22
$\begingroup$

In Einstein's theory of relativity, if motion is truly relative, then why would somebody in a rotating space station experience (artificial) gravity? I mean, I get why they experience gravity IF the space station is rotating, but what I do not get is how can you say it is rotating in the first place? Would you not need to define its motion in relation to another point in space?

So let's say we have two space stations and in our hypothetical universe they are the only two things that exist in the entire universe.

  • In space station 1, there is no sensation of gravity.
  • In space station 2, there is a sensation of gravity.

So we know that space station 2 is rotating and space station 1 is not because of the sensation of gravity we feel in station 2.

To an observer in station 2, they would feel gravity and see station 1 appearing to orbit around them. An observer in station 1 would feel no gravity and see station 2 spinning but staying in one spot. I think everyone would agree with all of those observations thus far.

But here is where it breaks down, I think, you have to have some fixed reference point in space to be able to say that station 2 is spinning and it is not station 1 orbiting in order to get the sensation of gravity on station 2. Newton seemed to have an answer to this because he said there was a fixed "at rest" but according to Einstein, I don't think there is a fixed at rest?

Does my question make sense? My confusion also applies to acceleration, it seems you have to have some fixed reference frame to say anything is accelerating at all. We know you would feel the effects of acceleration if in a space ship in deep space, but why if all motion is relative. I mean, how can you even say you are accelerating and it is not everything else in the universe accelerating and you are actually standing still without a fixed “at rest” point in space?

$\endgroup$
  • 2
    $\begingroup$ Jeremy Olson: "how can you say it is rotating in the first place?" -- Closely related ("the opposite question"): "What determines which frames are inertial frames?" (PSE/q/3193). "according to Einstein, I don't think there is a fixed at rest?" -- In the theory of relativity there are definitive methods for determining whether several participants are at rest to each other, or (only) rigid to the other (e.g.: rotating), or (even) moving wrt. each other. $\endgroup$ – user12262 Nov 26 '14 at 22:01
  • $\begingroup$ Jeremy Olson: "the sensation of gravity we feel" -- In the theory of relativity the measurements of geometric relations (such as, whether participants under consideration are at rest to each other, or instead merely rigid to each other) is not based on their possible "feelings", but on the judgement of coincidence (or else: sequence) of observations by each participant. Of course this approach does not deny the possibility or reality of such feelings; but it establishes a reference for measuring/comparing/distinguishing the "acuteness and trueness of feelings" of each participant. $\endgroup$ – user12262 Nov 26 '14 at 22:02
  • 4
    $\begingroup$ I'd like to stress one of the references being burried in the thread that user122262 linked to: The OP's question is more or less directly related to Mach's Principle. $\endgroup$ – balu Nov 26 '14 at 22:59
  • 1
    $\begingroup$ Here's a simpler version of this. If there is one spaceship in the universe, and you say it is rotating, what is it rotating with respect to? Einstein himself had a problem with that question. I'm no expert here, but I think he deferred to Mach's Principle, which roughly says that the rotation can be defined with respect to the average position of all the mass in the universe. As I understand things, General Relativity provides a mechanism for this. Experts will have to explain it to us. $\endgroup$ – garyp Nov 27 '14 at 0:15
  • 1
    $\begingroup$ Some historical background: Einstein was inspired by Mach's philosophy (which is not to say general relativity is necessarily 100% Machian), which in turn mirrored much of what Leibniz believed. Newton and Leibniz vehemently disagreed on a number of things (not just calculus), one of which is whether a notion of space exists independent of the matter in the universe. To counter Leibniz's "everything's relative" stance, Newton put forth his bucket thought experiment, which, as far as I know, none of his contemporaries had a good response to. $\endgroup$ – user10851 Nov 27 '14 at 1:08
33
$\begingroup$

Velocity is relative. There is no special reference frame that would be "at rest".

But acceleration is not and was never claimed to be. Reference frames in free fall are special and reference frames that are accelerating relative to the ones in free fall contain inertial forces (circular motion involves acceleration towards the centre; the corresponding inertial force is called centrifugal force).

So you can distinguish which of the stations is rotating and theory of relativity never claimed otherwise.

$\endgroup$
  • 1
    $\begingroup$ You have said -that- acceleration is not relative and -that- accelerating reference frames contain inertial forces. But what I am really asking is -why- do accelerating reference frames feel inertial forces when there is no fixed "at rest" by which to define that they are accelerating? $\endgroup$ – Jeremy Olson Dec 2 '14 at 5:56
  • $\begingroup$ @JeremyOlson Acceleration is dv/dt . You do not need a rest frame to observe that dv/dt is not zero, only two frames and a way to measure distance/angle between the two. For example the center of a rotating satellite versus the motion of the periphery. $\endgroup$ – anna v Dec 3 '14 at 11:49
  • $\begingroup$ But change in velocity is a change in the rate of speed which speed is the change in position over time, so change in position relative to what is the question? $\endgroup$ – Jeremy Olson Dec 3 '14 at 16:54
  • $\begingroup$ In other words, take the rotating disk, how do you know it is rotating if you are only comparing to itself? For instance, you take a point on that rotating frame at instant and then again at a later point in time it will still be in the same exact position in relation to every facet of the rotating frame as it was in the prior instant, so by that reckoning it has not moved. The only way to have "movement" and thus acceleration is to be able to say at first instant it was at point A then at later instant at point B. Point A and point B must be independent of the rotating frame. $\endgroup$ – Jeremy Olson Dec 3 '14 at 18:56
  • $\begingroup$ So coming back a few weeks later and taking a second read of the comment a few up that states, "You do not need a rest frame to observe that dv/dt is not zero, only two frames and a way to measure distance/angle between the two"...so are you saying that if you do not have 2 frames you will not experience the effects of acceleration? In other words, are you saying that in a hypothetical universe that only contained 1 frame of reference it would not be possible to accelerate, so no matter what you did, spin, thrust forward it wouldn't matter because it would always feel like you are not moving? $\endgroup$ – Jeremy Olson Jan 7 '15 at 22:40
11
$\begingroup$

Special relativity deals with "inertial" or "non-accelerating" frames. Physics in inertial frames are equivalent independent of their velocity and the velocity of inertial frames are relative. You are free to assume any inertial frame is stationary and all other frames are moving relative to it. Rotating frames are not inertial, they are accelerating frames and are therefore not relative. If a frame has no rotation then it must be maintaining orientation with the mean of the galaxies in the universe. This is an absolute. It has been conjectured, originally by Mach, that the distribution of matter in the universe sets the non-rotating frame.

General Relativity does deal with accelerating frames and discusses differences between rotating frames and gravitational accelerating frames, but that is another story.

$\endgroup$
  • $\begingroup$ Mach's theory does get to the heart of "why" the rotating space station would experience artificial gravity. This theory does make sense to me. $\endgroup$ – Jeremy Olson Dec 2 '14 at 6:02
10
$\begingroup$

In general relativity, angular motion actually does have some "relativity" to it as well. When you're in close proximity to a spinning object, you'll actually be dragged along with it. This is known as the Lense-Thirring effect, or just "frame-dragging". The most dramatic example is the ergosphere of a spinning black hole, a region where no object can remain stationary - it must rotate with the black hole. (similar to the event horizon from which objects cannot escape)

The fact that you feel a force under constant angular motion is because you're spinning relative to the background universe. (that statement may be somewhat controversial) If you were to take a large spherical mass shell, and set it spinning around yourself, you would actually begin to rotate with the shell but experience no centripetal force. If you added mass until the shell was nearly a black hole, your natural rest frame would have the same angular velocity as the shell relative to the rest of the universe.

$\endgroup$
  • $\begingroup$ Yeah, I think that makes sense...so basically an object is accellerating in relation to all the other matter in the universe. So by this theory, it seems to me if you only had 1 object in the entire universe, then it would not feel acceleration, inertia or anything like that...it would all be meaningless because you could not say it was moving, sitting still, or accelerating...do I have it right? $\endgroup$ – Jeremy Olson Nov 26 '14 at 23:51
  • 8
    $\begingroup$ That statement is controversial. What's preventing me from saying the universe is revolving around me? $\endgroup$ – Mehrdad Nov 27 '14 at 9:07
  • 1
    $\begingroup$ Special relativity is well-equipped to handle accelerating, non-inertial reference frames, while admittedly leaving out gravitational effects. See here. $\endgroup$ – Shivam Sarodia Nov 27 '14 at 23:58
  • $\begingroup$ @Draksis Thanks for pointing that out. As the easiest way to fix my answer, I've simply removed that statement. $\endgroup$ – adipy Nov 28 '14 at 14:56
  • $\begingroup$ Arguably, this question and Draksis' link provide the background needed to understand exactly why we are still looking for a way to reconcile gravity with GR. $\endgroup$ – dotancohen Nov 28 '14 at 17:28
5
$\begingroup$

General covariance applies only to freely falling observers -- once you invoke non-gravitational forces, like the inward pressure of the wall, the observer is no longer freely falling.

$\endgroup$
3
$\begingroup$

Easy way to distinguish between gravity and rotating space station: Throw a ball straight up in the air. If it comes straight down, gravity. If it moves away from you (behind your tangential velocity), it's a rotating space station.

$\endgroup$
  • 1
    $\begingroup$ An easier method might be to just place a straight, rigid beam on the floor. If you find the floor is concave, you're probably on a space station. The technique you describe would only work on a relatively small station (which would therefore require a rather large angular velocity to achieve 9.8m/s centrifugal acceleration) where you would probably notice the concavity of the floor just by looking at it. This phenomenon would also be observed on Earth, but the angular velocity of the planet is sufficiently small that atmospheric drag tends to overshadow its affect on ballistic objects. $\endgroup$ – bcrist Nov 29 '14 at 2:06
3
$\begingroup$

If the occupants of the space station were not aware of its design and could not look out a window then there is no way to tell if it is rotating or they are near a earth size planet that causes the gravity.

Orbiting around another space station will causes a sensation of gravity, and it seems you are contradicting yourself. If there is any rotational motion about of a body around its own axis or another space station then the occupants will feel a force you have defined as gravity. They can't feel it as in your space station 1 scenario.

That's how I see it as a non-pro.

As to acceleration, no you do not need a fixed reference point. In contrast to velocity which is defined as relative to the velocity of something else, acceleration is absolute in a sense that it is not relative to something else.

$\endgroup$
  • $\begingroup$ I remain open minded that I may be missing something here, because like I said, I am not an expert...but I do believe that even without windows if it was rotating you would know it because you would be stuck to the side of the wall (i.e. it would feel like gravity - I know this is not gravity, but it would feel like it to someone in the space station). $\endgroup$ – Jeremy Olson Nov 26 '14 at 20:45
  • $\begingroup$ The two space station example is not needed to illustrate my question, I just added it because I thought it helped me make my point...but maybe it just added more confusion. At any rate, the main question I have is why would you feel the effect of rotating if there is no fixed at rest to determine that you are rotating? $\endgroup$ – Jeremy Olson Nov 26 '14 at 20:48
  • $\begingroup$ or how can one explain the effects you would feel in an accelerating space ship when there is no fixed "at rest" in the universe, because in order for that object to be accelerating you have to compare it to some other frame of reference. $\endgroup$ – Jeremy Olson Nov 26 '14 at 20:51
  • $\begingroup$ acceleration is not relative to anything. you will feel it inside a black box, velocity not so. $\endgroup$ – Peter Nov 26 '14 at 20:57
  • 1
    $\begingroup$ because of inertia, your blood and everything will try resist the acceleration and be "left behind" giving a real real sensation. $\endgroup$ – Peter Nov 26 '14 at 21:12
0
$\begingroup$

The "reference" for acceleration, is its own previous "state". What this means is that, for linear acceleration, it is the initial point prior to the start of linear acceleration (at t = 0). And for angular acceleration, it is the imaginary line defined by the center of rotation and the initial position at t = 0 ($theta$ = 0).

$\endgroup$
  • $\begingroup$ I think I understand what you are saying, however, it still does not address the one simple question, how to determine if the state has changed. For instance, in acceleration it changed position in relation to the initial point/line, but the initial point/line in reference to what? If in reference to itself, then no, it didn't change position at all because that point/line continues to follow it along. So, it seems to me you absolutely must define that initial point in some way without reference to the accelerated objected itself. That means there would have to be some fixed reference. $\endgroup$ – Jeremy Olson Dec 3 '14 at 17:01
  • $\begingroup$ Let me clarify...not necessarily a "fixed" reference per se (I suppose it could be variable or fixed), but some outside reference other than the accelerated object itself would be necessary in order have a "change" in position. $\endgroup$ – Jeremy Olson Dec 3 '14 at 18:10

protected by Qmechanic Nov 28 '14 at 14:15

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.