Levi-Civita connection and pseudoRiemannian metric

[PeterDonis]

ds is not an arc length, s is.

This is just quibbling about terminology; $s = \int ds$, so $s = 0$ for a null curve, so a null curve has zero arc length.

ds=0 refers in GR to null vectors at a point's tangent space.

No, it doesn't. It's an infinitesimal that can be integrated to an arc length. See above.

Null geodesics for instance don't have arc length because since they can't be parametrized by arc length, they don't integrate to an arc length.

Incorrect. See above. You can do the integral above without having to parametrize the curve by arc length; *any* parametrization of the curve will work fine (including the implicit parametrization you have when you describe the curve in some coordinate chart).

 

[PeterDonis]

You can do the integral above without having to parametrize the curve by arc length; *any* parametrization of the curve will work fine (including the implicit parametrization you have when you describe the curve in some coordinate chart).

Since this appears not to be obvious, I'll give an explicit example. Consider the standard Minkowski coordinate chart on Minkowski spacetime, and the curve whose equation in that chart is $t = x$, from $(t, x) = (0, 0)$ (call this point A) to $(t, x) = (1, 1)$ (call this point B). The arc length along the curve is

$$
s = \int_A^B ds = \int_A^B \sqrt{\eta_{\mu \nu} dx^{\mu} dx^{\nu}} = \int_A^B \sqrt{- dt^2 + dx^2}
$$

The equation for the curve, $t = x$, immediately gives $dt = dx$, so we can eliminate one variable; it doesn't matter which, but for concreteness I'll eliminate $t$. Then we have

$$
s = \int_A^B \sqrt{- dx^2 + dx^2} = \int_0^1 dx \sqrt{ - 1 + 1 } = 0
$$

If we wanted to, we could treat $x$ as an affine parameter along the curve, and write (using capital letters for the coordinates now, since we are treating them as functions of the affine parameter $x$)

$$
s = \int_0^1 dx \sqrt{ - \left( \frac{dT}{dx} \right)^2 + \left( \frac{dX}{dx} \right)^2 } = \int_0^1 dx \sqrt{- 1 + 1 } = 0
$$

We get the same answer either way. So we can affinely parametrize this null curve even though it has zero arc length when we do the integral. We can't parametrize by arc length, but we don't need to to get a well-defined answer for the arc length.

 

[TrickyDicky]

Now we are just going in circles. As I said I'm not concerned with null curves in general here but just null geodesics. And to me vanishing arc length means the same as not having arc length so this is mostly about words now. In other words precisely the reason they can't be parametrized by arc length is that they don't have arc length(=it is zero).

Thanks a lot for your input anyway, Peter.

 

[PeterDonis]

to me vanishing arc length means the same as not having arc length

Of course you are at liberty to adopt whatever terminology you wish, but you should realize that this terminology is going to confuse a lot of people. "Not having arc length" to me, and I suspect to a lot of people, means that "arc length" as a concept is not well-defined. "Vanishing arc length" to me, and I suspect to a lot of people, means "arc length" is well-defined and its value is zero. Those are not at all the same thing, and your terminology makes it very unclear whether you mean the former (wrong) claim or the latter (correct) claim. I'm still not entirely certain which claim you are making.

 

[TrickyDicky]

I'm still not entirely certain which claim you are making.

A few subtleties that it is convenient to get out of the way to understand this discussion.

We have agreed that in general in GR the metric intergration of spacetime distances is not meaningful(specially for the light geodesic case), and this is a good thing to avoid problems with the metric space mathematical structure of manifolds. What it is run of the mill and quite routinely used in cosmology is to pick spacetimes where a clear cut can be made between the temporal and spatial parts, so that we can obtain times and distances without any topological problem. This is the case for spacetimes that have worldlines that are hypersurface orthogonal , i.e. FRW and static spacetimes.
In these cases one can concentrate on the purely time or length distances in the spatial slice hyperurface, it is even easier in the static case where one doesn't have to worry about the scale factor.
There is an equation in cosmometry that gives proper distance:
$$D_p=a_0 \int^r_0 \frac{dr}{\sqrt{1-kr^2}}=a_0 \int^{t_0}_t \frac{dt}{a}$$
That in the static case reduces to $D_p=t_0-t$ This is called light-travel distance or look-back distance although it is more frequently given as function of redshift z for the FRW spacetime.
This expression is obtained from the metric tensor, considering the null geodesic path with ds=0 followed by a light ray in a fixed direction(arbitrary due to isotropy) emitted at time t and coordinate r=0 and received at time t0 in r.

In this cosmological examples we can see how we can use both the radial and the time coordinates as affine parameters(for instance in the Schwarzschild case we could use the radial but not the time coordinate as affine parameter due to a factor in the time part of the metric and the radial coordinate can't be equated directly to a radial distance because obviously the spatial slice is non-euclidean).
And in the case with euclidean hypersurface the radial parameter is indeed a radial distance.
So these are very specific examples where distances are integrated and actually used in GR applied to cosmology, and only because for these spacetimes we can clearly separate a spatial slice from the spacetime manifold. And they all are perfectly compatible with the manifold topology.

I hope it is understood why I(coming from cosmology to GR) was puzzled when told distances are pointless in GR. This is true in a rather subtle way.

 

[PeterDonis]

We have agreed that in general in GR the metric integration of spacetime distances is not meaningful(specially for the light geodesic case)

It would really be helpful if you would stop using terminology that you have been told several times is ambiguous, especially when you are specifically responding to a statement of mine that was made because of that ambiguity.

If I take the statement "metric integration of spacetime distances" at face value, it refers to the integral $\int ds$, which I have already shown *is* meaningful, and therefore your statement quoted above is simply false.

If I take "metric integration of spacetime distances" to refer to integrating the distance function that is compatible with the underlying manifold topology (I'll call this distance function $D_T$), then your statement is true, but it's also irrelevant, because, as I've already said a number of times, that distance function has no physical meaning. In some cases, the physical distance function, the one that uses the metric tensor, and is found by computing $\int ds = \int \sqrt{g_{\mu \nu} dx^{\mu} dx^{\nu}}$ (I'll call this distance function $D_M$), happens to give the same answer as $D_T$; but that doesn't mean we're using $D_T$ to compute physical spacetime distances ("intervals" might be a better word, since the "distances" involved are not always spacelike).

Also, if you meant "metric" in the second sense (to refer to $D_T$ instead of $D_M$) in the quote above, then you're being inconsistent in your use of terminology, because later in your post you use "metric" explicitly to refer to $D_M$.

this is a good thing to avoid problems with the metric space mathematical structure of manifolds.

Which is a perfectly good mathematical structure, yes, but is irrelevant physically because we don't use $D_T$ to compute physical spacetime distances; we use $D_M$. Yes, $D_M$ does not give the same answers in all cases as $D_T$ (by the way, you have been concentrating on null curves, but this is also true of timelike curves; there is no way to get a negative squared interval from $D_T$). So what? That doesn't pose any problem for the physical interpretation of the theory, because, as I just said, we don't use $D_T$ to compute physical spacetime distances.

What it is run of the mill and quite routinely used in cosmology is to pick spacetimes where a clear cut can be made between the temporal and spatial parts, so that we can obtain times and distances without any topological problem.

Can you give any mainstream references that talk about this and why it's so important? I haven't seen anything like this in any of the reading I've done about cosmology. Of course there is lots of cosmology literature that I haven't read, so if you can point to some that talks about this point, please do.

There is an equation in cosmometry that gives proper distance:
$$D_p=a_0 \int^r_0 \frac{dr}{\sqrt{1-kr^2}}=a_0 \int^{t_0}_t \frac{dt}{a}$$

[STRIKE]Can you give a reference for the second equality? I've seen this expressed in terms of $r$, but not in terms of $t$. On first inspection, it does not look to me like the two expressions are equal.[/STRIKE] [Edit: Never mind, I worked it out.]

That in the static case reduces to $D_p=t_0-t$

Is there any relevant "static case" in cosmology? The universe is not static.

This expression is obtained from the metric tensor

Yes. But note that you are using "metric" here to refer to $D_M$, not $D_T$.

considering the null geodesic path with ds=0 followed by a light ray in a fixed direction(arbitrary due to isotropy) emitted at time t and coordinate r=0 and received at time t0 in r.

But the actual integral is not taken along that null geodesic, is it? Certainly the integral in terms of $r$ is not; it's taken along a spacelike geodesic from $0$ to $r$, at constant $t$. That's obvious because the integrand is just $dr \sqrt{g_{rr}}$.

In this cosmological examples we can see how we can use both the radial and the time coordinates as affine parameters

Yes, I've already agreed that coordinates can be used as affine parameters. That doesn't change any of the other things I've said.

So these are very specific examples where distances are integrated and actually used in GR applied to cosmology, and only because for these spacetimes we can clearly separate a spatial slice from the spacetime manifold.

Um, you can take a spacelike slice in any spacetime. See further comments below.

And they all are perfectly compatible with the manifold topology.

Well, of course they are, because they're all spacelike "distances". If we restrict attention to a spacelike slice, then the restriction of the physical metric tensor $g_{\mu \nu}$ to the slice obviously gives a Riemannian metric on the slice, and this metric obviously induces a topology that is compatible with the manifold topology (since that is true of any Riemannian metric on a manifold).

But you can cut a spacelike slice, and get a Riemannian metric on it which is compatible with the manifold topology, in any spacetime you like; so having a meaningful spacelike "distance" is in no way restricted to a particular class of spacetimes.

I hope it is understood why I(coming from cosmology to GR) was puzzled when told distances are pointless in GR. This is true in a rather subtle way.

Only for one meaning of the term "distance". If "distance" means a number obtained from $D_T$, or some distance function that induces the same topology on the manifold as $D_T$, then such "distances" can only be given physical meaning on spacelike slices (by choosing $D_T$ so it's equal to the restriction of $D_M$ to the slice); they have no physical meaning for null or timelike curves. But if "distance" means a number obtained from $D_M$, then such "distances" are physically meaningful in all cases. So it all depends on what you mean by "distance". As I pointed out at the start of this post, it would be really helpful if you would stop using terms that you have been told are ambiguous.

 

[TrickyDicky]

If I take the statement "metric integration of spacetime distances" at face value, it refers to the integral $\int ds$, which I have already shown *is* meaningful, and therefore your statement quoted above is simply false.

I was using meaningful in the sense of useful. Sorry about that.

The integral $\int ds$=0 is useless in GR for null geodesics. If you disagree with this we better let it go.

Is there any relevant "static case" in cosmology? The universe is not static.

Well, Einstein's universe is relevant for historical reasons, not physical, but it is instrumental because its metric is used as "unphysical metric" in conformal gravity(Penrose diagrams for black holes,etc).
I was not using it to imply the universe is static, static examples are in all books about cosmology as pedagogical tolos and to explain that our universo is not like that and why. No need for the condescent tone.

Yes. But note that you are using "metric" here to refer to $D_M$, not $D_T$.

Sure, no posible mistake because when I refer to this I use the word metric accompanied by tensor. And alone otherwise.

But the actual integral is not taken along that null geodesic, is it? Certainly the integral in terms of $r$ is not; it's taken along a spacelike geodesic from $0$ to $r$, at constant $t$. That's obvious because the integrand is just $dr \sqrt{g_{rr}}$.

Not strictly, no. In the sense that you only consider integrating null geodesics the integral of ds=0. But it is the only useful distance one can obtain from ds=0.
EDIT:It does integrate a null geodesic in the example. Using an affine parameter different from arc length. Just equate the line element to zero so you get the dt part equal to the dr part of the line element.

And that is what was creating my problems at the beginning of the thread. You see, I'm not really claiming I've discovered anything, certainly not something that disagrees with any mainstream notion. On the contrary, I'm just trying to understand it better myself with these posts. I'm saying I didn't know how to make consistent certain things(yes, I'm dumb and you people so gifted, so what?) in my head and now I can. That's all.

But you can cut a spacelike slice, and get a Riemannian metric on it which is compatible with the manifold topology, in any spacetime you like; so having a meaningful spacelike "distance" is in no way restricted to a particular class of spacetimes.

No. But I feel that I would only confuse you more if I tried to explain it now. I'll try later.

It would really be helpful if you would stop using terminology that you have been told several times is ambiguous, especially when you are specifically responding to a statement of mine that was made because of that ambiguity.

Only for one meaning of the term "distance". If "distance" means a number obtained from $D_T$, or some distance function that induces the same topology on the manifold as $D_T$, then such "distances" can only be given physical meaning on spacelike slices (by choosing $D_T$ so it's equal to the restriction of $D_M$ to the slice); they have no physical meaning for null or timelike curves. But if "distance" means a number obtained from $D_M$, then such "distances" are physically meaningful in all cases. So it all depends on what you mean by "distance". As I pointed out at the start of this post, it would be really helpful if you would stop using terms that you have been told are ambiguous.

See above how I distinguish it terminologically. I might have not used that distinction consistently, if so my apologies..
You seem to imply that the sorry fact that metric is a word confusingly used both for metric tensors and metrics as distance functions is my fault.

 

[PeterDonis]

The integral $\int ds$=0 is useless in GR for null geodesics. If you disagree with this we better let it go.

I wouldn't say it's completely useless, but I agree there's not much to it compared to the analogous computations for timelike and spacelike curves.

Well, Einstein's universe is relevant for historical reasons, not physical, but it is instrumental because its metric is used as "unphysical metric" in conformal gravity(Penrose diagrams for black holes,etc).

Ah, ok.

It does integrate a null geodesic in the example. Using an affine parameter different from arc length. Just equate the line element to zero so you get the dt part equal to the dr part of the line element.

That just establishes the second equality you give. It does not mean either integral is taken along a null geodesic. The first integral (in terms of $r$) is taken along a spacelike geodesic (all coordinates constant except $r$). The second is taken along a timelike geodesic (all coordinates constant except $t$). Integrating along the actual null geodesic path that the light takes would give zero, since $\int ds = 0$ along a null geodesic; you've already agreed to that multiple times.

I'm not really claiming I've discovered anything, certainly not something that disagrees with any mainstream notion. On the contrary, I'm just trying to understand it better myself with these posts. I'm saying I didn't know how to make consistent certain things(yes, I'm dumb and you people so gifted, so what?) in my head and now I can.

This does help to clarify where you are coming from.

You seem to imply that the sorry fact that metric is a word confusingly used both for metric tensors and metrics as distance functions is my fault.

No, just that it's better to avoid using the word "metric" since it's ambiguous. Adopting terminology like I did in my post, using $D_T$ and $D_M$ for the two "distance functions", makes things clearer. "Metric tensor" as a term is OK since, as you note, the term "tensor" implies $D_M$, not $D_T$. But even then, it's still a good idea, IMO, to avoid the word "metric" when referring to $D_T$, even if you don't ever use the term "metric tensor" to refer to $D_T$. In GR, the word "metric" is very closely associated to $D_M$, so using it to refer to anything else is highly likely to cause confusion.

 

[TrickyDicky]

The first integral (in terms of $r$) is taken along a spacelike geodesic (all coordinates constant except $r$). The second is taken along a timelike geodesic (all coordinates constant except $t$).

I don't understand why you say this. In both integrals neither r nor t are constant. the first in terms of r has t going from time at emmision to time at reception of the light ray, and the second in terms of t has r going from the origin set at the point of emmision(i.e. distant galaxy) to the point of reception of the light. This relies on the finiteness of c.

Integrating along the actual null geodesic path that the light takes would give zero, since $\int ds = 0$ along a null geodesic

But it looks like you are stressing this as if one couldn't always choose an afffine parameter diffferent from arc length for null geodesics.

 

[PeterDonis]

I don't understand why you say this. In both integrals neither r nor t are constant.

Are you sure? In the first integral, the integrand only contains $dr$, not $dt$, and is only a function of $r$, not $t$, and the limits of integration are values of $r$, not $t$. In the second integral, the integrand only contains $dt$, not $dr$, and is only a function of $t$, not $r$, and the limits of integration are values of $t$, not $r$. When I took calculus, I was taught that integrals like these are only over one variable; if there are any other variables involved (like $t$ in the first integral or $r$ in the second), they must be held constant. If you have an integral where two variables are both changing, it's a multiple integral, and neither of the integrals you wrote down are multiple integrals. (A multiple integral wouldn't make any sense in this example anyway; it would be computing an area, not a distance.)

the first in terms of r has t going from time at emmision to time at reception of the light ray

Then these values of $t$ should appear somewhere in the first integral. They don't.

the second in terms of t has r going from the origin set at the point of emmision(i.e. distant galaxy) to the point of reception of the light.

Then these values of $r$ should appear somewhere in the second integral. They don't.

But it looks like you are stressing this as if one couldn't always choose an afffine parameter diffferent from arc length for null geodesics.

I'm saying nothing whatever about an affine parameter. I made a simple statement that you have already agreed with multiple times: $\int ds = 0$ for any null geodesic, including the one the light follows in your example. To be explicit, I mean $\int ds = 0$ using $D_M$ as the distance function (I already wrote that out in a previous post, but it's worth saying again given all the confusion in this thread about different distance functions). That's the only relevant distance function for computing the integrals we're talking about. Given that, the statement I made is true regardless of how you choose to affinely parametrize the null geodesic.

Perhaps the issue here is that you are equating choosing an affine parameter with choosing a distance function. They're not the same. I can choose an affine parameter other than arc length along a null geodesic and still compute $\int ds = 0$ using $D_M$ as the distance function, not $D_T$.

 

[TrickyDicky]

Are you sure? In the first integral, the integrand only contains $dr$, not $dt$, and is only a function of $r$, not $t$, and the limits of integration are values of $r$, not $t$. In the second integral, the integrand only contains $dt$, not $dr$, and is only a function of $t$, not $r$, and the limits of integration are values of $t$, not $r$. When I took calculus, I was taught that integrals like these are only over one variable; if there are any other variables involved (like $t$ in the first integral or $r$ in the second), they must be held constant.
Then these values of $t$ should appear somewhere in the first integral. They don't.

Then these values of $r$ should appear somewhere in the second integral. They don't.

Ok,sure, but the expression with the integrals was an equation, it relates both integrals, that is obviously stablishing a relation between r and t that is clearly physical in the cosmological context, finiteness of c allows us to link the r=0 with time t and r with t0 for the light geodesic. My remarks were (I though it obvious) in he context of both integrals being equated. If you take the integrals independently and outside their physical context you are right.

 

[PeterDonis]

My remarks were (I though it obvious) in he context of both integrals being equated. If you take the integrals independently and outside their physical context you are right.

Agreed except for the "outside their physical context" part. Equating the integrals, and establishing a relationship between their limits of integration (all of which I agree with) doesn't make them integrals along the null curve. The fact that the $r$ integral, considered by itself, is actually an integral along a spacelike geodesic, and the $t$ integral, considered by itself, is actually an integral along a timelike geodesic, are part of the "physical context".

To put this point another way, the equality between the integrals is not just telling us something about the null geodesic that the light follows (all it tells us about that is that $ds = 0$ along that geodesic, since that's what makes the $r$ and $t$ integrals equal). The equality between the integrals is also telling us that a certain integral along a certain spacelike geodesic (an integral which has an obvious physical interpretation as the spatial "distance" between two comoving worldlines at the instant of comoving time $t$ that the light is received) gives the same numerical value as a certain integral along a certain timelike geodesic, when the integrands and limits of integration are related in the way you have shown (i.e., via the $ds = 0$ constraint imposed by the null geodesic the light follows). All that is part of the "physical context"--if nothing else, it's needed to justify calling the numerical value obtained a "distance".

 

[TrickyDicky]

Agreed except for the "outside their physical context" part. Equating the integrals, and establishing a relationship between their limits of integration (all of which I agree with) doesn't make them integrals along the null curve. The fact that the $r$ integral, considered by itself, is actually an integral along a spacelike geodesic, and the $t$ integral, considered by itself, is actually an integral along a timelike geodesic, are part of the "physical context".

To put this point another way, the equality between the integrals is not just telling us something about the null geodesic that the light follows (all it tells us about that is that $ds = 0$ along that geodesic, since that's what makes the $r$ and $t$ integrals equal). The equality between the integrals is also telling us that a certain integral along a certain spacelike geodesic (an integral which has an obvious physical interpretation as the spatial "distance" between two comoving worldlines at the instant of comoving time $t$ that the light is received) gives the same numerical value as a certain integral along a certain timelike geodesic, when the integrands and limits of integration are related in the way you have shown (i.e., via the $ds = 0$ constraint imposed by the null geodesic the light follows). All that is part of the "physical context"--if nothing else, it's needed to justify calling the numerical value obtained a "distance".

The first integral can be physically interpreted as a "spacelike=spatial" distance, yes, but not as integrating any physical path into a spacelike geodesic, nothing physical that we know of follows spacelike geodesics as it would have to go faster than light.

Similarly, the second integral can be interpreted as timelike in the sense that makes a temporal measure in coordinate time. But never as the distance integrated from a timelike geodesic that describes only material particles paths at less than c, as there is no proper time here, and nothing physical except a light ray could follow that path in that time according to its own clock.
That's why I said "outside their physical context".

 

[PeterDonis]

The first integral can be physically interpreted as a "spacelike=spatial" distance, yes, but not as integrating any physical path into a spacelike geodesic

That doesn't change the fact that our usual concept of "distance" is "arc length integrated along a spacelike geodesic". That's why the term "distance" is used to describe the first integral. It's a perfectly good physical quantity; not all physical quantities have to describe arc lengths along timelike or null curves.

nothing physical that we know of follows spacelike geodesics as it would have to go faster than light.

The second part is true, but the first part does not follow from it. "An object at an instant of time" is a perfectly good physical thing, and "the proper length of an object", which is what arc length along a spacelike geodesic describes, is a perfectly good physical quantity.

Similarly, the second integral can be interpreted as timelike in the sense that makes a temporal measure in coordinate time. But never as the distance integrated from a timelike geodesic that describes only material particles paths at less than c, as there is no proper time here

True; the integral actually gives the conformal time between emission and reception, which is not a "time" actually measured by any clock. I'll agree that conformal time is not a good example of a "physical quantity" for that reason.

nothing physical except a light ray could follow that path in that time according to its own clock.

A light ray can't do it either, since a light ray doesn't have a "clock"; the concept of "proper time" or "clock time" doesn't apply to a light ray. As far as I know, conformal time does not correspond to *any* actual physical observable that is "timelike". (The equality between integrals shows that conformal time can be equated to a "distance", but that just makes it superfluous, since we already have distance as a physical quantity.)

 

[TrickyDicky]

That doesn't change the fact that our usual concept of "distance" is "arc length integrated along a spacelike geodesic". That's why the term "distance" is used to describe the first integral. It's a perfectly good physical quantity; not all physical quantities have to describe arc lengths along timelike or null curves.

The second part is true, but the first part does not follow from it. "An object at an instant of time" is a perfectly good physical thing, and "the proper length of an object", which is what arc length along a spacelike geodesic describes, is a perfectly good physical quantity.

I wasn't disputing their being physical quantities, only your calling them integrals of spacelike and timelike geodesics.

A light ray can't do it either, since a light ray doesn't have a "clock"; the concept of "proper time" or "clock time" doesn't apply to a light ray.

That was my point, thus my problem with your calling the second integral that of a timelike geodesic.

 

[PeterDonis]

I wasn't disputing their being physical quantities, only your calling them integrals of spacelike and timelike geodesics.

Let me explain in more detail what I meant. Look at the integrals again:

$$
D_p = a_0 \int_0^r \frac{dr}{\sqrt{1 - k r^2}} = a_0 \int_t^{t_0} \frac{dt}{a(t)}
$$

The first integral has $r$ as the integration variable, and the limits of integration are $0$ to $r$. That corresponds to (meaning, has the obvious physical interpretation of) integrating along a spacelike geodesic where $r$ is the only coordinate changing: i.e., a radial (constant $\theta$ and $\phi$) spacelike geodesic lying in a surface of constant $t$.

The second integral has $t$ as the integration variable, and the limits of integration are $t$ to $t_0$, i.e., the coordinate time of emission to the coordinate time of reception. That corresponds to integrating along a timelike geodesic where $t$ is the only coordinate changing: i.e., a "comoving" worldline with constant $r$ (and $\theta$ and $\phi$, of course).

(Of course the integrand in the second integral is not just $dt$, so the integral doesn't give the elapsed proper time along the timelike geodesic. But you can integrate any function of $t$ you like along that geodesic; here we're integrating $1 / a(t)$ because that happens to be the integrand of interest for this problem. Similarly, we could integrate any function of $r$ we like along the spacelike geodesic used in the first integral above; the particular function of $r$ we actually integrate, which is $\sqrt{g_{rr}}$ and therefore gives us the actual proper distance along that geodesic, is the one that happens to be of interest for this problem.)

Now, just to confirm that I am not ignoring the role played by the null geodesic the light actually follows: here's an integral that computes the arc length along that geodesic, from the event of emission to the event of reception (I've squared the integral to make things simpler):

$$
0 = \int_e^0 ds^2 = \int_e^0 g_{\mu \nu} dx^{\mu} dx^{\nu} = \int_e^0 \left( - dt^2 + a(t)^2 \frac{dr}{1 - k r^2} \right) = \int_e^0 \left[ - \left( \frac{dt}{a(t)} \right)^2 + \left( \frac{dr}{\sqrt{1 - k r^2}} \right)^2 \right]
$$

Of course this looks a lot like the above; all we need to do is to move the $t$ integral to the LHS, right? No; it's not that simple. This is a single integral, not a multiple integral; so what is the single integration variable? Well, it's whatever affine parameter we choose along the null geodesic. To make things as easy as possible, let's choose $t$. Then we should be writing the integral like this:

$$
0 = \int_t^{t_0} dt^2 \left[ - \left( \frac{1}{a(t)} \right)^2 + \left( \frac{dr}{dt} \right)^2 \left( \frac{1}{\sqrt{1 - k r(t)^2}} \right)^2 \right]
$$

Of course we can still split this up into two pieces, and convert the second piece into an integral over $r$ by cancelling the $dt$ factors and converting the limits of integration. But what are we doing when we do that? We are taking a single integral in a single variable, and saying that it equals the sum of two integrals, each in a *different* variable, because once we split the integral up, and look at each piece by itself--which is what we have to do to get a "distance" $D_p$ out of all this--we are *changing* what the variables mean.

This is easy to see in the case of $r$; in the integral just above, $r$ is a function of $t$, but in the first integral (the one with $dr$ as the integration measure), $r$ is the integration variable, and is not a function of anything. If we were still trying to evaluate an integral along the null geodesic the light follows, that wouldn't make sense (note that, unlike $t$, we cannot choose $r$ as an affine parameter along the null geodesic).

But even in the case of the second integral (meaning the second one you originally gave, the one on the RHS of the equality at the top of this post), once we break it apart from the first integral and look at it by itself--which, again, is what we *have* to do in order to get a "distance" $D_p$--I would say that $t$ is no longer an affine parameter along the null geodesic, because we're no longer integrating along the null geodesic: we threw away a piece of the integral, and that changes its meaning. (See further comments below on this.)

Now I'll agree that, to an extent, all this is a matter of terminology, not physics; we appear to agree on the actual physics, we just disagree on the best words to use to describe it. But I would be very curious to see whether you can give any mainstream references that take your point of view: for example, that explicitly describe the first integral (the one in $dr$, on the LHS of the equality at the top of this post) as being taken along the null geodesic the light follows, rather than along a radial spacelike geodesic of constant $t$, which is the obvious physical interpretation.

That was my point, thus my problem with your calling the second integral that of a timelike geodesic.

If we take the second integral as it stands, it is not taken along the light ray's worldline; it is taken along the timelike geodesic (see above). So the fact that the concept of "proper time" doesn't apply along the light ray's worldline is irrelevant.

If we think of the second integral as being a piece of the third integral above (the one taken along the null geodesic the light actually follows), then yes, we shouldn't use the term "time" or "timelike" to refer to it. But the same argument would also say that we shouldn't use the term "distance" to refer to the first integral, if we view it as a piece of the third integral, because "distance" implies spacelike, and the null geodesic the light follows is not spacelike any more than it's timelike.

Finally, the specific phrase of yours that prompted my comment was "nothing physical except a light ray could follow that path in that time according to its own clock". If that phrase was your way of saying that a null geodesic is not timelike so the concept of "proper time" doesn't apply, it was a very unclear way of saying it; it doesn't make sense to use the phrase "according to its own clock" about something that can't even have a "clock". If you would say clearly what you actually mean to begin with, you would get far fewer comments about what you say being wrong or misleading.

 

[TrickyDicky]

Thanks. I appretiate your post and I agree on the substantial. I'll try to find a reference and if I find it will post it.

 

[TrickyDicky]

Narlikar's "An introduction to cosmology", end of page 112 and beginning of 113. There is a preview in google books.

 

[PeterDonis]

Narlikar's "An introduction to cosmology", end of page 112 and beginning of 113. There is a preview in google books.

Yes, this basically runs through a similar derivation to the one I posted; it uses the integral along the null geodesic path of the light to derive an equality between two other integrals (the ones that appear added together in the integral along the null geodesic).

But I don't see this passage committing to any physical interpretation of the $r$ integral. The discussion you are referencing is trying to get the answer to a different question:

At what time should a light wave leave $G_1$ in order to arrive at $r = 0$ at the present time $t = t_0$?

(p. 112, next to last paragraph)

In other words, the point here is to find "how long" the light travels, not "how far" it travels! The equality of the integrals is just used to give a way of finding the emission time $t$ (Narlikar calls it $t_1$) in terms of known quantities: the function $a(t)$ (Narlikar calls it $S$), the constant $k$, and the (presumed known) radius $r$ of the point of emission.

And to the extent anything is given a physical interpretation, it appears to be an interpretation of the $t$ integral as giving a time, which, to me, implies interpreting it as being taken along a timelike geodesic, for the same reasons that I gave in my previous post with reference to the derivation I gave there (but see further comment below). I agree there isn't anything explicit about this in the book, but there isn't anything explicit the other way either, because, again, the derivation is open to the same interpretational issues that I brought up in my previous post (i.e., splitting the integral into two pieces changes the variables and the interpretation of the pieces).

There is one other difference between this passage and the derivation I gave in my previous post: the time $t$ that is being solved for is one of the limits of integration of the $t$ integral, not the result of the integral (whereas in my derivation, the distance being solved for was the result of the $r$ integral). So instead of determining both integration limits and then solving the $t$ integral, Narlikar is determining the value that the $t$ integral has to produce, and using that to determine the lower limit of integration (the upper limit $t_0$ being presumed known). So no physical interpretation of the actual *value* of the $t$ integral is necessary; all that's necessary is to interpret the limits of integration as times. But "no physical interpretation is necessary" isn't the same as "the physical interpretation is as an integral of something along the null geodesic". (Also, this doesn't change the fact that the $t$ integral is only taken over a single variable, and its range can't be along the null geodesic because we've taken away a piece of the integral.)

 

[TrickyDicky]

In the text it is literally stated "we need to calculate the null geodesic from G1 to us". Calculations based on the integration of the null geodesic equations are routinely used in studies of look-back, light-travel distances, time delay or frequency shifts of light.
Hairsplitting whether thy are about "how far" or "how long" or whether each integral by itself is the integral of a null geodesic is a purely terminologic discussion that is of not much utility in practice. My point all along was that any calculation based on integrals of the equation ds=0 would be done in terms of an affine parameter different from arc length s or proper time tau because both of those parameters gave vanishing results upon integration that were useless to calculate anything(besides the fact that a vanishing distance between distinct points is not a feature one finds in a metric space with the manifold topology).

 

[martinbn]

Just out of curiosity, what is the point of this thread?

 

[PeterDonis]

In the text it is literally stated "we need to calculate the null geodesic from G1 to us".

Which, taken in context, means "find an expression for $\int ds = 0$ along the null geodesic, and then split it up into pieces that have different physical interpretations". What the text does *not* say is anything along the lines of "these separate pieces, even though they have obvious physical interpretations as integrals along timelike or spacelike geodesics, are actually integrals of some function other than $ds = 0$ along the null geodesic the light follows".

Calculations based on the integration of the null geodesic equations

I'm not sure what you mean by "integration of the null geodesic equations". Finding an expression for $\int ds = 0$ is not the same as integrating the geodesic equation; the $\int ds = 0$ expression can be written down purely from the FRW line element. The geodesic equation itself doesn't appear anywhere in the integrand.

Hairsplitting whether thy are about "how far" or "how long" or whether each integral by itself is the integral of a null geodesic is a purely terminologic discussion that is of not much utility in practice.

If you mean you don't need to go into all these details in order to get a numerical answer, of course you're right. But the same can be said of hairsplitting about whether what we are doing can be interpreted as calculating a "distance" along the null geodesic that's different from the physical $ds = 0$ distance (i.e., whether we can interpret what we're doing as calculating $D_T$ instead of $D_M$), as opposed to splitting up the $\int ds = 0$ expression into pieces with obvious physical interpretations as integrals along timelike or spacelike geodesics. None of these interpretations matter if all you're concerned with is getting a numerical answer for $D_p$. They only matter if you have particular preferences for how to justify interpreting $D_p$ as a "distance".

 

[TrickyDicky]

Which, taken in context, means "find an expression for $\int ds = 0$ along the null geodesic, and then split it up into pieces that have different physical interpretations". What the text does *not* say is anything along the lines of "these separate pieces, even though they have obvious physical interpretations as integrals along timelike or spacelike geodesics, are actually integrals of some function other than $ds = 0$ along the null geodesic the light follows".

I am saying(like you I think) that they are integrals of $ds = 0$ along the null geodesic the light follows.

I'm not sure what you mean by "integration of the null geodesic equations". Finding an expression for $\int ds = 0$ is not the same as integrating the geodesic equation; the $\int ds = 0$ expression can be written down purely from the FRW line element. The geodesic equation itself doesn't appear anywhere in the integrand.

Right, the sentence is read:integration of the "null geodesic" equations (ds=0), not integration of the null "geodesic equations".

If you mean you don't need to go into all these details in order to get a numerical answer, of course you're right. But the same can be said of hairsplitting about whether what we are doing can be interpreted as calculating a "distance" along the null geodesic that's different from the physical $ds = 0$ distance (i.e., whether we can interpret what we're doing as calculating $D_T$ instead of $D_M$), as opposed to splitting up the $\int ds = 0$ expression into pieces with obvious physical interpretations as integrals along timelike or spacelike geodesics. None of these interpretations matter if all you're concerned with is getting a numerical answer for $D_p$. They only matter if you have particular preferences for how to justify interpreting $D_p$ as a "distance".

Fair enough.
Since there is some curiosity about the point of this thread and just in case you are interested in what goes on beyond the purely computational side of it, the chapter "On the structure of space-time" page 178(in google books preview) in the book "Space, time and geometry" is what inspired my OP.

 

[martinbn]

Do you mean the book by P. Suppes? I must say that I have absolutely no idea how that page of the book led you to this question!

 

[TrickyDicky]

Do you mean the book by P. Suppes? I must say that I have absolutely no idea how that page of the book led you to this question!

Yes, that book. But that is just the first page in the essay, he mentions metric spaces and how this definition applies to relativity and classical space-times in pge 180.