In your case, dividing the geodesic by is meaningless. To further prove that, do the same operation for the general geodesic: .
The expression for time dilation. Now what was your point again?What did you obtain?
The operation i performed (actually it was grav who first performed it, i merely responded to your unfounded criticism of this specific operation) in the first case succintly suffices to derive the entire framework of SR whilst making its most important aspects as clear as possible, first that the 3-space for an inertial lorentz observer is euclidean (by the statement of the pythagorean theorem) and secondly that the speed of light is an invariant for that family of observers. In the second case the same operation was used to derive time dilation which is a cornerstone of the experimental confirmation of SR. If you think that an operation that is easily shown to have applications in both theoretical understanding and experimental verification of a physical theory is "meaningless", then what exactly do you think "meaningful physics" actually is?the operation you performed is meaningless in the realm of physics.
Also the mere fact that an equation involving two degrees of freedom cannot be reduced to a single solution without more information is a simple mathematical fact that has no bearing on physics or anything else. It is equivalent to claiming that statements of the form are meaningless in physics because we cannot figure out the value for without information on the value of .
Last edited by caveman1917; 2012-Jul-27 at 09:55 PM. Reason: layout of equations
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The last few days I have been trying to transform between the coordinate systems of the local observer and the distant observer. If I figure that many local observers are at a distant r on the surface of a planet, then the distant observer will measure the planet as spherical, and the local observers, all observing the same thing from any position and since the contraction is radial only, might determine their planet as spherical also. At what radius though, I'm not sure, since the inferred radial distance is greater but the circumference is the same as the distant observer measures. If the tube is perfectly straight in free space and is brought down to the planet, local observers should still measure the tube as straight, right? If they then lay the tube level on the ground, each end will rise a slight distance from the ground. After finding that distance that the local observers measure in the radial direction for the end of the tube from the ground of a spherical planet, I then contracted that radial distance for what the distant observer measures between the ends of the tube and the ground, bringing the ends closer to the ground, while the center of the tube remained on the ground at the same place on the surface. The distant observer, then, will measure a slight bend in the tube this way that is the same as what was found earlier that it would need to be, but only half the necessary value, so there's still something missing.
If I simply extend a single local observer's local coordinate system globally, so that if the local observer is located at some point upon the y axis in the radial direction, say, then all distances in the y direction are 1 / sqrt(1 - r_s / r) greater than the distant observer measures and all distances along the x axis are the same. This would make the planet look like an ellipse to that local observer with him at the peak. If I then place the straight tube with the center at the peak, then transform in the same way back to the distant observer's coordinate system, then there is no bend. But then, if we're dealing with single local observers at this point, then there should be another at the end of the tube to measure that distance, but would then observe the planet and tube differently than the first local observer that was at the center of the tube, so might not say the tube is straight, hard to say. My latest attempts were to start with the distant observer's coordinate system, already knowing what the bend should be, then trying to transform that back to what the local observer would measure that would make the tube appear straight, but the results are so far indefinite. So I'm stuck, help please.
A tube that is resting with its center on the ground is effectively a tube where a constant force is applied to its center, whether it will bend and how much then also depends on the material the tube is made of.
Wouldn't it be better to take a step back for a moment and clearly explain what it is exactly that you're trying to test? Just "the equivalence principle in the schwarzschild metric" is a bit vague. Given the centrality of the field and the curvature effects it would be difficult to do the analysis with extended objects. Perhaps it would be easier to do it in an appropriate limit, say . In that case you can use rindler for your local observer and consider only a single radial line, which gets you closer to what the equivalence principle actually is.
Also did you take into account that the tube would not only bend, but by tidal effects be shorter than it is in free space?
Would it help to use Christoffel symbols for your purpose? I just found this 'Catalogue of spacetimes' http://arxiv.org/abs/0904.4184 on arxiv that displays them ready for use for a range of coordinate systems. A cool document imho.
The path of a geodesic can be simulated with linear steps like this:
where the nabla is calculated using the Christoffel symbols. Of course you need higher order extrapolation to get accurate results after a few iterations, but I guess you know that part better than I do.
Another advantage of the Christoffel symbols is that you can almost directly read off the behavior of the photon and the elevator mentioned in your OP. The photon follows a lightlike geodesic, the edges of the elevator--or a rigid tube--should follow a spacelike geodesic. In the tangential direction the outcome should then be that the path of the photon bends twice as much as the elevator edge.
If that doesn't reestablish your confidence in the equivalence principle in the Schwarzschild metric space, I hope at least you can make use of that catalogue.
The tube is immensely strong and is designed in free space to be perfectly straight. Then the tube is brought down to the surface of the planet and laid level with its center on the ground. Let's say that according to the distant observer, the center lies at y = r and the tube extends a distance x. We know that for a distance x that the tube extends along the x axis, in order to accomodate the equivalence principle for what the distant observer measures, the end of the tube must bend slightly downward a distance
to first order. The coordinates of the center of the tube, then, is (0, r) and the ends are at (x, r - Δy) and (-x, r - Δy). I want to know what those distant coordinates translate to locally, such that the tube should be locally viewed as straight. With an infinitesimal length of the tube, we have
and the difference in angle from the center of the tube to the center of the planet to the end of the tube becomes
and the difference in radius becomes
but I'm having trouble transforming those to the coordinates in the local observer's frame.
I figured its straight length would be the same according to the local observer, and since it bends according to the distant observer, the length would be along the bend, but otherwise about the same length. But to be sure, I'm trying to avoid that issue now by just taking an arbitrary dx and dy that the distant observer measures and transforming those coordinates to that of the local observer to find out what the local observer will measure, which will then give the locally measured length and all.Also did you take into account that the tube would not only bend, but by tidal effects be shorter than it is in free space?
and if we figure in the tangent direction with no drop along y, then finding for the radius r' that the local observer measures, with about r' + dr' along the hypotenuse for infinitesimal , then we get
So it looks like the local observer uses the inferred distance for the radius according to his local radially directed ruler for his coordinate system. I'll see if I can verify that another way also, maybe using the angle.
Yay, I finally got it. The tangent speed from which the metric was derived in an earlier post was
The numerator is the tangent distance which the distant observer and local observer will measure the same, so
(primed for the local observer's measurements)
where , so
The angle we're finding for is that between the radial direction and along the line from the center of the planet to the end of the tube, with some very small x'/r' and x/r, which the local observer will measure as
with the radial distance r' and the length of the straight tube x' in the tangent direction. For the distant observer, we have
where the radial distance to the end of the tube is r-y and the tangent distance x, where the tube bends some distance y in the radial direction.
Since both agree upon the tangent distance, then x' = x, and we have
Applying the quadratic formula, we get
and taking the negative value since the positive would give us 2 r, we have
which to first order gives us
precisely as it would need to be for the equivalence principle to work out. Cool.
Last edited by grav; 2012-Jul-30 at 01:28 AM.
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