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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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 at 17:02
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I think the analysis being performed in this Q and A is based on a simple misunderstanding of relativity. The Principle of Relativity (referred to above as the 1st Postulate) states that the laws of physics are valid in any inertial frame of reference locally in that frame. This means that if the observer moving at $0.866c$ (w.r.t the Solar System) does an experiment in a laboratory on his own spacecraft, where everything in the laboratory is stationary with respect to him, all normal physical laws of motion will hold good. Every result in his lab will agree with the results in our labs on Earth. If his entire solar system around him is speeding through the galaxy at $0.866c$ (e.g. like this), Kepler's 3rd Law will hold for that system just as well as for ours.

The Principle of Relativity does not say that far away events and objects will look the same regardless of your state of motion. In fact, the groundbreaking aspect of the theory of relativity was that, in order for the first paragraph to be true, events and objects moving relative to the viewer must look different. For example, in Einstein's original 1905 paper, he shows that a rigid sphere (Sec. 4), seen from a stationary point of view, will be an ellipsoid compressed in the direction of motion when seen from a moving frame. This will apply to the shape of the planetary orbits in the original example (especially if the orbital speeds of the planets are much less than the speed of light, as is the case), as well as the shapes of the planets themselves. Likewise, light which is a certain frequency when viewed in the same frame as the source will be blue-shifted and higher-intensity when the source is moving toward the observer (Sec. 7).

All of these effects stem from an observer viewing an object or event that is far away and/or moving relative to him. But Special Relativity guarantees that for events in your immediate vicinity and stationary with respect to you, the laws of physics do not depend on your rate of motion (with respect to something else). Thus, there is no preferred frame of "absolute rest," because everything is at absolute rest with its immediate environment.

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  • $\begingroup$ "According to his/her wristwatch the observer measures the earth's orbital period around the sun to be 730.5 days, correct?" Yes. "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" Only if the orbit is in $yz$ plane and the traveler's line of motion is $x$, but that is possible. "the earth will spiral into the sun" No, because Kepler's 3rd Law is valid if you count time $t$ in the frame of the Earth (not $t'$ in the frame of the traveler). $\endgroup$
    – RC_23
    Jan 16 at 2:08
  • $\begingroup$ What you are saying is similar to observing that muon particles decay much more slowly when they are observed as incoming cosmic rays traveling at 99.99%+ light speed. In the frame of the Earth, the cosmic muon's decay time may be on the order of 100s of $\mu$s, which appears to violate the Standard Model – and indeed it does, unless we account for the motion of the muon and its frame's time dilation relative to ours. $\endgroup$
    – RC_23
    Jan 16 at 2:14
  • $\begingroup$ So in a sense yes, you're correct that Kepler's law does not hold for an observer moving relative to the solar system, but Special Relativity never claims that it does. It claims that if you uniformly boost the Solar System to $0.01c$ or $0.5c$ or $0.99c$ with respect to say the galactic center or the Cosmic Microwave Background, the inhabitants of Earth will observe no difference in Kepler's law for their own planets. $\endgroup$
    – RC_23
    Jan 16 at 2:16
  • $\begingroup$ I'm not sure what your goal is here. Do you think that because a space rock has flown past the Solar System, the Earth should have burnt up in the Sun by now? Clearly that has not happened. $\endgroup$
    – RC_23
    Jan 16 at 13:46
  • $\begingroup$ From the traveler's perspective, the Earth will take 700+ days to complete its orbit at the current distance. Kepler's 3rd law will not hold for him. It does not have to hold for him for SR to be true. I don't see the problem. $\endgroup$
    – RC_23
    Jan 16 at 17:52
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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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