Special Relativity Jim Wheeler Physics Advanced Mechanics Postulates

Special Relativity Jim Wheeler Physics: Advanced Mechanics

Postulates of Special Relativity 1. Spacetime is homogeneous and isotropic 2. All inertial frames are equivalent 3. 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 = 2. 998 x 108 m/s = 1 light second / second This is particularly striking, given the equivalence of inertial frames.

A picture of spacetime

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

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

What we all agree on more light (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 Suppose some observers move in straight lines Observers move slower than light. Postulate 1: Spacetime is homogeneous and isotropic World line of an observer

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

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

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

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

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

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

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

Observe Alice +2 s (0, 2) 0 s The distance of a remote event 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. -2 s (Postulate 3: Constancy of the speed of light)

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

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

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

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

Spacetime coordinates 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, 0) (0, -2) (0, -1) -2 s (0, 1) (0, 2) (0, 3)

Spacetime coordinates 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, 0) (0, -2) (0, -1) -2 s (0, 1) (0, 2) (0, 3)

Spacetime coordinates 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, 0) (0, -2) (0, -1) -2 s (0, 1) (0, 2) (0, 3)

Spacetime coordinates 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, 0) (0, -2) (0, -1) -2 s (0, 1) (0, 2) (0, 3)

Spacetime coordinates 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, 0) (0, -2) (0, -1) -2 s (0, 1) (0, 2) (0, 3)

Spacetime coordinates 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, -2) (0, -1) (-2, 0) (0, 1) (0, 2) (0, 3)

Spacetime coordinates 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, -2) (0, -1) (-1, 0) (-2, 0) (0, 1) (0, 2) (0, 3)

Spacetime coordinates 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, -2) (0, -1) (-1, 0) (-2, 0) (0, 1) (0, 2) (0, 3)

Spacetime coordinates 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, -2) (0, -1) (-1, 0) (-2, 0) (0, 1) (0, 2) (0, 3)

Spacetime coordinates (5, 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, -2) (0, -1) (-1, 0) (-2, 0) (0, 1) (0, 2) (0, 3)

Spacetime coordinates (5, 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. (0, -3) (2, 0) (1, 0) (0, -2) (0, -1) (-1, 0) (-2, 0) (0, 1) (0, 2) (0, 3) (0, 4)

Spacetime coordinates (5, 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. (0, -4) (2, 0) (1, 0) (0, -3) (0, -2) (0, 0) (0, -1) (-1, 0) (-2, 0) (0, 1) (0, 2) (0, 3) (0, 4)

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

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

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

Alice’s coordinate axes t Alice x

Alice and Bill’s coordinate axes t t’ Alice Bill 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. y x

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Return to Alice’s coordinate axes Alice

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

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. t Alice Events along the x-axis no longer line up. x

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

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’ Bill t Alice x x’

What do Alice and Bill agree on? t’ Bill t Alice x x’

What do Alice and Bill agree on? t’ Bill t Alice The light cone x x’

What do Alice and Bill agree on? t’ Bill t Alice The light cone Timelike, null, and spacelike separations of events. x x’

Do Alice and Bill agree on anything else? t’ Bill t Alice The light cone Timelike, null, and spacelike separations of events. x x’

Spacetime is a vector space! t’ Bill t Alice x x’

t’ Bill t Alice Does a spacetime vector have an invariant length? s x x’

In Euclidean space, the Pythagorean theorem gives an invariant length. y L 2 = Dx 2 + Dy 2 Dy Dx x

y The invariant length can be expressed as a quadratic expression in terms of the coordinates. (x 2, y 2) L 2 = (x 2 - x 1)2 + (y 2 - y 1)2 Dy = y 2 - y 1 (x 1, y 1) Dx = x 2 - x 1 x

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

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

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 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.

The Lorentz transformation 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Find the cycle time, Dt’, of the moving clock L v. Dt’

Find the cycle time, Dt’, of the moving clock L L 2 + (v. Dt’/2)2 v. Dt’ = c. Dt’/2

Find the cycle time, Dt’, of the moving clock L 2 + (v. Dt’/2)2 = c. Dt’/2 L 2 + v 2 (Dt’)2 /4 = c 2 (Dt’) 2 /4 (c 2 - v 2)(Dt’) 2 = 4 L 2 Dt’ = 2 L/(c 2 - v 2)1/2

Find the cycle time, Dt’, of the moving clock Since Dt = 2 L/c, Dt’ = Dt 1 - v 2 /c 2

Now substitute into the linear Lorentz transformation Let g = 1 1 - v 2 /c 2 Then Dt’ = t’ 2 - t’ 1 = d(v)(x 2 - x 1) + a(-v)(t 2 - t 1) = 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/c 2

Velocity The Lorentz transformation must be of the form: x’ = gx + bt t’ = bx/c 2 + 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/c 2 + g dt So: a dx + b dt v = dx’/dt’ = b dx/c 2 + g dt = b g

The Lorentz transformation must be of the form: x’ = g(x + vt) t’ = g (t + vx/c 2) 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/c 2) We already know the form of the invariant for Euclidean space: L 2 = x 2 + y 2 + z 2 This must still be invariant when t = 0.

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

The invariant interval Substitute the Lorentz transformation: s’ 2 = f t’ 2 + x’ 2 + y’ 2 + z’ 2 + gt’ (x’ + y’ + z’) = f g 2 (t + vx/c 2)2 + (g 2(x + vt)2 + y 2 + z 2) + g g 2 (t + vx/c 2)(x + vt + y + z) = g 2 [ f (t 2 + 2 vxt/c 2 + v 2 x 2/c 4 ) + x 2 + 2 xvt + v 2 t 2] + y 2 + z 2 + g 2 g (xt + vx 2/c 2 + vt 2 + v 2 tx/c 2 + ty + vxy/c 2 + tz + vxz/c 2]

The invariant interval Now compare: s 2 = f t 2 + x 2 + y 2 + z 2 + g t (x + y + z) s’ 2 = g 2 [ f (t 2 + 2 vxt/c 2 + v 2 x 2/c 4 ) + x 2 + 2 xvt + v 2 t 2] + y 2 + z 2 + g 2 g (xt + vx 2/c 2 + vt 2 + v 2 tx/c 2 + ty + vxy/c 2 + tz + vxz/c 2] Since s has no xy, xz or yz terms, we require g = 0.

The invariant interval With g = 0, s’ 2 = g 2 [ f (t 2 + 2 vxt/c 2 + v 2 x 2/c 4 ) + x 2 + 2 xvt + v 2 t 2] + y 2 + z 2 s’ 2 = g 2 [ (f + v 2)t 2 + (fv 2/c 4 + 1)x 2 + 2(f/c 2 + 1)vxt] + y 2 + z 2 Then 0 = s’ 2 - s 2 = [ (g 2 f + g 2 v 2 -f )t 2 + (g 2 fv 2/c 4 + g 2 - 1)x 2 + 2(f/c 2 + 1)vxt

The invariant interval This gives the three conditions 0 = (g 2 f + g 2 v 2 -f ) 0 = (g 2 fv 2/c 4 + g 2 - 1) 0 = f/c 2 + 1 All three equations are solved by the single condition f = - c 2 The invariant interval finally takes the form s 2 = - c 2 t 2 + x 2 + y 2 + z 2

The invariant interval s 2 = - c 2 t 2 + x 2 + y 2 + z 2 s is called the proper length

The invariant interval It is also convenient to define the proper time, t, by c 2 t 2 = c 2 t 2 - x 2 - y 2 - z 2 Both s and t are Lorentz invariant. We use whichever is convenient.

Event separation and the invariant interval s 2 = - c 2 t 2 + x 2 + y 2 + z 2 For events lying in the x-t plane, we can set y=z=0 Then, with c = 1, we write s 2 = - t 2 + x 2 t 2 = t 2 - x 2

Timelike separated events have s < 0 Whenever events are timelike separated, it is possible for an observer to be present at both events. s 2 = - t 2 + x 2 < 0 t 2 = t 2 - x 2 > 0

Spacelike separated events have s > 0 t Whenever events are spacelike separated, they are simultaneous for some observer. x s 2 = - t 2 + x 2 > 0 t 2 = t 2 - x 2 < 0

Lightlike separated events have s = 0 Whenever events are lightlike separated, light can travel from one to the other. s 2 = - t 2 + x 2 = 0 t 2 = t 2 - x 2 = 0

Some paths are longer than others

Some paths are longer than others t Alice B This is due to the familiar “triangle inequality” property of a vector space Let’s check… C A x

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

Some paths are longer than others t (6, 0) Alice B Now compute the invariant length of each vector. t. A 2 = (3 -0)2 - (2 -0)2 = 5 (3, 2) C t. B 2 = (6 -3)2 - (0 -2)2 = 5 A t. C 2 = (6 -0)2 - (0 -0)2 = 36 x (0, 0)

Some paths are longer than others t (6, 0) Alice B The triangle inequality holds with the inequality reversed: t. A = t. B = 5 = 2. 24 (3, 2) C t. C = 6 A t. C > t. A + t. B x (0, 0)

Some paths are longer than others t (6, 0) Alice B t. C > t. A + t. B The change from < to > is because of the minus sign in the invariant interval. (3, 2) C A x (0, 0)

Some paths are longer than others t (6, 0) Alice B t. C > t. A + t. B Now consider the physical interpretation (3, 2) C A x (0, 0)

Some paths are longer than others t (6, 0) t. C > t. A + t. B C is Alice’s world t. Path C is just the elapsed Alice sees Path B as the During his return, Sid’s world line of the friend clock advances another Breturning at 2/3 the 2. 24 years speed of light. (3, 2) line. In reference time onher Alice’s watch. C frame, ages she by is at rest. Alice 6 years. Alice sees two Pathlight A asyears the Sid covers A world line of friend Sid in 3 years. Atarest in his moving to the at 2. 24 2/3 own frame, Sidright ages the speed of light. years. Alice ages 6 years Sid ages 4. 48 years x (0, 0)

The result is independent of frame t Alice (6, 0) (3, 2) Alice ages 6 years Sid ages 4. 48 years x (0, 0)

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

Look in Carlos’ frame t’ Carlos t (6, 0) Alice (6 g, - 1. 5 g) B (3, 2) Locate the key events in Carlos’ frame g= (3. 5 g, 2. 75 g) A (16/15)1/2 x (0, 0) x’

Compute the proper times t’ Carlos t (6, 0) t. A 2 = (3. 5 g)2 - (2. 75 g)2 = (196 - 121)/15 =5 Alice (6 g, 1. 5 g) B t. B 2 = (2. 5 g)2 - (1. 25)2 = (100 - 25)/15 =5 g = (16/15)1/2 (3, 2) (3. 5 g, 2. 75 g) t. C 2 = (6 g)2 - (1. 5 g)2 = 36 g 2 - 2. 25 g 2 = 33. 75 x 16/15 = 36 A x (0, 0) x’

Compute the proper times t’ Carlos t (6, 0) Alice (6 g, 1. 5 g) t A 2 = 5 B g = (16/15)1/2 t B 2 = 5 (3, 2) t. C 2 = 36 (3. 5 g, 2. 75 g) Now consider the physical interpretation. A x (0, 0) x’

Compute the proper times t’ Carlos t (6, 0) Alice (6 g, 1. 5 g) B In Carlos’ frame Alice ages 6 years while Sid ages 4. 48 years. g = (16/15)1/2 (3, 2) (3. 5 g, 2. 75 g) A x (0, 0) x’

The result is independent of frame t’ t (6, 0) Alice (6 g, 1. 5 g) B Carlos In Carlos’ frame: Alice ages 6 years Sid ages 4. 48 years. (3, 2) (3. 5 g, 2. 75 g) In Alice’s frame: Alice ages 6 years Sid ages 4. 48 years. A x (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!