I picked this up on the net:

Einstein came to realize the principle of equivalence, and it states that an accelerated system is completely physically equivalent to a system inside a gravitational field.

When I am being accelerated I will pick up speed over time. This will have a noticeable effect, e.g. there will be an increasing blue-shift in the light of stars which lie in the direction of acceleration. This will not happen if I am just in a gravitational field and I could well distinguish acceleration from gravity.

I suspect that being completely physically equivalent may be a bit of an overstatement and is only true as long as I don't look outside?

Or is the solution, that I cannot really know, whether those stars are being accelerated too? So when I feel the gravity and I see a constant blue shift, I might conclude that I am being accelerated, but all these stars are being accelerated the same as me?

Or is there a way to interpret gravity as acceleration, which also leads to an increasing red/blue shift?

  • 1
    $\begingroup$ LOCALLY physically equivalent is the right terminology. By the way, I'm not sure your example is correct, but I don't have the time and energy right now to look into it. $\endgroup$ – Danu Nov 25 '13 at 20:21

First, it turns out that there are no uniform gravitational fields so the equivalence principle holds only locally.

But, for the sake of argument, let's assume that a uniform gravitational field can exist.

Now, consider the situation where an astronaut is in a rocket and the rocket's accelerometer reads a constant, non-zero value.

According to the equivalence principle, the following two perspectives are equivalent and thus indistinguishable by experiment:

(1) There is no gravitational field and the rocket is absolutely accelerating (due to some type of motor)

(2) The rocket is stationary (due to the same motor) in a uniform gravitational field

Note that in case 2, the stars are freely falling in the uniform gravitational field and thus, their speed relative to the stationary rocket increases with time just as in the first case.

So, it is not the case that you can distinguish the two perspectives. In both cases, the stars "in front" of the rocket will become more and more blue-shifted with time.


The stars you see at night are blue-shifted partly because they gain energy as they fall down Earth's potential well. Sorry, but Einstein wins this round :)

  • $\begingroup$ (They are red-shifted because of Hubble :) $\endgroup$ – lionelbrits Nov 25 '13 at 20:23
  • $\begingroup$ But the blue shift does not increase over time. This makes me believe I am in a gravitational field and not being accelerated. $\endgroup$ – Martin Drautzburg Nov 25 '13 at 20:25
  • $\begingroup$ @MartinDrautzburg Does assuming these other stars are moving away address this issue? $\endgroup$ – BMS Nov 25 '13 at 21:38
  • $\begingroup$ The equivalence principle only holds locally. It says nothing about "over time". $\endgroup$ – lionelbrits Nov 25 '13 at 23:14

What if I don't look at other stars, but the microwave background radiation, the "echo of the Big Bang", as it is sometimes called. Doesn't that give me a clue about my "absolute" motion, i.e. relative to "the universe"?

  • $\begingroup$ Sorry, didn't mean to post this as an answer to the original question, rather a contribution to the discussion below. I'm new to stackexchange and still learning... $\endgroup$ – Tom Nov 26 '13 at 18:30
  • $\begingroup$ This may be worth discussing, but is off topic. My original question was about distinguishing gravity from acceleration, by the fact that acceleration produces speed and gravity (apparently) does not. It was not about finding a system of rest. $\endgroup$ – Martin Drautzburg Nov 28 '13 at 21:52
  • $\begingroup$ I admit. The correct answer to your original question, BTW, is the one up there (Alfred Centauri). $\endgroup$ – Tom Nov 29 '13 at 19:00


The condition of locality conventionally only requires a locally constant gravitational field during the time the experiment runs. For a stationary rocket in the gravitational field of e.g. the earth, this condition is perfectly satisfied, without requiring the gravitational field to be uniform throughout space. Clearly, the stars are not accelerating towards the earth with g, yet the condition of a uniform field is satisfied locally, as required by the equivalence principle. Neither will a stationary rocket observe any progressive blue-shift of e.g. spectral lines of earth's atmospheric emissions (that you can consider well within the local uniform region). So I think the only consistent way to rescue the equivalence principle here is to actually disallow any external references (as it is usually done).

  • $\begingroup$ "One can also start with a space that has no gravitational field. A material point in this space, when sufficiently distant from other masses, behaves free of acceleration relative to an inertial system K. However, if one introduces a uniformly accelerated coordinate system K’ relative to K (uniformly accelerated parallel translation), … we can also view K’ as an admissible system (at rest) and attribute the acceleration of masses relative to K’ to a static gravitational field that fills the entire space that is under consideration. mathpages.com/home/kmath622/kmath622.htm $\endgroup$ – Alfred Centauri Dec 9 '13 at 2:46
  • $\begingroup$ A local gravitational field necessarily requires an anisotropic mass distribution with regard to this point. It is not possible to have the same anisotropy throughout entire space. Neither is it admissible that K' observes a different mass distribution to K (if K' would reverse the acceleration within a fraction of a second, that would be equivalent to the mass distribution of the whole universe shifting around within a fraction of a second). $\endgroup$ – Thomas Dec 9 '13 at 21:44

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