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I realize there are answers to my question many places across the internet and many places on this site, but from what I have seen, most of them either involve terms I don't know or assume you can use a test mass or "accelerometer," which I think does not apply in the case I outline below (if it does, please explain how the accelerometer would work). After what I deemed enough time trying to understand various internet explanations, I decided to ask my own question. Please answer the question in a manner that satisfies the situation below, and please refrain from references to equations or devices that I, someone who has only taken a year or two of physics, would not be well acquainted with.

Situation: Someone is in a closed lab in which every particle, including those that make up the lab, experiences the same constant force. (I don't believe it is necessary to articulate exactly what is causing that force, but if it is please explain why.)

To clarify: I am looking for a conceptual explanation about the extent to which acceleration is relative. It seems to me that, in the situation above, there is no way to tell if you are accelerating, but I keep reading that the acceleration of a mass is not relative.

My problem with the answers I have seen is that they often don't speak in terms I understand, and when they do they are often unsatisfying or contradict other answers. For example, in this question: Is acceleration relative?

The first answer doesn't explain what is wrong with viewing an accelerating reference frame.

The second answer is close to what I am looking for, but seems to disagree with everything else I have seen, and simply mentions the idea of "all the mass in the universe" without explaining why that matters.

The third seems to take it as a fundamental assumption from Newton's laws. To me this is unsatisfying. It seems to apply that, for any object, there is some real acceleration, but there may be no way of finding that acceleration. Basically, the argument seems purely platonic. I would hope that an unsupported assumption isn't all there is to the answer.

An example of an answer I don't understand is the first answer for this question: Why acceleration is not relative in General Relativity?

The answer talks about local experiments, which I don't think would work in the situation above, then goes on to talk about Minkowski metrics, which are completely foreign to me.

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    $\begingroup$ For your title question, have you read this possible duplicate : physics.stackexchange.com/q/22803. It might be one of the explanations you mentioned but you might cite it, if so. $\endgroup$
    – user163104
    Jul 31, 2017 at 21:56
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    $\begingroup$ Sir: from your last paragraph it is unclear as to what is the question. $\endgroup$ Jul 31, 2017 at 22:42
  • $\begingroup$ It's not clear what you're asking, can you please elaborate? $\endgroup$
    – Omar Nagib
    Jul 31, 2017 at 23:18
  • $\begingroup$ @ZeroTheHero Added clarification. $\endgroup$ Aug 1, 2017 at 0:36

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Your example situation does not make sense. If every particle in the lab experiences the same force then every particle will accelerate at a different rate depending on its mass. The lab will be torn apart.

Neglecting that example, velocity is relative to the frame of reference. It is the same in all frames which are at rest relative to each other.

In the same way acceleration is relative to the frame of reference. It is the same in all frames which move with constant velocity relative to each other (which includes being at rest).

We could extend this to 'jerk', the rate of change of acceleration, and all higher derivatives of position. 'Jerk' is the same in all frames of reference which move with constant acceleration relative to each other.


It seems to me that, in the situation above, there is no way to tell if you are accelerating, but I keep reading that the acceleration of a mass is not relative.

Your example scenario seems to be asking if it is possible (in a closed, window-less laboratory) to distinguish between uniform acceleration in empty space and being at rest in a gravitational field. This is Einstein's Principle of Equivalence. The answer is No, that is not possible. Those two motions are equivalent. Likewise it is not possible to distinguish between free-fall acceleration in a gravitational field and zero acceleration in empty space.

However, it is possible to distinguish between two different magnitudes of acceleration in empty space, or being at rest in two different magnitudes of gravity. As Ben Crowell explains in a comment to Why acceleration is not relative in General Relativity?

what GR says is detectable is acceleration relative to a free-falling frame of reference.

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  • $\begingroup$ For the example I meant to imply that the force affects every infinitesimal unit of mass the same way, like gravity. I used the word particle to imply an infinitesimal mass. I realize now that is not accurate. My bad. $\endgroup$ Aug 1, 2017 at 15:21
  • $\begingroup$ OK. I think I understand what you mean now. ... Another difficulty is what you mean by "I keep reading that the acceleration of a mass is not relative." Can you provide an example of where this is written? $\endgroup$ Aug 1, 2017 at 16:01
  • $\begingroup$ By mass I just meant object, like in the question and first answer of physics.stackexchange.com/questions/22803/… $\endgroup$ Aug 1, 2017 at 17:35
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    $\begingroup$ That question is only another user repeating what he has heard without providing any authority for it, such as a textbook. It is pointless for us to try to justify rumours. This user even admits that it does not make sense to his/her present supervisor. dmckee's answer ends by saying The magnitude and direction measured depend on the frame of the observer, which is often what is meant when people say "it's relative". Magnitude and direction are all there is to acceleration, so he is saying that acceleration is relative. ... $\endgroup$ Aug 1, 2017 at 17:49
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    $\begingroup$ ... His example refers only to frames which have constant relative velocity, concluding that they measure the same acceleration, as stated in my answer. $\endgroup$ Aug 1, 2017 at 17:53

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