# A speed equivalent of gravitational acceleration?

1. Mar 28, 2015

### friend

Things space contract and time dilate in SR. And also things space contract and time dilate near heavy objects in GR, right? Particles are accelerated towards heavy objects in GR. My question is does the the acceleration due to gravity produce a frame of reference equal to some speed which would produce the same space contraction and time dilation in SR? The difference would be that the contraction in SR would only be in one direction whereas the contraction in GR would be in all directions. But in GR it would not matter in which direction you approached the point near the heavy object. However you got to the point near the mass, you still end up in an accelerated frame while traveling in that direction so that all directions are contracted in GR.

If space contraction due to gravitation corresponds to a space contraction of speed, then what speed does a gravitational acceleration correspond to? And what is being accelerated, some kind of hypothetical test particle, or maybe light itself, or maybe space itself?

2. Mar 28, 2015

### Staff: Mentor

These are not the same thing. Length contraction and time dilation in SR are frame-dependent; two observers moving relative to each other each see the other's clock running slow and the other's lengths contracted. Gravitational time dilation doesn't work this way; two observers at different altitudes in a gravity well will both agree that the higher observer's clock runs faster.

(What happens to space in a gravity well is more complicated and isn't really properly described as "space contraction". It's better described as the geometry of space being non-Euclidean.)

No.

Not really. In a gravity well, the non-Euclideanness of space is only in the radial direction. There is no effect in the tangential directions.

3. Mar 29, 2015

### A.T.

I don't get this. Isn't the non-Euclideaness detected by comparing the distances in both directions, and comparing their relation to the Euclidean value? How can you attribute a non-Euclidean relation to either of the directions?

4. Mar 29, 2015

### friend

To me this is like saying that acceleration has nothing to do with velocity. Of course they are related. You can't have velocity without acceleration, and you can't have acceleration without velocity. The only question I have is how are the effects of gravitational acceleration related to the effects of velocity. That seems like a straight forward question.

5. Mar 29, 2015

### A.T.

Of course you can, by choosing a reference frame where the velocity is non-zero.
What effects? Coordinate acceleration due to gravity is the effect.

6. Mar 29, 2015

### friend

I suppose you can always do any calculation you'd like to imagine what a faster or slower reference frame might observe. But that's only a calculation. For real particles, no one particle can have a different velocity than another except that there has been an acceleration for at least one particle at some point in its history. This is because all particles start from a single point.

7. Mar 29, 2015

### DrGreg

That's not true. Particles are being created all over the Universe as a result of other high-energy particles colliding with each other.

8. Mar 29, 2015

### Staff: Mentor

No, the non-Euclideanness is detected by looking at the difference in the areas of two 2-spheres centered on the gravitating body, with slightly different radius. Then you compare the actual distance between them along a radial line, with the expected distance calculated from Euclidean geometry. In Schwarzschild spacetime, the actual radial distance is larger, indicating that space is non-Euclidean in the radial direction. But there's no corresponding non-Euclideanness in the tangential directions: for example, the relationship between the area of a given 2-sphere and the circumference of great circles is exactly what is expected from Euclidean geometry.

(Also, you have to look at the areas of the 2-spheres--or the circumferences of great circles--to see the non-Euclideanness. Just looking at local distances won't show it; those distances will always be consistent with Euclidean geometry in a local frame. The non-Euclideanness of space is a global property, not a local one. And, of course, it depends on a particular slicing of spacetime into space and time, the one given by standard Schwarzschild coordinates. In Painleve coordinates, for example, the spacelike slices of constant coordinate time are Euclidean.)

9. Mar 29, 2015

### A.T.

You mean different Schwarzschild coordinate, which is based on the circumference and its Euclidean ratio to radius? By using this circumferential/Euclidean measure as reference, you will of course conclude that it is the radial direction that deviates from Euclidean geometry .

You could just as well say: In Schwarzschild spacetime, the change in circumference (or sphere surface area) for a given change of the actual radial distance is smaller than the Euclidean value, so its the circumferential direction that is to blame for the non-Euclideanness.

All I see, is that the ratio of actual radial distance change to circumference/area change is different than the Euclidean value. But I don't see how you can say which of the two is the one deviating from Euclidean geometry.

Now you are talking about the 2D geometry of a single spherical slice, not about the geometry of the 3D space. If you take only a radial line (1D slice), you will also find no non-Euclideanness within it. So how can you say, that it's the radial line, and not the sphere that is responsible for the deviation from Euclidean geometry of the 3D space.

10. Mar 29, 2015

### Staff: Mentor

Yes.

The spacetime is spherically symmetric, so the sphere surface areas (and their circumferences) are invariant geometric features of the spacetime. The radial distances are not; they depend on the choice of coordinates--or, to put it in coordinate-free terms, they depend on what spacelike slice you choose to measure the radial distances in. So there is no invariant notion of "radial distance" that you can use as a standard of reference the way you can use the surface areas of the spheres.

Or, to put it yet another way: consider the difference between spacelike slices of constant Schwarzschild coordinate time, and spacelike slices of constant Painleve coordinate time. The former are non-Euclidean; the latter are Euclidean. What is the difference? It can't be any difference in areas of the 2-spheres, because those are the same in both slices--they have to be, by spherical symmetry. So the difference must be in the radial distances.

11. Mar 29, 2015

### friend

So if we talk about particles being created and annihilated with so much energy/velocity, wouldn't that be infinite acceleration, but acceleration never the less?

12. Mar 29, 2015

### DrGreg

No. As soon as these particles come into existence, they already have velocities relative to each other.

13. Mar 29, 2015

### A.T.

I don't understand why the actual radial distance (not Schwarzschild coordiante) is less of an invariant geometric feature, than the circumferences (or surface areas)

If the effects of curvature are only radial and not tangential, how come there is a geodetic effect for a gyroscope in a circular orbit?
http://en.wikipedia.org/wiki/Geodetic_effect

14. Mar 29, 2015

### Staff: Mentor

One way to say it is, as I've already said, that the radial distance between two 2-spheres of given areas depends on how you slice spacetime up into space and time.

Another way to say it is that, as I said before, the spacetime has spherical symmetry. That means the 2-spheres are submanifolds formed from orbits of the 3 spacelike Killing vector fields associated with spherical symmetry. There is no radial Killing vector field in this spacetime.

If that's still too abstract, consider this hypothetical scenario: we have a compact object that is emitting radiation. To make the effect I'll describe as pronounced as possible, this object will be at the very limit of compactness allowed for a static object: the radius of its surface will be 9/8 of the Schwarzschild radius associated with its mass. It emits spherically symmetric radiation with some total luminosity $L$. Then it is simple to show that the flux of this radiation does not obey the inverse square law if we use actual radial distance.

The emitted flux of radiation is just $L$ divided by the area of a 2-sphere at radial coordinate $(9/8) r_s$, where $r_s$ is the Schwarzschild radius. To make things simple, we will scale everything so that $r_s = 1$. Then the emitted flux is $F_e = L / 4 \pi r_e^2 = 16 L / 81 \pi$.

At any 2-sphere at a larger radial coordinate, the flux will be $L$ divided by the area of that 2-sphere; so the flux declines as the inverse square of the radial coordinate. So, for example, at $r = 9/4$, twice the radial coordinate of the surface, the flux will be 1/4 of the above value: $F_o = L / 4 \pi r_o^2 = 4 L / 81 \pi$.

However, the actual radial distance between these two 2-spheres is considerably larger than $r_o - r_e = 9/8$. The radial distance is given by

$$D = \int_{r_e}^{r_o} \sqrt{\frac{r}{r - 1}} dr = \sqrt{r ( r - 1 )} + \ln \left( \sqrt{r} + \sqrt{r - 1} \right) \bigg|_{r_e}^{r_o}$$

where we have plugged in $r_s = 1$. Evaluating this for $r_e = 9/8$ and $r_o = 9/4$ gives $D \approx 1.918$, as compared with $r_o - r_e = 1.125$. So the flux declines with actual radial distance more slowly than the inverse square. Since the flux is total luminosity divided by surface area, it gives a physical realization of the spherical symmetry, i.e., of the fact that the 2-spheres and their areas are the "right" things to use as references.

I didn't say they were. I only said the "non-Euclideanness" of space is described as radial distances being different from the Euclidean expectation. That's by no means the only effect of spacetime curvature.