Derivation of expansion scalar for FRW spacetime - weird observation

[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.

 

[PeterDonis]

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.

Isn't this obvious from the definition I gave? I said:

u_{a} = g_{ab} u^{b}

This is just the definition of a 1-form corresponding to a vector; put another way, it is the definition of "lowering an index" in tensor algebra. If that is what you mean by the directions being "connected", then yes, I agree. If not, then you'll have to explain what you mean by the directions being "connected".

 

[TrickyDicky]

Isn't this obvious from the definition I gave? I said:

u_{a} = g_{ab} u^{b}

This is just the definition of a 1-form corresponding to a vector; put another way, it is the definition of "lowering an index" in tensor algebra. If that is what you mean by the directions being "connected", then yes, I agree. If not, then you'll have to explain what you mean by the directions being "connected".

Yes, that is what I mean. Basically, in flat space, a covector and its dual vector represent the same object and even have the same components-Added:in orthonormal basis of course- (not so in curved spaces).
Now doyou agree this object, in this case the 4-velocity, belongs to the local Minkowski space any Lorentzian manifold resemble at infinitesimal scales?

 

[PeterDonis]

Now doyou agree this object, in this case the 4-velocity, belongs to the local Minkowski space any Lorentzian manifold resemble at infinitesimal scales?

Yes, if by "local Minkowski space" you mean the tangent space:

http://en.wikipedia.org/wiki/Tangent_space

There is a tangent space at every event in the manifold (the spacetime, in this case), and vectors and 1-forms (and tensors and all other geometric objects) at each event "live" in the tangent space at that event.

 

[Ben Niehoff]

If all we are discussing now are the mathematical consequences of sign conventions, this thread is become quite silly. Are there any physical consequences you are concerned about, Tricky?

 

[TrickyDicky]

So when you agreed previously that a change of sign in g00 has as a "side effect" (or whatever you want to call it) the change in direction of the 4-velocity covector, I tend to think that it does the same with the 4-velocity vector, in the sense that we are ultimately talking about the same geometrical object although in abstract terms the covector belongs to the cotangent vector space, it happens to coincide in the minkowskian setting with the tangent vector space.

 

[TrickyDicky]

If all we are discussing now are the mathematical consequences of sign conventions, this thread is become quite silly.

Wel, maybe is silly, though that is a very subjective appreciation, perhaps not everybody finds it so.

Are there any physical consequences you are concerned about, Tricky?

Not really, I always said my point was mathematical, about coordinates.
Are there any you are concerned about now?

 

[PeterDonis]

So when you agreed previously that a change of sign in g00 has as a "side effect" (or whatever you want to call it) the change in direction of the 4-velocity covector, I tend to think that it does the same with the 4-velocity vector

You are mistaken as to what I agreed to. What I said was that the change in the metric sign convention had as a side effect the change in *sign* of the 4-velocity covector, when expressed in component form. I did *not* say, and in fact explicitly denied, that the change in sign of the 4-velocity covector, as a result of the change in metric sign convention, was a change in the *direction* of the covector. The latter is a statement about physics; the former is a statement about sign conventions, i.e., about math only. That's why I was careful to describe explicitly what a "change in direction" would mean: it would mean the 4-vector and its covector would describe an observer moving in the -t direction instead of the +t direction. Just to make absolutely sure it's clear what I mean, I'll spell it all out explicitly:

* The 4-vector u^{a} = (1, 0, 0, 0) always describes an observer moving in the +t direction. I explained why in a previous post.

* The covector u_{a} = (-1, 0, 0, 0) also describes an observer moving in the +t direction, *if* the metric sign convention is (-+++). That was the sign convention I used in the OP. So with this metric sign convention, the covector (-1, 0, 0, 0) describes the same geometric object as the vector (1, 0, 0, 0).

* If we *change* the metric sign convention to (+---), then the covector that describes an observer moving in the +t direction is now u_{a} = (1, 0, 0, 0); in other words, the components of the covector flip sign due to the change in metric sign convention. But the components of the 4-vector do *not* flip sign; see the point above. So with the (+---) metric sign convention, the covector (1, 0, 0, 0) describes the same geometric object as the vector (1, 0, 0, 0), i.e., they both describe an observer moving in the +t direction. We have not actually changed the covector itself; we have only changed the sign convention used to represent it.

* If, on the other hand, we keep the metric sign convention as (-+++), but we want to describe an observer moving in the -t direction instead of the +t direction, then *both* the vector and its covector change sign: we have u^{a} = (-1, 0, 0, 0) and u_{a} = (1, 0, 0, 0) both representing the same geometric object; but it's a *different* geometric object from the one we were representing before, because it describes an observer moving in the opposite direction in time (-t instead of +t).

* Finally, if we are describing an observer moving in the -t direction, and we change the metric sign convention to (+---), then once again the sign of the covector flips but the sign of the vector does not; the observer moving in the -t direction is now described by u^{a} = (-1, 0, 0, 0) and u_{a} = (-1, 0, 0, 0), so both of these now represent the same geometric object (which is the same one we were representing in the previous item, just represented with a different sign convention).

in abstract terms the covector belongs to the cotangent vector space, it happens to coincide in the minkowskian setting with the tangent vector space.

Yes, I glossed over this in my previous post but it's good to point out. There are actually two spaces at each event, the tangent and the cotangent space. Things with "upstairs" indexes, like vectors and contravariant tensors, live in the tangent space. Things with "downstairs" indexes, like covectors (1-forms) and covariant tensors, live in the cotangent space. If you have a metric available, you can switch between spaces by raising and lowering indexes, so the metric basically gives you an isomorphism between them, which is why we can speak of a 1-form that "corresponds" to a vector, for example. But they are still two distinct spaces.

 

[TrickyDicky]

You are mistaken as to what I agreed to. What I said was that the change in the metric sign convention had as a side effect the change in *sign* of the 4-velocity covector, when expressed in component form. I did *not* say, and in fact explicitly denied, that the change in sign of the 4-velocity covector, as a result of the change in metric sign convention, was a change in the *direction* of the covector. The latter is a statement about physics; the former is a statement about sign conventions, i.e., about math only. That's why I was careful to describe explicitly what a "change in direction" would mean: it would mean the 4-vector and its covector would describe an observer moving in the -t direction instead of the +t direction. Just to make absolutely sure it's clear what I mean, I'll spell it all out explicitly:

* The 4-vector u^{a} = (1, 0, 0, 0) always describes an observer moving in the +t direction. I explained why in a previous post.

* The covector u_{a} = (-1, 0, 0, 0) also describes an observer moving in the +t direction, *if* the metric sign convention is (-+++). That was the sign convention I used in the OP. So with this metric sign convention, the covector (-1, 0, 0, 0) describes the same geometric object as the vector (1, 0, 0, 0).

* If we *change* the metric sign convention to (+---), then the covector that describes an observer moving in the +t direction is now u_{a} = (1, 0, 0, 0); in other words, the components of the covector flip sign due to the change in metric sign convention. But the components of the 4-vector do *not* flip sign; see the point above. So with the (+---) metric sign convention, the covector (1, 0, 0, 0) describes the same geometric object as the vector (1, 0, 0, 0), i.e., they both describe an observer moving in the +t direction. We have not actually changed the covector itself; we have only changed the sign convention used to represent it.

* If, on the other hand, we keep the metric sign convention as (-+++), but we want to describe an observer moving in the -t direction instead of the +t direction, then *both* the vector and its covector change sign: we have u^{a} = (-1, 0, 0, 0) and u_{a} = (1, 0, 0, 0) both representing the same geometric object; but it's a *different* geometric object from the one we were representing before, because it describes an observer moving in the opposite direction in time (-t instead of +t).

* Finally, if we are describing an observer moving in the -t direction, and we change the metric sign convention to (+---), then once again the sign of the covector flips but the sign of the vector does not; the observer moving in the -t direction is now described by u^{a} = (-1, 0, 0, 0) and u_{a} = (-1, 0, 0, 0), so both of these now represent the same geometric object (which is the same one we were representing in the previous item, just represented with a different sign convention).

So, I guess the sign in this particular case has nothing to do with direction of the covector. How odd. I stand corrected.

BTW, you are fixing the convention for the vector and letting it change for the covector, but I think I'll leave it here

 

[PeterDonis]

you are fixing the convention for the vector and letting it change for the covector

In this particular case (the 4-velocity vector), the sign of the vector is not a matter of convention; it's dictated by the definition of 4-velocity. See the last part of my post #32.

 

[TrickyDicky]

In this particular case (the 4-velocity vector), the sign of the vector is not a matter of convention; it's dictated by the definition of 4-velocity. See the last part of my post #32.

Yes, you are fixing beforehand (by convention) the sign of dt as positive.

 

[PeterDonis]

Yes, you are fixing beforehand (by convention) the sign of dt as positive.

No, I'm saying that the sign of dt is determined by the sign of t, which implies that, for an observer moving in the +t direction, dt is positive. If you want to call that a "convention", I suppose I can't stop you, but calling it that kind of implies that the alternative is at least sensible. In the case of the sign convention for the metric, obviously both alternatives are sensible. I have a hard time seeing how calling dt negative for an observer moving in the +t direction is sensible, since it flies in the face of the definition of the coordinate t.

 

[TrickyDicky]

Peter, would you care to look at this? http://en.wikipedia.org/wiki/4-velocity

The last heading, where it says Interpretation.

 

[PeterDonis]

The last heading, where it says Interpretation.

Yes, this page is consistent with what I've been saying. In units where c = 1, the Wiki page is saying that

u^{a} u_{a} = -1

if the metric signature is (-+++), and

u^{a} u_{a} = +1

if the metric signature is (+---). That's consistent with what I said in post #42.

Also, further up the page, it gives the same definition of 4-velocity that I gave:

u^{a} = \frac{dx^{a}}{d\tau}

And it later splits it into time and space components:

u^{a} = (\gamma, u^{i})

Since the relativistic \gamma factor is always >= 1, this is consistent with what I was saying about the sign of u^0 = t and hence dt. For a "comoving" observer, \gamma = 1 and u^{i} = 0, so u^{a} = (1, 0, 0, 0).

 

[TrickyDicky]

So is there any circumstance when u^{a} = (-1, 0, 0, 0)?

 

[PeterDonis]

As I said in earlier posts, the 4-velocity will be (-1, 0, 0, 0) if the observer is moving in the -t direction. The Wiki page doesn't go into that, but here's how I would see it being represented in the math.

The 4-velocity, once again, is defined as:

u^{a} = \frac{dx^{a}}{d\tau}

The Wiki page immediately writes down:

x^{0} = t = \gamma \tau

and hence:

u^{0} = \gamma

But notice that there is a hidden assumption there: that the direction of increasing t is the *same* as the direction of increasing \tau. In other words, the Wiki page assumes that the observer is moving in the +t direction. For an observer moving in the -t direction, the two equations above would instead read:

x^{0} = t = - \gamma \tau

and:

u^{0} = - \gamma

For the "comoving" case, this leads to u^{a} = (-1, 0, 0, 0), which is what I wrote in post #42 for the case of an observer moving in the -t direction.

I see on re-reading post #46 that I didn't make this part of it clear; I was focused on the sign of t and dt instead of the sign of \tau. Apologies if that caused confusion.

 

[TrickyDicky]

I see on re-reading post #46 that I didn't make this part of it clear; I was focused on the sign of t and dt instead of the sign of \tau. Apologies if that caused confusion.

No need to apologize, thanks I'll think about it.

 

[TrickyDicky]

Ugh, I have to admit my #6 is both silly and wrong, I simply slipped-up and well, I'm a bit stubborn at times, anyway thank you Peter and Ben for your patience.

 

[Ben Niehoff]

No problem. Glad you see it now.

 

[PeterDonis]

No worries.