# Derivation of expansion scalar for FRW spacetime - weird observation

1. Oct 19, 2011

### Staff: Mentor

Derivation of expansion scalar for FRW spacetime -- weird observation

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.

2. Oct 19, 2011

### Ben Niehoff

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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!).

3. Oct 19, 2011

### Staff: Mentor

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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.

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

4. Oct 20, 2011

### TrickyDicky

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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).

5. Oct 20, 2011

### Staff: Mentor

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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:

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.

Last edited: Oct 20, 2011
6. Oct 20, 2011

### TrickyDicky

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

I surely didn't imply such "flipping".

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.

7. Oct 20, 2011

### Ben Niehoff

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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

8. Oct 20, 2011

### Staff: Mentor

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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.

9. Oct 20, 2011

### TrickyDicky

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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

10. Oct 20, 2011

### Ben Niehoff

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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:

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

11. Oct 20, 2011

### TrickyDicky

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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

12. Oct 20, 2011

### Ben Niehoff

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

None of this is relevant to anything.

13. Oct 20, 2011

### TrickyDicky

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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.

14. Oct 20, 2011

### Ben Niehoff

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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

15. Oct 20, 2011

### TrickyDicky

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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

16. Oct 20, 2011

### Ben Niehoff

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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.

17. Oct 20, 2011

### TrickyDicky

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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

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

18. Oct 20, 2011

### Ben Niehoff

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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 $x^2 - a = 0$?

19. Oct 20, 2011

### TrickyDicky

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

Really, no need for snide remarks.

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

20. Oct 20, 2011

### Staff: Mentor

Re: Derivation of expansion scalar for FRW spacetime -- weird observation

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.