Understanding Einstein Field Equations Through Tensor Calculus

[Passionflower]

Interesting discussion, it seems a lot of the argument has to do with terminology and perhaps an understanding of the matter. I will try to summarize in order to organize my own thoughts about it, and hopefully do not instead add more confusion.

If I understand JDoolin's view correctly he considers the 'real' distance being the cord length, e.g. the length of the cord if we span this cord using an embedding dimension, between two points (we assume we slice time and space in a 'sensible' way):

Certainly, if someone told me the distance between two cities, I would expect them to be talking about the distance along a particular road or set of roads. But where are you taking this? The real distance between the two points is the one connecting the two cities right through the surface of the planet.

That is certainly a point of view but is there any physics associated with it? According to GR it seems not, assuming this is right then my question to JDoolin is what is the point in calling this the "real distance"?

I think the main argument to call a space curved (again assuming a 'sensible' slicing of time and space) is that it is not physically possible to travel on the 'cord distance'.

Let's for instance take an example using the Schwarzschild solution:

Suppose we have 8 space stations forming the vertices of a cube around a planet. All stations keep a fixed distance D from the surface of a planet. Then if they start sending out light signals to each other, or when small space crafts travel from one to another all will have to conclude that the space around this planet is not Euclidean. One could argue and say, 'well they did not travel on the cord distance', however this seems a moot point as such traveling is not physically possible.

The reason we can make this distinction, though, is that we know the Earth's surface is a 2-dimensional manifold that is embedded in a 3-dimensional space that is (as far as we can tell) Euclidean

I agree that the surface of the Earth is a 2-dimensional manifold however I disagree that space is Euclidean.

However, there is no guarantee that we will always be able to find a higher-dimensional Euclidean space in which a given manifold is embedded.

As far as I understand one can always embed a given manifold in higher dimensions, whether that makes sense is a different question. How do you conclude there is no guarantee?

For example, it is possible (though not likely, given the current state of evidence) that the 3-dimensional manifold that we call "space" is actually not Euclidean

As I said before we already know space is not Euclidean.

For instance in the Schwarzschild solution, which is one of the simplest solutions, we could 'push' all the curvature into the 'time' dimension. For instance consider the GP coordinate chart. But this 'pushing' only works for the Schwarzschild solution, generally it is impossible to do so.

Generally when there is mass-energy spacetime will be curved and if this mass-energy is distributed we cannot push all this curvature into the 'time' dimension only.

 

[PeterDonis]

As far as I understand one can always embed a given manifold in higher dimensions, whether that makes sense is a different question. How do you conclude there is no guarantee?

Mathematically, we can always find an embedding, yes. I don't know for sure if, mathematically, we can always find an embedding in a higher-dimensional *Euclidean* manifold. But in any case, that wasn't what I was referring to; I was referring to the fact that, *physically*, there's no guarantee that there will always be physical phenomena that *require* the higher-dimensional manifold to exist. In the case of the Earth's surface, we know the higher-dimensional manifold exists, because we can directly experience it, we can run various physical experiments that show it exists, etc. But, for example, if the universe as a whole turns out to be spatially curved (see below), there may not be any physics that requires it to be embedded in a higher-dimensional manifold (in fact, according to our current understanding, there isn't).

As I said before we already know space is not Euclidean.

For instance in the Schwarzschild solution, which is one of the simplest solutions, we could 'push' all the curvature into the 'time' dimension. For instance consider the GP coordinate chart. But this 'pushing' only works for the Schwarzschild solution, generally it is impossible to do so.

Generally when there is mass-energy spacetime will be curved and if this mass-energy is distributed we cannot push all this curvature into the 'time' dimension only.

The Schwarzschild solution doesn't describe the universe as a whole; it was the universe as a whole that I was referring to when I said that the 3-dimensional space we experience may be flat (that's consistent with all the evidence we currently have), but we don't know for sure. In terms of solutions to the EFE, I was referring to the various FRW models that, depending on various parameters, can be spatially positively curved, flat, or negatively curved. Our current best-fit model is a spatially flat one; all the curvature is in the time dimension, just as it is for GP coordinates in Schwarzschild spacetime. But the error bars are large enough to allow the small possibility that there might be a very tiny spatial curvature.

 

[Passionflower]

The Schwarzschild solution doesn't describe the universe as a whole; it was the universe as a whole that I was referring to when I said that the 3-dimensional space we experience may be flat (that's consistent with all the evidence we currently have), but we don't know for sure. In terms of solutions to the EFE, I was referring to the various FRW models that, depending on various parameters, can be spatially positively curved, flat, or negatively curved. Our current best-fit model is a spatially flat one; all the curvature is in the time dimension, just as it is for GP coordinates in Schwarzschild spacetime. But the error bars are large enough to allow the small possibility that there might be a very tiny spatial curvature.

Sorry Peter but you are mistaken. Perhaps you are thinking of conformally flat.

If you would have read what I wrote accurately you would have seen that I chose the Schwarzschild solution as an example where one actually can consider space flat.

By the way the term Euclidean in GR has two meanings, it could mean flat but it can also be used as the counter to Minkowskian.

A Riemannian manifold can always be embedded in Euclidean higher dimensions. However spacetime is not really a Riemannian manifold because it is locally Minkowskian. Often we speak of a pseudo-Riemannian or Lorentzian manifold to make the distinction. But also Lorentzian manifolds can be embedded in higher dimensions. However, as far as I know, we cannot embed a Lorentzian manifold into a Euclidean higher dimension.

 

[PeterDonis]

Sorry Peter but you are mistaken. Perhaps you are thinking of conformally flat.

No, I meant "spatially flat, for an appropriate value of the parameters". The FRW metric is:

$$ds^{2} = -dt^{2} + a \left( t \right)^{2} \left[ \frac{1}{1 - k r^{2}} dr^{2} + r^{2} \left( d\theta^{2} + sin^2 \theta d\phi^{2} \right) \right]$$

(From the Wikipedia page on the FRW metric--the URL doesn't seem to translate right, but it's easily accessible via Google, or of course there are plenty of other references. I'm using units where c = 1, and using s instead of $\tau$ on the LHS because the sign convention is spacelike.)

If k = 0 this metric is obviously spatially flat; all the curvature is contained in the function a(t) for the scale factor. Our current best-fit model has k = 0.

By the way the term Euclidean in GR has two meanings, it could mean flat but it can also be used as the counter to Minkowskian.

A Riemannian manifold can always be embedded in Euclidean higher dimensions. However spacetime is not really a Riemannian manifold because it is locally Minkowskian. Often we speak of a pseudo-Riemannian or Lorentzian manifold to make the distinction. But also Lorentzian manifolds can be embedded in higher dimensions. However, as far as I know, we cannot embed a Lorentzian manifold into a Euclidean higher dimension.

You're right, I was speaking loosely. A more precise way of saying what I was trying to say is that I'm not sure if a given curved Lorentzian manifold can always be embedded into a higher-dimensional Minkowski space.

 

[Passionflower]

No, I meant "spatially flat, for an appropriate value of the parameters". The FRW metric is:

$$ds^{2} = -dt^{2} + a \left( t \right)^{2} \left[ \frac{1}{1 - k r^{2}} dr^{2} + r^{2} \left( d\theta^{2} + sin^2 \theta d\phi^{2} \right) \right]$$

(From the Wikipedia page on the FRW metric--the URL doesn't seem to translate right, but it's easily accessible via Google, or of course there are plenty of other references. I'm using units where c = 1, and using s instead of $\tau$ on the LHS because the sign convention is spacelike.)

If k = 0 this metric is obviously spatially flat; all the curvature is contained in the function a(t) for the scale factor. Our current best-fit model has k = 0.

The scaling factor obviously influences the spatial part so I honestly cannot understand how you maintain it is spatially flat.

If you are drawing triangles on a rubber sheet that is stretching in time would you maintain that sheet is Euclidean?

 

[PeterDonis]

The scaling factor obviously influences the spatial part so I honestly cannot understand how you maintain it is spatially flat.

"Spatially flat" means that any slice of constant time t has a flat spatial metric. For t = constant the scale factor a(t) is also constant, so any slice of constant t has the same scale factor everywhere; that's why a(t) is factored out in the metric formula I gave. The factor inside the brackets is the spatial metric at any constant time t, and is obviously flat if k = 0.

If you are drawing triangles on a rubber sheet that is stretching in time would you maintain that sheet is Euclidean?

If the sheet stretches in such a way that all distances change by exactly the same factor, then yes, the sheet at any given constant time t is Euclidean. That's what the FRW metric with k = 0 describes.

 

[Passionflower]

"Spatially flat" means that any slice of constant time t has a flat spatial metric.

I am not sure if you are trying to move the goal posts by now considering constant time slices, but at any rate there seems there is not much of a point in continuing this discussion.

If we only consider the spatial section it is flat but when multiplied by the expansion factor it is no longer flat but instead conformally flat.

See for instance: "An Introduction to General Relativity" - Reyder: formula (10, 37) and accompanying text.

 

[PeterDonis]

I am not sure if you are trying to move the goal posts by now considering constant time slices, but at any rate there seems there is not much of a point in continuing this discussion.

I haven't moved the goal posts, since I was using the term "spatially flat" to refer to constant-time slices from the beginning. My understanding is that that is standard usage; I don't have the text you referred to and it didn't come up on my quick search of Google Books, but when I have more time I'll look at the texts I have to check my usage.

You referred to the Painleve chart, in which the Schwarzschild geometry is spatially flat. Can you explain how that case differs from the case of the FRW metric with k = 0? I understand that the Painleve chart doesn't have a time-varying scale factor; it's a static metric, whereas the FRW metric is not. But if that's the key difference, then you're basically saying that "spatial flatness" requires a static metric, which seems much too restrictive to me. (Perhaps the reference you gave discusses this, but as I said, I don't have it. I don't remember any similar discussion in MTW or Wald, but when I have time I'll check them.)

 

[Passionflower]

You referred to the Painleve chart, in which the Schwarzschild geometry is spatially flat. Can you explain how that case differs from the case of the FRW metric with k = 0? I understand that the Painleve chart doesn't have a time-varying scale factor; it's a static metric, whereas the FRW metric is not. But if that's the key difference, then you're basically saying that "spatial flatness" requires a static metric, which seems much too restrictive to me.

That is indeed the key difference.

Good question, can we only in case of static curved spacetimes push all the curvature into the time dimension? I think the answer is yes, could someone confirm this is the case for the Schwarzschild interior solution?

 

[PeterDonis]

That is indeed the key difference.

Ok, then it looks to me like there are two questions, one of physics (the one you posted), and the other of terminology. I haven't cracked my textbooks yet, but a Google search for the phrase "spatially flat Friedmann" (see link below) turned up links to a whole slew of journal articles that use the term "spatially flat" to refer to the Friedmann model with k = 0, as I did (and I'm pretty sure that MTW, at least, uses it the same way).

Google search link: http://www.google.com/search?hl=en&...mann"&btnG=Search&aq=f&aqi=&aql=&oq=&gs_rfai=

So I think there is also a question of usage: which of the following should the term "spatially flat" be used to refer to?

(A) Any metric with flat spatial slices (i.e., slices of constant time);

or

(B) Only metrics with flat spatial slices *and* a time-independent scale factor.

It looks to me like the literature is not consistent on this point. I'm OK with either usage; I adopted usage (A) only because I thought it was standard.

Regarding the physics question, on thinking it over, I would think that at the very least a metric would have to be stationary to have a time-independent scale factor. But I'm not sure that a metric that was stationary but not static (e.g., the Kerr metric) could have flat spatial slices. So I would guess that you are correct that only a static metric can be spatially flat in the (B) sense.

 

[pervect]

But another question would be to ask "what direction is the next city" One person (comfortable with driving) points along the road that curves several times before getting to the city. Another person (comfortable with non-Euclidian geometry)-points straight, tangentially along the curve of the Earth to a point on the horizon that passes several thousand feet above the city. A third person (comfortable with Euclidian geometry) points straight, through the planet toward the city. Which person is correct? All three people are comfortable in their particular mental framework. But it can still be asked "which way are they really pointing" The first is actually pointing along a road. The second is actually pointing at a distant star. The third is actually pointing toward the city.

While the metaphor of the geometry of a curved surface for a non-Euclidean geometry is a good one, it is ultimately a visual aid. What I would like you to at least consider (and you seem to be doing your best to avoid doing it from what I can tell) is the thought that actual, physical, clocks and rulers, the best we can make, will actually not behave according to the rules of a Euclidean geometry, but will behave according to different rules, the rules of an underlying non-Euclidean geometry. So that "reality" is actually non-Euclidean - which we should actually already know from SR, as we know that the geometry of space-time is locally Lorentzian, not Euclidean.

As far as other tests go - while we don't have any direct tests, if one were to insist that the space-like hyperslice of constant Schwarzschild time around our sun have a Euclidean geometry, one would get different results for the path that light would take compared to GR, which assumes that it does not. And the actual deflection of light has been measured, and been found to agree with the predictions of GR.

 

[Passionflower]

But I'm not sure that a metric that was stationary but not static (e.g., the Kerr metric) could have flat spatial slices. So I would guess that you are correct that only a static metric can be spatially flat in the (B) sense.

No, the Kerr metric cannot have all its curvature pushed into the time dimension. In this context see the Doran chart which is as 'close' as we can get.

 

[JDoolin]

I haven't read all of the comments here just yet; (have to get back to work) but I think where I'm going with this is that while "space-time" is non-euclidian "space" by itself, actually *is* Euclidian; that is, when you consider the momentarily comoving reference frame of any given observer at any given instant, it can be represented perfectly well by an infinite three-dimensional cartesian coordinate system.

Another question to pursue is exactly why and how Special Relativity is non-Euclidian. It is not because parallel lines cross, but because right-angles are not all congruent.

As for Fredrik saying that he has no trouble with someone saying that planets orbiting stars are "traveling in straight lines" I really don't understand. In what way is this a straight line? Do you mean it is traveling in a straight line at any given instant? Or do you mean that there is some global transformation you can do to make it look like a straight line? Or do you mean you want to change the definition of straight line so that it means geodesics?

 

[PeterDonis]

I haven't read all of the comments here just yet; (have to get back to work) but I think where I'm going with this is that while "space-time" is non-euclidian "space" by itself, actually *is* Euclidian; that is, when you consider the momentarily comoving reference frame of any given observer at any given instant, it can be represented perfectly well by an infinite three-dimensional cartesian coordinate system.

Take away the word "infinite" and substitute "appropriately limited in extent in space and time". (And technically, of course, the momentarily comoving frame has to be Minkowski, not cartesian, when you include time--"cartesian" refers to the spatial coordinates only.)

In certain special cases, and as long as we limit the accuracy of our measurements, we can usefully model space as Euclidean on a larger scale. The example I was thinking of earlier was that, on a cosmological scale (i.e., over distances of tens to hundreds of millions of light-years and larger), the "space" of the universe as a whole appears to be Euclidean, at least according to our current best-fit model. But obviously that's only an average on large scales (and, as Passionflower indicated, there are also subtleties because the scale factor changes with time). Another example would be our ordinary experience here on Earth: we can usefully model, for example, the underlying 3-dimensional space in which the Earth sits as Euclidean (for example, if we're digging railroad tunnels under a mountain, or looking at the paths of earthquake shock waves through the Earth's interior) only because the accuracy of measurement we need in such cases is loose enough that we can ignore the small relativistic corrections. But those corrections are there: light is bent slightly when passing the Sun (by about 1.75 seconds of arc for light just grazing the limb of the Sun on its way to Earth), indicating that space itself around the Sun is not exactly Euclidean (technically, only half of the observed effect is due to "space curvature", the other half is due to "time curvature"). If we could measure accurately enough, we would see the same effect for light passing close to the Earth, for example from a spacecraft some distance away.

These effects are *not* the same as, for example, using geodesics over the Earth's surface instead of straight lines through its interior, say to determine the great-circle course of an airplane. In the 3-dimensional space of our experience, for those cases where the geodesics we observe are not exactly Euclidean (such as the paths of light rays bent by the Sun), there aren't any paths corresponding to "straight lines through the interior" that we can use as our standard for lines that are "really" straight. The *only* paths are the geodesics we observe, and we can tell by simple geometric measurements that they are *not* exactly Euclidean straight lines.

 

[JDoolin]

(technically, only half of the observed effect is due to "space curvature", the other half is due to "time curvature").

That sounds vaguely quantifiable. Where's this come from?

 

[Passionflower]

That sounds vaguely quantifiable. Where's this come from?

JDoolin has got a point.

Peter, could you give us the formula on which you base this?

 

[PeterDonis]

That sounds vaguely quantifiable. Where's this come from?

It seems to be a fairly common observation in relativity textbooks; it's based on the fact that a "naive" Newtonian calculation of the bending of light by the Sun gives an answer that's half the general relativistic value. I don't have time right now to dig up "official" references, but this page:

http://www.mathpages.com/rr/s6-03/6-03.htm

has a pretty good, if somewhat brief, discussion.

Edit: I should also note that I was using language loosely in referring to the "Newtonian" part of the bending of light as "space curvature" and the rest as "time curvature". The full general relativistic deflection is what will show up in any measurements, even if they are "purely spatial" in nature. For example, suppose we sent two satellites out and positioned them in such a way that they could send out laser pulses to Earth that would appear to come from opposite sides of the Sun and just grazing the Sun's surface. If we then measured the three angles of the triangle formed by the two pulses and a third pulse sent between the satellites at the same time as the first two pulses were emitted (we assume that this third pulse's path is far enough away from the Sun that any relativistic effects on the path due to the Sun's gravity are unmeasurable), we would find that the sum of the angles was larger than 180 degrees by the full GR deflection amount, summed for both laser pulses.

 

[Passionflower]

Edit: I should also note that I was using language loosely in referring to the "Newtonian" part of the bending of light as "space curvature" and the rest as "time curvature". The full general relativistic deflection is what will show up in any measurements, even if they are "purely spatial" in nature. For example, suppose we sent two satellites out and positioned them in such a way that they could send out laser pulses to Earth that would appear to come from opposite sides of the Sun and just grazing the Sun's surface. If we then measured the three angles of the triangle formed by the two pulses and a third pulse sent between the satellites at the same time as the first two pulses were emitted (we assume that this third pulse's path is far enough away from the Sun that any relativistic effects on the path due to the Sun's gravity are unmeasurable), we would find that the sum of the angles was larger than 180 degrees by the full GR deflection amount, summed for both laser pulses.

So are you no longer saying that the space and time curvature is each exactly 50% in this scenario? If you are still saying that would it be fair for me to ask you to back that up with math and show where exactly we have 50% spatial and 50% temporal curvature?

 

[PeterDonis]

So are you no longer saying that the space and time curvature is each exactly 50% in this scenario?

My comment about loose use of language was basically retracting that statement, yes, but it should be noted that that does not mean there are no timing effects involved in the scenario. The laser pulses passing close to the Sun would be time delayed (the "Shapiro time delay") as well as arriving at Earth at a slightly increased angle.

Edit: I should also note that I remember reading in at least some sources (as I noted in an earlier post, I haven't had time to check references for specific quotes) that the existence of the Shapiro time delay is the reason for attributing the fact that the GR result for the angle of light bending is twice the Newtonian one to the presence of "time curvature" as well as "space curvature" (hence the 50-50 split). However, as I said, I agree that's a loose use of language.

 

[PeterDonis]

...the existence of the Shapiro time delay is the reason for attributing the fact that the GR result for the angle of light bending is twice the Newtonian one to the presence of "time curvature" as well as "space curvature" (hence the 50-50 split).

Hmm...I may have been remembering it backwards. In going back through the huge collection of bookmarks in my browser, I noticed a link to the Gravity Probe B site:

http://einstein.stanford.edu/SPACETIME/spacetime3.html

This has the following quote in the section on light bending by the Sun:

If Mercury's perihelion shift led to the acceptance of general relativity among Einstein's peers, then light deflection made him famous with the public. He had already found in 1911 that the equivalence principle implies some light deflection, since a beam of light sent horizontally across a room will appear to bend toward the floor if the room is accelerating upwards. (Similar arguments had in fact been proposed on purely Newtonian grounds by Henry Cavendish in 1784 and Johann Georg von Soldner in 1803.) In 1915, however, Einstein realized that space curvature doubles the size of the effect, and that it might be possible to detect it by observing the bending of light from background stars around the sun during a solar eclipse.

So in this interpretation, the doubling of the light-bending effect is due to the incorporation of *space* curvature in GR.

 

[Passionflower]

So in this interpretation, the doubling of the light-bending effect is due to the incorporation of *space* curvature in GR.

Yes that is often said but forgive me for not taking this as gospel. I am not claiming it is wrong but I rather see formulas that actually show that. You think that is unreasonable?

For starters in light bending calculations the r coordinate is actually used as a physical distance so spatially speaking there is not much going on as space is now assumed spatially flat (e.g. rho is assumed equal to r). On the other hand light actually decelerates when attracted by a gravitational field and my intuition would say that is the reason for the 'double whammy' not spatial curvature.

Surely I must be wrong, for my understanding is little, and greater minds on this website will obviously have no trouble at all to demonstrate I am wrong and lack understanding by using the appropriate formulas. So I remain.

 

[PeterDonis]

Yes that is often said but forgive me for not taking this as gospel. I am not claiming it is wrong but I rather see formulas that actually show that.

I'm not sure I take it as gospel either. I can see the general line of argument: the Newtonian calculation gives an answer D for the deflection, but in Newtonian physics space is assumed to be flat; GR allows space to be curved and gets the answer 2D for the deflection. However, I don't know that there's an equation that cleanly shows the GR deflection as a sum of two terms, one for "space curvature" and one for "Newtonian effect". In the derivation on the page I linked to earlier, the factor of 2 in GR comes in as a multiplier, not a separate term added to the Newtonian one. (Also, it's a matter of opinion whether the Newtonian effect is aptly described as "time curvature"; the Gravity Probe B quote I gave notes that Einstein's original argument for light bending was based on the equivalence principle, which is supposed to hold in local inertial frames that look like small pieces of *flat* spacetime.)

 

[PeterDonis]

So I would guess that you are correct that only a static metric can be spatially flat in the (B) sense.

On further thinking, I'm no longer sure this is right, but I could easily be misunderstanding something. I've started a new thread about this question here:

https://www.physicsforums.com/showthread.php?t=446589