Banned
What is the speed of light according to the Schwarzschild metric, but in any direction, not only radial or tangential separately?
I mean the dependence of c on the angle, which is measured relative to the radius, like this:
+---------> c
..+ / - angle
....+
-----M
Order of Kilopi
Banned
Not necessarily. The speed of light depends on many factors.
Only the locally measured average speed - two-way - is constant.
Radial speed of light in gravity is: 1 - 2a/r;
and tangential: 1 - a/r; a = GM/c^2.
Established Member
Isn't the speed of light constant for all viewers/reference frames?
Established Member
The speed of light is constant if and only if:Originally Posted by xylophobe
-you are in flat space(no gravity)
-you are in vacuum(no medium)
-you are standing next to it(locally)
The locally is the big part. If you're falling through a black hole, you measure the speed of light in all directions as the same. But that's you, in the tiny piece of space you occupy. If you're past the event horizon, you look back and see stars. You measure that starlight and get...c. Shine a light at them and for you, it leaves at the speed of....drum roll....the speed of light. But for us outside the black hole, that's impossible, since your light never leaves the black hole. The two frames can not be treated as the speed of light being c everywhere.
Do not mess with a black hole!
Order of Kilopi
That's the coordinate speed of light, which is indeed anisotropic in standard schwarzschild coordinates. Not that this means anything, we can make the coordinate speed of light anything we want, even in flat spacetime. We can also make it isotropic if we want by using the surprisingly named isotropic coordinates.
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You don't need the first if you have the third and you don't need the third if you have the first.
Established Member
Perhaps it would be simpler to ask in what circumstances would the speed of light be measured to be anything other than c?
From caveman1917's last reply, it would seem that you can be in a vacuum and measure the speed of light to be something other than c if there is gravity present, but you are not measuring the local speed of light.
I have also been told (I think it was Grant or Ken) that the speed of light is not constant in an accelerating frame of reference, and there was some mention of Rindler coordinates (probably Grant then!), but again this involves the coordinate speed of light.
Can anyone explain further?
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I think the best way to put it is light follows null paths. In flat spacetime, a null path resolves to a speed of 'c' locally and globally. In curved spacetime and/or non-inertial reference frames, the coordinate speed of a null path is not globally 'c'. For example, in Rindler coordinates, the coordinate speed of light will be c*(1 + gz/c^2), for an observer at the origin, accelerating at 'g' in the z direction (when that expression goes to zero, one has found the Rindler horizon -- it is sometimes common to take z = 0 to be that horizon and write a slightly different expression). But every local observer will, using his own ruler and clock, measure the speed of light passing by him to be 'c' always.
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Thank you publius.
Would there be any similar horizon to the Rindler horizon for a frame at rest in relation to a gravitational field, such as an observer standing on the surface of the Earth? I only ask this due to the equivalence principle.
Banned
Very funny.
So we know that the speed of light is isotropic, or not?
There are quite significant differences between the radial and tangential velocity.
a = GM/c ^ 2;
For the Earth: a / R = 7e-10
Thus on the earth's surface horizontal speed of light should be greater than the vertical by: c * 7e-10 = 21 cm/s
Sun gives: 3m/s, measured on Earth (close to the Sun is 600 m/s).
The Milky Way: 300 m/s
I think it is possible to detect such an anisotropy, for example with the famous Lunar Laser.
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No, there isn't actually. Remember the EP is local. You can show that show for a local observer stationary in Schwarszchild, the metric is the same as Rindler for whatever 'g' that observer is feeling in a local neighborhood. Or perhaps I should better phrase it as Schwarzschild and Rindler reduce to the same thing in a small enough local neighborhood about the observer. But globally, things can of course be very different.Thank you publius.
Would there be any similar horizon to the Rindler horizon for a frame at rest in relation to a gravitational field, such as an observer standing on the surface of the Earth? I only ask this due to the equivalence principle.
THe exterior spacetime for a spherically symmetric body is just Schwarzschild, but the interior metric inside the body is something different and there is no horizon anywhere. Now, if an observer undergoes a coordinate acceleration, there will likely be some Rindler horizon like thing in those coordinates. Consider an observer accelerating away from the body at some far distance away where the field is very weak. That's very nearly flat spacetime, and Rindler is the metric for an accelerating observer in flat spacetime, so we know it must reduce to something close to Rindler globally there.
Established Member
You need to be more precise in your language.
We know that the local speed of light is constant and so isotropic. See for example the Michelson–Morley experiment in 1887 (andf later experiments) which mesaured the speed of light in various directions and found it to be isotropic.
The coordinate speed of light depends funnily enough on the system of coordinates that you use!
In Schwarzschild coordinates, the coordinate speed of light varies radially.
In Kruskal–Szekeres coordinates, the coordinate speed of light is constant.
Your calculation is thus wrong because you do not state the coordinate system that you use.
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You are not going to measure anything different from c via a local measurement because that's the way local rulers and clocks work. Your local ruler and clock just cancel out whatever factors in the metric make the coordinate speed different from 'c'. However, you can *infer* the coordinate speed via various measurements such as time of flight vs your own notion of coordinate distance. The Shapiro delay is an example of such a measurement and the results for the solar system confirm the predictions of GR to the accuracy of the measurements.
Coordinates are just coordinates and coordinate speeds are just that. There is nothing physical about them.
Order of Kilopi
It's a somewhat meaningless question, the coordinate speed of light is exactly what you choose it to be.
It's like asking "what is the color of a ball when i can choose the color?", it'll be exactly the color you choose it to be. If you choose coordinates in which the speed of light will be anisotropic, then the coordinate speed of light will be anisotropic. If you choose coordinates in which it will be isotropic (check the wiki article on isotropic coordinates), then it will be isotropic.
Suppose i'm flat spacetime. Suppose i choose my unit rulers such that my unit ruler in the x direction is twice as long as the other directions, then the coordinate speed of light in those coordinates will be anisotropic, it will be twice as slow in the x direction as in the other directions. The anisotropy you see in schwarzschild coordinates has no more meaning than this example, it's just that we find it convenient to choose the radial coordinate to be "different" than the tangential coordinate.
Banned
I am afraid that these are only conventions proposed by Mr. Einstein.
You claim that the metrics are not applicable in practice - in the real world?
Banned
I ask for real time and distance measurements, such as the NASA performed with the Pioneers.
In fact, this anisotropy in the Schwarzschild metric is unavoidable - necessary, because it is a consequence of the non-zero curvature of the spherical space, which determines the gravitational acceleration.
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Michelson measured the difference of twa-way speeds of light, and horizontally, which are the same - directly from the wave equation (hence it is invariant under Lorentz transformation).
Kruskal-Szekeres coordinates are a mix of various physical quantities (time, with the distance), so it is completely nonphysical - fully four-dimensional structure.
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??? A metric is a metric. A given spacetime has invariant properties. The form of the metric varies with the coordinates chosen, but the invariant properties remain the same. Coordinate speeds are not invariant. However, a null path is invariant. That is, the nullness of the path, which is the path light follows, is an invariant. Light follows null paths -- how that path is coordinatized is arbitrary.
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A crackpot poster on Usenet (now Google Groups) sci.astro
shortly before I joined BAUT suggested a method of measuring
the one-way speed of light that looked like it might work. His
suggestion involved sending a light signal to a satellite beyond
the Moon, but I think the Moon was not essential. As proposed
it would be terribly expensive. A much less expensive version
seems highly likely to be completely equivalent. However, I
don't remember the details, and don't have a link. I'm afraid
I'd have to look through a huge number of long posts to find it.
A second, easily-done experiment was proposed for a very
similar purpose. It involved timing a light pulse in an optical
fiber laid in a straight line with a looped section which could
be moved along the length of the fiber, with a high-resolution
oscillioscope to determine whether the travel time of the light
pulse changes with the postion of the looped section. I was
disappointed that the crackpot didn't carry it out.
-- Jeff, in Minneapolis
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
Banned
I don't understand this frivolity in the treatment of metrics.
What are the properties of space-time, and what the speed of light is invariant and with respect to what?
In relation to what, the light paths are invariant?
I think, since these paths of light and the speed of light are invariant (somehow - at all), so surely there must be a metric, in which it will be very satisfied, because we measure something, and these metrics are used to do just that.
Banned
Rather superfluous.
Two-way speed of light is independent of direction, so just realize this relationship geometrically.
We obtain the equation of an ellipse.
Or directly from the SR: contraction along the v + time dilation, so what comes out?
Ellipse of course: c1 + c2 = 2c = const
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As I think I understand it, the speed of light is in theory invariant as measured locally under conditions in which special relativity can be applied. When we introduce gravitational spacetime warps and look at it globally, it becomes much more complicated, and I don't think a few short answers in a forum like this will bring about real understanding. I certainly am too rusty on modern physics to attempt it.
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I suppose you didn't look up the article on isotropic coordinates? It specifically shows an isotropic coordinate chart of the schwarzschild metric in which the speed of light is isotropic, so in fact it is everything but unavoidable.
Banned
Unfortunately, the speed of light can not be both isotropic and anisotropic.
I suspect the operation of substitution, of a new variable in the equations, requires also the conversion of boundary conditions.
These metrics are identical with the mathematical and practical point of view.
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I think you might be confused about the difference between a metric and a coordinate (matrix) representation of a metric. Most people will however simply say "the schwarzschild metric" rather than "the schwarzschild coordinate chart of the schwarzschild metric", which might be a source of confusion.
Intuitively, a metric is an abstract concept that contains the invariant (ie physical) properties of a spacetime manifold, a coordinate representation of a metric is a way of making this abstract notion concrete so that you can actually work with it. There are many ways to do that, which is equivalent to choosing a coordinate basis. The coordinate speed of light is not a function of a metric itself, but a function of a chosen coordinate representation of a metric. Since you are (somewhat*) free to choose your coordinate basis, you are free to choose your coordinate speed of light any way you please.
The coordinate light speed anisotropy you see is in the schwarzschild coordinate representation of the schwarzschild metric, it is however isotropic in the isotropic representation of the schwarzschild metric. They are both the exact same schwarzschild metric.
[*] timelike, null and spacelike intervals must stay that way
Banned
Okay, I said once that someone lost something in these metrics, and hence the confusion.
I have such a question: with respect to what we measure the curvature of space?
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With respect to itself. For manifolds of dimension 2 and higher you can measure a so-called intrinsic curvature that does not relate to anything else.
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Free light follows null geodesics, integral of ds^2 = 0. That's the meaning of null. That nullness is invariant. However, how one coordinatizes such a null path is not invariant.
Consider a simple example, a flat plane, where we have a simple positive definite manifold. If we coordinatize with a simple Cartesian x-y coordinate system, our metric is ds^2 = dx^2 + dy^2. Now, if go to polar coordinates, our metric is
ds^2 = dr^2 + (r dO)^2. Note the metric takes different forms with different coordinates. But it is the same manifold, the invariants are the same. We could go to any other of a myriad of coordinates, each with different ds^2 expression, but they represent the same invariant manifold, the flat Euclidian plane.
Banned
It is rather impossible - in principle.
I guess this is the secret postulate of non-Euclidean geometry.
The consequence is the ambiguity of concepts, metrics, etc.
And finally, it is possible to calculate the speed that I mentioned at the beginning of a subject?
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