Consider a scientist who is accelerating at a constant rate along the x direction, in flat minkovski space. The relevant coordinate transformations are $$\begin{cases}t'=t \\ x'=x+\frac{1}{2}at^2 && a=\text{const} \\ y'=y \\ z'=z\end{cases}$$

The question is, do you expect the curvature tensor to vanish in the primed coordinates?

One argument is that the curvature tensor is indeed a tensor. Since the space is flat, the curvature tensor will vanish for a stationary observer and hence will vanish in all coordinate systems. This makes sense because the curvature of space-time is an intrinsic property of the manifold and doesn't care about which reference frame (RF) we measure it in.

But when i try to apply the equivalence principle (EP) to answer the same question i wind up contradicting myself. Which is worrying because it likely means that my basic understanding of the concept is flawed.

The EP says that the scientist cannot decide whether his laboratory is uniformly accelerating in flat space or if his laboratory is fixed on the surface of his home planet, who's surface conveniently results in an acceleration of 'a' outward from the planet (recalling that a freely falling observer is now the definition of unaccelerated). So actually, the primed coordinates aren't those of someone who is accelerating uniformly in flat space, they're equivalently the coordinates of someone fixed in a gravitational field, the space of which is curved and so has a non zero curvature tensor.

Please correct where my logic fails, it's something quite basic and fundamental so i would really like to straighten it out. Thanks.

  • $\begingroup$ In SR, there is no possibility of constant coordinate acceleration since no physical object can accelerate to c (or beyond). $\endgroup$ Jun 15, 2017 at 1:57
  • $\begingroup$ I see your point. How can the question be amended to make sense then? I guess I could let the acceleration be a suitable function of v', but that would complicate the concept I am trying to explore here more than I want to. $\endgroup$ Jun 15, 2017 at 2:11

2 Answers 2


Your coordinate transformations are wrong. They are obviously wrong because for large enough $t$ the velocity will exceed the speed of light. The tranformations are actually:

$$\begin{cases} t'=\frac{1}{a}\sinh at \\ x'=\frac{1}{a}\cosh at \\ y'=y \\ z'=z \end{cases}$$

These are known as the Rindler coordinates and give the spacetime geometry for an observer with constant acceleration $a$ in the $x$ direction. The geometry is described by the Rindler metric:

$$ ds^2 = -\left(1 + \frac{a}{c^2}x \right)^2 c^2 dt^2 + dx^2 + dy^2 + dz^2 $$

If you take this metric and calculate the Riemann tensor then you'll find it comes out zero, just as it does for the Minkowski metric. I speak from experience, having done exactly this calculation to convince myself it was true :-)

An observer can measure their proper acceleration by measuring their acceleration relative to a freely falling object. Note that this is a local measurement i.e. it is done at the observer's position. If you take the Rindler metric and calculate the proper acceleration from it the result comes out as $a$, so it's just the acceleration we started with.

The equivalence principle tells us that the observer cannot tell whether their proper acceleration is due to gravity or motion. For example an observer in a gravitational field, e.g. me sitting here on the surface of the Earth, can also calculate a proper acceleration. If you're interested this calculation is described in What is the weight equation through general relativity?.

To go into a bit more detail, the proper acceleration is the norm of the four-acceleration, and the four-acceleration is given by:

$$ A^\alpha = \frac{d^2x^\alpha}{d\tau^2} + \Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu $$

There are two parts to this. The first part $d^2x^\alpha/d\tau^2$ is just the coordinate acceleration as in Newtonian mechanics i.e. how your position changes with time. The second part $\Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu$ is due to the curvature of the coordinates. The symbols $\Gamma^\alpha{}_{\mu\nu}$ are the Christoffel symbols and they tell us about the curvature in our coordinate system.

Take your example of an accelerating observer and suppose I am an inertial observer watching you accelerate. In my coordinates spacetime is flat, so the Christoffel symbols are all zero, but I see your position changing. So I calculate your four-acceleration to be:

$$ A^\alpha{}_\text{me} = \frac{d^2x^\alpha}{d\tau^2} + 0 $$

In your coordinates you are stationary at the origin so your coordinates aren't changing, but the Christoffel symbols are now non-zero, so you calculate your four-acceleration to be:

$$ A^\alpha{}_\text{you} = 0 + \Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu $$

The calculation looks different for the two of us, but when we take the norm of the four-acceleration we both get the same answer of $a$. So you consider yourself to be in a gravitational field with gravitational acceleration $a$ while I consider you to be in flat spacetime with an inertial acceleration $a$.

  • $\begingroup$ Thank for the thorough answer. I will do the same as you and do all the calculations myself.. and for the record, this was a question from a past exam at my school, so i took the coordinate transformations directly from there. $\endgroup$ Jun 15, 2017 at 6:22
  • 2
    $\begingroup$ @NormalsNotFar: If you Taylor expand $\cosh x$ you get $\cosh x = 1 +\tfrac{1}{2}x^2 + O(x^4)$. So dropping the higher order terms we get $\tfrac{1}{a}\cosh at \approx \tfrac{1}{a} + \tfrac{1}{2}at^2$. That's why your expression for $x'$ is a good approximation for small times. $\endgroup$ Jun 15, 2017 at 6:29
  • $\begingroup$ @JohnRennie would it be legitimate to do the calculation with that cosh approximation and the approximation for sinh? $\endgroup$
    – GlennGould
    Dec 3, 2021 at 6:25
  • $\begingroup$ @GlennGould Only for speeds much less than the speed of light i.e. in the non-relativistic limit. $\endgroup$ Dec 3, 2021 at 7:01
  • $\begingroup$ @JohnRennie interesting. $\endgroup$
    – GlennGould
    Dec 3, 2021 at 7:04

No, no need to ammend it, it's a good question, and you are almost right.

See for instance the Wikipedia article on proper acceleration. In a small enough lab you can't tell if it's constantly accelerated or in a gravitational field. If the lab is bigger, at some size you'll start seeing tidal a forces from gravity, which is different than constant acceleration. In the article it talk about real acceleration is acceleration wrt a freely falling reference frame

Fact is that there is a curvature non-zero tensor if there are tidal forces. Not otherwise. Straight acceleration has the same effect as a constant gravitational field.

Second, there is a wheel known coordinate frame for an 'accelerated observer', in the sense you meant it, i.e. an accelerated reference frame in Minkowski spacetime. In fact because of the limitation on c, in that reference frame (i.e., if you were a constantly accelerated observer), you can only see part of spacetime, i.e. there is a horizon. The coordinate frame is called the Rindler coordinates. The observer at rest in that coordinate system, i.e., constantly accelerated in Minkowski spacetime, sees radiation coming from the horizon. The equation is similar to Hawking radiation from the horizon of a black hole. In fact, the geometry close and outside a black hole horizon, with an observer and observer at rest in that frame can be approximated by the Rindler coordinates. This is pretty amazing and is another example of how Equivalence Principle gets it right, if you interpret correctly. For more see the Wikipedia article on Rindler coordinates.

  • $\begingroup$ Minor comment to the post (v1): In the future please link to desktop Wikipedia pages rather than mobile Wikipedia pages. $\endgroup$
    – Qmechanic
    Jun 15, 2017 at 4:21

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