# Swartzchild Anisotropy

1. Apr 13, 2010

### utesfan100

Given:

1) Any observer in an inertial reference frame will observe the speed of light at their location to be the invariant speed of the Lorentz transforms.

2) The Swartzchild metric requires a transformation to avoid having an anisotropy in the speed of light, with different values in the radial and circumferential directions.

Does this mean that the Swartzchild metric does not define a locally inertial reference frame? Is the anisotropy an artifact of the distortion imposed on the space-time by defining r to be a geometric radius (as defined by the circumference around the singularity at r)?

With what ruler is the circumference measured in the Swartzchild metric anyways? It appears to me that the Swartzchild ruler is calibrated to measure like a ruler at infinity in the circumferential direction rather than the local proper length.

What, if any, is the physical meaning of r in the isotropic coordinates? Is the factor of four in the horizon radius merely a calibration issue? What is the difference in calibration?

How does the difference in surface area in the two representations of the same object effect black hole entropy, which is proportional to the area of the event horizon? My gut says that since we changed coordinates a change in our measure of entropy should not be unexpected, but a change in temperature at infinity would be harder for me to justify. I am not sure these two are incompatible.

Last edited: Apr 13, 2010
2. Apr 14, 2010

### starthaus

It is always a bad sign when a thread starts with botching the name of a very famous physicist. It is Schwarzschild, not Swartzchild .

Secondly, the Schwarzschild metric does not "require ANY transformation to avoid an anisotropy in the speed of light", whatever what migh mean. In fact, the Schwarzschild metric implies isotropic light speed. You can prove it by yourself, in a couple of lines of algebra. Start with :

$$ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2 (d \phi)^2$$

Make $$d \phi =0$$ (light moving along r direction) and you get the isotropic solution for light speed

$$\frac {dr}{dt}= \alpha$$

Last edited: Apr 14, 2010
3. Apr 14, 2010

### utesfan100

Why, then, do people use the transformarion
$$r = r_1 \left(1+\frac{GM}{2c^2 r_1}\right)^{2}$$
to form the metric
$$c^2 {d \tau}^{2} = c^2 dt^2 \left(1-\frac{GM}{2c^2 r_1}\right)^{2}/\left(1+\frac{GM}{2c^2 r_1}\right)^{2} - \left(1+\frac{GM}{2c^2 r_1}\right)^{4}\left(dr_1^2 + r_1^2 d\theta^2 + r_1^2 \sin^2\theta \, d\varphi^2\right)$$

How do you explain the anisotropy of the index of refration in a gravitational field derived in exact form at this link?

http://arxiv.org/abs/0902.0728
The index of refraction was found to be:
n = c/v = (r/rs)^2/[(r/rs-1)^3/2*sqrt(r/rs-sin^2(phi))]

This clearly indicates that the Schwarzschild metric had an anisotropy in the azimuthal angle of its direction, as Wikipedia states was first pointed out by A.S. Eddington decades ago.

http://en.wikipedia.org/wiki/Schwar...c.29_formulations_of_the_Schwarzschild_metric

EDIT:
P.S. Hokt on fonix wurkt fur mi!

4. Apr 14, 2010

### starthaus

I suggest that you start by studying the basics before you get to reading advanced papers on arxiv. As I explained, for light propagating along the radial coordinate r, $$d \theta=d \phi =0$$. What happens when you plug this into the Schwarzschild solution? What values do you get for the speed of light?

Last edited: Apr 14, 2010
5. Apr 14, 2010

### starthaus

...because, in the general case, when you assume light moving also in the $$\phi$$ and $$\theta$$ directions, not only along the radial direction $$r$$ it is very easy to see, that the speed of light is STILL the same in all directions but now varies from point to point along the transformed radial direction due to the presence of the term in $$r_1$$ in the formula. The transformation from polar to Cartesian coordinates shown on the wiki page cannot and does not get rid of the anisotropy in this case. Technically, the polar to cartesian transformation on wiki exchanges the anisotropy between the radial and angular direction observed by Eddington for a non-homogeneity along the transformed radial direction $$r_1$$..

Schwarzschild is not pronounced Swartzchild :-)

Last edited: Apr 14, 2010
6. Apr 14, 2010

### utesfan100

A different value than when I set r=const and vary $$d \theta$$.
EDIT: I did not see your second post. It addresses this issue. Thank you.

Last edited: Apr 14, 2010
7. Apr 14, 2010

### utesfan100

Yes, the speed of light in the transformed coordinates varies with the radial distance.

Am i mistaken to think that in the proper frames of reference these revert back to c? Does that not imply that niether of the metrics we are discussing above are in a local inertial frame?

Since these metrics are not in the local proper frame some form of calibration to rulers and clocks must be applied for a local observer to measure these coordinates.

How do I find the form of such calibrations?

8. Apr 14, 2010

### starthaus

What do you mean by "proper frames of reference"? This is GR, there is no such thing as frames of reference in GR, so the question makes no sense (this is why I didn't answer the first time). What we can say is that, for very small vicinities, the local speed of light is isotropic and equal to c. Indeed, if we look at my simplified post:

$$\frac{dr}{dt}=\alpha=1-2m \frac {1}{r}$$

If you take a small vicinity of radius $$\epsilon$$ and you calculate the difference between the light speeds at the radial distances $$r$$ and $$r+\epsilon$$ respectively, what can you conclude?

The Schwarzschild solution to EFE applies to gravitational fields, i.e. to curved spacetime. There are no "inertial frames" to talk about.

I have no idea what you are asking.

9. Apr 14, 2010

### Mentz114

It helps to understand the situation if the observer frame is specified. For instance in the Schwarzschild case we could define a freely falling observer who has a local Minkowski frame, or an observer at rest wrt to the source and so on. There is a formalism for this, see for instance Fermi coordinates or Painleve coordinates. There is a good intro here

en.wikipedia.org/wiki/Frame_fields_in_general_relativity

10. Apr 14, 2010

### utesfan100

The other answers above helped me significantly understand what I am asking.

Suppose I am an observer in a Schwarzschild like region of a solution to the EFE. Also suppose I have prior knowledge of my instantanious possition and velocity relative to the field. (Obviously my mass would alter the field, but if the field is strong enough this is negligable.)

What transformations would I need to convert my local space-time measurments to the Schwarzschild coordinates?

How does this change for the isotropic coordinates?

11. Apr 14, 2010

### Ich

But there are http://en.wikipedia.org/wiki/Local_inertial_coordinates" [Broken].
What you are talking about, utesfan100, are Riemannian http://en.wikipedia.org/wiki/Normal_coordinates" [Broken]. I don't know offhand how to construct these generally, but in the Schwarzschild case it's relatively easy to transform to approximated Rindler coordinates, and then to inertial coordinates.

Last edited by a moderator: May 4, 2017
12. Apr 14, 2010

### yuiop

This is generally true and desirable. The Schwarzschild metric in its original form has this property.
I think this is the transformation you are talking about: http://en.wikipedia.org/wiki/Schwar...c.29_formulations_of_the_Schwarzschild_metric

In Schwarzschild coordinates the local speed of light is isotropic (c in all directions), but the coordinate speed of light from the point of view of an observer at infinity is anisotropic (slower vertically than it is horizontally and slower with reducing r).

In order to make the coordinate velocity of light isotropic from the point of view of the observer at infinity, Eddington created an isotropic formulation of the Schwarzschild metric. In practice this would mean a vertical ruler would be a different length to a horizontal ruler, if you rotated one of them, so they lay side by side. It also means that the local speed of light in Eddington's version is not isotropic. Since the speed of light should be constant and isotropic locally for any observer in any coordinate or gravity system, this casts doubt on whether Eddington's modifications are valid.

The original Schwarzschild metric defines a locally inertial reference frame. The Eddington isotropic version does not.
A normal physical ruler using proper length is used to measure the Schwarzschild circumference. In other words a short ruler calibrated using light in flat space far from a gravitational force is transported down and used to make the circumference measurement locally. In practice it is not necessary to calibrate the ruler in flat space and transport. Any sufficiently local region in curved space is locally flat or Minkowskian. We could calibrate a ruler locally deep in a gravitational well using light and an ideal clock and calibrate another ruler far from the gravitational source using the same method and when the two rulers are brought together they will match. The Schwarzschild radius is inferred from the circumference/(2pi). The radius can not measured directly by a ruler.

Last edited: Apr 14, 2010
13. Apr 14, 2010

### starthaus

Yes, this is a very good article. I have not encountered the term of "frame" before, I always knew these as "vector bases" (which is what they are).

14. Apr 14, 2010

### Ich

No, you heavily misunderstood something here. The whole purpose of isotropic coordinates is to be, well, isotropic. This applies to the coordinate speed of light and thus, necessarily, to the space coordinates. It is also explicit in the formulation of the metric.
No.
It defines locally an anisotropic accelerating frame with scaled r- and t- coordinates.

15. Apr 14, 2010

### yuiop

I do not see how what you are saying differs in substance from what I said. I basically said (paraphrased) that the purpose of isotropic coordinates is to make the coordinate speed of light isotropic, which is essentially what you said.

The local speed of light in Schwarzschild coordinates IS already isotropic in a very local perhaps infinitesimal region. The cost of making the coordinate speed of light isotropic is that the local speed of light is no longer isotropic. You can not have both.
You are right to point out that a local stationary observer in Schwarzschild spacetime, is in an anisotropic accelerated frame. I was thinking of a free falling local observer in Schwarzschild spacetime who sees the local region as flat or Minkowskian. Thanks for the correction. Having said that, I think that even in the accelerating frame of the local stationary observer in Schwarzschild coordinates, the speed of light is approximately isotropic in an infinitesimally local region.

16. Apr 14, 2010

### Altabeh

What are you attempting to say here? When talking about a "local" speed, it always happens to be the speed of the object in a region which is "very" small so as to be called local!

This is trivial, I think!

17. Apr 14, 2010

### Ich

No, it isn't. ds!=dr, but ds=rd\theta, for example. The r coordinate is scaled, while tangential coordinates correspond to proper length.
Just to make sure I understand you correctly: Whatever you're referring to by "speed of light" as opposed to "coordinate speed of light" should be the measured value? Which is, of course, isotropic? Or how would a change of coordinates influence this "speed of light"?
You can have both, by using isotropic coordinates. Really, that's what they're made for.
No, not when the observer uses Schwarzschild coordinates.

18. Apr 14, 2010

### utesfan100

Let's see what I have learned so far ...

Mostly I have gained an appreciation for the difference between a projection of a geometrical object and its true shape.

True.

Semi-true. Any set of coordinates will not be true to the curved space-time, except maybe at a few finite points. These are chosen at infinity for all cases discussed here, and are distorted relative to any inertial tangent frame at any other point.

The coefficients of the metric essentially describe how we have deformed the actual curved shape to fit it to our flat chart.

Yes. A fixed point with finite radius in the chart must have a local acceleration, requiring a non-inertial tangent plane and an inertial tangent plane can't be at rest with this metric.

Yes.

It is not really a ruler but a map making rule. Since the origin is a singularity we can't use a true radial distance like latitude from the North Pole. This frame appears to use the equivalent of projections to a plane along concentric cylinders.

This appears to use a conformal mapping that preserves local angles, distorting r.

New Question: What would a gnomonic-like projection of this geometry look like? This would distort space to preserve the straightness of geodesics. Gnomonic projection is defined only for spheres, can this concept even be generalized to other geometries?

http://en.wikipedia.org/wiki/Gnomonic_projection

A black hole with a pointy red hat might be an interesting :)

I am not certain this question is coherent anymore.

Last edited: Apr 14, 2010
19. Apr 18, 2010

### yuiop

Hi Ich, I have had a closer look at isotropic Schwarzschild coordinates (something I probably should have done earlier) and concede you are right. Thanks again Ich. It appears the isotropic coordinates use a different concept of radius. In normal Schwarzschild coordinates the radius of a ring centred on the gravitational body is in agreement with Euclidean geometry (i.e circumference = 2*pi*r) while isotropic coordinates do not have a Euclidean geometry.

20. Apr 18, 2010

### Ich

You can always define r as circumference/(2pi). That doesn't make geometry euclidean, because by doing so you have to scale r such that, say, 1 "coordinate meter" is 1.2 "real meters". (*) It is this scaling that makes the coordinates anisotropic, because r has a different scale than the tangential coordinates, so 1 coordinate meter in r-direction is different from 1 coordinate meter in theta direction.
In isotropic coordinates, you scale all three directions with the same factor. So 1 coordinate meter is n real meters, independent of direction. But now the circumference is not 2*pi*r.

* in Schwarzschild coordinates, the scaling factor is
$$\sqrt{\frac{1}{1-\frac{2M}{r}}}$$