Derivation of expansion scalar for FRW spacetime - weird observation

[PeterDonis]

Derivation of expansion scalar for FRW spacetime -- weird observation

In a recent thread...

https://www.physicsforums.com/showpost.php?p=3567386&postcount=137

...I posted a formula for the expansion scalar for the congruence of "comoving" observers in FRW spacetime. When I posted, I didn't have any references available, and I didn't have time to explicitly compute the answer, so I just wrote down what looked right to me based on memory and on the physical meaning of the expansion scalar.

Later, when I decided to check myself by computing the expansion explicitly, I observed something weird. I'll briefly summarize the computation and then discuss the weirdness.

We are working in standard FRW coordinates, in which the metric coefficients are (leaving out non-diagonal terms since they're all zero):

$$g_{00} = -1$$

$$g_{ii} = a^{2} h_{ii}$$

where the scale factor a is a function of the time coordinate, $x^{0} = t$, only, and the exact form of $h_{ii}$ is not needed for this problem (though of course it can be easily read off the FRW line element).

We will also need the inverse metric, which is

$$g^{00} = -1$$

$$g^{ii} = \frac{1}{a^{2}} h^{ii}$$

The congruence of worldlines of "comoving" observers in these coordinates consists of all worldlines with 4-velocity $u^{a} = (1, 0, 0, 0)$ at every event. The 1-form corresponding to this 4-velocity is then $u_{a} = g_{ab} u^{b} = (-1, 0, 0, 0)$.

The kinematic decomposition, which will give us the expansion, is given, for example, here:

http://en.wikipedia.org/wiki/Congruence_(general_relativity)

(This page uses X for the 4-velocity, which I am calling u.) For this congruence, the expansion scalar is the only interesting part of the decomposition, since the shear, vorticity, and acceleration are all zero. So we have the expansion tensor given by:

$$\theta_{ab} = u_{a;b} = u_{a,b} + \Gamma^{c}_{ab} u_{c}$$

Since the partial derivatives of the 4-velocity are all zero, the only terms of interest are the Christoffel symbol terms. Computing those, we find (considering only indices that give rise to nonzero terms):

$$\theta_{ii} = \frac{1}{2} g_{ii,0} = a \frac{da}{dt} h_{ii}$$

The expansion scalar is just the trace of the above:

$$\theta = g^{ii} \theta_{ii} = \frac{1}{a^{2}} h^{ii} a \frac{da}{dt} h_{ii} = \frac{3}{a} \frac{da}{dt}$$

(Btw, in my original post I left out the factor of 3. However, one could argue that the actual quantity of interest is 1/3 the trace of the expansion tensor, to account for there being 3 spatial dimensions. That's not the weirdness I want to talk about here, though.)

Now for the weird observation. In the computation above, you will notice that I took covariant derivatives of the spatial components of the 4-velocity (more precisely, of its corresponding 1-form), even though they are zero at every event! That seems weird; the temptation is great to say that those indices ought not to even appear, because they don't appear in the 4-velocity. But of course, if we do that, we don't get the right answer; we get that the expansion is identically zero.

The only way I can make sense of this is to think of the connection coefficients as acting to keep the spatial components of the 4-velocity zero from event to event, even though the spacetime is dynamic. Thus, what we are evaluating when we evaluate the covariant derivatives above is how much the connection coefficients have to do to keep the 4-velocity pointing purely in the "time" direction at each event. I'm curious, though, if anyone else has thought this to be weird, and if this hand-waving explanation makes sense to others besides me.

 

[Ben Niehoff]

This is hardly surprising. After all, presumably the connection coefficients have

$$\Gamma^0_{ab} \neq 0$$

Think of a simpler example of a set of observers all traveling on the meridians of a sphere, from the south pole to the north pole. In standard spherical coordinates, these observers have a velocity that points solely in the $-\theta$ direction. Suppose the observers are all synchronized so that they form a circle; then obviously this circle will change size as they travel up the sphere.

The $\phi$ component of their velocities is always zero. However, their velocity vectors are not parallel. Hence the covariant derivative in the $\phi$ direction will not vanish (not even the covariant derivative of the $\phi$ component!).

 

[PeterDonis]

This is hardly surprising. After all, presumably the connection coefficients have

$$\Gamma^0_{ab} \neq 0$$

Yes, I saw that when I wrote the thing down in index notation and realized that those coefficients had to be included to get a nonzero answer. It just seemed weird that they corresponded to taking the covariant derivative of something that looked like it was identically zero.

Think of a simpler example of a set of observers all traveling on the meridians of a sphere, from the south pole to the north pole...

Thanks! This is a good simpler example that illustrates the principle involved.

 

[TrickyDicky]

Now for the weird observation. In the computation above, you will notice that I took covariant derivatives of the spatial components of the 4-velocity (more precisely, of its corresponding 1-form), even though they are zero at every event! That seems weird; the temptation is great to say that those indices ought not to even appear, because they don't appear in the 4-velocity. But of course, if we do that, we don't get the right answer; we get that the expansion is identically zero.

The only way I can make sense of this is to think of the connection coefficients as acting to keep the spatial components of the 4-velocity zero from event to event, even though the spacetime is dynamic. Thus, what we are evaluating when we evaluate the covariant derivatives above is how much the connection coefficients have to do to keep the 4-velocity pointing purely in the "time" direction at each event. I'm curious, though, if anyone else has thought this to be weird, and if this hand-waving explanation makes sense to others besides me.

This is the weirdness that triggered my starting the thread "worldlines congruence...", thanks for putting it in much clearer mathematical terms than I managed in the descriptive way:sometimes trying to treat things only in descriptive or conceptual terms further confuses instead of make it easier.

Precisely evaluting "how much the connection coefficients have to do to keep the 4-velocity pointing purely in the "time" direction at each event" leads us to that problematic in terms of coordinate transformations "direction of time" or time signature coordinate dependency.
It is maybe interesting to note that the 4-velocity vector is a "special relativistic" object that is not defined in GR for "not local" distances (it is only defined for the "comoving rest frame" where it has magnitude c and time direction).

 

[PeterDonis]

It is maybe interesting to note that the 4-velocity vector is a "special relativistic" object that is not defined in GR for "not local" distances (it is only defined for the "comoving rest frame" where it has magnitude c and time direction).

I think you mean that in a curved spacetime there is no unique way to compare 4-velocity vectors at different events; in fact, *any* vector in a curved spacetime is only uniquely defined at one specific event, and in order to compare it with vectors at any other event you have to specify how it is "transported" from one event to the other, and there isn't in general a unique way to do that. (In the specific example I gave above, the 4-velocity of a given "comoving" observer's worldline is the tangent vector to that worldline at each event, and the worldlines are geodesics, so they parallel transport their tangent vectors along themselves, which specifies the method of transport and hence how to compare 4-velocities at different events. The connection coefficients then tell you what parallel transport "does" to the 4-velocity vectors as they are "moved" along each worldline.)

However, there is a limit to the weirdness. For example, the non-uniqueness of "transport" of vectors does *not* imply that the time direction of vectors might be "flipped" purely by this kind of effect, as you seem to be implying here:

Precisely evaluting "how much the connection coefficients have to do to keep the 4-velocity pointing purely in the "time" direction at each event" leads us to that problematic in terms of coordinate transformations "direction of time" or time signature coordinate dependency.

As I said in the other thread, the light cones are invariant; they don't depend on the coordinates, which means they aren't affected by all this weirdness with connection coefficients. So if a vector is timelike and points in a particular "time direction", which we call the "future" (meaning it points into a specific half of the light cone), at one event, then no matter how we transport it to other events in the spacetime, it will still be a timelike vector pointing into the same half of the light cone (again, assuming the spacetime meets the very general conditions I described in the other thread). Put another way, there's a limit to how much the connection coefficients can be forced to do to keep a timelike vector pointing in the future time direction; the coefficients will always be able to count on only having to "move the vector around" within one specific half of the light cone at each event.

The weirdness is that, in the course of moving around the timelike vector, the coefficients also have to move around the corresponding spacelike vectors, in order to keep the spatial components of the 4-velocity equal to zero. But the spacelike vectors only get moved around *outside* the light cone (in the spacelike region), just as the timelike vector only gets moved around inside the future half of the light cone.

 

[TrickyDicky]

However, there is a limit to the weirdness. For example, the non-uniqueness of "transport" of vectors does *not* imply that the time direction of vectors might be "flipped" purely by this kind of effect, as you seem to be implying here

I surely didn't imply such "flipping".

As I said in the other thread, the light cones are invariant; they don't depend on the coordinates, which means they aren't affected by all this weirdness with connection coefficients. So if a vector is timelike and points in a particular "time direction", which we call the "future" (meaning it points into a specific half of the light cone), at one event, then no matter how we transport it to other events in the spacetime, it will still be a timelike vector pointing into the same half of the light cone (again, assuming the spacetime meets the very general conditions I described in the other thread). Put another way, there's a limit to how much the connection coefficients can be forced to do to keep a timelike vector pointing in the future time direction; the coefficients will always be able to count on only having to "move the vector around" within one specific half of the light cone at each event.

The weirdness is that, in the course of moving around the timelike vector, the coefficients also have to move around the corresponding spacelike vectors, in order to keep the spatial components of the 4-velocity equal to zero. But the spacelike vectors only get moved around *outside* the light cone (in the spacelike region), just as the timelike vector only gets moved around inside the future half of the light cone.

It's much simpler than that, we all know Lorentzian metrics have this little coordinate convention issue with signs, and the sign in the g00 component is what defines the time direction. If something that depends on convention is not a coordinate dependent type of issue, I don't know what could be; this type of coordinate dependency is used in every other instance where it happens to declare the property it defines as not invariant and therefore unphysical.

Not that any conclusion needs to be drawn from this evidence but sometimes it seems like people are denying the evident or disguising it as weird.

 

[Ben Niehoff]

It's much simpler than that, we all know Lorentzian metrics have this little coordinate convention issue with signs, and the sign in the g00 component is what defines the time direction. If something that depends on convention is not a coordinate dependent type of issue, I don't know what could be; this type of coordinate dependency is used in every other instance where it happens to declare the property it defines as not invariant and therefore unphysical.

The sign of g00 has nothing to do with the time direction. Not sure what you're going on about.

 

[PeterDonis]

the sign in the g00 component is what defines the time direction.

To amplify what Ben said, the sign of g_00 tells you whether the x^0 coordinate is timelike or spacelike (roughly speaking), but it doesn't tell you anything about the "time direction"; g_00 multiplies the *square* of the differential dx^0, so information about the sign of x^0 is not contained in the metric.

Looking at the computation I posted in the OP, the sign convention for the time coordinate appears in the factor da/dt, which isn't in the metric (that only contains a itself), but is in the connection coefficients. So in a time asymmetric spacetime, it looks like the sign convention for the time coordinate will show up in the connection, but not the metric. But as I said before, that doesn't affect the actual physics; it just changes the words we would use to describe it ("universe contracting in the direction of time we remember" instead of "universe expanding in the direction of time we anticipate"). Which makes sense since the connection itself is not a generally covariant quantity.

 

[TrickyDicky]

The sign of g00 has nothing to do with the time direction.

Would you say it has anything to do with the time coordinate at all?

 

[Ben Niehoff]

Would you say it has anything to do with the time coordinate at all?

Generically speaking, no. If the metric is diagonal, then the one term with a "minus" (in -+++ conventions) tells you which coordinate is timelike. But it doesn't tell you which direction time flows.

For more complicated metrics, the answer is not so simple. It's not even necessary that any specific coordinate refer to "time". Take null coordinates, for example:

$$ds^2 = -2 du \, dv + dx^2 + dy^2$$

 

[TrickyDicky]

We are not speaking generally about any metric and any g00, we are centering on the general relativity case.


When constructing the one-form from the 4-velocity vector we use the metric and the sign of g00 determines the sign of the time component of the one-form that is then used to compute the expansion tensor


$u^{a} = (1, 0, 0, 0)$ $u_{a} = g_{ab} u^{b} = (-1, 0, 0, 0)$.

with $g_{00} = -1$ but if we choose $g_{00} = 1$ then the sign of the one-form changes

 

[Ben Niehoff]

We are not speaking generally about any metric and any g00, we are centering on the general relativity case.

I am talking about GR.

When constructing the one-form from the 4-velocity vector we use the metric and the sign of g00 determines the sign of the time component of the one-form that is then used to compute the expansion tensor


$u^{a} = (1, 0, 0, 0)$ $u_{a} = g_{ab} u^{b} = (-1, 0, 0, 0)$.

with $g_{00} = -1$ but if we choose $g_{00} = 1$ then the sign of the one-form changes

None of this is relevant to anything.

 

[TrickyDicky]

I am talking about GR.



None of this is relevant to anything.

If you say so.

But I just gave you a very simple way the sign of g00 determines "time direction" of 4-velocity. An that is all I claimed in post #6.

 

[Ben Niehoff]

But I just gave you a very simple way the sign of g00 determines "time direction" of 4-velocity. An that is all I claimed in post #6.

No, you wrote something about choosing sign conventions, which has nothing to do with physics. Physics doesn't care whether you use (-+++) or (+---).

 

[TrickyDicky]

No, you wrote something about choosing sign conventions, which has nothing to do with physics. Physics doesn't care whether you use (-+++) or (+---).

That's precisely my point. You are stressing here "time direction" is unphysical.

 

[Ben Niehoff]

That's precisely my point. You are stressing here "time direction" is unphysical.

You're conflating terminology. Take flat Minkowski space:

$$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$

Since this metric is diagonal, we can conclude that the coordinate t is timelike.

But the metric does not give us information about whether time flows in the t direction or in the -t direction.

Now take flat Minkowski space again:

$$ds^2 = -2 \; du \; dv + dx^2 + dy^2$$

None of these coordinates are timelike. In fact $g_{00} = 0$. Hence the sign of g00, generically speaking, has nothing to do with anything.

 

[TrickyDicky]

You're conflating terminology. Take flat Minkowski space:

$$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$

Since this metric is diagonal, we can conclude that the coordinate t is timelike.

But the metric does not give us information about whether time flows in the t direction or in the -t direction.

Yuo have already picked direction by choosing this convention instead of (+,-,-,-)

Now take flat Minkowski space again:

$$ds^2 = -2 \; du \; dv + dx^2 + dy^2$$

None of these coordinates are timelike. In fact $g_{00} = 0$.

What does that tell you about the distinction timelike-spacelike-null in SR?

 

[Ben Niehoff]

I'm not sure what your mental block is, but you're now making some very basic conceptual errors.

How many solutions are there to the equation $x^2 - a = 0$?

 

[TrickyDicky]

I'm not sure what your mental block is, but you're now making some very basic conceptual errors.

Really, no need for snide remarks.

How many solutions are there to the equation $x^2 - a = 0$?

How does this have any connection with what it's being discussed?

 

[PeterDonis]

When constructing the one-form from the 4-velocity vector we use the metric and the sign of g00 determines the sign of the time component of the one-form that is then used to compute the expansion tensor

$u^{a} = (1, 0, 0, 0)$ $u_{a} = g_{ab} u^{b} = (-1, 0, 0, 0)$.

with $g_{00} = -1$ but if we choose $g_{00} = 1$ then the sign of the one-form changes

If by "changing the sign of g_00" you mean writing the metric such that the 0 coordinate is no longer timelike, I'm not sure how that's relevant to FRW spacetime. Anyway, I don't think that's what you meant.

If by "changing the sign of g_00" you mean changing from a (-+++) sign convention to a (+---) sign convention for the metric, you've got to incorporate all the sign changes in order to judge whether any physics is affected. Let me run through the same computation as is in the OP, with the opposite sign convention for the metric, so we can see how it goes. (This will also make clear how changing the sign convention for g_00 is different from changing the sign of the t coordinate.)

The metric is:

$$g_{00} = 1$$

$$g_{ii} = - a^{2} h_{ii}$$

The inverse metric is:

$$g^{00} = 1$$

$$g^{ii} = - \frac{1}{a^{2}} h^{ii}$$

The 4-velocity and its 1-form are, as you note:

$$u^{a} = (1, 0, 0, 0)$$

$$u_{a} = (1, 0, 0, 0)$$

The equation for the expansion tensor $\theta_{ab}$ is the same, but since the sign of u_a has changed, let's dig a little deeper into the connection coefficient term, $\Gamma^{c}_{ab} u_{c}$. The connection coefficient, in terms of derivatives of the metric, is:

$$\Gamma^{c}_{ab} = \frac{1}{2} g^{cd} \left( g_{da,b} + g_{db, a} - g_{ab, d} \right)$$

Since only u_0 is nonzero, we are only interested in the c = 0 term of the above, and since the metric is diagonal that means only d = 0 appears. Since the 00 metric components are constant, only the last term survives, and only spatial indices survive for a, b, so we have

$$\Gamma^{0}_{ii} = \frac{1}{2} g^{00} g_{ii,0}$$

and therefore

$$\theta_{ii} = \Gamma^{0}_{ii} u_{0} = \frac{1}{2} g^{00} g_{ii,0} u_{0}$$

As you can see, the sign changes in u_0 and g^00 cancel each other out here, so we end up with the same equation as before for the expansion tensor; when we work it out for just the nonzero terms we get:

$$\theta_{ii} = \frac{1}{2} g_{ii,0} = - a \frac{da}{dt} h_{ii}$$

The trace then becomes:

$$\theta = \left( - a^{2} h_{ii} \right) \left( - a \frac{da}{dt} h_{ii} \right) = \frac{3}{a} \frac{da}{dt}$$

Again we see that the sign changes cancel each other out, so we end up with the same equation as before for the expansion. So changing the sign convention for the metric doesn't change anything about the physics; the sign of the expansion is still determined by the sign of da/dt, so we can change it by changing the sign of t, but not by changing the sign of g_00.

 

[TrickyDicky]

Peter I have not mentioned the expansion scalar yet, you have the habit of answering about things I have not said, that is usually called straw man fallacy.

 

[Ben Niehoff]

$$x^2 - a = 0$$

How does this have any connection with what it's being discussed?

The quantity $dt^2$ appears in the metric, and yet you have claimed that the metric tells us what is the sign of $dt$.

 

[TrickyDicky]

The quantity $dt^2$ appears in the metric, and yet you have claimed that the metric tells us what is the sign of $dt$.

Nope, that is not what I have claimed, I've only referred to the sign of the component g00, that has to do with $dt^2$.

 

[TrickyDicky]

Really, this is very obvious and well known stuff. I'd recommend you to read the last sentence of #6 again.

It's bedtime for me, bye.

 

[PeterDonis]

Peter I have not mentioned the expansion scalar yet, you have the habit of answering about things I have not said, that is usually called straw man fallacy.

I know you didn't, but you are making a claim (or at least you appear to be making a claim) about how the sign convention for the metric affects physical observables, or at least the sign of physical observables that depend on the direction of time, and the expansion scalar is such a physical observable. Also, since I had already posted the computation with one metric sign convention, posting it with the other was an easy comparison to make. If you think what I posted is not relevant to what you've been saying, then I don't understand what point you've been trying to make. What I posted makes it clear that changing the metric sign convention doesn't change any physics, and in particular it doesn't change the sign of physical observables that depend on the direction of time; and what I posted also makes it clear that changing the metric sign convention is completely independent of changing the sign of the time coordinate.

 

[Ben Niehoff]

The quantity $dt^2$ appears in the metric, and yet you have claimed that the metric tells us what is the sign of $dt$.

Nope, that is not what I have claimed, I've only referred to the sign of the component g00, that has to do with $dt^2$.

Right here you claim that the sign of g00 can be used to determine the direction of time; i.e., whether time flows in the +dt or -dt direction:

But I just gave you a very simple way the sign of g00 determines "time direction" of 4-velocity. An that is all I claimed in post #6.

If you meant something otherwise, then you have been using terminology incorrectly (as I have previously pointed out).

Clearly since the metric contains only dt^2, it doesn't care whether time flows in the +dt or the -dt direction. Hence information about the direction of time is not contained in the metric.

In general, for any manifold with metric (not just in GR), the metric and the orientation are completely independent pieces of information that don't care about each other.

Finally, there is what you actually said in #6:

It's much simpler than that, we all know Lorentzian metrics have this little coordinate convention issue with signs, and the sign in the g00 component is what defines the time direction. If something that depends on convention is not a coordinate dependent type of issue, I don't know what could be; this type of coordinate dependency is used in every other instance where it happens to declare the property it defines as not invariant and therefore unphysical.

As has been pointed out in several different ways, the metric signature convention is irrelevant. It has nothing at all to do with physics, and furthermore it is not even special to Lorentzian manifolds! In Riemannian manifolds, one can just as easily use a (----) convention. It doesn't mean anything at all!

 

[pervect]

I agree with Ben.

Consider for example a black hole and a white hole. The metric for the two is identical (in fact, you can think of the fully extended Schwarzschild metric as two asymptotically flat space-times, one of which has the black hole solution and the other which has the white hole solution).

This did get me wondering about something else, but I should probably open a different thread rather than hijack this one.

But because the exact same metric represents either a white hole or a black hole, it should be obvious that the metric does not determine the direction of time.

 

[TrickyDicky]

Right here you claim that the sign of g00 can be used to determine the direction of time; i.e., whether time flows in the +dt or -dt direction

And I showed how this sign influences the 4-velocity one-form. So I'm only referring here to the time direction of comoving observers, not to the general dt of the line element. I'm talking about a local property. There not even a unique way to compare 4-velocity vectors at different events.
Edit: Ben if you read things in context you'd have noticed from post #1 Peter referred to "comoving observers".

Clearly since the metric contains only dt^2, it doesn't care whether time flows in the +dt or the -dt direction. Hence information about the direction of time is not contained in the metric.

And I'm not saying otherwise, I explained there is a previous choice of coordinate convention, that's where the information is contained.

it is not even special to Lorentzian manifolds! In Riemannian manifolds, one can just as easily use a (----) convention.

In Riemannian manifolds this choice doesn't single out timelike from spacelike. There is a reason we need Lorentzian metrics to deal with time.

 

[PeterDonis]

And I showed how this sign influences the 4-velocity one-form.

Did you also notice that the sign does *not* affect the sign of the 4-velocity *vector*? A change in the direction of time would affect the vector. The 1-form is just an alternate representation of the vector, defined so that the contraction of the vector with the 1-form is always 1. That definition is what makes the sign of the 1-form change when you change the sign convention of the metric. It has nothing whatever to do with the physics.

So I'm only referring here to the time direction of comoving observers, not to the general dt of the line element. I'm talking about a local property.

The line element *is* local. The coordinate differentials represent very small changes in coordinates in the local neighborhood of an event. And since coordinate time in standard FRW coordinates is the same as proper time for comoving observers, the dt in the line element is equal to the differential of proper time for comoving observers, so the sign of dt does reflect the time direction of comoving observers. (Also, as I've said several times before, since the sign of dt picks out which half of the light cone is the future half, and that choice is preserved by a local Lorentz transformation, any timelike observer who can relate his motion to that of a comoving observer at the same event via a Lorentz transformation will see the same direction of time, meaning their proper time will have the same sign as dt, even if not the same magnitude. So the sign of dt plus local Lorentz invariance is enough to define the direction of time for *all* timelike observers, not just comoving ones.)

 

[PeterDonis]

A change in the direction of time would affect the vector.

Just to be completely precise about this: the 4-velocity vector describes the congruence of "comoving" worldlines, so the sign of that vector describes which direction of time the observers moving along those worldlines are moving in, relative to the sign of the time coordinate (i.e., u^0 positive means the observers "forward" direction of time is the +t direction, u^0 negative means it's the -t direction).

 

[TrickyDicky]

Did you also notice that the sign does *not* affect the sign of the 4-velocity *vector*? A change in the direction of time would affect the vector. The 1-form is just an alternate representation of the vector, defined so that the contraction of the vector with the 1-form is always 1. That definition is what makes the sign of the 1-form change when you change the sign convention of the metric. It has nothing whatever to do with the physics.

Let's see if we reach some understanding, for some reason you guys seem highly emotional and defensive about this.
The 4-vector is affected exactly in the same way as the one-form is, because it is actually a Minkowskian vector (remember equivalence principle and GR being locally minkowkian at the limit where t and r tend to zero) and if the direction of the covector is changed, its dual space vector is changed similarly.
BTW, this is not physics, it's just math, and so far I have not claimed any physical consequence from the purely mathematical coordinate related issues I'm pointing out, I only remembered what these coordinate dependence considerations usually mean for reference.

The line element *is* local. The coordinate differentials represent very small changes in coordinates in the local neighborhood of an event. And since coordinate time in standard FRW coordinates is the same as proper time for comoving observers, the dt in the line element is equal to the differential of proper time for comoving observers, so the sign of dt does reflect the time direction of comoving observers. (Also, as I've said several times before, since the sign of dt picks out which half of the light cone is the future half, and that choice is preserved by a local Lorentz transformation, any timelike observer who can relate his motion to that of a comoving observer at the same event via a Lorentz transformation will see the same direction of time, meaning their proper time will have the same sign as dt, even if not the same magnitude. So the sign of dt plus local Lorentz invariance is enough to define the direction of time for *all* timelike observers, not just comoving ones.)

Peter, you do a much better job than I do at supporting and extending what I mean. Thanks. No irony whatsoever here.
You are absolutely right in the quoted paragraph. I was using the term local in a strictest way at the tangential point the vector represents.

 

[PeterDonis]

if the direction of the covector is changed, its dual space vector is changed similarly.

This sentence, as you state it, is true, but it doesn't mean what you seem to think it means.

Let's start from the basics. We have a 4-velocity vector $u^{a} = (1, 0, 0, 0)$ that describes a "comoving" observer at a given event. Just by looking at that 4-vector, you can see that the observer must be moving in the +t direction (because the 0 component is 1, not -1). (If you are thinking of questioning this, see below.)

Corresponding to this 4-vector, we have a 1-form or covector $u_{a} = g_{ab} u^{b}$. This is purely a definition and doesn't specify any physics; the 1-form is defined this way to ensure that the contraction of the 4-vector and its corresponding 1-form equals the norm of the 4-vector: $u_{a} u^{a} = g_{ab} u^{a} u^{b}$. (Btw, I may have mis-stated this before; the norm of the 4-velocity is not always +1, since its sign depends on the sign of timelike intervals; if we are using a (-+++) metric sign convention, then timelike intervals have negative norm, so the norm of the 4-velocity is -1. The formula I just gave makes this obvious, of course.)

Supose I want to "change the direction" of the covector, in the sense that I want it to describe an observer moving in the -t direction, not the +t direction. What does that mean? It means I want to flip the sign of $u_{a}$, without changing anything else except what has to change just as a result of flipping that sign. In particular, I want to flip the sign *without* changing the sign convention of the metric or the sign of the t coordinate (since those changes would change what the mathematical sign of the covector means physically). As you can see from the above, if I flip the sign of $u_{a}$ and don't change anything else, then the sign of $u^{a}$ has to flip as well, because of the definition of $u_{a}$ in terms of $u^{a}$. In other words, changing the direction of the covector (which is equivalent, physically, to changing the direction of the corresponding vector--if both were previously describing observers moving in the +t direction, they are now both describing observers moving in the -t direction) will change the sign of *both* $u_{a}$ and $u^{a}$.

Compare the above with what I posted before. What I posted was a description of what happens when you change the sign convention of the metric, without changing anything else except what has to change just as a result of flipping that sign. As you saw, in that case, the sign of the 1-form $u_{a}$ changes but the sign of the vector $u^{a}$ does *not* change. That means that, whatever changing the sign convention of the metric amounts to, it *can't* amount to changing the actual, physical direction of the 1-form, since the sign of the 4-vector corresponding to it does not change, and we showed above that for the actual physical direction to change, *both* signs have to change.

So what *does* changing the metric sign convention do? Obviously, it changes which sign of the 1-form describes observers moving in the +t direction. And with regard to 4-vectors and 1-forms, that's all it does; as I showed in previous posts, it doesn't affect any actual physics, just how the 1-forms mathematically describe the physics.

(Note that, as I noted above, a positive sign on the 4-vector must always describe observers moving in the +t direction; fundamentally, this is because coordinate differences and their derivatives are 4-vectors. Specifically, the coordinates of a "comoving" observer are described by a 4-vector $x^{a}$, which is the coordinate difference between a given event on that observer's worldline and the "origin" event of the coordinate chart. The 4-velocity is then $u^{a} = \frac{dx^{a}}{d\tau}$, where $\tau$ is the observer's proper time. Since for "comoving" observers, coordinate time is the same as their proper time, we have $u^{a} = \frac{dx^{a}}{dt}$; and for "comoving" observers, the spatial coordinates are all constant, so this equation leads direction to the 4-vector $u^{a} = (1, 0, 0, 0)$ that I gave above for an observer moving in the +t direction. So an observer moving in the +t direction must have a positive sign for the 4-velocity vector. For non-comoving observers, the math looks more complicated, but the final result is the same; basically, we use the chain rule to write $u^{a} = \frac{dx^{a}}{d\tau} = \frac{dx^{a}}{dt} \frac{dt}{d\tau}$, and we note that local Lorentz invariance requires $\frac{dt}{d\tau}$ to be positive for *any* observer whose proper time flows in the +t direction.)

 

[TrickyDicky]

in that case, the sign of the 1-form $u_{a}$ changes but the sign of the vector $u^{a}$ does *not* change. That means that, whatever changing the sign convention of the metric amounts to, it *can't* amount to changing the actual, physical direction of the 1-form, since the sign of the 4-vector corresponding to it does not change, and we showed above that for the actual physical direction to change, *both* signs have to change.

I think your post is basically right except we disagree in the quoted part.
What you showed in your previous post is that changing the one-form sign doesn't affect the expansion scalar due to the cancelation of signs you explained. This is evident, I'm not arguing this.
It seems to me the sentence you say is true does mean what I think it means, at least mathematically. You are entitled to think otherwise, this is no big deal.

 

[PeterDonis]

I think your post is basically right except we disagree in the quoted part.

I don't understand. The statement of mine you quoted, which you say you disagree with, is just math. If you change the metric sign convention without changing any other sign conventions, you change the sign of the 1-form, *without* changing the sign of the corresponding 4-vector. That's required by the definition of the 1-form. Are you disagreeing with that? If so, what exactly do you think happens when you change the metric sign convention without changing any other sign conventions?

The other part of what you quoted just says that, if you want to change the actual, physical meaning of the 1-form (for example, by having it describe an observer moving in the -t direction instead of the +t direction), without changing anything else, you have to change the sign of *both* the 1-form and its corresponding 4-vector. Again, this is required by the definition of the 1-form. Are you disagreeing with that? If so, how exactly do you propose to change the 1-form to describe an observer moving in the -t direction instead of the +t direction, without changing any sign conventions, and not change the sign of the corresponding 4-vector?

What you showed in your previous post is that changing the one-form sign doesn't affect the expansion scalar due to the cancelation of signs you explained.

You are mis-stating it. What I showed in my previous post is that changing the *metric sign convention* doesn't change the sign of the expansion scalar. The change of sign of the 1-form was a *side effect* of the change in metric sign convention. For the argument in my previous post, which you say you agree with, to go through, it *must* be the case that, as I said above, changing the metric sign convention changes the sign of the 1-form *without* changing the sign of the corresponding 4-vector. So again, if you think you are disagreeing with me, what, exactly, do you think happens when you change the metric sign convention, without changing any other sign conventions?

 

[TrickyDicky]

Ok, let's go one thing at a time.

Do you agree the direction of the covector velocity is "connected" to the vector velocity direction and viceversa? say, like is the case with the basis of a covector wrt the basis of the vector.