x− vty′ = y

z′ = z

Inverse transformation:

t = t′

x = x′ + vt′

y = y′

z = z′

The Lorentz transformation:

t′ =t− vx/c2√1− v2/c2

x′ =x− vt√1− v2/c2

y′ = y

z′ = z

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Inverse transformation:

t =t′ + vx′/c2√

1− v2/c2

x =x′ + vt′√1− v2/c2

y = y′

z = z′

Notice that in the limit that v/c → 0, but v remains finite, the Lorentz transformationsapproach the Galilean transformation. So, only when v is comparable to c are the effects ofspecial relativity revealed.

Derive time dilation from the Lorentz transformations:Two events, #1 at (t1, x1) and #2 at (t2, x2), with occur at the same place (x1 = x2) in thet, x coordinate system. Thus the proper time between the events is

∆T0 = t2 − t1.

The time between the events in the primed coordinate system is

∆T ′ = t′2 − t′1

=t2 − vx2/c2√

1− v2/c2− t1 − vx1/c2√

1− v2/c2

=t2 − vx2/c2 − t1 + vx1/c

2√1− v2/c2

=t2 − t1√1− v2/c2

because x1 = x2

=∆T0√

1− v2/c2,

which is the correct expression for time dilation.

Derive length contraction from the Lorentz transformations:A stick is at rest in the unprimed t, x coordinate system. Two events, #1 at (t1, x1) and #2at (t2, x2), occur at different times t1 6= t2 but at either end of the stick. And because thestick is at rest in the unprimed coordinate system, the proper length of the stick is

L0 = x2 − x1

It has been arranged that these same two events occur at the same time in the primedcoordinate system, t′1 = t′2. So, in the primed coordinates the length of the stick is measured

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to be L′ = x′2 − x′1. Now, the proper length of the stick is

L0 = x2 − x1

=x′2 − vt′2√1− v2/c2

− x′1 − vt′1√1− v2/c2

=x′2 − vt′2 − x′1 + vt′1√

1− v2/c2

=x′2 − x′1√1− v2/c2

because t′1 = t′2

=L′√

1− v2/c2and, finally

L′ = L0

√1− v2/c2

which is the correct expression for length contraction.

Consider the Galilean addition of velocities:With the Lorentz transformations in hand, we can now see how velocities are viewed fromdifferent coordinate systems. First consider the Galilean addition of velocities. Consider abird flying along, and note two nearby events along the bird’s path—perhaps the events aretwo flaps of the bird’s wings. These events are noted to occur at (t1, x1, y1) and (t2, x2, y2) inthe unprimed coordinate system, and at (t′1, x

′1, y

′1) and (t′2, x

′2, y

′2) in the primed coordinate

system. The components of the speed of the bird in the unprimed system are

ux =x2 − x1t2 − t1

=∆x

∆tuy =

y2 − y1t2 − t1

=∆y

∆t

and in the primed coordinates

u′x =x′2 − x′1t′2 − t′1

=∆x′

∆t′u′y =

y′2 − y′1t′2 − t′1

=∆y′

∆t′

Now use the Lorentz transformations to relate these different speeds:

ux =x2 − x1t2 − t1

=∆x

∆t

=(∆x′ + v∆t′)/

√1− v2/c2

(∆t′ + v∆x′/c2)/√

1− v2/c2

=∆x′/∆t′ + v

1 + v∆x′/∆t′c2and, finally

ux =u′x + v

1 + vu′x/c2,

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and

uy =y2 − y1t2 − t1

=∆y

∆t

=∆y′

(∆t′ + v∆x′/c2)/√

1− v2/c2

=∆y′/∆t′

(1 + v∆x′/∆t′c2)/√

1− v2/c2and, finally

uy =u′y√

1− v2/c2

1 + vu′x/c2.

If v and u′x and u′y are all less than c then ux, uy, and also√u2x + u2y are all less than c. Try

this out with v = u′x = u′y = 9c/10.