- #1

jk22

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$$\theta(t)=\arccos(\sin\alpha\cos\omega t)),$$

$$\phi(t)=\arctan(\tan(\omega t)/\cos\alpha),$$

$$t=\tau$$

not solution to the geodetic equation ?

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- #1

jk22

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$$\theta(t)=\arccos(\sin\alpha\cos\omega t)),$$

$$\phi(t)=\arctan(\tan(\omega t)/\cos\alpha),$$

$$t=\tau$$

not solution to the geodetic equation ?

- #2

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One obvious issue is that ##t=\tau## is not possible for finite ##r##, assuming that ##\tau## is supposed to be the proper time on the geodesic. If you simply intended to rescale your time unit then that has implications for the modulus of your four velocity and hence the form of your geodesic equations.

- #3

jk22

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I supposed a circle was a solution so that time flows equally, that's my mistake

- #4

jk22

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I supposed the equation is scale invariant and I got ##\ddot{t}=0## I'll look again

- #5

PeterDonis

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In the Schwarzschild metric case why is :

...

not solution to the geodetic equation ?

What "geodesic equation" (I assume that's what you mean) are you trying to solve? Please write it down explicitly or give a reference. There isn't just one of them.

- #6

jk22

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The solution I wanted to check is a circular uniform motion, it is a solution if ##\theta=\pi/2## but if the circle is rotated by an angle ##\alpha## around the y-axis I got that it is no more a solution. Maybe I made a calculus mistake, but if not the question is :

Is the Schwarzschild metric spherically symmetric, or should one use isotropic coordinates to have this rotated circular trajectory a solution ?

- #7

PeterDonis

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The solution I wanted to check is a circular uniform motion

Do you mean a free-fall circular orbit? That will only be a solution to the geodesic equation for particular combinations of parameters; not all circular motion in Schwarzschild spacetime is free-fall geodesic motion.

Is the Schwarzschild metric spherically symmetric

Yes.

should one use isotropic coordinates to have this rotated circular trajectory a solution ?

Whether a particular worldline is a geodesic is independent of any choice of coordinates.

- #8

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The solution I wanted to check is a circular uniform motion, it is a solution if ##\theta=\pi/2## but if the circle is rotated by an angle ##\alpha## around the y-axis I got that it is no more a solution. Maybe I made a calculus mistake, but if not the question is :

Is the Schwarzschild metric spherically symmetric, or should one use isotropic coordinates to have this rotated circular trajectory a solution ?

The simplest approach is to show by symmetry arguments that motion stays in a plane and then take that plane to be defined by ##\theta = \frac \pi 2##. Otherwise, the more general solution gets unnecessarily complicated.

- #9

kent davidge

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What you mean by that? Even the most general solution is simple in the case of the sphere.the more general solution gets unnecessarily complicated

- #10

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The equations are messier, especially for trans-polar orbits that go through the ##\theta## coordinate singularities. And unnecessarily so, since you can just solve for equatorial orbits and then rotate coordinates.What you mean by that? Even the most general solution is simple in the case of the sphere.

- #11

jk22

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At least in the Newtonian case the OP relations are a solution.

So I was asking myself if the EFE were symmetric but not the geodesic equation ?

- #12

PeterDonis

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So I was asking myself if the EFE were symmetric but not the geodesic equation ?

What do you mean by "symmetric"? How would the geodesic equation be symmetric or not symmetric?

- #13

jk22

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So is it correct to deduce that either the equation shall be recomputed in the rotated coordinates,

Or the background spacetime given by the EFE is absolute and hence there were preferred directions ?

- #14

PeterDonis

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By symmetric I mean the symmetry of the problem.

The symmetry of the problem obviously applies to all equations involved in the problem, so it would apply equally to the solution of the EFE and to the geodesic equation.

We can check the circular uniform motion is a solution of the geodesic equation at but the same motion seen by another rotated non moving observer is not.

No. Circular uniform motion with the appropriate orbital parameters (angular velocity matched up correctly with radial coordinate) is always a solution of the geodesic equation. It's just mathematically much harder to demonstrate this if you insist on choosing the coordinates so the circular orbit is not in the ##\theta = \pi / 2## plane. But you don't have to go to all that trouble precisely because of the spherical symmetry: you can

- #15

PeterDonis

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is it correct to deduce that either the equation shall be recomputed in the rotated coordinates

Obviously if you change the coordinates you have to recompute everything that depends on the coordinates. That includes the particular terms and numerical factors that appear in the geodesic equation.

- #16

jk22

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Can This argument of letting ##\theta=\pi/2## also be done to ##\phi=0##, giving a free fall case admitting the initial speed is zero, but r is not constant ?

- #17

PeterDonis

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Can This argument of letting ##\theta=\pi/2## also be done to ##\phi=0##, giving a free fall case admitting the initial speed is zero, but r is not constant ?

If you mean, is setting ##\theta = \pi / 2## and ##\phi = 0## a valid way to treat purely radial geodesic motion, yes, it is.

- #18

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...and the equations can be solved analytically.If you mean, is setting ##\theta = \pi / 2## and ##\phi = 0## a valid way to treat purely radial geodesic motion, yes, it is.

- #19

jk22

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- #20

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The geodesic equation is invariant under coordinate change. However, when you actually try to compute a geodesic you need to adopt a coordinate system, and the computations you make will differ. But the same is true of the EFE. For example you can write down the Schwarzschild metric in Cartesian-like ##t,x,y,z## coordinates and it does not have the same form as the familiar spherical polars version.

- #21

jk22

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Since x space is flat then the geodesics are straight lines in it : $$\frac{d^2 x^\mu}{d\tau^2}=\frac{d^2\xi^\mu_p}{d\tau^2}(y^\nu)=0=\frac{d}{d\tau}(\frac{\partial\xi_p^\mu}{\partial y^\alpha}\frac{d y^\alpha}{d\tau})=\frac{\partial^2\xi_p^\mu}{\partial y^\alpha\partial y^\beta}\frac{d y^\alpha}{d\tau}\frac{d y^\beta}{d\tau}+\frac{\partial\xi^\mu}{\partial y^\alpha}\frac{d^2 y^\alpha}{d\tau^2}

\Rightarrow \frac{d^2 y^\nu}{d\tau^2}=-\underbrace{\frac{\partial y^\nu}{\partial\xi^\mu}\frac{\partial^2\xi^\mu}{\partial y^\alpha\partial y^\beta}}_{-\Gamma^\nu_{\alpha\beta}}\frac{dy^\alpha}{d\tau}\frac{dy^\beta}{d\tau}$$ ?

Then ##\xi_p^\mu## is weird since it is a coordinate change that creates curvature out of flat space ?

- #22

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- #23

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It should be obvious that your conclusion is wrong. Quite apart from @vanhees71's comment, consider two objects in the same freefall orbits, but opposite directions. They encounter each other twice per orbit. How could two straight lines do that?

More generally your ##x^\mu## coordinates are poorly constrained. You have defined them along a particular geodesic, but that does not say anything about their behaviour off the geodesic. Your equation ##0=\frac{dx^\mu}{d\tau}## only applies along your chosen geodesic. Furthermore, ##x^\mu=\xi^\mu_p(y^\nu)## makes no sense - you have different numbers of free indices. Presumably you meant ##x^\mu=\xi^\mu_\nu(y^\nu)##.

- #24

PeterDonis

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The fact that the rotated circular motion is not a solution

Any circular orbit

contrary to the EFE the geodesic equation is not invariant under coordinates change ?

No. It is. But you need to do the math correctly. Incorrect math will of course give you incorrect answers.

- #25

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https://www.physicsforums.com/threads/twin-paradox-for-freely-falling-observers.991709/

- #26

jk22

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It should be obvious that your conclusion is wrong. Quite apart from @vanhees71's comment, consider two objects in the same freefall orbits, but opposite directions. They encounter each other twice per orbit. How could two straight lines do that?

More generally your ##x^\mu## coordinates are poorly constrained. You have defined them along a particular geodesic, but that does not say anything about their behaviour off the geodesic. Your equation ##0=\frac{dx^\mu}{d\tau}## only applies along your chosen geodesic. Furthermore, ##x^\mu=\xi^\mu_p(y^\nu)## makes no sense - you have different numbers of free indices. Presumably you meant ##x^\mu=\xi^\mu_\nu(y^\nu)##.

Yes that's the point, either x is of dimension 5 and y dimension 4, representing the embedding, or both are of dimension 4 then it was said ##\xi_p(y^\nu)## is a "local" coordinates transformation, meaning p is a point at which the function is determined but not derived. This permits to induce curvature but it's a bit sloppy. It were like writing ##f(x)=_x g(x)## and ##f'(x):=_x g'(x)##.

For example ##f(x)=_x\cos(x)## and hence ##f'(x)=-_x\sin(x)##

Does this scripture I saw in an old course of the TU Vienna make sense ?

- #27

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You do not address my point that geodesics in curved spaces can cross multiple times but flat space ones cannot. You have addressed my point about mismatched indices by mismatching them in a different way. Furthermore, you don't explain how an apparently index-free function ##\xi## can convert a four-index coordinate into a five-index one. Either it needs an upper and a lower index, or else you are using yet more new notation without explaining it.

- #28

PeterDonis

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