It would appear that curved spacetime and the first postulate of special relativity are mutually exclusive.

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 distant observer measures the earth's orbital period around the sun to be 730.5 days inasmuch as $\gamma=2$, correct?

Because of the observer's position and direction of travel relative to the solar system, he/she 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.

In the distant observer's inertial frame the radial force acting on the earth to accelerate it away from the sun is calculated to be diminished by $\frac{1}\gamma$

$$\frac{\gamma m(\frac{v}\gamma)^2}r=\frac{m v^2}{\gamma r}$$ where $r$ is the same in both the solar system's rest frame and the distant observer's rest frame as previously specified. This result is in accordance with the calculation of any mechanical 4-force perpendicular to the direction of relativistic velocity.

But the spacetime curvature in the vicinity of the sun is always identical for all inertial frames. We know this to be empirically true from the precession of Mercury's perihelion and the deflection of starlight during eclipses; spacetime curvature in the vicinity of a gravitating body is unchanged for all inertial observers no matter their velocity with respect to the gravitating body.

An inconsistency now arises from the distant observer's point of view because the earth no longer has enough radial acceleration to counteract the spacetime curvature it is moving through. So the distant observer calculates that the earth will spiral into the sun in short order.

Does curved spacetime permit/require the violation of the first postulate? Does it reduce the laws of mechanics to mere approximations which are only true in the non-relativistic limit? Which physical laws actually do survive intact from one inertial frame to the next pursuant to the first postulate?

  • 1
    $\begingroup$ "An inconsistency now arises from the distant observer's point of view because the earth no longer has enough radial acceleration to counteract the spacetime curvature it is moving through. So the distant observer calculates that the earth will spiral into the sun in short order." What makes you think this is true? I fail to see where this point arises. $\endgroup$
    – Zack
    Apr 27, 2021 at 3:45
  • $\begingroup$ Zack, thanks for your question. The best way for you to appreciate the situation is work out the equation you think the distant observer would use to demonstrate that the earth would not spiral into the sun, since the earth possesses only half the radial acceleration in the distant observer's frame compared to the rest frame of the solar system. $\endgroup$
    – Pat Dolan
    Apr 27, 2021 at 6:30

4 Answers 4


One of the first things a person must get rid of when coming to GR from special relativity is the existence of a global reference frame for a particular observer. In general relativity you cannot ask a question "what is the state of a distant object relative to me?", because the question is ill-defined. You can talk about what you see and what you feel, and things that are happening right at the spot where you are. You can ask "if I take two people, synchronize their clocks, and then send one of them on a spaceship flight, what are they going to see when they both meet?". But you cannot ask "is the clock of a distant moving observer going to be slowed down or accelerated relative to me". You can can compare things that are only right next to you. You can though ask "will I see that moving clock slowed down or accelerated", which is a question about light coming into your eyes from that clock.

So, first, there is no "force acting on a planets relative to a distant observer" because a term "force relative to the distance observer" does not exist in general relativity. There is also no "curvature relative to a distance observer". But second, which is probably even more important, in General relativity there are no gravitational forces! Gravity is not a force in general relativity. Free falling body moves without acceleration! These are people on the ground who are moving with acceleration $g$ upwards due to the ground pushing them up.

Overall, I would recommend to read several first introductory chapters of Geroch's lecture notes

  • $\begingroup$ Thanks for your answer Pavlo. It's difficult to completely follow your argument. But you may have drifted into the relativistic epistemological fallacy. The observer can really, really, really know what the orbital velocity of the earth is in his frame once the relativistic doppler is corrected. That orbital velocity is no more an illusion than is the slower rate the observer measures for Big Ben as seen through his telescope. There can be no doubt that the earth is moving through the identical spacetime curvature with only half the radial acceleration of the solar system's rest frame. $\endgroup$
    – Pat Dolan
    Apr 27, 2021 at 7:01
  • $\begingroup$ In general relativity there is no global "his frame of reference", similar to how there is no correct flat map of the Earth surface since the sheet is flat and the earth is a sphere. In general relativity "reference frame" is just coordinates that one puts onto a curved spacetime. They do not bare any physical significance, but just help to do calculations. It is a very different philosophy from special relativity, where reference frames are "real". In general relativity they are not real, but just arbitrarily picked coordinates $\endgroup$
    – Pavlo. B.
    Apr 27, 2021 at 7:23
  • $\begingroup$ I believe, conventional teaching of special relativity with it's emphasis on "paradoxes" rather harms proper understanding of physics. When you study general relativity properly, all the paradoxes and weird things one used to have in Special relativity suddenly disappear and start looking super trivial. I highly recommend that reference at the end of my answer $\endgroup$
    – Pavlo. B.
    Apr 27, 2021 at 7:33

I assume that by "first postulate" you are referring to

The laws of physics are invariant (that is, identical) in all inertial frames of reference (that is, frames of reference with no acceleration).

This remains true in general relativity, but needs some modification.

The difference between special relativity and general relativity is that, in special relativity, the spacetime metric is given by $$ ds^2 = dt^2 - dx^2 - dy^2 - dz^2 $$ and is called "flat." In general relativity, it is given by a general metric $$ ds^2 = g_{\mu \nu} dx^\mu dx^\nu. $$ This is the only difference between the two.

In particular, the Lorentz transformations are symmetries of the flat metric (they are "Killing vectors") but an arbitrary $g_{\mu \nu}$ will usually have no symmetries. So does general relativity break special relativity?

Well, no. The thing is, at any particular point, an observer can set up a local system of coordinates which are locally Minkowski. These are sometimes called "freefalling coordinates," naturally used by an observer who is travelling in spacetime along a geodesic. (You see, "inertial" in general relativity means "freefalling" or "on a geodesic.") Locally, the Lorentz transformations will still be symmetries of the metric. A second freefalling observer, who passes close to the first one, will use their own coordinates that differ by one of these local Lorentz transformations.

However, these freefalling coordinates and Lorentz transformations lose their meaning when you look far away from these two observers.

However, that is irrelevant to the first postulate! It just says the "laws" are observer independent. For instance, both of them will think that planets and such move in geodesics around the sun. Sure, the coordinates they use to describe these processes will be different, but both will agree on the value of physical quantities like proper time elapsed for any process. In general relativity, there is a notion of general covariance, where people can compute things in any coordinate system they want, and convert easily between the two to find that the actual "laws" are unchanged. Therefore the first postulate remains entirely true. Your conclusion "An inconsistency now arises from the distant observer's point of view because the earth no longer has enough radial acceleration to counteract the spacetime curvature it is moving through" is not correct. If you don't pull out the full equations of general relativity you are bound to make incorrect conclusions about it. In fact, because they're covariant, it really is impossible to make any coordinate change and derive a physical inconsistency.


You need to take the Schwarzschild metric (in the Sun's frame): $$ g_{tt} = -\left(1-\frac{r_s} r\right)$$ $$ g_{rr} = \left(1-\frac{r_s} r\right)^{-1}$$ $$ g_{\theta\theta} = 1$$ $$ g_{\phi\phi} =\sin^2{\theta}$$ and transform it into the distant observer's frame, and see if the geodesic equation:

$$ \frac{d^2x^{\mu}}{ds^2}+\Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{ds}\frac{dx^{\beta}}{ds}$$

does (or doesn't) describe the Earth's trajectory.

...or use the weak field gravioelectromagentic analog to the Lorentz force law:

$$ {\bf F}= m({\bf E}_g+{\bf v}\times4{\bf B}_g)$$

where ${\bf E}_g$ is Newtonian gravity and ${\bf B}_g$ is the magnetic field analog satisfying:

$$\nabla\times{\bf B}_g=-\frac{4\pi G}{c^2}{\bf J}_g+\frac 1{c^2}\frac{\partial {\bf E}_g}{\partial t} $$

The mass current density is:

$${\bf J}_g = \rho{\bf v} $$


For a spherical mass like the sun, the Schwarzschild metric applies:

enter image description here

See this Wikipedia article. Here $\tau$ stands for the proper time, i.e. the time as measured by a clock following the Earth. This time is independent of coordinates so no matter from where you look, the metric will always be the same. Which means that also the curvature is independent from where you look. Note also that the factor in front of $dt^2$ is dependent only on radial distances. These don't change if you approach the solar system "perpendicularly".
Which time then sees the person inside the rocket to be going slower than what he/she reads on his clock on the wall? Well, this is indeed the clock on Earth. but this is a special relativistic effect. People on Earth see the clock inside of the rocket go slower too. This doesn't mean that their clock is actually ticking faster than the clock inside the rocket. What matters for curvature is only the proper time, not the time as seen from where ever.


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