Special Relativity Jim Wheeler Physics: Advanced Mechanics.

Special Relativity Jim Wheeler Physics: Advanced Mechanics

Postulates of Special Relativity
Spacetime is homogeneous and isotropic All inertial frames are equivalent The paths and speed of light are universal

Postulate 1: Homogeneity and Isotropy
We assume that, in empty space, any two points are equivalent and any two directions are equivalent. In particular, this means that there exists a class of inertial frames which move with constant relative velocities. Further, the spatial origins of these frames may be represented by straight lines in a spacetime diagram.

Postulate 2: Inertial Frames
The equivalence of inertial frames implies that there is no absolute motion or absolute rest. Only relative velocities can be of physical relevance. One consequence is that there is no such thing as “vertical” or “horizontal” in a spacetime diagram.

Postulate 3: The speed of light
In agreement with Maxwell’s theory of electromagnetism, and in conflict with Newton’s laws of mechanics, we assume the speed of light in vacuum relative to any observer in an inertial frame takes the same value, c = x 108 m/s = 1 light second / second This is particularly striking, given the equivalence of inertial frames.

A picture of spacetime

How we think about it Time flows sort of upward
Space goes side to side How we think about it

What we all agree on Light Postulate 3:
(moving right) Postulate 3: Constancy of the speed of light

What we all agree on more light Postulate 3:
(moving left) Postulate 3: Constancy of the speed of light

We can choose our units so that light moves at 45 degrees.
(e.g., light seconds & seconds) Lotsa light Postulate 3: Every observer sees light move the same distance in a given amount of time

An observer moving along in spacetime
Observers move slower than light. Suppose some observers move in straight lines Postulate 1: Spacetime is homogeneous and isotropic World line of an observer

Postulate 2: Equivalence of inertial frames
Observers move along in spacetime Postulate 2: Equivalence of inertial frames

How might an observer label points in spacetime?
An observer moving along in spacetime How might an observer label points in spacetime?

Some observers have watches
An observer marking time They can mark progress along their world line. A watch gives a perfectly good way to label points along an observer’s world line. Some observers have watches

Some observers have watches
An observer marking time Postulate 1: Homogeneity and isotropy. We assume “good” clocks give uniform spacing. Some observers have watches

How might an observer label other points in spacetime?
Observer A How might an observer label other points in spacetime?

Observer Al How might an observer label other points in spacetime? They can send out light signals.

Observer Ali There needs to be dust or something, so some light comes back. How might an observer label other points in spacetime? They can send out light signals.

Labeling points in spacetime
Observer Alic +2 s Suppose they send out a signal two seconds before noon… …and the signal reflects and returns at two seconds after noon. Labeling points in spacetime Noon = 0 s dust -2 s

The time of a remote event.
Observer Alice +2 s The observer assumes the reflection occurred at noon. (0, x) The time of a remote event. 0 s -2 s

The distance of a remote event
Observe Alice +2 s The light took 2 seconds to go out, and two seconds to come back. The dust must be 2 light seconds away at t = 0. (Postulate 3: Constancy of the speed of light) (0, 2) The distance of a remote event 0 s -2 s

Spacetime coordinates
O serve Alice +2 s (0 s, 2 ls) Spacetime coordinates 0 s -2 s

serve Alice +2 s The observer can send out other signals, at various times, in various directions. (0, 2) Spacetime coordinates 0 s -2 s

serv Alice +2 s (0, 2) 0 s Sometimes there will be dust in just the right places. -2 s

ser Alice +2 s (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) We assign coordinates to each point where a reflection occurs. -2 s

se Alice In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. +2 s (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) -2 s

s Alice In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. +2 s (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) -2 s

Alice In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. +2 s (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) -2 s

Alice In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. (2, 0) (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) (-2, 0)

Alice In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. (2, 0) (1, 0) (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) (-1, 0) (-2, 0)

Alice (3, 0) In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. (2, 0) (1, 0) (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) (-1, 0) (-2, 0)

Alice (4, 0) (3, 0) In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. (2, 0) (1, 0) (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) (-1, 0) (-2, 0)

Alice (5, 0) (4, 0) (3, 0) In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. (2, 0) (1, 0) (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) (-1, 0) (-2, 0)

Alice (5, 0) (4, 0) (3, 0) In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. (2, 0) (1, 0) (0, 4) (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -3) (0, -2) (-1, 0) (-2, 0)

Alice (5, 0) (4, 0) (3, 0) In this way we find all points that are labeled by t = 0, but different values of x. This is what we mean by the x-axis. (2, 0) (1, 0) (0, 4) (0, 3) (0, 2) (0, 1) (0, 0) (0, -1) (0, -2) (0, -3) (0, -4) (-1, 0) (-2, 0)

Alice (5, 0) (4, 0) (3, 0) We assign coordinates to other points in the same way. (2, 0) (1, 2) (1, 1) (1, 3) (1, -2) (1, -1) (1, -3) (1, 4) (1, -4) (1, 0) (0, 2) (0, 1) (0, 3) (0, -2) (0, -1) (0, -3) (0, 4) (0, -4) (0, 0) (-1, 0) (-2, 0)

Alice (5, 0) (4, 0) (3, 0) (2, 0) (1, 0) (0, 2) (0, 1) (0, 3) (0, -2) (0, -1) (0, -3) (0, 4) (0, -4) (0, 0) (-1, 0) We now have coordinate labels for each point in spacetime. (-2, 0)

Alice (5, 0) (4, 0) (3, 0) (2, 0) (1, 0) (0, 2) (0, 1) (0, 3) (0, -2) (0, -1) (0, -3) (0, 4) (0, -4) (0, 0) (-1, 0) Notice that the paths of light move one light second per second (-2, 0)

Alice’s coordinate axes

Alice and Bill’s coordinate axes
Since the coordinate system is constructed using only the postulates, similar coordinates can be constructed for any inertial frame x x’

Spacetime terminology

Events Alice Generic points in spacetime are called events.
An event is characterized by both time and place

The light cone Alice

The light cone Alice The t and x axes make equal angles with light paths. j j

The light cone t Alice If we add another spatial dimension the light cone really looks like a cone. x y

The light cone t Alice Think of the light cone as the surface of an expanding sphere of light. x y

Forward light cone Alice t
The forward light cone includes all places that can receive light from Alice’s origin. x y

The past light cone Alice t
The past light cone includes all places that can send light to Alice’s origin. x y

Timelike separated events
Whenever events are timelike separated, it is possible for an observer to be present at both events.

Spacelike separated events
Whenever events are spacelike separated, they are simultaneous for some observer. t x

Lightlike separated events
Whenever events are lightlike separated, light can travel from one to the other.

Return to Alice’s coordinate axes

We want to show that spacetime is a vector space
Alice Consider an event

Coordinates for the event
Alice (3, 2)

Can we associate a vector with this point?
Alice (3, 2) Vectors must add linearly. u

Consider a pair of these “vectors”
Alice (3, 2) u (1,-3) v

Their sum is the “vector” at the event (4,-1)
Alice (4,-1) (3, 2) u+v u (1,-3) v

The sum is the sum of the components, as required
Alice This works because we have assumed that Alice’s coordinates are uniform. Postulate 1: Spacetime is homogeneous and isotropic.

If clocks don’t tick uniformly, we need to reassess what happens
If clocks don’t tick uniformly, we need to reassess what happens. Events along the x-axis no longer line up. t Alice x

Like inertial frames in Newtonian dynamics, we are restricted to special “inertial frames” or “frames of reference” Alice Both Newton’s 2nd Law and Special Relativity allow generalizations to arbitrary coordinates. As we have seen, the definition of vectors becomes more subtle.

For now, we assume that there exists a set of observers with uniform clocks.
Alice (3,2) (1,-3) For these observers, the components of spacetime events add as vectors.

Multiple observers

Consider two observers . . .

Alice . . . t Alice x

. . . and Bill t t’ Bill Alice x x’

What do Alice and Bill agree on?
x x’

The light cone x x’

The light cone Timelike, null, and spacelike separations of events. x x’

Do Alice and Bill agree on anything else?
The light cone Timelike, null, and spacelike separations of events. x x’

Spacetime is a vector space!
Bill Alice x x’

Does a spacetime vector have an invariant length?
Bill Alice s x x’

In Euclidean space, the Pythagorean theorem gives an invariant length.
L2 = Dx2 + Dy2 Dy Dx x

y The invariant length can be expressed as a quadratic expression in terms of the coordinates. (x2, y2) L2 = (x2 - x1)2 + (y2 - y1)2 Dy = y2 - y1 (x1, y1) Dx = x2 - x1 x

t t’ Bill Alice Here’s a plan: 1. Relate the (x,t) components of vectors to the (x’,t’) components. 2. Try to find an invariant quadratic form. s x x’

How are the coordinates (x, t) and (x’, t’) related?
Bill Alice x x’

How are the coordinates (x, t) and (x’, t’) related?
The coordinates (t, x) for Alice and the coordinates (t’, x’) for Bill are the components of the same spacetime vector. They must therefore be related by a linear transformation. x’ = ax + bt t’ = dx + et

The coordinates transform linearly
The coefficients a,b,c and d may be functions of velocity. x’ = a(v)x + b(v)t t’ = d(v)x + e(v)t The inverse transformation must simply replace v by -v. x = a(-v)x’ + b(-v)t’ t = d(-v)x’ + e(-v)t’

The Galilean transformation
In Newtonian physics, the result is the Galilean transformation x’ = x + vt t’ = t and it’s inverse, x = x’ - vt’ t = t’ For special relativity, the linear map is called the Lorentz transformation

The Lorentz transformation
First, consider the special form of the inverse transform. x = a(-v)x’ + b(-v)t’ t = d(-v)x’ + e(-v)t’

x = a(-v)x’ + b(-v)t’ = e(v)x’ - b(v)t’
The inverse transform In general, the inverse of a linear transformation is given by: x = l (e(v)x’ - b(v)t’) t = l (-d(v)x’ + a(v)t’) where l is the inverse determinant, l = 1/(ae -bd). By adjusting the units in the two systems we can impose l = 1. Therefore, x = a(-v)x’ + b(-v)t’ = e(v)x’ - b(v)t’ t = d(-v)x’ + e(-v)t’ = -d(v)x’ + a(v)t’ Equating like terms: a(-v) = e(v) b(-v) = - b(v) d(-v) = - d(v)

The Lorentz transform x’ = a(v)x + b(v)t t’ = -d(v)x + a(-v)t
We may now write the Lorentz transformation in the form: x’ = a(v)x + b(v)t t’ = -d(v)x + a(-v)t where b(v) and d(v) are antisymmetric.

We need to consider clocks in detail to derive the full Lorentz transformation

Let’s study a simple clock.

Imagine a pulse of light bouncing between a pair of mirrors.

A simple clock. One full cycle of the clock takes time Dt = 2L/c
L meters v = c

Now let’s watch the same clock as it moves past us with velocity v
L meters v

This time, the light appears to us to travel further.

Find the cycle time, Dt’, of the moving clock
vDt’

L2 + (vDt’/2)2 = cDt’/2 vDt’

L2 + (vDt’/2)2 = cDt’/2 L2 + v 2 (Dt’)2 /4 = c 2 (Dt’) 2 /4 (c v2)(Dt’) 2 = 4L2 Dt’ = 2L/(c v2)1/2

Since Dt = 2L/c, Dt Dt’ = 1 - v2 /c2

Now substitute into the linear Lorentz transformation
1 g = Let 1 - v2 /c2 Then Dt’ = t’2 - t’1 = d(v)(x2 - x1) + a(-v)(t2 - t1) = a(-v) Dt Since Dt’ = g Dt we have a(v) = a(-v) = g

Constancy of the speed of light
Now imagine an expanding sphere of light. In Alice’s frame, the x-coordinate of the sphere is given by x = ct. In Bill’s frame, the x’-coordinate of the sphere is given by x’ = ct’. Substitute these conditions into the Lorentz transformation equations.

An expanding sphere of light
Set x = ct and x’ = ct’: ct’ = gct + bt = (gc + b)t t’ = dct + gt = (dc + g)t Then combining, c(dc + g) = gc + b d= b/c2

Velocity The Lorentz transformation must be of the form: x’ = gx + bt t’ = bx/c2 + gt An object at rest in Alice’s frame has dx/dt = 0. In Bill’s frame, the same object moves with velocity dx’/dt’ = v. Therefore, differentiating: dx’ = a dx + b dt dt’ = b dx/c2 + g dt So: v = dx’/dt’ = a dx + b dt b dx/c2 + g dt b = g

The Lorentz transformation must be of the form: x’ = g(x + vt) t’ = g (t + vx/c2) We assume that the relative motion of the frames is in the x direction. This also leads to y’ = y z’ = z With a bit of algebra, it is not too hard to find the form of the transformation for an arbitrary relative velocity.

The invariant interval
We now seek a quadratic form that is independent of the observer’s frame of reference. x’ = g(x + vt) y’ = y z’ = z t’ = g (t + vx/c2) We already know the form of the invariant for Euclidean space: L2 = x2 + y2 + z2 This must still be invariant when t = 0.

The invariant interval must therefore be of the form: s2 = f t2 + x2 + y2 + z2 + gt (x + y + z) Invariance requires s2 = s’2 = f t’2 + x’2 + y’2 + z’2 + gt’ (x’ + y’ + z’)

Substitute the Lorentz transformation: s’2 = f t’2 + x’2 + y’2 + z’2 + gt’ (x’ + y’ + z’) = f g2 (t + vx/c2)2 + (g2(x + vt)2 + y2 + z2) + g g2 (t + vx/c2)(x + vt + y + z) = g2 [ f (t vxt/c2 + v2x2/c4 ) + x2 + 2xvt + v2t2] + y2 + z2 + g2 g (xt + vx2/c2 + vt2 + v2tx/c2 + ty + vxy/c2 + tz + vxz/c2]

Now compare: s2 = f t2 + x2 + y2 + z2 + g t (x + y + z) s’2 = g2 [ f (t vxt/c2 + v2x2/c4 ) + x2 + 2xvt + v2t2] + y2 + z2 + g2 g (xt + vx2/c2 + vt2 + v2tx/c2 + ty + vxy/c2 + tz + vxz/c2] Since s has no xy, xz or yz terms, we require g = 0.

With g = 0, s’2 = g2 [ f (t vxt/c2 + v2x2/c4 ) + x2 + 2xvt + v2t2] + y2 + z2 s’2 = g2 [ (f + v2)t2 + (fv2/c )x2 + 2(f/c2 + 1)vxt] + y2 + z2 Then 0 = s’2 - s2 = [ (g2f + g2v2 -f )t2 + (g2fv2/c4 + g2 - 1)x2 + 2(f/c2 + 1)vxt

This gives the three conditions 0 = (g2f + g2v2 -f ) 0 = (g2fv2/c4 + g2 - 1) 0 = f/c2 + 1 All three equations are solved by the single condition f = - c2 The invariant interval finally takes the form s2 = - c2t2 + x2 + y2 + z2

s2 = - c2t2 + x2 + y2 + z2 s is called the proper length

It is also convenient to define the proper time, t, by c2t2 = c2t2 - x2 - y2 - z2 Both s and t are Lorentz invariant. We use whichever is convenient.

Event separation and the invariant interval
s2 = - c2t2 + x2 + y2 + z2 For events lying in the x-t plane, we can set y = z = 0 Then, with c = 1, we write s2 = - t2 + x2 t2 = t2 - x2

Timelike separated events have s < 0
Whenever events are timelike separated, it is possible for an observer to be present at both events. s2 = - t2 + x2 < 0 t2 = t2 - x2 > 0

Spacelike separated events have s > 0
Whenever events are spacelike separated, they are simultaneous for some observer. t x s2 = - t2 + x2 > 0 t2 = t2 - x2 < 0

Lightlike separated events have s = 0
Whenever events are lightlike separated, light can travel from one to the other. s2 = - t2 + x2 = 0 t2 = t x2 = 0

Some paths are longer than others

Alice B This is due to the familiar “triangle inequality” property of a vector space Let’s check… C A x

Alice (6,0) B Assign coordinates to all relevant events. Notice that it doesn’t matter which frame of reference we choose! (3,2) C A x (0,0)

Alice (6,0) B Now compute the invariant length of each vector. tA2 = (3-0)2 - (2-0)2 = 5 tB2 = (6-3)2 - (0-2)2 = 5 tC2 = (6-0)2 - (0-0)2 = 36 (3,2) C A x (0,0)

Alice (6,0) B The triangle inequality holds with the inequality reversed: (3,2) C tA = tB = 5 = 2.24 tC = 6 A tC > tA + tB x (0,0)

Alice (6,0) B tC > tA + tB The change from < to > is because of the minus sign in the invariant interval. (3,2) C A x (0,0)

Alice (6,0) B tC > tA + tB (3,2) Now consider the physical interpretation C A x (0,0)

Alice t (6,0) Alice sees Path B as the world line of the friend returning at 2/3 the speed of light. During his return, Sid’s clock advances another 2.24 years B tC > tA + tB tC is just the elapsed time on Alice’s watch. Alice ages by 6 years. Path C is Alice’s world line. In her reference frame, she is at rest. (3,2) C Sid covers two light years in 3 years. At rest in his own frame, Sid ages 2.24 years. Alice sees Path A as the world line of a friend Sid moving to the right at 2/3 the speed of light. A Alice ages 6 years Sid ages 4.48 years x (0,0)

The result is independent of frame
Alice t (6,0) (3,2) Alice ages 6 years Sid ages 4.48 years x (0,0)

Let Carlos move to the left at .25c relative to Alice
Look in Carlos’ frame t’ t Alice Carlos (6,0) B Let Carlos move to the left at .25c relative to Alice (4,-1) (3,2) His x’ axis is symmetrically placed Carlos moves to x = -1 by time t = 4 A x (0,0) x’

Locate the key events in Carlos’ frame g = (16/15)1/2
Look in Carlos’ frame t’ t Alice Carlos (6,0) (6g,- 1.5g) B Locate the key events in Carlos’ frame g = (16/15)1/2 (3,2) (3.5g,2.75g) A x (0,0) (0,0) x’

Compute the proper times
Alice Carlos (6,0) (6g, 1.5g) tA2 = (3.5g)2 - (2.75g) = ( )/ = 5 tB2 = (2.5g)2 - (1.25) = ( )/ =5 tC2 = (6g)2 - (1.5g) = 36g g = x 16/ = 36 B g = (16/15)1/2 (3,2) (3.5g,2.75g) A x (0,0) (0,0) x’

Alice Carlos (6,0) (6g, 1.5g) tA2 = 5 tB2 = 5 tC2 = 36 B g = (16/15)1/2 (3,2) (3.5g,2.75g) Now consider the physical interpretation. A x (0,0) (0,0) x’

Alice Carlos (6,0) (6g, 1.5g) In Carlos’ frame Alice ages 6 years while Sid ages 4.48 years. B g = (16/15)1/2 (3,2) (3.5g,2.75g) A x (0,0) (0,0) x’

The result is independent of frame
Alice (6,0) (6g, 1.5g) B Carlos In Carlos’ frame: Alice ages 6 years Sid ages 4.48 years. In Alice’s frame: Alice ages 6 years Sid ages 4.48 years. (3,2) (3.5g,2.75g) A x (0,0) (0,0) x’

Physical properties may always be characterized by invariant quantities.
2.24 yr 6 yr 2.24 yr Compute invariant spacetime quantities in whichever reference frame is most convenient.

Lotsa light!

Special Relativity Jim Wheeler Physics: Advanced Mechanics.

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