Lesson 13 Web

Lesson 13: More About Special Relativity

Introduction

Special relativity first appeared in chapter 11 as a set of Lorentz transformations, four-vectors, and a metric that turned the interval into a quadratic form. Lesson 12 then gave that arena a name: Minkowski spacetime is a smooth manifold, and the objects we called four-vectors and one-forms are the rank-one tensors that live on it. What we have not yet done is stay with a single observer and read the geometry of that observer’s history.

That history is a world-line, a timelike curve parameterized by proper time. Once the curve is in hand, the familiar questions of elementary relativity—how velocities add, what an accelerating clock does, how one observer looks at another—become statements about the tangent, the curvature, and the twisting of that curve in spacetime. The same constructions that Frenet and Serret wrote for a curve in Euclidean three-space have Minkowski analogues. The unit tangent is the four-velocity. The curvature of the world-line is the magnitude of the four-acceleration. The normal, binormal, osculating plane, and rectifying plane organize the instantaneous rest space of the traveler. Torsion measures how that rest space is twisting. The moving trihedron they form is a frame that an accelerated observer can actually carry.

We will begin with the geometry of world-lines and recover velocity addition and acceleration from it, including the relation between four-acceleration and the curvature of the path. That leads to the reference frame of an observer and then to the Frenet–Serret frame in spacetime, with its unit tangent, the tangent line, the normal plane, curvature, the principal unit normal (with a short pause on analytic curves and points of inflection), the principal normal line, the osculating plane, the binormal and binormal line, the rectifying plane, the moving trihedron, torsion, and the Frenet equations themselves.

Two tensor tools make the bookkeeping clean. The Levi-Civita tensor in spacetime turns pairs of vectors into the planes they span and lets us decompose two-forms into “electric” and “magnetic” pieces relative to an observer. Along the way we have to ask what happens when the local frame itself is changing. Differentiation in spacetime then splits into three operations we will need again in general relativity: the absolute derivative along a world-line, the derivative with respect to an observer, and the Fermi–Walker derivative that transports a vector without the extra rotation that a spinning observer would see.

None of this requires curvature of spacetime. The manifold is still flat. What is curved is the world-line of an accelerated traveler, and that is enough to make frames, simultaneity, and “what I measure” into geometric questions. The last section does the same constructions in Mathematica so that the frames, curvatures, and Fermi–Walker transport can be computed rather than only written.

By the end of the lesson the informal four-velocity and four-acceleration of earlier chapters will have become the first two legs of a spacetime Frenet frame, and the derivative along an observer’s history will be a well-defined operator rather than a slogan. That is the language we will carry into fields, continua, and, later, gravity.

The Geometry of World-Lines

Consider a spacetime manifold M. Assume that there is an open interval, I, of the real line. We can establish a mapping from the open interval, I, to the manifold, M, and call it β. We assume β to be a smooth mapping, β:I→M. For every number in our open interval, λ∈I, there is a tangent vector at the point, β(λ)∈M. This mapping forms a curve in our manifold. If, for every point λ the tangent vector is timelike, then the curve is also timelike. Similarly for null and spacelike tangent vectors, the curves will be null and spacelike, respectively.

In what follows the arguments are—as such arguments always are—circular. This is based on the section Light-Cone Structure from Lesson 11. The world line of a particle is a time-like curve. The world line of a light beam is a null curve. We work throughout with signature (−,+,+,+).

We can reparameterize the curve by stating that λ=λ(λ'), thus replacing the parameter λ by the new parameter λ’. So the tangent vectors and are related by

(13.1)

Given that p and q are events along the world line we can write

(13.2)

The two sides are equal by (13.1). Note that this integral depends on only three things: the world line, p, and q. It is independent of the parametrization we have used. This integral is the length of the timelike world line between the events. This length of the world line is the elapsed proper time between events along the world line.

Let’s say we have two twins that begin in a spacetime event and proceed along two different world lines extending from event p and ultimately arriving at event q.

Figure 13.1 The twin paradox represented by world lines.

One world line is longer than the other. That means that one twin will experience a longer worldline, and thus will be older, than the other when they arrive at q. This is called the twin paradox.

From (13.1) we can always choose a parameter—given a timelike curve—whose tangent vector is a unit, . Such a parameter is unique within a constant. By the integral in (13.2) we can see that the proper time between events is just the difference in time parameters of the points. The world line is said to be parameterized by proper time. The unit tangent vector to the curve is the four-velocity of the particle at that location on the world line.

By these arguments, when we speak of an observer we are not attributing intellect, or will, to some particle, instead we are speaking of a world line.

Given an event p and a nearby event q. The displacement vector can be denoted . How do we find ? Think about it. There have to be two parts, right? One part is the spatial distance between the events. The other has to be orthogonal to the four-velocity representing how much earlier or later it is. This temporal displacement is given by . The spatial displacement can be given by the sum , so the elapsed time is then

(13.3)

Exercise 13.1: Derive this result from the projection .

The remainder is orthogonal to the four-velocity , , and the spatial distance is

(13.4)

Exercise 13.2: Derive this result.

The spacetime interval is

(13.5)

Exercise 13.3: Derive this result.

Each observer can decompose spacetime into space and time. The squared spacetime interval between events will always be the same, no matter the observer. This gives us a peculiar situation, where the spacetime interval between events is independent of the observer, but the decomposition of spacetime into space and time is not.

If q lies on the world-line of a particle, the speed of that particle relative to the observer is

(13.6)

Exercise 13.4: Derive this result from (13.3) and (13.4).

Let be the observer’s four-velocity and the particle’s four-velocity. The four-momentum of the particle is . So, for the energy, we write -, or

(13.7)

where v is the speed (13.6) of the particle relative to that observer. If the observer is the particle itself, then U=v and E=m, the rest energy.

Exercise 13.5: Derive this result.

Velocity Addition

Say the we are in the frame F moving with speed u and we see a particle moving with some speed v relative to F. We can further state that this particle is moving in the negative direction with respect to F. We can write, for some arbitrary constant d,

(13.8)

Exercise 13.6: Derive this result.

The question is, what is the speed with respect to the frame of the particle, say F'?

Exercise 13.7: What would this be in a non-relativistic frame?

To answer this question we first need to look at the matrix representation of the Lorentz transformation from Lesson 11 (11.13)

We use this to operate on our position vector to transform it into the new position vector for the F' frame,

(13.9)

or, in terms of coordinates

(13.10)

So we can write

(13.11)

and

(13.12)

Exercise 13.8: Derive these results.

So, the speed in F' is

(13.13)

This is the famous velocity addition formula.

Exercise 13.9: Show that this works when u and v are 4-vectors and not just speeds.

If we define the quantity

(13.14)

we can call it the rapidity. We can apply this to the Lorentz transformations, and write

(13.15)

Exercise 13.10: Show that this to be true.

We can then rewrite the velocity transformation law

(13.16)

Exercise 13.11: Show that this to be true.

Acceleration

Suppose you are on a building and you fall off the roof. This constitutes one event, say p. At some later point on the world line you reach event q, you hit the Earth. The world line might look like Figure 13.2.

Figure 13.2 The spacetime diagram for a falling body.

Let’s say our observer is carrying an accelerometer. In this way an observer can always measure a vector that is orthogonal to their four-velocity at every event along their world line, . This vector is called the four-acceleration along the world line at the location of the event. What the device actually displays is the magnitude of the proper acceleration, the acceleration you feel.

We can examine the parts of the world line and see what the accelerations are at various times. We will write the acceleration for each part. It is important not to confuse that reading with “the acceleration due to gravity” in the Newtonian sense. An accelerometer does not measure gravitational pull. It measures whatever non-gravitational force is acting on you, written as an acceleration. Before you leave the roof, the roof of the building pushes up on you. The accelerometer reads a proper acceleration of magnitude g, directed away from the Earth.

Figure 13.3 The acceleration you experience as the building supports you against the pull of gravity.

Once you begin to fall, you no longer experience your own weight, so the acceleration you experience is 0.

Figure 13.4 The acceleration you experience as you fall.

When you hit the ground the ground stops you in a short interval. The accelerometer reads a large spike.

Figure 13.5 The large acceleration you experience as you hit the Earth.

After that you again experience the acceleration due to gravity as you are on the ground.

Figure 13.6 Feeling the Earth support you against gravity again.

In a sense you only experience the acceleration due to gravity when something is pushing you up, the building or the ground in this case.

The same device makes the contrast with electromagnetism clear. In an electric field a charged body accelerates. If the test mass inside the accelerometer is neutral, the accelerometer still records only the proper acceleration of its carrier. If that test mass were charged as well, and had the same charge-to-mass ratio as you, it would fall with you and the needle would stay at zero. That is why an accelerometer is built with neutral masses: it is not a detector of electric fields, and it is not a detector of gravity either. It detects pushes and pulls that are not universal.

An observer who “feels an acceleration” is being acted on by a non-gravitational force. That is the content of the equivalence principle in this setting. In a small closed room you cannot tell, by local mechanical experiments, whether the room is at rest in a gravitational field or accelerating in empty space. Free fall in a gravitational field is locally indistinguishable from sitting still in the absence of one.

The precise statement is not that gravity is an acceleration, and not that gravity is “an artifact of acceleration.” The precise statement is that the locally measurable acceleration is proper acceleration, and gravity does not contribute to it. In the language we will use later, a freely falling world-line has vanishing four-acceleration; a world-line held at rest on the Earth does not.

Acceleration and World-Line Curvature

If we examine a world line, how do we link it to the presence of an acceleration? At every point along the world line we have an acceleration vector, orthogonal to the four-velocity at that point.

Return to the falling observer of Figure 13.2. While the observer stands on the roof or on the ground, the world-line is straight in an inertial frame attached to the Earth. During the fall it is also nearly straight (free fall). At the impact event (q) the path bends sharply. That bend is the geometric mark of a large proper acceleration. A true kink would mean a jump in the tangent itself, as in an ideal impulse. A smooth but tight bend means a large finite acceleration. In either case, the more the world-line turns away from its tangent, the larger the acceleration.

We can measure this bending. If the world line is a curve, c, then we can say that c:I→M that is timelike and parameterized by length. There is a corresponding tangent vector to this curve. We can define a quantity,

(13.17)

On a general spacetime the same object is the absolute derivative along the world-line,

This is called the curvature vector of the world line. Its magnitude is the curvature of the curve.

Exercise 13.12: Prove that the curvature is orthogonal to the tangent vector .

Acceleration is the curvature of a world line, in other words . There are three things that make this at least plausible. First, both and are orthogonal to . Second, we can see the curvature and acceleration in Figure 13.2

where there is a kink at q that corresponds to a large acceleration. Third, is the derivative of , and this seems very acceleration-like.

I have demonstrated that this argument is plausible. There seems no way to present it conclusively. In fact, there exist theories—and I will not present them here—where this argument does not hold. We will treat it as a postulate. It does have one strong advantage, there need be no additional tensor field on M to link curvature to acceleration.

Hyperplanes

Start with a space of dimension n and cut it by one linear equation. Fix a nonzero covector and a constant c. The points that satisfy

(13.18)

form a flat set of dimension n−1. If c=0 the set passes through the origin and is a vector subspace. If c≠0 it is a parallel copy of that subspace.

In a 2-dimensional plane that set is a line. In ordinary 3-space it is an ordinary plane. In 4-dimensional spacetime it is a flat 3-dimensional slice. In every dimension the same object is called a hyperplane.

The vector dual to is orthogonal to every displacement that remains inside the set. It is a normal to the slice.

At an event p on a world-line with four-velocity , the points that satisfy

(13.19)

are the instantaneous rest space of the observer at p. That rest space is a hyperplane. The four-acceleration lies in it, which is why the world-line bends in the observer’s rest space and not along the four-velocity.

Four-Momentum and Four-Forces

The instantaneous rest space of an observer at an event is the hyperplane orthogonal to the four-velocity . The curvature vector lies in that hyperplane where the world-line is turning in the observer’s rest space, not along the four-velocity.

Let a particle of mass m travel on that world-line. Its four-momentum is

(13.20)

The four-force is the absolute derivative of this momentum along the world-line,

(13.21)

In inertial coordinates the same statement is

(13.22)

Because is orthogonal to , the four-force is orthogonal to the four-velocity as well,

(13.23)

That is the relativistic form of the fact that a force changes the momentum but does not change the rest mass, the part of along would change m.

Every non-gravitational force enters the motion in this way. Gravity does not. A free particle has , hence , hence vanishing world-line curvature. Its history is a timelike geodesic.

That is the geometric meaning of the falling observer. Standing on the roof you are not on a geodesic. The floor supplies a four-force, the accelerometer reads a nonzero proper acceleration, and the world-line is curved. In free fall there is no four-force, the accelerometer reads zero, and the world-line is straight in the inertial sense. Hitting the Earth is a brief, violent four-force and a sharp bend in the world-line.

The Reference Frame of an Observer

As an observer moves along a world-line, each proper time τ has its own instantaneous rest space: the hyperplane through that event orthogonal to the four-velocity. Call that hyperplane . The collection of these hyperplanes, one for each τ, is the family of rest spaces of the observer.

Figure 13.7 A system of hyperplanes along the world line of an observer.

Each single rest space is a three-dimensional Euclidean vector space once an origin is chosen at the observer’s location. The rest spaces at different τ are different slices of spacetime. Taken together they are what we will mean by the space of the observer, written O.

We then represent the location of an event P in the local frame by assigning the coordinates in the space having the orthonormal basis . We can then establish the mapping where .

The mapping φ also induces an isomorphism between the local rest frames of O and , such that, ; in other words the isomorphism sends the local basis at the proper time τ to the Euclidean basis. Another way of looking at this it to say that the spacelike vectors of the local frames of O are held fixed in while they evolve with τ in .

We can then say that a vector field along a world line is considered fixed with respect to O iff . A vector field where is fixed with respect to O iff turns out not to depend on τ.

From this we can conclude that a particle is fixed with respect to the observer iff the spatial coordinates are the same for all events on the world line.

With this idea we return to our view of motion as composed of translation and rotation. We have already discussed, to some extent translation. What about rotation?

In some respect we need to connect this idea to the evolution of a local frame along the world line of our observer. To do this we will need to study another derivative, this time of the basis vectors, with respect to the proper time along that world line. We will call this the 4-rotation of the local frame.

We seek to find the derivative of a basis vector with respect to τ. Assuming we find such, it will also be a vector. Since we are dealing with a unique basis of spacetime there must be a set of four real functions where,

(13.24)

For some vector field, , fixed with respect to O, we can see the time-rate of change,

(13.25)

Exercise 13.13: Explain this result.

We could write this in component-free notation

(13.26)

Here Ω is some mapping. In fact this is a special kind of mapping. We will define it as a structure-preserving mapping from our local rest frame to itself. This is called an endomorphism. So Ω is an endomorphism operating on the vector .

If we allow our vector field to be the 4-velocity , then we write

(13.27)

If we accept this as a definition of 4-acceleration, we write

(13.28)

If we consider the duality of the metric tensor, we can associate a 2-form to Ω. We can justify this by saying that for any vector in , say , then is also a vector in . If we have another vector, , also in , then forms a scalar; it is this we will use to define the 2-form ,

(13.29)

Exercise 13.14: Explain this result.

In component form, we write

(13.30)

Exercise 13.15: Explain this result.

From this we can conclude

(13.31)

If we adopt a special symbol for the Minkowski metric (and we have been calling it by the same symbol as the generic metric tensor), we choose the traditional , where

(13.32)

Now, if we take the derivative

(13.33)

or,

(13.34)

We can rewrite this

(13.35)

We can apply (13.24) to get

(13.36)

Exercise 13.16: Derive this result.

From this we conclude that the 2-form is antisymmetric.

Exercise 13.17: Show that is not antisymmetric.

The Frenet-Serret Frame

We have been discussing basis vectors for the local reference frame of an observer. That raises a natural question, “Is there a preferred basis, fixed by the world-line itself rather than by an arbitrary choice of axes?”

There is a standard answer for ordinary curves in Euclidean 3-space. Along a smooth curve in one can form the unit tangent, the principal unit normal, and the binormal. Those three vectors are determined by the curve alone. They are a moving basis adapted to the path, not imported from a fixed set of laboratory axes.

The same construction can be carried out for a timelike world-line in spacetime. The unit tangent is already the four-velocity. What remains is to build the rest of the frame from the way that tangent turns and twists. That frame is the Frenet–Serret frame of the observer.

The Unit Tangent Vector in Spacetime

In the classical theory of space curves the natural parameter is arc length s. Let a curve C in be given by a position vector . Differentiating with respect to s asks how that position changes when we move a little way along the curve.

Figure 13.8 A small change in the position vector for some trajectory.

The derivative is the limit of the difference quotient

(13.37)

The quotient itself is a secant of C. In the limit the secant becomes the tangent.

Figure 10.9 The secant and velocity of the parameterized curve.

It turns out that the vector is a unit vector by our parameterization by arc length, since we are assuming a natural representation.

Exercise 10.18: Can you prove that this derivative is unique?

We call this derivative the unit tangent vector, .

What would this be when we transition from the three-dimensional to the four-dimensional spacetime manifold M? First of all, we are no longer parameterizing by s, instead we parameterize with the proper time τ. So, what is the derivative of the position with respect to proper time? It is the 4-velocity . So the analogy of the unit tangent vector is 4-velocity. We then write,

(13.38)

So the unit tangent of the world-line is the four-velocity. We take it as the first leg of the local frame,

(13.39)

That is the first of the four unit vectors of the Frenet–Serret frame in spacetime.

The Tangent Line in Spacetime

In the classical theory of space curves the tangent line at a point of C is the straight line that passes through that point and is parallel to the unit tangent. If

(13.40)

then the tangent line is the set of points

(13.41)

Here x is distance along the line, not the original arc-length parameter of C.

The same idea applies to a world-line in spacetime. At an event the four-velocity is . The tangent line is the straight world-line through that event in the direction of ,

(13.42)

The parameter λ is proper time along this straight line, not along the original world-line, unless the original world-line is already inertial.

If the observer is inertial, the world-line is its tangent line. If the observer is accelerating, the world-line peels away from this line, and the rate at which it peels away is the four-acceleration.

The Normal Plane

In the classical theory of space curves the normal plane at a point of C is the plane that passes through that point and is orthogonal to the tangent line. If and , its equation is

(13.43)

Every displacement that stays in this plane is perpendicular to .

The spacetime analog is not a plane but a hyperplane. At an event with four-velocity , the set of nearby events that satisfy

(13.44)

is the instantaneous rest space of the observer. That is the same hyperplane already introduced. It plays the role of the normal plane: it is the slice through the event orthogonal to the unit tangent of the world-line.

Exercise 13.19: Start from the Euclidean equation of the normal plane and obtain the spacetime equation above. Why is the result a hyperplane rather than a plane?

Curvature in Spacetime

Returning to classical space curve theory we next define the curvature of the curve. Since the space curve C is of the class , where m≥2, then is of the class , where n≥1. Thus we can take it’s derivative with respect to arc length, . The vector defined by this is called the curvature vector .

Exercise 13.20: Can you prove that the derivative of a unit vector function is orthogonal to the unit vector?

By this principle, we can state that the curvature vector is orthogonal to the unit tangent vector. The magnitude of the curvature vector is the curvature.

(13.45)

The inverse of this is called the radius of curvature,

(13.46)

Defining this in words, we can say that the curvature is the change of the unit tangent vector as you proceed down the curve.

The same construction applies to a timelike world-line parameterized by proper time. The unit tangent is the four-velocity , so the curvature vector is the four-acceleration,

(13.47)

For spacetime it is clear that the derivative of 4-velocity is the 4-acceleration that we introduced earlier. Thus,

(13.48)

This makes sense from the argument we made when saying that acceleration was the curvature of the world line. Similarly the radius of curvature is then,

(13.49)

This is the same identification made earlier, where the world-line bends at a rate set by the acceleration the observer feels. An inertial world-line has κ=0 and no finite radius of curvature; it is already straight.

The Principal Unit Normal Vector

Returning to classical space curve theory, if we take and divide it by the curvature κ (or multiply it by the radius of curvature), we get the principal unit normal vector,

(13.50)

Note that a straight line has no definite principal unit normal vector (just to be clear, we are not saying that there is a zero vector, we are saying that it is undefined), since for a straight line κ=0.

We can rewrite (13.41)

(13.51)

if we take the scalar product with , we get

(13.52)

(13.53)

So, let’s see what happens at a point of inflection. From what we said about straight lines we could conclude that the principal unit normal vector vanishes at a point of inflection. Let’s see if our intuition is correct. We define the inflection point as the point described on the curve C located at the head of the position vector . Being a point of inflection we write,

(13.54)

Being that the derivative is non-zero away from the point of inflection, then the function is of the class , where k>2. We see that even if the function is of the class that the derivative, and hence the curvature, can be zero at a point—so the principal unit normal vector may not exist everywhere.

So, what about analytic functions? “What is an analytic function?” you ask. Well, here we go.

Analytic Functions

A function f(x) that can be represented by a power series of the form

(13.55)

with a positive radius, R, of convergence, , for all x in the neighborhood of the point is an analytic function. For a vector function we write this,

(13.56)

or equivalently,

(13.57)

This is a normal Taylor expansion with remainder 0, since the series converges very fast. These facts can be proven, but their proof is beyond the scope of this work (though it makes a nice project for those of you who are analytically inclined).

The Principal Unit Normal Vector for A Point of Inflection Along Analytic Curves

Let k be the lowest order derivative greater than first that does not vanish at the point , . We assume that the curve parameterized by s is not straight, so such a derivative exists.

Exercise 13.21: Why is there such a k?

We can write the Taylor expansion about this point of inflection,

(13.58)

Then the unit tangent vector will be the derivative of this,

(13.59)

If we look at this long enough we will realize that we can factor out ,

(13.60)

This gives us a new series, we will call ,

(13.61)

So we can rewrite (13.51)

(13.62)

Problem 13.1: Prove that is continuous.

Since is continuous, then there must exist a neighborhood around where for some s in the neighborhood. Let’s see what happens when we play a little manipulation game with (13.53)

(13.63)

this gives us a new unit vector, I will not hold you in suspense—it is the Principle Unit Normal vector,

(13.64)

So we can rewrite (13.54)

(13.65)

To make this consistent with our curvature, then , so the curvature vector can be written

(13.66)

This will be true for any s in the neighborhood of , including by continuity.

From this we can write the following theorem.

Theorem 13.1: Any analytic curve that is not a straight line will have a definite and continuous principal unit normal vector in the neighborhood of a point of inflection.

So our intuition was almost correct, but not quite! We need to be careful in our investigations. Many textbooks on classical differential geometry (almost all of them) fail to discuss this issue.

How does this relate to our spacetime analogy for the Principal Unit Normal Vector? If we look at

and recall our discussion of curvature, this will give us

(13.67)

The Principle Normal Line

In the principal normal line at is the straight line through that point parallel to the principal unit normal,

(13.68)

Along a world-line the analogous line through the event is

(13.69)

where is the principal unit normal already defined. The line is spacelike, thus it lies in the instantaneous rest space.

Exercise 13.22: Write this spacetime equation by direct analogy with the Euclidean principal normal line. Where does the construction fail if ?

The Osculating Plane

The osculating plane of a space curve at is the plane through that point spanned by the unit tangent and the principal unit normal. It is the plane in which the curve is turning. The name means that the plane has contact with the curve: to second order, C lies in this plane.

It is not the plane perpendicular to . That later plane is the rectifying plane. The osculating plane is perpendicular to the binormal . Its equation is the triple product

(13.70)

Along a world-line the same span is two-dimensional rather than three.

Now, a 2-flat is a flat two-dimensional slice of a higher-dimensional space: an ordinary plane, but possibly sitting inside 3-space or spacetime rather than being “the whole space.” More precisely, it is an affine two-dimensional subspace. Through a point p you pick two linearly independent vectors and . The 2-flat is every point you can reach by

(13.71)

with real numbers α and β. No bending: the slice is as flat as a plane in .

At the event the osculating plane is the timelike 2-flat generated by the four-velocity and the principal unit normal,

(13.72)

The world-line and its curvature vector both lie in this 2-flat. If there is no principal normal and the osculating plane is not defined.

Exercise 13.23: Obtain the spacetime description from the Euclidean triple-product equation. Why is the result a 2-flat rather than a 3-dimensional hyperplane?

The Levi-Civita Tensor in Spacetime

The Levi-Civita tensor on spacetime is the totally antisymmetric rank-4 tensor . Swapping any two arguments changes its sign. If two arguments are equal, the tensor vanishes.

Fix an oriented local frame whose time leg is future-pointing and whose spatial triad is right-handed. In that frame

Say we choose any local frame where is pointing to the future. If the spatial basis vectors form a right-handed set, then the covariant components in this frame become

(13.73)

The same orientation rule will be used for every later tetrad, including the Frenet–Serret frame.

If α, β, γ, and δ are not all different

(13.74)

Exercise 13.36: Show this to be true due to total antisymmetry.

In an oriented orthonormal frame the components are therefore

(13.75)

Exercise 13.37: Taking an even permutation to mean an even number of swaps, how many even permutations of four distinct labels are there, and what are they?

Exercise 13.38: How many odd permutations are there, and what are they?

Raising all four indices with produces one minus sign, so

(13.76)

and, as numerical arrays in this frame,

(13.77)

With this tensor we can finish the spacetime Frenet frame. Given and , and given a third unit vector orthogonal to both, the last leg is

(13.78)

That is the four-dimensional stand-in for the Euclidean vector product that will define the binormal.

The Binormal

In the binormal is the remaining leg of the right-handed triad. It is the vector product of the unit tangent and the principal unit normal

(13.79)

It is a unit vector, orthogonal to both and , and therefore orthogonal to the osculating plane.

That formula does not copy into spacetime as it stands. The vector product of two vectors in four dimensions is not a vector; it is a 2-form. Equivalently, the orthogonal complement of span{u,n} is two-dimensional, so two unit vectors, not one, remain to be chosen.

The component in the time-like direction is the 4-velocity

(13.80)

What does carry over is the first spatial leg of the frame, with the principal unit normal already defined,

(13.81)

The next Frenet vector is fixed only after we see how itself turns. Differentiate along the world-line and throw away the pieces along and . If that remainder is nonzero, its unit vector is the analog of the binormal. The last leg then completes an orthonormal tetrad, for instance by

(13.82)

If is taken to be the next Frenet vector—the unit direction of the part of dn/dτ orthogonal to and —then is the spacetime analog of the binormal and completes the frame. Equivalently one may fix first and recover by

(13.83)

The Euclidean formula is the special case in which the ambient space is already three-dimensional, so one vector product finishes the basis. In spacetime the same job takes three vectors and .

The construction fails if : there is no principal normal, and no binormal to define from it.

The Binormal Line

In the binormal line at is the straight line through that point parallel to the binormal,

(13.84)

It is orthogonal to the osculating plane.

Along a world-line let be the spacetime analog of the binormal constructed in the last section. The corresponding line through the event is

(13.85)

This line is spacelike. It lies in the instantaneous rest space and is orthogonal to the osculating 2-flat spanned by and .

Exercise 13.24: Write this spacetime equation by direct analogy with the Euclidean binormal line. What fails if ?

The Rectifying Plane

In the rectifying plane at is the plane through that point spanned by the unit tangent and the binormal. It is perpendicular to the principal normal. If is a point of the plane,

(13.86)

That single equation is enough in three dimensions, where the set orthogonal to is already a plane.

Along a world-line the same idea produces a 2-flat, not a hyperplane. At the spacetime rectifying plane is the timelike 2-flat spanned by the four-velocity and the binormal analog

(13.87)

The single condition is not enough. In four dimensions that equation cuts out a whole hyperplane, which still contains . The rectifying 2-flat is the subset of that hyperplane orthogonal to as well,

(13.88)

Exercise 13.25: Obtain the spacetime description from the Euclidean equation of the rectifying plane. Why does one orthogonality condition suffice in but not in spacetime?

The Moving Trihedron

The ordered right-handed triple {,,} is called the moving trihedron in .

Of course, in spacetime we have four dimensions, so not only do we not have complete information about we need a fourth unit vector.

Torsion

Start again in . Differentiate the binormal

(13.89)

With the first term vanishes, so

(13.90)

Exercise 13.26: Write the analog of this identity for a world-line, using in place of the 3-dimensional vector product.

Since is a unit vector, then is orthogonal to . This implies that is parallel to the rectifying plane, and is thus parallel to both the unit tangent vector and principal unit normal vectors. We can thus write it as a linear combination of these vectors. We will introduce term-wise coefficients σ(s) and t(s), where t is called the torsion. It measures the rate at which the curve leaves its osculating plane. Something like this,

(13.91)

Now pass to a world-line, with proper time as the parameter and signature (−,+,+,+). The first two Frenet vectors are and , with

(13.92)

The same orthogonality argument as above gives

(13.93)

but the signs change with the metric. Differentiating u ⋅ n=0 yields σ=a, not σ=−a. So

(13.94)

Exercise 13.27: Derive (13.94), including the coefficient of u.

Torsion still means departure from the osculating 2-flat. To see the order at which that happens, fix an event at τ=0 and write the displacement to a nearby event as a Taylor series. Let be the curvature at and set . Then

(13.95)

and

(13.96)

Exercise 13.28: Explain this change of parameter.

The first three τ-derivatives at are

(13.97)

(13.98)

(13.99)

Exercise 13.29: Obtain these from and (13.94).

The expansion therefore reads

(13.100)

with every coefficient evaluated at .

Exercise 13.30: Derive this series.

Up to order the world-line remains in the osculating 2-flat of and . Torsion first appears at order .

If that torsion is nonzero, is defined. Its derivative is orthogonal to , so

(13.101)

where is a new unit vector orthogonal to , , and .

Exercise 13.31: Show that α=0.

Exercise 13.32: Show that β=-t.

Hence

(13.102)

The new coefficient is the second torsion. It is defined whenever is not collinear with . If , the world-line stays in the 3-flat spanned by , , . That 3-flat is the osculating hyperplane. The second torsion measures departure from it and first appears at order . If , then is defined. Its derivative is orthogonal to , so

(13.103)

Exercise 13.33: Show that α=0.

Exercise 13.34: Show that β=0.

Exercise 13.34: Show that .

Hence

(13.104)

There is no third torsion, the ambient space is only four-dimensional, and the frame is now complete.

The Frenet Equations

So, returning to the theory of classical space curves in , we can write a set of three equations,

(13.105)

These equations are called the Frenet-Serret equations. These form the foundation of the differential geometry of curves.

Note that we can write the system as a matrix,

(13.106)

or just a matrix of its coefficients,

(13.107)

Our equations for a world line in spacetime will then be,

(13.108)

We can write these in matrix form,

(13.109)

Decomposing 2-Forms

Suppose that we have a generic 2-form, A on a local tangent space R. Then we can write . If we consider some timelike vector, for instance an observer’s 4-velocity , then there will be a unique 1-form (the dual space of R) and a unique tangent vector such that

(13.110)

with the conditions

(13.111)

and

(13.112)

The first two terms are the wedge product . The last term is already antisymmetric in its last two slots. The whole of A remains a 2-form; the Levi-Civita tensor does not destroy antisymmetry.

We write a generic tensor product of 1-forms where we can write .

Exercise 13.35: Is this end result a scalar?

We can use this make (13.110) more precise,

(13.113)

If we look at this result we will note that this is no longer an antisymmetric 2-form. Why not? The rank 4 antisymmetric tensor ε removes the antisymmetry.

We can then write the definition of the 1-form

(13.114)

From this we can write then .

Exercise 13.36 Show that .

We can now write the antisymmetric 2-form

(13.115)

If we choose the vector as an argument,

(13.116)

we can sort of isolate parts of this

What is the action of the 2-form on the hyperplane normal to ?

Problem 13.2: Can you show that the local rest space is Euclidean?

If we assume that the local rest space is Euclidean, , then we can establish a spatial orthonormal basis . From this we can define the numbers,

(13.117)

along with the vector

(13.118)

Exercise 13.37: Show that .

Exercise 13.38: Show that is the scalar triple product of the vectors , , and .

Since ε is an antisymmetric 4-form, we can define

(13.119)

Here is an antisymmetric 3-form on . If we have a right-handed basis , then . We can then define the vector product,

(13.120)

If we examine this closely, we can see that is the vector of associated by the metric duality to the 1-form that maps where .

We can perform a similar analysis for , where it is a vector in O associated by the metric duality of the 1-form that maps O→E where .

We can then define the scalar triple product,

(13.121)

What About the Variation of the Local Frame?

We can apply the decomposition of an antisymmetric 2-form to our local frame. Here we apply this to

(13.122)

If we denote the vector from the decomposition by , then the decomposition can be written,

(13.123)

If we consider two vectors and , we can write

(13.124)

or we can expand this,

(13.125)

By considering the definition of the vector product, we can write the endomorphism,

(13.126)

If we consider the basis vectors of the local frames within O, we can write,

(13.127)

We can examine the parts of this,

The Fermi-Walker terms form what we call the Fermi-Walker Tensor,

(13.128)

The spatial rotation term forms what we call the spatial rotation tensor,

(13.129)

These are in direct analogy to the finite displacement tensor of a rigid body and the finite rotation tensor of a rigid body. In fact, the combination is reminiscent of the strain tensor for a deformable body where we have the translation, rigid rotation, and extension of the rotated axes.

We define an accelerated observer, iff for O . Similarly we define a rotating observer, iff for O .

We can thus define an inertial observer as one whose local frame is neither accelerated nor rotated. In other words,

(13.130)

Differentiation in Spacetime

The Absolute Derivative Along a World Line

If we have a vector field along a world line of our observer, then we can write the absolute derivative of the vector field as

(13.131)

Of course, we need to write the components of in the local frame defined by , so we actually have, . Thus the actual derivative is

(13.132)

Now we can apply (13.131)

(13.133)

The Derivative with Respect to an Observer

We can also determine the derivative of the vector field with respect to the observer traveling along the world line. We write this derivative,

(13.134)

In this way we see that is the variation of the vector field along the world line. This is entirely the result of variations of the vector field components within the local frame.

Recall that a particle is said to be fixed in a local frame if it is the spatial coordinates with respect to O are the same for all events on the world line. In a similar way we can define a fixed vector field with respect to an observer as one where .