Any coordinate system in GR?
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From: Edward Green on 21 Aug 2006 19:13 The offhand claim is sometimes made that GR, which I take it means the formal machinery of GR, treats all coordinate systems equally. Is this true? Aside from questions about smoothness, it seems to me there is at least some other implicit condition on permissible coordinate systems. Consider a 1-dimensional manifold, a real coordinate x, and the metric ds = |dx|. Next, stretch this manifold by a postive factor a(x), also stretching the coordinates (so that points on the manifold are still labeled by the same numbers). Calling the new, stretched, coordinates u, the most natural transformation law for the metric would be ds = a(u)|du| Now, instead of stretching the manifold, compress the coordinates by this same function a( ), so that dx = a(u)du. The most natural transformation law for the metric would again be ds = a(u)|du| So, if we allow "any coordinate system", we would apparently be unable to distinguish the (physical) case of the stretched manifold from the (unphysical) case of the compressed coordinates. It may be objected that by compressing the coordinates, we have implicitly changed physical units, and this is not allowed. But that's a new rule, isn't it? What's a "unit"? We need additional structure to specify what we mean by this; we must equip our manifold with some physics -- a local property establishing the natural scale of the coordinates. This will insure that we don't arbitrarily distort the coordinates, and any distortion can be attributed to the manifold. Am I right or wrong?
From: Bill Hobba on 21 Aug 2006 22:42 "Edward Green" <spamspamspam3(a)netzero.com> wrote in message news:1156201991.957020.198840(a)h48g2000cwc.googlegroups.com... > The offhand claim is sometimes made that GR, which I take it means the > formal machinery of GR, treats all coordinate systems equally. Is this > true? Aside from questions about smoothness, it seems to me there is > at least some other implicit condition on permissible coordinate > systems. > > Consider a 1-dimensional manifold, a real coordinate x, and the metric > ds = |dx|. Next, stretch this manifold by a postive factor a(x), also > stretching the coordinates (so that points on the manifold are still > labeled by the same numbers). Calling the new, stretched, coordinates > u, the most natural transformation law for the metric would be > > ds = a(u)|du| > > Now, instead of stretching the manifold, compress the coordinates by > this same function a( ), so that dx = a(u)du. The most natural > transformation law for the metric would again be > > ds = a(u)|du| > > So, if we allow "any coordinate system", we would apparently be > unable to distinguish the (physical) case of the stretched manifold > from the (unphysical) case of the compressed coordinates. First why is one unphysical and the other not? Secondly since all rulers are compressed or stretched by the same amount nothing will change. But that is not really the claim of GR - it does not claim like SR does for inertial frames that if we take the same experiment and shift it to an other frame then exactly the same result will occur - shift it to an accelerated frame and it will be different. Here lies the crucial difference - SR by being restricted to inertial frames has that property - GR does not. In both cases the laws of physics are still the same. GR obeys the principle of general invariance (not to be confused with the principle of covariance). As pointed out by Kretchmann (and eventually agreed by Einstein) any law can be put into covariant from so the principle of general covariance contains no actual physics. Its physics lies in the fact that when in such a form all absolute terms (ie things like the speed of light, planks constant etc) remain unchanged and any other terms must be dynamical - technically this is called general invariance. But you will find a lot of articles confuse one with the other - but as long as you understand what is happening no problems will arise - at least I have never found any. This immediately implies that the metric tensor for example is not an absolute term so must be dynamical - which is a cornerstone of GR - in fact all by itself it pretty much implies the EFE's. For a detailed discussion see http://www.pitt.edu/~jdnorton/papers/decades.pdf#search=%22general%20covariance%20vs%20invarience%22 Thanks Bill > > It may be objected that by compressing the coordinates, we have > implicitly changed physical units, and this is not allowed. But that's > a new rule, isn't it? What's a "unit"? We need additional structure > to specify what we mean by this; we must equip our manifold with some > physics -- a local property establishing the natural scale of the > coordinates. This will insure that we don't arbitrarily distort the > coordinates, and any distortion can be attributed to the manifold. > > Am I right or wrong? >
From: Edward Green on 22 Aug 2006 00:56 Bill Hobba wrote: > "Edward Green" <spamspamspam3(a)netzero.com> wrote ... <stretching the manifold vs. compressing the coordinate system> > First why is one unphysical and the other not? Well, that depends on whether it is "physical" to stretch the manifold. In a simple minded analogy, the one dimensional manifold might be represented by a rubber band: stretching the rubber band is a physically distinct situation from compressing the coordinates -- one might be able to determine the state of strain by local measurements, or at least determine the relative strain for two locations. In GR the "stretch" might correspond to a gravitational field vs. flat spacetime. Not _everything_ is an artifact of the coordinate system. > Secondly since all rulers > are compressed or stretched by the same amount nothing will change. This might be the case, but I think you assume too much. It may be the case that "nothing will change" locally -- meaning we cannot tell the difference by local measurements alone -- but that we can very definitely tell that something is changing over paths. This is the situation in GR, I believe. I made my "stretching" dependent on position BTW -- not the very simplest situation one could imagine -- to try to avoid the impression that this was solely about an arbitrary choice of units. As a silly illustration of principle about a possible physical difference between distortion of a manifold and distortion of coordinates, imagine an elastic manifold which turns blue beyond a certain strain state. Now,eitehr stretching the manifold in strata, or, on the other, inversely compressing the coordinates, we either see blue striations or we don't -- physically distinct situations! > But > that is not really the claim of GR - it does not claim like SR does for > inertial frames that if we take the same experiment and shift it to an other > frame then exactly the same result will occur - shift it to an accelerated > frame and it will be different. Here lies the crucial difference - SR by > being restricted to inertial frames has that property - GR does not. In both > cases the laws of physics are still the same. > > GR obeys the principle of general invariance (not to be confused with the > principle of covariance). As pointed out by Kretchmann (and eventually > agreed by Einstein) any law can be put into covariant from so the principle > of general covariance contains no actual physics. Its physics lies in the > fact that when in such a form all absolute terms (ie things like the speed > of light, planks constant etc) remain unchanged and any other terms must be > dynamical - technically this is called general invariance. But you will > find a lot of articles confuse one with the other - but as long as you > understand what is happening no problems will arise - at least I have never > found any. This immediately implies that the metric tensor for example is > not an absolute term so must be dynamical - which is a cornerstone of GR - > in fact all by itself it pretty much implies the EFE's. Well, you have confused me, though I claim I am alive to the kinds of issues you raise, and you have failed to persuade me, if that was your purpose, that my question is somehow misguided. However, I thank you very cordially for your serious answer, and especially your detailed reference, available free through the miracle of pdf, and I certainly must be compelled to study it. That post was rejected by the moderator of s.p.r., BTW, where I only repaired because that's where the experts are, and the cranks aren't. The reason given was "mainly mathematical, and containing elementary errors". I've been led to believe that s.p.r. was waiting with open arms for the disillusioned amateur to escape the trials and tribulations of trolls and orcs, provided he only asked respectful questions. I hardly went ranting about how GR is wrong, only mentioned a common aside which I suspected was wrong, and if not, could the error in my ointment please be strained out? I may be wrong -- my usual refrain -- but I'm not _that_ obviously wrong. Your thoughtful answer shames this miscreant, but unfortunately does not remove him from his position.
From: Ken S. Tucker on 22 Aug 2006 04:34 Basically s.p.r. has no expertise in GR so although your question is well posed and penetrating it exceeded the moderators understanding. Myself I'd be interested in possible well informed opinions that occasionally post to s.p.r. , that said I'll try. Edward Green wrote: > The offhand claim is sometimes made that GR, which I take it means the > formal machinery of GR, treats all coordinate systems equally. Is this > true? Pardon my apparent evasiveness... GR treats all Frames of Reference's (FoR's) equally. AE's law, G_uv = T_uv constrains the metrical geometry, to solutions of the tensor g_uv satisfying that law, that is why G_uv=T_uv is a called a field equation. For example a simplistic solution follows from G_uv=0 called the Schwarzchild Solution, that I'm sure you're aware of, that provides metrics like, g_00 = 1 - 2m/r , g_rr = 1/g_00 etc. The reason for CS independence is because g_uv is a tensor and can be properly transformed via the usual procedure, g'_ab = (&x^u/&x'_a) (&x^v/&x'_b) g_uv to any other CS, using light-years, millimetres polar, elliptical, cylindrical as you please, and in time, seconds or dog-years, meaning the physical reality is independent of units. > Aside from questions about smoothness, it seems to me there is > at least some other implicit condition on permissible coordinate > systems. Yes, the ultimate choice/solution of the metric is conditioned by the field equation, but you can use inches or cubits. > Consider a 1-dimensional manifold, a real coordinate x, and the metric > ds = |dx|. Next, stretch this manifold by a postive factor a(x), also > stretching the coordinates (so that points on the manifold are still > labeled by the same numbers). Calling the new, stretched, coordinates > u, the most natural transformation law for the metric would be > > ds = a(u)|du| Sure that's a good example, use a tapered elastic. Lightly stretched mark points against a ruler then stretch it and you'll end up generating a logrithm ruler, that's familiar. > Now, instead of stretching the manifold, compress the coordinates by > this same function a( ), so that dx = a(u)du. The most natural > transformation law for the metric would again be > > ds = a(u)|du| > > So, if we allow "any coordinate system", we would apparently be > unable to distinguish the (physical) case of the stretched manifold > from the (unphysical) case of the compressed coordinates. By "any CS" you can still use feet or metres, polar or cartesian. > It may be objected that by compressing the coordinates, we have > implicitly changed physical units, and this is not allowed. But that's > a new rule, isn't it? What's a "unit"? We need additional structure > to specify what we mean by this; we must equip our manifold with some > physics -- a local property establishing the natural scale of the > coordinates. This will insure that we don't arbitrarily distort the > coordinates, and any distortion can be attributed to the manifold. > Am I right or wrong? I think you're right when you focus on "unit" specifically Planck's constant "h" as used here, E = h*f. Think back in Newtonian terms, light frequency "f" and energy "E" were unaffected by gravitation. For example, one could imagine setting up a laser based cartesian CS, X,Y,Z from distant laser's like from stars. From the standpoint of Newton, placing a mass into the X,Y,Z Grid will NOT affect the Grid. However, light energy is subject to the conservation of mass-energy law, (in hindsight...duh) and hence Newtons X,Y,Z Grid needed to be modified to AE's x,y,z flexible grid, which, as it turns out, flexs the x,y,z grid to maintain the scalar invariance of "h". Famous confirmations are the Einstein Shift due to gravitation and the deflection of light, each mods the Newtonian Grid to the Einsteinian grid. Best Regards Ken S. Tucker (kst)
From: hilbertsnutsack on 22 Aug 2006 09:09
I replied to this post last night through mathforum.org, and for some reason my reply still hasn't made it to google groups. Here it is again, apologies if it now appears twice. ******************************************* I'm not entirely sure I understand your question, but I will have a go at answering it; let me know if I am barking up the wrong tree. > The offhand claim is sometimes made that GR, which I take it means the >formal machinery of GR, treats all coordinate systems equally. Is this >true? For some definition of "all coordinate systems", yes. More precisely, spacetime is assumed to be equipped with a differentiable structure, i.e. an equivalence class of local coordinate charts. What this means is that, given some coordinate chart x that maps part of spacetime to a subset, say U, of R^n, defining a new coordinate system y by y=f(x), where f is a smooth bijection with smooth inverse from U to some other subset V of R^n, y will be a valid coordinate system. The equations of GR are constructed in such a way that they will look the same in both the x and y coordinate systems. >Consider a 1-dimensional manifold, a real coordinate x, and the metric >ds = |dx| OK. For simplicity let's take the manifold to be R^1. >Next, stretch this manifold by a postive factor a(x), also >stretching the coordinates (so that points on the manifold are still >labeled by the same numbers) I am not sure that I understand what you mean by "stretch this manifold"... it seems like it should mean "change the metric on the manifold by ds |-> a(x)ds, is this what you had in mind? >Now, instead of stretching the manifold, compress the coordinates by >this same function a( ), so that dx = a(u)du. The most natural >transformation law for the metric would again be >ds = a(u)|du| Yes. It seems that what you have shown, if I have understood what you are doing, is that the "stretched" manifold is isometrically equivalent to the original manifold, i.e. there is a diffeomorphism between the two which preserves the metric. >So, if we allow "any coordinate system", we would apparently be >unable to distinguish the (physical) case of the stretched manifold >from the (unphysical) case of the compressed coordinates. >From a purely mathematical standpoint, there is no difference, since the spaces are equivalent considered as manifolds with metrics. From a physics standpoint, we can tell the difference if there is anything in this spacetime: suppose there are two planets in the original manifold at x=0 and x=1, then if the space is stretched by some supernatural being changing the metric but leaving the planets fixed at x=0 and x=1, the inhabitants of this universe would be able to tell the difference, since, for example, the metric determines how the laws of electromagnetism look in a given coordinate system, and if the metric gets bigger, then the difference in x-values of two adjacent atoms in ruler gets smaller. Therefore if the inhabitants use a ruler to measure the distance between the planets before and after the stretching, they will get different answers. On the other hand if you simply change to using u-coordinates then the ruler and planets will both appear different, but the length between the planets, as measured by the ruler, will stay the same. Does that help at all? |