Derivation of expansion scalar for FRW spacetime - weird observation

[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.