Characterizing GR Traj in Minkowski Space

[sophiatev]

TL;DR Summary

Why do our notions of "timelike", "spacelike", and "null" trajectories remain the same in GR?

In Minkowski space, with line element $$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$ (and $c = 1$) we take spacelike trajectories to have $ds^2 > 0$, null trajectories to have $ds^2 = 0$, and timelike trajectories to have $ds^2 < 0$. This makes sense given our definition of the line element, since a particle moving at the speed of light should have $ds^2 = 0$ at every point along the path, a particle moving slower should have $ds^2 < 0$, and a particle moving faster should have $ds^2 > 0$. But in general relativity, where the line-element can take other forms, why do we still take these statements to be true? Does it have something to do with the fact that the metric still has Lorentzian signature (why is this true, by the way)?

My initial thought was that we can transform to the Minkowski metric at every point in spacetime, and since these characterizations hold in Minkowski and the line-element is invariant under coordinate transformations, they hold generally. But strictly speaking the line-element is not located at one point, but extends from one point to another. It is infinitesimally small, and I know that we can take the first-derivative of the metric to vanish at the point, which maybe somehow helps? Regardless, I'm not sure my argument holds.

 

[PeterDonis]

Does it have something to do with the fact that the metric still has Lorentzian signature

Yes. That is in fact why the timelike/spacelike/null distinction remains valid in curved spacetime: because the spacetime still has Lorentzian signature everywhere.

(why is this true, by the way)?

Because that model works; it makes accurate predictions. That's the only answer science can give to a question like this.

the line-element

Focusing on the specific form of the line element can be misleading, because there are many ways to choose coordinates, even in flat Minkowski spacetime, that can make it not at all apparent from the form of the line element what vectors are timelike, spacelike, or null, or even what the metric signature is.

 

[Ibix]

Does it have something to do with the fact that the metric still has Lorentzian signature (why is this true, by the way)?

Peter is correct that we justify these assumptions post hoc: if we assume x we predict y, and y matches experiment. However, it's worth noting that there's an application of the correspondence principle here - SR is well tested, and a relativistic gravity theory must reduce to it in cases where gravity is not relevant. For a geometric relativistic gravity theory that means locally Lorentz.

Mathematically, once you've got a locally Lorentzian metric, you'll find that the inner product of any two vectors can be positive, zero or negative, so the notion of spacelike, null, and timelike remains useful. And physically we find that "light can never be overtaken" remains true even when gravity is relevant, so it seems like a useful distinction.

 

[sophiatev]

Yes. That is in fact why the timelike/spacelike/null distinction remains valid in curved spacetime: because the spacetime still has Lorentzian signature everywhere.

Correct me if I'm wrong, but from my understanding "Lorentzian signature" simply means that when we put the metric into its canonical form, there is exactly one minus in the signature (and all the rest of the eigenvalues are positive). Is it simply a matter of convention that we always take the minus to correspond to the "timelike" coordinate? I'm thinking specifically of the case of the Schwarzschild metric, where once we pass the event horizon the $r$ coordinate is the one with the negative coefficient in $ds^2$ rather than the $t$ coordinate. In that case, "timelike progression" is along the direction of decreasing $r$ now rather than $t$, even though we still take $r$ to correspond to radial distance (I think). Would a Lorentzian signature imply that there always exists trajectories in spacetime with a negative $ds^2$, and we simply choose all massive particles to move along those trajectories because "it works"?

Mathematically, once you've got a locally Lorentzian metric, you'll find that the inner product of any two vectors can be positive, zero or negative, so the notion of spacelike, null, and timelike remains useful. And physically we find that "light can never be overtaken" remains true even when gravity is relevant, so it seems like a useful distinction.

That seems to be what Ibix is implying here, at least (though again correct me if I'm wrong).

 

[PeterDonis]

from my understanding "Lorentzian signature" simply means that when we put the metric into its canonical form, there is exactly one minus in the signature (and all the rest of the eigenvalues are positive)

What do you mean by "canonical form"? If you mean "pick a local inertial chart so that the metric takes its Minkowski form at your chosen point", then there could be one minus or three minuses in the signature, depending on your choice of signature convention. The "odd one out" (the signature sign that is different from the other three) is the timelike one.

I'm thinking specifically of the case of the Schwarzschild metric, where once we pass the event horizon the $r$ coordinate is the one with the negative coefficient in $ds^2$ rather than the $t$ coordinate.

If you are taking the Schwarzschild metric in Schwarzschild coordinates to be an example of "canonical form" (which then would mean that any chart in which the metric is diagonal would count as a "canonical form"--note that the Schwarzschild chart is not the only chart in which this is true for the Schwarzschild metric), then you have to add to your specification that it does not matter which coordinate is the "odd one out" (the one whose signature sign is different from the other three). As long as one is, the metric is Lorentzian. It doesn't even have to be the same coordinate on different patches, as the Schwarzschild example illustrates. (Note that the two cases, outside and inside the event horizon, are two disconnected patches; Schwarzschild coordinates are not valid at the horizon, so these coordinates cannot connect the exterior and interior regions. That is why it's fine for the "odd one out" coordinate to be $t$ outside and $r$ inside, without affecting the signature.)

In that case, "timelike progression" is along the direction of decreasing $r$ now rather than $t$

Yes.

even though we still take $r$ to correspond to radial distance (I think).

No. $r$ is an "areal radius", but changes in $r$ are not the same as changes in radial distance. Inside the horizon, there isn't even a well-defined concept of "radial distance" at all.

 

[sophiatev]

What do you mean by "canonical form"?

Sorry, I should have been more specific. I'm using Carroll's definition from An Introduction to General Relativity. I'm not sure what a good way to do this is so I will just post images of the relevant pages (pg. 73-74)

Above is his discussion of what "canonical form" means.

If you mean "pick a local inertial chart so that the metric takes its Minkowski form at your chosen point", then there could be one minus or three minuses in the signature, depending on your choice of signature convention.

Later he goes on to describe putting the metric into canonical form, which seems to agree with what you said here (with specific choice of one negative and three positive in the signature)

If you are taking the Schwarzschild metric in Schwarzschild coordinates to be an example of "canonical form"

So using this definition, no, I should not have associated the Schwarzschild metric with the "canonical form".

then you have to add to your specification that it does not matter which coordinate is the "odd one out" (the one whose signature sign is different from the other three). As long as one is, the metric is Lorentzian.

Although in Carroll's definition, it does not seem to matter which coordinate is the negative one in the signature when the metric is in "canonical form" for the signature to be characterized as Lorentzian. However, in our specific case, he does seem to imply that when we put the metric into "canonical form" locally, we can make it specifically Minkowski to zeroth order and have the first derivative vanish.

Thank you for clarifying more about the specific case of Schwarzschild coordinates. I have two more questions. Let's say we "put the metric into canonical form" at some point P past the event horizon, which according to Carroll means equation 2.47 is satisfied. Does that mean that the coordinate mapping between the Minkowski coordinates and Schwarzschild coordinates at that point P is such that the areal radius coordinate $r$ in Schwarzschild is now related to the time coordinate $t$ in Minkowski? Since $r$ is now the "timelike" coordinate?

Would a Lorentzian signature imply that there always exists trajectories in spacetime with a negative ds2, and we simply choose all massive particles to move along those trajectories because "it works"?

And is my understanding here correct?

 

[PeterDonis]

I'm using Carroll's definition from An Introduction to General Relativity.

Ok, that corresponds to the first option I described, choosing a local inertial chart so the metric is in Minkowski form at a chosen point.

using this definition, no, I should not have associated the Schwarzschild metric with the "canonical form".

Correct.

In Carroll's definition, it does not seem to matter which coordinate is the negative one in the signature when the metric is in "canonical form" for the signature to be characterized as Lorentzian.

More precisely, it doesn't matter whether you choose one positive and three negative, or one negative and three positive. Which of these you choose is usually called a choice of "signature convention" in the literature; the "one positive and three negative" choice is called the "timelike signature convention", and the "one negative and three positive" choice is called the "spacelike signature convention". This choice makes no difference to the physics.

he does seem to imply that when we put the metric into "canonical form" locally, we can make it specifically Minkowski to zeroth order and have the first derivative vanish.

Yes. The technical name for this choice of coordinates in the literature is "Riemann normal coordinates"; another common name is "local inertial coordinates".

Let's say we "put the metric into canonical form" at some point P past the event horizon, which according to Carroll means equation 2.47 is satisfied. Does that mean that the coordinate mapping between the Minkowski coordinates and Schwarzschild coordinates at that point P is such that the areal radius coordinate in Schwarzschild is now related to the time coordinate in Minkowski?

Sort of. A more precise way of saying it is that, at any point inside the horizon, we can find a choice of local inertial coordinates such that the Schwarzschild basis vector $\partial / \partial r$ points in the same direction as the local inertial basis vector $\partial / \partial T$ (I use capital letters to make clear the distinction between the two charts), and the Schwarzschild basis vector $\partial / \partial t$ points in the same direction as the local inertial basis vector $\partial / \partial X$. (The $Y$ and $Z$ basis vectors of the local inertial chart then lie in the tangential plane; in fact we can ignore those two spacetime dimensions and just consider the Schwarzschild $t-r$ plane and the local inertial $T-X$ plane.)

 

[PeterDonis]

Would a Lorentzian signature imply that there always exists trajectories in spacetime with a negative $ds^2$, and we simply choose all massive particles to move along those trajectories because "it works"?

The key thing is not whether the sign of $ds^2$ for a timelike trajectory is negative (since we can flip the sign by our choice of signature convention); it is that we have, heuristically, one coordinate's worth of trajectories with one sign of $ds^2$, and three coordinates' worth of trajectories with the other sign of $ds^2$. The former is the set of trajectories that massive particles are found to travel on.

 

[Ibix]

So using this definition, no, I should not have associated the Schwarzschild metric with the "canonical form".

Careful - I think you mean the Schwarzschild metric in Schwarzschild coordinates is not in canonical form. It can, of course, be put into canonical form.

Although in Carroll's definition, it does not seem to matter which coordinate is the negative one in the signature when the metric is in "canonical form" for the signature to be characterized as Lorentzian.

You are always free to change the order of coordinates - so while it's conventional to write the Schwarzschild metric as $\mathrm{diag}(g_{tt},g_{rr},g_{\theta\theta},g_{\phi\phi})$ there's no obligation to do so. As long as you keep track of which component is which, you can do whatever you want (although since the conventional $t r \theta \phi$ order of spherical coordinates is one of the very few conventions in GR that everyone seems to agree on, it would probably be a bad idea). But if you transform some metric into local Minkowski ones and get (-1,+1,-1,-1) you can just rearrange.

And is my understanding here correct?

Kind of. The point about the one different sign (both the difference and that there's only one of it) is that this is the ultimate source of a difference between a "time" direction and "space" directions, at least in the model.

Our experience is that we may choose not to move in space (by choosing to use our own, possibly non-inertial, rest frame), but we cannot choose not to advance in time. In the model, that means that we must be moving in a timelike direction (the tangent vector to our worldlines must always be timelike), or else we could model ourselves stopping in time or traveling in the opposite direction from someone - all the things you can do in space, not time.

 

[sophiatev]

Our experience is that we may choose not to move in space (by choosing to use our own, possibly non-inertial, rest frame), but we cannot choose not to advance in time.

it is that we have, heuristically, one coordinate's worth of trajectories with one sign of ds2, and three coordinates' worth of trajectories with the other sign of ds2. The former is the set of trajectories that massive particles are found to travel on.

I think I see what you guys mean. The idea is that because "experimentally" we know that we must advance in time, we want the notion of "time-progression", which I guess in this case relates to "timelike", to be distinct from other types of progression (namely spatial). So we can accomplish this either by having the timelike coordinate be the negative one in the signature and all others positive, or vice versa. And it must be the case that at every point in spacetime, we can find coordinates (Minkowski in this case, which I think is an even stronger statement related to the equivalence principle) where we have one coordinate of opposite sign from the other three. And that distinct coordinate would determine what "timelike progression" looks like for a (massive) particle at that point? (More specifically, the tangent vector of that "distinct coordinate", call it $\partial / \partial t$, is what the particle would move along if it were to only move in "time" and not "space"?).

And physically we find that "light can never be overtaken" remains true even when gravity is relevant, so it seems like a useful distinction.

And as you seem to suggest here, it sounds like this notion of one coordinate having opposite sign from the other three is also "useful" in the sense that it gives us a bound. When a particle moves at the speed of light, the quantities of opposite sign are equal (or I guess more precisely for Minkowski $dt^2 = dx^2 + dy^2 + dz^2$) and thus $ds^2 = 0$, giving us in essence an "upper bound" on particle trajectories.

 

[vanhees71]

Correct me if I'm wrong, but from my understanding "Lorentzian signature" simply means that when we put the metric into its canonical form, there is exactly one minus in the signature (and all the rest of the eigenvalues are positive). Is it simply a matter of convention that we always take the minus to correspond to the "timelike" coordinate?

Yes, it's convention. You can as well use the west-coast convention, where the signature is $(+---)$. The only important thing is that its one eigenvalue of the matrix representing the pseudo-metric form that has different sign from the three others. Only then you can establish a causality structure on spacetime as usual in relativsitic physics.

 

[PeterDonis]

that distinct coordinate would determine what "timelike progression" looks like for a (massive) particle at that point?

For one particular massive particle, the one whose 4-velocity points in the same direction as the timelike basis vector. But one can do a local Lorentz transformation to change the direction of the timelike basis vector to be anywhere in the future light cone. In other words, there is not just one (future-pointing) timelike direction in spacetime at a given point; there is a whole future light cone's worth of them.

 

[PeterDonis]

this notion of one coordinate having opposite sign from the other three is also "useful" in the sense that it gives us a bound

The "bound" you refer to is the light cone, yes (as I said in my previous post just now).

Btw, looking at the light cone gives another way of distinguishing timelike from spacelike vectors. The interior of the light cone of a given point in spacetime divides into two disconnected halves, the future half and the past half; therefore so do the timelike vectors at a given point. There is no continuous way to rotate (in the spacetime sense of "rotation") a future-directed timelike vector to a past-directed timelike vector. The spacelike vectors, on the other hand, form a single connected set; one can continuously rotate (in the spacetime sense) any spacelike vector into any other spacelike vector.

 

[sophiatev]

There is no continuous way to rotate (in the spacetime sense of "rotation") a future-directed timelike vector to a past-directed timelike vector. The spacelike vectors, on the other hand, form a single connected set; one can continuously rotate (in the spacetime sense) any spacelike vector into any other spacelike vector.

This is very interesting. I often wondered about what makes the "future" and "past" lightcones distinguishable - I guess this is one answer to that question. Thank you very much for all your help in general in this thread (as well as @vanhees71 and @Ibix), I feel I have a deeper understanding of the topic now.

 

[vanhees71]

I think the fact that there's a "orientation of time" is one of the assumptions we make. As @PeterDonis points out that's consistent with the Lorentz structure of the relativistic spacetime manifold, because the part of the Poincare group that's connected continuously with the identity is the proper orthochronous Poincare group, and as high-energy particle physics tells us, that's the spacetime symmetry group that is realized in Nature exactly, while the weak interaction breaks all the discrete symmetries (space reflection, time reversal, charge conjugation and all combinations of these, i.e., P, T, C, CP, PT, and CT).

I'd call this postulate, often not made explicitly in textbooks, the assumption of a "causal arrow of time". From it you can derive the other usually quoted arrows of time by analyzing the dynamical equations. E.g., the socalled "electrodynamic arrow of time" of classical electrodynamics, where you consider only the retarded solutions of the Maxwell equations ("Jefimenko equations" or equivalently the retarded potentials in the Lorenz gauge), i.e., if you have a compact source of charges and currents you consider the outgoing-wave solutions. The time-reversal of this solution is of course also a solution of Maxwell's equations, but for all practical purposes impossible to realize experimentally, because you'd have to prepare sources over a wide region far from the place of your compact sources to create a precisely incoming wave, being exactly absorbed by the compact sources.

The most prominent arrow of time, related to our everyday experience is the "thermodynamical arrow of time", i.e., the observation that the forward direction of time is the one, where the entropy of a macroscopic system is not decreasing, i.e., the dynamics drives the system (at least on average) towards the state of maximum entropy, which is the state of thermal equilibrium, given the constraints the system is underlying due to conservation laws. This direction of time can be shown to be the same as the postulated causal arrow of time, and it comes at the point in the derivation of the Boltzmann equation from the underlying microscopic dynamics (i.e., the Liouville equation for classical mechanics or the Kadanoff-Baym equations for many-body qunantum theory) where you "coarse grain" your description, by "throwing away irrelevant information". E.g., in the standard derivation of the Boltzmann equation from classical point-particle mechanics you simplify your collision term for two-body scattering by assuming that the incoming particles are uncorrelated, i.e., you substitute the two-particle distribution function by the product of the corresponding two one-particle distribution function to close your equation. At this moment you throw away the information of the two-particle correlations, and this leads to the H-theorem, i.e., that in the "forward time evolution" ("forward" in the sense of the causal arrow of time) the entropy is never decreasing. The intuitive meaning is again the same as for the electrodynamic arrow of time: For macroscopic objects it's often impossible to prepare the precise time reversed state. E.g., if your coffee cup is falling down from the table and shattering in zillions of pieces of the floor, which is a common experience of everyday life, it is practically impossible to time-reverse this process, i.e., put all the pieces on the floor precisely in motion in such a way that the cup resembles unbroken on the table. Though this is in principle possible, because of time-reversal invariance (neglecting the weak interaction, which is not, but that's always justified for such macroscopic phenomena), it is so unlikely to happen "spontaneously" that for all practical purposes it indeed never happens.

 

[Ibix]

I think the fact that there's a "orientation of time" is one of the assumptions we make.

I think it depends what the OP means. Light cones, or equivalently the Lorentzian signature of the metric, is enough to explain why I could overtake you in the corridor then turn around and pass you going the other way, but I can never pass you so your watch ticks the opposite way to mine. It isn't enough to determine which light cone is the future one and which the past without additional input, though, as you say.

 

[vanhees71]

It's of course also the meaning of "time-reversal symmetry". Experience of course shows that you cannot achieve it in an active way, i.e., you cannot simply "run time in the opposite direction", while of course you can, e.g., repeat the same experiment at another place (spatial translation), rotate the entire equipment (spatial rotations), repeat the experiment at a later time (time translation), or do it in a boosted frame of reference (Lorentz boost). Time-reversal symmetry rather means to repeat an experiment with initial conditions, where all positions are the same as the final state of a previously done experiment but with all momenta pointing to the opposite direction (speaking for a mechanical system for convenience; of course the same holds for any other system, where the various quantities like em. field components, charge and current densities, electric and magnetic moments,etc., have a determined behavior under "time reversal"). Then "time-reversal symmetry" means that you should end up after the corresponding time in the time-reversed initial state of the previous experiment. Then it becomes clear that for the thermodynamic arrow of time, it's practically impossible to achieve this, because you'd have to "time-reverse" the final state of one experiment precisely to set up the time-reversed evolution, which is impossible due to the impossibility of this formidable task for macroscopic systems consisting of $10^{24}$ atoms/molecules.

 

[PeterDonis]

I think the fact that there's a "orientation of time" is one of the assumptions we make.

More precisely, the disconnected structure of the two timelike light cone interiors is a geometric fact in any Lorentzian spacetime, but the choice of which half to call the "future" half and which half to call the "past" half is a choice we make when constructing our model; we could choose to construct the model either way.

Note, also, that in some spacetimes (ones that are not time orientable), it is impossible to make a continuous choice throughout the spacetime of which half of the light cone is the "future" half and which half is the "past" half.