1
$\begingroup$

Laws of physics cannot distinguish one inertial frame from another which is equivalent to the statement of Poincaré invariance.

What is the meaning of general coordinate invariance? General coordinate transformation does include transformations form one inertial frame to a non-inertial frame. And general coordinate invariance mean laws of physics are invariant under such transformations. Does it therefore imply that it is also impossible to distinguish between an inertial (I) and a non-inertial (NI) observer?

But Newton's first law can distinguish between inertial and non-inertial frame of reference. Then what is the meaning of general coordinate invariance then?

EDIT: Though I'm familiar with Poincaré invariance and special relativity, my knowledge of general relativity is still in its embryonic stage. Hence, an answer in simple non-technical language will be helpful.

$\endgroup$
  • $\begingroup$ I'm assuming you wanted something more than this? $\endgroup$ – Yogi DMT Oct 14 '16 at 13:01
  • $\begingroup$ Poincare invariance doesn't allow us to distinguish one inertial observer from another. Right? Then my simple question is does general coordinate invariance imply it is also impossible to distinguish between an inertial (I) and a non-inertial (NI) observer? If yes then there is a confusion. Newton's law seems to be able to distinguish between a I and NI observer. If no, what does it say about the invariance of physical laws under transforming from one inertial frame to another? $\endgroup$ – SRS Oct 14 '16 at 13:12
  • $\begingroup$ I think what you're asking is at the heart of the issue with reconciling relativity with QM and the "arrow of time". Special relativity shows us that inertial and non-inertial frames are not symmetrical (twin paradox) despite QM wanting the time elapsed to be invariant. $\endgroup$ – Yogi DMT Oct 14 '16 at 13:24
  • $\begingroup$ General coordinate transformation does include transformations form one inertial frame to a non-inertial frame. And general coordinate invariance mean laws of physics are invariant under such transformations. Are these two statements correct? $\endgroup$ – SRS Oct 14 '16 at 13:27
  • 1
    $\begingroup$ Your first sentence says "Laws of physics cannot distinguish one inertial frame from another", your second sentence says, "But the laws of physics ... can distinguish between inertial and non-inertial frame of reference". ???? $\endgroup$ – DilithiumMatrix Oct 14 '16 at 14:54
2
$\begingroup$

General coordinate invariance does not mean that you can't distinguish inertial from non-inertial frames: you can. What it means is that you can write the laws of physics in such a way that they have the same form in any good coordinate system, where 'good' means something like 'related to another good coordinate system by a non-singular differentiable transformation' (I realise this definition looks circular: it's not really, since you get to bootstrap the whole thing with some known-to-be-good sets of coordinates when you define your manifold).

So, for instance, imagine a rotating Cartesian coordinate system (for ordinary Newtonian mechanics in flat space). Well, the laws of physics look all funny here: things don't move in straight lines in the absence of forces according to the coordinate system and so on. But we can introduce various fudge factors 'fictitious forces' and so on and get things to make sense.

Now imagine a coordinate system (again for Newtonian mechanics) anchored in a car which is being driven about madly. Things are now even worse because the motion is not regular: the fudge factors are now quite complicated. But it turns out that, with a little mathematical sophistication, we can write the laws of physics in such a way that they will work in the car's frame too: we can systematically describe the fudge factors so they can be expressed in any coordinate system we like so long as it's differentiably related to a good one.

All of that seems like a lot of work for no real gain: we can just find some nice simple coordinate system and transform back to it, can't we?

Well, no, we can't. We can so long as there is a nice simple coordinate system. And in flat spacetime there always is (this is pretty much the definition of flat spacetime in fact). But we're not going to be dealing with flat spacetime, we're going to be dealing with spacetime which has curvature. And, for such a spacetime, there is no globally 'good' coordinate system: indeed, there may well not be any global coordinate system at all, we may need a chart of several coordinate systems each of which cover just some patch of the thing we are interested in (consider $S^2$ (the two-dimensional surface of a sphere) as the canonical example of that: you need at least two coordinate systems in its chart).

So general covariance, aka general coordinate invariance is sorting out a way of writing things down so that they work in any good coordinate system, not just in some privileged set of particularly simple ones.

Sorry for the wordy reply.

$\endgroup$
0
$\begingroup$

The trick is that when you tranforrm between two coordinate systems which are not uniformly moving with respect to each other you should at the same time transform also the value of the gravitational force. Since the latter is felt universally by all matter, the equations of motion then come out right in both coordinate systems.

This is the content of the famous "lift" example: if you sit in a lift that is in free falling motion in, say, the earth gravitational field, all experiments you can make will give just the same results as they would in a good old Newtonian inertial frame.

However, the correct statement is not "it is impossible to distinguish between an inertial and a non-inertial observer" but rather "there is no such thing as inertial observers" (= no experiment can single out a class of distinguished observers). Compare to the analogous statements for Newtonian mechanics: "it is impossible to distinguish between a static and a non-static observer" versus "there is no such thing as static observers".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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