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Consider a distant observer traveling at .866 c relative to the solar system along the line that is co-linear with the sun's axis of rotation. According to his/her wristwatch the observer measures the earth's orbital period around the sun to be 730.5 days, correct?

But the observer also measures the major and minor axes of the earth's orbit around the sun to be identical to its major and minor axes in the solar system's rest frame, where the orbital period is only 365.25 days.

So it appears as if Kepler's 3rd law of planetary motion is only valid in the rest frame of the solar system. Does this violate the first postulate of special relativity?

If so then how can Kepler's 3rd law be made frame-invariant?

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  • $\begingroup$ I don't know if this helps, but there is an article that derives the Schwartzschild solution from Kepler''s third law. This is a result in GR, which assumes SR. Maybe the discussion there will put you mind at ease? mathpages.com/rr/s5-05/5-05.htm $\endgroup$ – m4r35n357 Sep 18 at 8:06
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It appears as if Kepler's 3rd Law of Planetary Motion is only valid in the rest frame of the solar system. Does this violate the First Postulate of Special Relativity?

No. It just means that Kepler’s laws are not actually laws of physics. Instead, they are approximations to the laws of physics in the non-relativistic limit

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  • $\begingroup$ Dale, thanks for your answer. Can you please give an example of a law that Einstein meant the First Postulate to apply to? $\endgroup$ – Pat Dolan Sep 18 at 1:29
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    $\begingroup$ I don’t presume to know Einstein’s mind, but Maxwell’s equations are one example of a law that satisfies the first postulate $\endgroup$ – Dale Sep 18 at 1:37
  • $\begingroup$ Dale, if the first postulate remains unviolated by the example then would you conclude that it is possible for identically shaped orbits with identical rest masses to have different orbital periods in different frames? If not, then where might the problem lie in the example? $\endgroup$ – Pat Dolan Sep 18 at 16:34
  • $\begingroup$ Sorry, I don’t understand your comment. The first postulate has nothing to do with the example because Keplers laws are not actually laws of physics. But yes, it is possible for the same orbit to have different periods in different reference frames $\endgroup$ – Dale Sep 18 at 16:41
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    $\begingroup$ Curvature is a tensor. The curvature itself is the same in all frames, but it’s components will differ $\endgroup$ – Dale Sep 18 at 17:02
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I agree with the first answer. Kepler’s 3rd law of planetary motion appears to permit different orbital periods in different inertial frames for identical elliptical orbits (same major & minor axes) and identical rest masses. But I disagree that the reason for this is because the first postulate does not apply to pseudo-laws of physics. The same first postulate-defying behavior is predicted when the problem is transferred to a situation where a bonafide law of physics is involved, to which the first postulate is applicable.

Consider the mechanical force responsible for keeping the earth in orbit around the sun

$force = m$ω$^2r$

where m is the mass of the earth, ω and r are the average angular velocity and orbital radius.

The relativistic expression of the above force when generalizes to all inertial frames is known as a 4-force and takes the form

4-$force = m$ω$^2r/$γ

when the 4-force is perpendicular to the relative velocity, as is the case in the example. The same 4-force expression can be derived by applying the Lorentz transforms to all the quantities m, ω, r that make up the non-relativistic expression. See https://www.researchgate.net/publication/254235022_Derivations_of_Relativistic_Force_Transformation_Equations

The distant observer traveling at .866 c relative to the solar system will calculate a γ of 2. Therefore the distant observer will calculate a 4-force only half the magnitude of the same 4-force calculated in the solar system’s rest frame. But the spacetime curvature, or the gravity if you like, in the vicinity of the sun-earth system remains unchanged in both frames. This again provokes a first postulate issue between the two frames.

From the distant observer’s point of view, as the situation stands the earth does not possess enough orbital energy to withstand the spacetime curvature in the vicinity of the sun. The earth's rest frame orbit is not sustainable in the observer's frame because it is orbiting too slowly. The orbit will need to change shape. But that would also violate the first postulate.

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  • $\begingroup$ “The same first postulate-defying behavior is predicted when the problem is transferred to a situation where a bonafide law of physics is involved”. This is incorrect. Your example is also not correct since there is no actual law of physics that treats gravitation as a four force. The actual law of physics is the Einstein field equation which is fully compatible with the first postulate. In fact it is compatible with a more general postulate. $\endgroup$ – Dale Sep 20 at 1:57
  • $\begingroup$ Dale, thanks again for your close attention to this question. I believe you have confused the mechanical 4-force I employ, mv^2/(r*gamma) for the gravitational pseudo-force GMm/r^2, which I agree is not a 4-force. Please reconsider your comment in light of this correction. I remain interested in your thoughts. You might also provide an actual calculation proving the point you are trying to make. Thanks again. $\endgroup$ – Pat Dolan Sep 20 at 2:49
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    $\begingroup$ Sure. The calculation is pretty straightforward. The gravitational four force is 0 (that is what it means for free fall to be a geodesic). It is 0 in every reference frame (inertial or not) and regardless of the curvature, so that is consistent with the first postulate. $\endgroup$ – Dale Sep 20 at 3:11
  • $\begingroup$ I thought we both agreed that there is no gravitational 4-force. "...since there is no actual law of physics that treats gravitation as a four force." How then can it be equal to zero? What is not zero in any frame is the spacetime curvature in the vicinity of the sun. Unless the sun's spacetime curvature is also frame-dependent then it appears the contradiction stands. $\endgroup$ – Pat Dolan Sep 20 at 3:17
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    $\begingroup$ “No gravitational four force” and “gravitational four force is always 0” are two ways of saying the same thing. “No money in the bank“ and “a bank balance of 0” is the same. Etc. For curvature the law of physics that determines it is: $G_{\mu\nu}+\Lambda g_{\mu\nu}=\kappa T_{\mu\nu}$. (Curvature is the left side of this equation). This is the law of physics for determining curvature in any reference frame and in any spacetime. So it satisfies the first postulate. Indeed, any law of physics that can be written in tensor form satisfies it. Anyway, we are getting the “chat” warning, so I’m done $\endgroup$ – Dale Sep 20 at 3:28
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While Kepler's third law is known to be true in the rest frame of the solar system, it can be demonstrated to be invalid when applied to the solar system by observers in other inertial frames of reference. This conflicts with special relativity's first postulate which requires that the laws of physics be valid and take the same form in all inertial frames of reference.

To a first approximation Kepler's third law of planetary motion states

$$\frac{R^3}{T^2}=K$$

where $K$ has the numerical value $7.5x10^{-6}$ when the orbital period $T$ is measured in $days$ and the semi-major axis of the orbit $R$ is measured in astronomical units $AU$.

The earth's orbital period in the solar system's rest frame is

$T=365\,days$

The Lorentz time dilation for the earth's orbital period between the solar system's rest frame and all other inertial frames is

$T'=\biggl(\frac{365\,days}{\gamma}\biggr)$

The semi-major axis (and the semi-minor axis) of the earth's orbit is identical for all inertial frames where the observer's velocity vector is collinear with the sun's axis of rotation

$R'=R=1AU$

So the relativistic expression of Kepler's third law when the observer's velocity vector is collinear with the sun's axis of rotation is

$$\frac{1AU^3}{\biggl(\frac{365\,days}{\gamma}\biggr)^2}=7.5\times10^{-6}\frac{AU^3}{days^2}$$

From this if follows that

$133333\,days^2 =\biggl(\frac{365\,days}{\gamma}\biggr)^2$

$365\,days = \frac{365\,days}{\gamma}$

$365\,days-365\,days\sqrt{1-\frac{v^2}{c^2}}=0$

The last equation is only valid in the rest frame of the solar system where $v=0$. The equation is invalid in all other inertial frames which is a violation of the first postulate of special relativity.

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