The Lorentz Transformations and a Few Concerns

[Passionflower]

You might want to use the conceptual frame to say that light speed differ comparing A-B to B-A in a accelerating frame, but I don't see it that way. The only thing differing to me is how gravity acts upon their paths, relative the detectors.

So even if an experiment would demonstrate it you would still believe light speed is constant?

locally both A-B and B-A will give 'c'. So yes, to me Peter is definitely defining it correctly.

If it wasn't this way, light would not be a constant.

What is locally when A and B are for instance 1000 kilometers removed?

Locally the speed of light is c but locally the world is flat as well.

 

[juanrga]

The derivation of the Lorentz transformations is based on the homogeneity[of space and time] and the isotropy of space.
Could one derive the same transformations wrt space which is not homogeneous or[not] isotropic?
You may consider a few chunks of dielectric strewn here and there. I am assuming for the sake of simplicity that they are at rest in some inertial frame. Such a distribution is not possible without introducing gravitational effects. Running of clocks is affected by gravity.Does the anisotropy of space itself have any effect on them[running of clocks]?

[Incidentally isotropy of space is connected with clocks in the derivation of the Lorentz transformation.Clocks placed symmetrically wrt the x-axis[and lying on the y-z plane as example] should record the same time. Otherwise isotropy of spece gets violated.This idea is commonly used in the derivations.You may consider the one given in "Introduction to Relativity" by Robert Resnick ]
Again the Lorentz transformations are embedded in [present in] Maxwell's equations. But they are the vacuum equations---homogeneity of space[and time] and isotropy are in due consideration.

The Lorentz Transformations are of course correct--only in the context of the homogeneity[of space and time] and isotropy of space. They are extremely useful, like frctionless planes we studied in our childhood days.Frictionless planes helped us in understanding mechanics--but it is extremely difficult to realize them in practice.

Effectively the LT is only valid in a specific context, as Robert Resnick remarks. Precisely when you introduce a second body, you cannot consider that homogeneity is valid everywhere. The properties around particle 1 in the direction where particle 2 is placed are not the same than the properties in the direction where the particle 2 is not (charge distribution is not homogeneous for instance).

This has important consequences. As Jackson emphasizes in his book, one cannot built a Lagrangian for a charged two-body system that was truly Lorentz invariant.

 

[yoron]

Well, theoretically it is defined so, flat that is, in the same way we also use 'point particles', but in general there are no straight lines naturally in nature, as I know? And in 'space' you will find 'gravity', everywhere even though we also define a 'flat space'. And, it's not a theoretical definition that you won't be able to measure lights speed in a vacuum as other than 'c' locally, again as I know it, it's a experimental one even though rather counter intuitive. And the last point stands, if you want to define 'c' as a 'variable' you can't simultaneously define it as a constant. So I believe that to be correct.

"If general relativity is correct, then the constancy of the speed of light in inertial frames is a tautology from the geometry of spacetime. The causal structure of the universe is determined by the geometry of "null vectors". Travelling at the speed c means following world-lines tangent to these null vectors. The use of c as a conversion between units of metres and seconds, as in the SI definition of the metre, is fully justified on theoretical grounds as well as practical terms, because c is not merely the speed of light, it is a fundamental feature of the geometry of spacetime"

Is The Speed of Light Constant.

 

[JDoolin]

Sorry for the delay. This was a long post, and took some thought. Thanks.

Ah, I see I left out a key word: "freely falling". Neither of the objects in question are freely falling.

You asked how, in a curved spacetime with gravity present, two worldlines with the same constant θ and φ, and values of r that differ by a small amount dr, at a given instant of coordinate time t = 0, will continue to remain separated by that same amount dr for all times t. I gave the correct answer, and apparently you knew the correct answer. But you are also right; you cannot have freely falling objects maintain the same constant dr for all times t. That's the essential flaw of using geodesics as a coordinate system.

I meant that the sign convention that spacelike squared intervals are positive, and timelike ones are negative, is just a convention; you can flip it around, so that timelike squared intervals are positive and spacelike ones are negative, without affecting the physics, provided you change all the formulas appropriately.

If you like. But regardless of how you do it, ds^2=-c^2 d\tau^2 and if ds is real then dtau is imaginary. And if dtau is real, then ds is imaginary. There's only one degree of freedom here in the space-time interval, and the interval is either space-like or time-like or zero.

As it stands, this statement is at best wrong, and at worst not even wrong. Flatness or curvature is an invariant feature of the underlying geometry, and it is there and can be talked about without even assigning coordinates. If you do assign coordinates, then you figure out whether the underlying geometry is flat or curved by computing the Riemann curvature tensor in those coordinates, using the proper form of the metric in those coordinates for the manifold in question.

When you do this using standard Schwarzschild coordinates on Schwarzschild spacetime, using the expression for the metric on Schwarzschild spacetime in those coordinates, which you have written down several times now, you find that the Riemann curvature tensor is non-zero: i.e., the spacetime is curved.

If you do this using the same Latin and Greek letters, t, r, θ, and φ, for coordinates on Minkowski spacetime, using the expression for the metric on Minkowski spacetime in those coordinates, you find that the Riemann curvature tensor is zero; the spacetime is flat.

It's important to recognize that "curvature" as I've just described it is *intrinsic* curvature of the manifold, since that's what we physically observe. See further comments below.

Okay. It appears we both agree that , t, r, θ, and φ can represent coordinates on the Minkowski spacetime. And those coordinates are "flat" according to the Riemann curvature tensor. So it appears that where we disagree is that I am claiming that the (t, r, θ, φ) in the Schwarzschild metric have exactly the same global geometric meaning as the (t, r, θ, φ) in the Minkowski spacetime.

Yes, it is, because if no body ever follows a perfect "straight line" as you are using the term, how can such a line possibly exist? There's no physical observable that picks it out.

Right, there is no physical observable that picks it out. But we don't need a physical observable. Regarding straight lines, one should NOT ask "What direction does the light ray actually take around the planet." but rather one SHOULD ask, "What direction would the ray have gone if the mass had not been there." There is a clear and unambiguous answer.

In other words, I am supposed to believe that you can draw lines that go outside our universe, taking the "straight" route between two events instead of the "curved" route inside our universe that physical objects actually take. Can you see why I'm a bit skeptical?

Sure, it is an odd way of saying it. But if you imagine the path of a satellite if the Earth did not exist, you are essentially imagining a line that exists "outside the universe." There is still no difficulty imagining this line and superimposing it inside our universe.

Does the mapping preserve the metric? No. You admit as much later on when you try to claim that the "real" distance to Alpha Centauri remains the same if a black hole comes between there and Earth, even though every actual physical object has to travel a greater distance because of curvature. If the mapping doesn't preserve the metric, then it's physically meaningless.
Check out the "river model" of black holes:

http://arxiv.org/abs/gr-qc/0411060

Inside the horizon, freely falling objects do go faster than light relative to the "river bed", which is basically what you mean by the flat geometry that everything is supposedly mapped to. But as the authors of the paper make clear, the "river bed" is not physical and has no physical meaning; it's only used to help with visualization. Also, the authors make clear that the ability to construct a model with a flat "river bed", even just for visualization, is a special feature of Schwarzschild geometry, and cannot be done in general for a curved spacetime.

I'd like to see the original papers of Gullstrand and Painleve, but I have a few comments. (1) The conservation of energy requirement uses a nonrelativistic calculation of kinetic energy. (2) Space does not flow like a river into a black hole. Objects flow toward it, or in their reference frame, the black hole moves toward them. (3) Whether or not the authors believe the "river bed" has physical meaning, it does represent a global Minkowski Reference frame comoving with the black hole. And like I've been saying earlier, although the black hole prevents objects from maintaining straight lines in that reference frame, nothing prevents us from imagining the straight lines, and superimposing those straight lines in the physical space. (4) Using the negative escape velocity as the "speed of space" certainly does not form a Lorentz Reference frame.

Show me a physical object that doesn't have to follow all of the curves.
Yes. If you disagree, again, show me a physical object that doesn't have to stay within the curved manifold. The distant bodies do affect the observations indirectly, by affecting the curvature, but we can measure the curvature without even knowing the distant bodies are there, as long as we measure the curvature over a region that's not too small for our measurement accuracy.

Again, here is what I was talking about with the "tyranny of physical observables" While all physical objects (that we know of) will follow the geodesic curves, we can also describe the curve that would exist if a certain mass were not there. We can draw a line on a page that is straighter than any geodesic. We can envision a path through space that is straighter than any geodesic. When we KNOW what a straight line looks like, and we KNOW those straight lines can be mapped right into the real space.

Velocity relative to what?
Relative to what?

My velocity is changing relative to everything else in the universe. There is another way of looking at it. From my perspective, I did not change velocity, but everything else in the universe rotated and changed velocity relative to me. In fact, that is precisely what the Lorentz Transformation and rotation transformations do. They take local parameters( my change in angle and speed) and transform the coordinates of every object (event) in the universe. Effectively, as I circle around the planet, there are two ways of looking at it: Either (a) I am continually changing velocity, or (b) I am continually changing my momentarily comoving reference frame. I actually think that the second way may be a more constructive way of looking at it.

Relative to what? You keep saying all these things when you have basically admitted that there is *no* physical observable to refer them to. It just "looks to you" like one set of lines is straighter than another. Whatever this is, it isn't physics.
I suppose I should note here, once again, that we are talking about paths in *spacetime*, not space. But you can also "draw a tangent line" in spacetime, so I'm ok with what you are visualizing here at this point. However:

If you can see that the tangent lines of any given geodesic actually have a meaningful distant meaning, and are NOT just "abstract vector spaces" existing only "at" a point, then you are well on your way to appreciating the overlying flat geometry.

And you will find that the lines are *not* "straight", in the sense that they will violate the theorems of Euclidean geometry--specifically, the ones that depend on the parallel postulate. The metric in these coordinates will *not* be the Euclidean (in space) or Minkowski (in spacetime) metric.

Well, let's get specific, and logical, about what I mean by that. I'm saying the overlying geometry represents "What would happen if the masses weren't there" I think we all agree that if no masses were present in the universe, we have a Minkowski metric. But the point is, we can construct a Minkowski metric (or at least we can get closer and closer to it) simply by imagining the tangents to the geodesics.

You appear to believe that the metric doesn't matter: that you can project the curved manifold into a flat manifold, like projecting the Earth's surface onto a flat projection like a Mercator projection, distorting all the distances (and times, in spacetime), and everything will still be just fine. You are simply ignoring actual physical observables when they don't match up with your mathematical scheme. Whatever this is, it isn't physics.

(1) No, I believe the metric does matter. (2) Yes, I do believe you can project the curved manifold onto a flat manifold AND vice versa; you can project the flat manifold onto the curved manifold, not unlike a mercator projection, distorting all the distances and times--essentially that is what the Schwarzschild metric does. dtau and dt are NOT the same. ds and dr are not the same. (3) No, I am not ignoring actual physical observables; I am merely trying to place the physical observables in a global context. (4) You HAVE to deal with the flat coordinates because that's where we can make comparisons. If we actually existed in the curved coordinates, then the clock rates in the valley and the clock rates on the mountain would be the same.

I am not saying any of the things you think I am saying. I am saying this: in GR, the theory of *physics*, we *define* "straight" motion as freely falling motion. That is our *physical* standard for mapping out the geometry of spacetime. When we look at the behavior of freely falling geodesic worldlines, we find that the geometry they map out is intrinsically curved, and we have an equation, the Einstein Field Equation, that relates that curvature to the presence of matter-energy. In other words, what we call "gravity" is a manifestation of spacetime curvature, *in this theory of physics*.

If you want to put together an alternate theory of physics that uses a flat background and goes through all the contortions to reproduce the predictions of GR, go ahead. But there's no point in wrangling over the meanings of words. If you don't like putting the labels "flat" or "curved" where I am putting them, fine, substitute your own labels. The physics remains the same. You can't change the physics by asserting that a certain set of coordinates is "flat" because you say so, even though you have to distort all the distances and times to make things fit into those coordinates, so that your model loses all connection to what we actually observe.

Okay, other words besides flat that could be substituted include "Euclidean" "Cartesian" or "Minkowski." The point is that if you don't acknowledge an overlying flat coordinate system, then how can you have an underlying curved coordinate system? If you can imagine a tangent-line parallel to the curve of a geodesic at a point, that tangent line should go on forever. It should not be a "tangent space defined only at a point." These tangent lines parallel to the curve of the geodesic are paths that the object WOULD take if the gravitating mass weren't there. When we imagine paths of objects in a space where no matter exists, we are imagining a Minkowski space, where parallel lines remain parallel. The Minkowski space is an overlying coordinate system. And specifically, the Schwarzschild metric (t,r,theta,phi) are spherical coordinates in that global flat Minkowski Euclidean space. They are the coordinates that represent straight lines geodesics would follow if the central mass were not there.

Where have I said the rock won't hit the ground? You are reading an awful lot into what I'm saying, that I haven't said. I've never said the rock isn't moving relative to the ground.

Directly above in this very post, you told me that I was not allowed to use a clock sitting on the floor and a clock sitting on a table to describe two objects that maintained a constant distance r. You said you "left out a key word 'freely-falling.'" The ground does not follow a geodesic, nor do objects sitting on the ground.

And you would substitute what? Apparently a bunch of unobservables that throw away the observables.

No, don't throw away the observables, but put them in a global context. I go upstairs and look at the clock and see that it is going at the same rate as my watch, I look at the height and it's the same length as my tie.. Then I go downstairs and see that the clock and my watch are going at the same rate, and its height is the same length as my tie. However, when I look at both clocks at the same time, I find that the downstairs clock is going slower than the upper clock, and the lower clock is actually shorter than the upper clock. You can't get the entire picture of what is going on without looking at nonlocal observables. There is a global reference frame where the upper clock is actually taller and faster than the lower clock. It's not a meaningless set of Greek and Latin letters, but a global flat coordinate system.

As the ultimate *basis* for distances and times, yes. Obviously you can build accounts of distances and times over extended regions on this basis. If I want to know how much proper time will elapse along the Earth's worldline from now until Christmas, I can calculate it based on local clock and ruler readings.

And it makes for pretty good data why? Because the spacetime around Earth is ALMOST flat.

No, I claim the rock's path is not curved because it feels no force. Again, that is a *definition* I adopt because I am using a particular theory of physics. You are free to adopt a different definition, as long as you contort your model appropriately. But you'll have to expect some skepticism when you explicitly state that the distances and times we actually measure are not the "real" ones.

I am trying to give definitions and draw distinctions. I already pointed out that at each point in space, there are two important vector spaces; one made up of lengths and times of the physical observables. And another made up of distance and direction of the overlying flat coordinate space. However, you are arguing that there is only one vector space. Once you acknowledge that there are two vector spaces, then you can also draw the distinctions that I am making, and make sense of what I'm saying.

Really? Give some examples, please. I certainly agree there are ways of obtaining observables for distant objects without stretching rulers from here to there. But to relate any of those observables to a "distance" requires a theoretical model that gives you the relationship. The theoretical relationships used in actual cosmology are based on GR, so they are based on distances that "follow every curve" of the manifold.

Well, the fact is that we don't see very many situations where light is significantly curved. If there is a big enough mass between us and the distant star to deflect the light, then either the star is eclipsed, by the mass in front of it, or there is a visible gravitational lensing effect. Yes, the light coming from there to here does follow every curve in the manifold, but throughout the region in between, the manifold is ALMOST flat, and when I say flat, that is in comparison to the overlying Minkowski coordinates. Only in regions of intense gravitational fields, in regions where the light would be eclipsed anyway, do you find a significant variation from flatness.

 

[PeterDonis]

You asked how, in a curved spacetime with gravity present, two worldlines with the same constant θ and φ, and values of r that differ by a small amount dr, at a given instant of coordinate time t = 0, will continue to remain separated by that same amount dr for all times t. I gave the correct answer, and apparently you knew the correct answer. But you are also right; you cannot have freely falling objects maintain the same constant dr for all times t. That's the essential flaw of using geodesics as a coordinate system.

No, that's the essential flaw in assuming that SR is globally valid, i.e., that spacetime is flat. It isn't flat. If we don't have agreement on this point, then any further discussion is useless unless you can give an actual *physical* observable that picks out your "flat" lines in spacetime instead of the "curved" ones. I emphasize that this isn't just about geodesics; there are no "flat" lines in your sense in a curved spacetime, geodesic or otherwise. None. If you think there are, show me one. It doesn't have to be a geodesic; any "flat" line will do.

So it appears that where we disagree is that I am claiming that the (t, r, θ, φ) in the Schwarzschild metric have exactly the same global geometric meaning as the (t, r, θ, φ) in the Minkowski spacetime. Right, there is no physical observable that picks it out. But we don't need a physical observable. Regarding straight lines, one should NOT ask "What direction does the light ray actually take around the planet." but rather one SHOULD ask, "What direction would the ray have gone if the mass had not been there." There is a clear and unambiguous answer. Sure, it is an odd way of saying it. But if you imagine the path of a satellite if the Earth did not exist, you are essentially imagining a line that exists "outside the universe." There is still no difficulty imagining this line and superimposing it inside our universe.

See further comments below on the "river model". Also, even if there is a (strained) way of doing this in one particular spacetime, that's a far cry from having a way of doing it in *any* spacetime. Finally, as I've said before, and as I point out again below, even if you can construct such a "flat background", when you project the actual, physical, curved spacetime into it, the metric gets distorted, so the laws of physics, as projected into the flat background, are *not* the laws of SR. So even where what you say is possible, it doesn't mean what you think it means.

(2) Space does not flow like a river into a black hole. Objects flow toward it, or in their reference frame, the black hole moves toward them.

In one model, yes. The river model is another model. Since both models make exactly the same physical predictions, they are physically equivalent.

(3) Whether or not the authors believe the "river bed" has physical meaning, it does represent a global Minkowski Reference frame comoving with the black hole.

A "frame" that nevertheless violates the laws of SR, because objects inside the EH move inward faster than light in this "frame".

And like I've been saying earlier, although the black hole prevents objects from maintaining straight lines in that reference frame, nothing prevents us from imagining the straight lines, and superimposing those straight lines in the physical space.

And distorting the metric so that the laws of physics in the "flat background" are not the laws of SR. I just gave one example of that above.

(4) Using the negative escape velocity as the "speed of space" certainly does not form a Lorentz Reference frame.

Yes, exactly. The "flat background" is *not* a global Lorentz Reference frame. In a curved spacetime there is no such thing. There just isn't.

Again, here is what I was talking about with the "tyranny of physical observables" While all physical objects (that we know of) will follow the geodesic curves, we can also describe the curve that would exist if a certain mass were not there. We can draw a line on a page that is straighter than any geodesic. We can envision a path through space that is straighter than any geodesic. When we KNOW what a straight line looks like, and we KNOW those straight lines can be mapped right into the real space.

Only by distorting the metric and thereby distorting the physical laws in this "flat" background. You can complain all you want about a "tyranny of physical observables", but you do remember that we are discussing physics here, right? It's a little weird to come into a physics forum and complain that people want theories to match physical observables.

My velocity is changing relative to everything else in the universe. There is another way of looking at it. From my perspective, I did not change velocity, but everything else in the universe rotated and changed velocity relative to me.

Sure, you can always construct your own "local" reference frame around any event on your worldline. It will only cover a small local patch of spacetime around that event, though. More precisely, the coordinate values in that frame will only obey the laws of SR, to some given accuracy of measurement, within a small local patch around that event.

In fact, that is precisely what the Lorentz Transformation and rotation transformations do.

In that small local patch, within the given accuracy of measurement, yes. Outside that small local patch, no. You can certainly try to extend your local coordinate patch, but you will find that the coordinate values no longer work right--they no longer predict the correct observational results--when you plug them into the SR formulas. The Lorentz Transformation is just one example of that. You can, if you insist, work out what changes in coordinate values a Lorentz Transformation in your local patch around a given event induces in the region outside that patch, but again, you will find that those changes don't obey the laws of SR.

They take local parameters( my change in angle and speed) and transform the coordinates of every object (event) in the universe. Effectively, as I circle around the planet, there are two ways of looking at it: Either (a) I am continually changing velocity, or (b) I am continually changing my momentarily comoving reference frame. I actually think that the second way may be a more constructive way of looking at it.

Yes, I would agree. And if you are in free fall, and yet you are continually changing your MCIF, then you are not obeying the laws of SR. The laws of SR require that an object in free fall, feeling no force, can never change its MCIF.

If you can see that the tangent lines of any given geodesic actually have a meaningful distant meaning, and are NOT just "abstract vector spaces" existing only "at" a point, then you are well on your way to appreciating the overlying flat geometry.

Sorry, this train never left the station, because its driver (you) fails to see that there is no unique way to define the "same" tangent vector at different events in a curved manifold. If I have a tangent vector T at event A, and I want to compare it with a tangent vector V at event B, I have to first "transport" T to event B (or V to event A, but I'll stick with moving T for this example). But what vector at B I end up with when I transport T from A to B depends on the path I take between those events; that's one way of expressing what it means for a manifold to be curved. So there is no unique way to compare T and V, because there is no unique way to transport either one to the other's "location" to compare them.

I won't comment right now on the rest of your post because I think the above captures the fundamental point at issue. I don't think we disagree on the actual math (except that the last point I raised, about the non-uniqueness of transporting vectors from event to event, is very important, so if you're not sure about it, we should resolve it), or the actual physical observables; I think the disagreement is about "interpretation" of an aspect of the math that you think is very important physically, but I think is irrelevant physically, though it can help sometimes with visualization.

 

[JDoolin]

I think we are agreed then, on the fundamental point of issue. So let me make that point once more, hopefully with a little more clarity. As concisely as possible, my point is that the Minkowski metric and the Schwarzschild metric overlap. The Schwarzschild metric describes physical observables as they really are (with the central gravitating body actually present), and the Minkowski metric describes how things WOULD BE (if the central gravitating body weren't present).

More precisely still, if I have a world-curve of an accelerating clock defined parametrically (t,r(t), θ(t), φ(t)) I would be able to find the age of that clock by integrating the metric ds^2 =- \left ( 1 - \frac{2 G M}{c^2 r} \right )c^2 dt^2 + \left ( 1-\frac{2 G M}{c^2 r} \right )^{-1} dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2) over that curve.

I could also find how old that clock WOULD have been without the central gravitating body following the same curve (t,r(t), θ(t), φ(t)) but integrating the metric ds^2 =- c^2 dt^2 + dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2) This is also the metric you would get simply by setting M=0 in the Schwarzschild metric.

We have these two different metrics; one answering the question of "what would the observable be if the mass weren't there" and one answering "what is the observable with the mass there." They overlap. The coordinates (t,r(t), θ(t), φ(t)) don't change their locations, it is only the question of whether you have a central mass.

Whether the mass is there or not, we can define a straight lines with "priveleged" status, being the path the object would have taken if the mass weren't there. That is not an ambiguous direction in Schwarzschild, but it is well defined whether the central mass is there or not.

 

[Passionflower]

You are comparing apples and oranges as the geodesics of the two metrics are completely different.

 

[PeterDonis]

I think we are agreed then, on the fundamental point of issue. So let me make that point once more, hopefully with a little more clarity. As concisely as possible, my point is that the Minkowski metric and the Schwarzschild metric overlap. The Schwarzschild metric describes physical observables as they really are (with the central gravitating body actually present), and the Minkowski metric describes how things WOULD BE (if the central gravitating body weren't present).

Um, no, we don't agree on this point. We appear to agree on the actual physical observables, since you say the Schwarzschild metric describes them. But we do *not* agree that the Minkowski metric describes "how things WOULD BE" as you appear to be using the term; IMO that statement has no physical meaning at all. I would say that the Minkowski metric describes the physical observables in a *different spacetime*, a flat one. See further comments below.

More precisely still, if I have a world-curve of an accelerating clock defined parametrically (t,r(t), θ(t), φ(t)) I would be able to find the age of that clock by integrating the metric ds^2 =- \left ( 1 - \frac{2 G M}{c^2 r} \right )c^2 dt^2 + \left ( 1-\frac{2 G M}{c^2 r} \right )^{-1} dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2) over that curve.

Ok so far.

I could also find how old that clock WOULD have been without the central gravitating body following the same curve (t,r(t), θ(t), φ(t)) but integrating the metric ds^2 =- c^2 dt^2 + dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)

Mathematically, yes, I see what you are doing. You are constructing a distorted version of the spacetime where the same events have the same coordinate labels but the metric is different. Again, mathematically you can do this (at least outside the horizon--see just below), but physically it has no meaning. You can capture all the physics in the actual, curved spacetime without bringing this extra math in at all. Also, there is no physical justification for saying that the same 4-tuple of coordinate values labels "the same event" in both cases, since the metric is changed. See further comments below.

It's worth noting that your correspondence only works above the horizon. At r = 2M the Schwarzschild coordinates are singular so you can't label points on the horizon using these coordinates. And inside the horizon, if you were to set up Schwarzschild coordinates, the "age of the clock" you would calculate by your formula above would be imaginary. (More precisely, the squared interval ds^2 would be positive--spacelike--instead of negative--timelike--for a set of parametrized coordinate values where the "Minkowski" squared interval was negative--timelike.)

It's also worth noting that Schwarzschild spacetime is just one of many curved spacetimes in GR. Your "flat correspondence" scheme would not work in most of the others; in FRW spacetime, for example. Even in Kerr spacetime outside the horizon I'm not sure it would work.

We have these two different metrics; one answering the question of "what would the observable be if the mass weren't there" and one answering "what is the observable with the mass there." They overlap. The coordinates (t,r(t), θ(t), φ(t)) don't change their locations, it is only the question of whether you have a central mass.

No, you can't say that events labeled by the same 4-tuple of coordinate values "don't change their locations", because in the actual, physical spacetime with the central mass present, the metric *changes*, therefore the "locations" of events labeled by the same 4-tuple of coordinates change, since "location" is defined by the metric. You can calculate your "Minkowski" interval, but you can't just call that the "location" of a point by fiat; at least not if you want to do physics. Physically, "location" is determined by the metric.

Whether the mass is there or not, we can define a straight lines with "priveleged" status, being the path the object would have taken if the mass weren't there. That is not an ambiguous direction in Schwarzschild, but it is well defined whether the central mass is there or not.

Mathematically, you can "define" straight lines like this, but the definition has no physical meaning. Again, all the physics is captured by the actual, curved metric. There is *no* additional physics being added by your definition of "straight" lines. So physically, in the actual curved spacetime, those lines have no meaning.

If you disagree, then please tell me what actual physics is left out by the curved spacetime description, but is included in the "flat" description. That should be interesting since you've already agreed that the Schwarzschild metric describes all the actual physical observables.

If, on the other hand, you agree that the actual curved metric captures all the physical observables, then you can talk all you want about your mathematically defined "straight" lines, and I can simply ignore all your talk and compute physical observables using the actual curved metric, just as I always have. So why are you bothering to talk about them? Why should anyone else even pay attention?

(Btw, it's not that I don't see how you are picking the "straight lines" out; I do. You are saying that lines of constant r, theta, phi in Schwarzschild coordinates are "straight lines", even though they are not geodesics--objects following these worldlines are accelerated. But physically, nothing is added to my understanding of these lines by picking them out as the "straight" ones in this way; indeed, adding this definition *complicates* matters, since now I have to remember that your "straight lines" are not geodesics, and furthermore that the proper acceleration along one of your "straight lines" *varies* with radial coordinate r. I don't have to deal with any of this in the standard GR picture. Furthermore, I don't even need your criterion--pick out the lines that "would be straight" with the same coordinate values if the metric were Minkowski--to pick out the lines of constant r, theta, phi in Schwarzschild coordinates as being "special"; they are already picked out by the time translation symmetry of the spacetime with the *curved* metric, again without having to introduce your "flat" metric at all. So again, there is *no* physics that is added by your description.)

 

[DrGreg]

As concisely as possible, my point is that the Minkowski metric and the Schwarzschild metric overlap. The Schwarzschild metric describes physical observables as they really are (with the central gravitating body actually present), and the Minkowski metric describes how things WOULD BE (if the central gravitating body weren't present).

Whether the mass is there or not, we can define a straight lines with "priveleged" status, being the path the object would have taken if the mass weren't there.

Instead of considering 4D spacetime, let's consider a 2D analogy, the curved surface of the Earth. For simplicity, let's assume the Earth is a perfect sphere, and choose units to make the radius of the Earth equal 1. The metric for the Earth's surface in terms of latitude and longitude is<br /> ds^2 = d\theta^2 + \sin^2 \theta \,\, d\phi^2<br />Would you be happy for me to say that I can also use the same latitude and longitude to describe what the Earth would be like if it were flat, with a metric<br /> ds^2 = d\theta^2 + d\phi^2 \,\, \mbox{?}<br />Would you be happy to say that the equation<br /> \theta = A\phi + B<br />describes a geodesic on the flat Earth and therefore the same equation defines a "straight line with 'privileged' status" on the surface of the curved Earth?

 

[JDoolin]

Peter, it seems to me we are interpreting the meaning of the Schwarzschild metric differently. Whereas I see the coordinates remaining in place as the central mass becomes bigger, you see them moving around somehow to adjust to the "physical observables.

Let me ask a hypothetical question that may further clarify the difference between my interpretation and yours.

Let's say we have a nonrotating planet with a high gravity. And lots and lots of identical meter-sticks. Tall buildings are constructed on the planet that sretch from the ground all the way out into space, and the meter-sticks are placed end to end from the ground up.

Now a photograph is taken from a long distance away; so far away that the small-angle formula applies, i.e. theta=sin(theta)=tan(theta) (ask me if you don't know what I mean.)

Now the meter-sticks measure physical observables, but in the photograph, will it appear that each meter-stick is the same length, or will the meter-sticks near the surface of the planet appear shorter than the meter-sticks further away, due to the effects of the Schwarzschild metric?


The way I interpret the metric, the meter sticks close to the planet will appear shorter in the photograph, even though each meter-stick is "locally" identical. They all measure off the same "physically observable" one meter. However, every observer in the system will agree (if they make careful eough measurements by sight) that the meter-sticks closer to the planet are indeed shorter than the meter-sticks above.

(Conversely, I think we both agree that if an observer climbs up and down the towers, physically checking the length of each meter stick with one he carries, he will find that all of the meter-sticks are the same length, it's just when he makes the comparison by sight. i.e. nonlocal observation of the metric, where the difference in lengths can be observed.)

It seems to me that your interpretation is different; am I right?

 

[PeterDonis]

Peter, it seems to me we are interpreting the meaning of the Schwarzschild metric differently. Whereas I see the coordinates remaining in place as the central mass becomes bigger, you see them moving around somehow to adjust to the "physical observables.

What does "the coordinates remaining in place" even mean? Coordinates are numbers used to label events. If you mean "an event labeled by a given 4-tuple of coordinates remains in place", what does *that* even mean? The only physical interpretation I can give this is that the "location" of an event labeled by a given 4-tuple is "the same" whether a gravitating body is there or not, and since "location" is determined by the metric, this is obviously false; or perhaps a better way to say it would be that any interpretation of the term "location" that has the "location" of an event labeled by a given 4-tuple being "the same" when the metric changes is not a good interpretation for doing physics. You could salvage it somewhat by steadfastly refusing to draw *any* deductions that you would normally draw from the phrase "the location is the same"--for example, you could refuse to deduce that two "locations" which remain the same will be the same physical distance apart (since they won't if the metric changes). But I don't see the point; the usefulness of the term "location" is completely thrown away, so why use it at all?

Let's say we have a nonrotating planet with a high gravity. And lots and lots of identical meter-sticks. Tall buildings are constructed on the planet that sretch from the ground all the way out into space, and the meter-sticks are placed end to end from the ground up.

Now a photograph is taken from a long distance away; so far away that the small-angle formula applies, i.e. theta=sin(theta)=tan(theta) (ask me if you don't know what I mean.)

Now the meter-sticks measure physical observables, but in the photograph, will it appear that each meter-stick is the same length, or will the meter-sticks near the surface of the planet appear shorter than the meter-sticks further away, due to the effects of the Schwarzschild metric?

There was a *long* thread about this recently. The thread started with a different question, but the "meaning" of the Schwarzschild metric as regards radial measures of length came up, starting more or less with this post:

https://www.physicsforums.com/showpost.php?p=3557020&postcount=138

The way I interpret the metric, the meter sticks close to the planet will appear shorter in the photograph, even though each meter-stick is "locally" identical. They all measure off the same "physically observable" one meter. However, every observer in the system will agree (if they make careful eough measurements by sight) that the meter-sticks closer to the planet are indeed shorter than the meter-sticks above.

No, they will agree that the meter sticks *look* shorter from far away. But locally, each meter stick still measures one meter; if I am far away from the planet, and I have a meter stick with me, which checks out as being exactly the same length as all other meter sticks, and I take it down to the surface of the planet, it will be the same length as the meter sticks there; I can put it side by side with any meter stick and they will exactly overlap. Or, I can take a meter stick from the surface of the planet, that "looks shorter" as seen from far away, put it in my rocket and bring it far away to the location of the telescope through which that meter stick "looked shorter", and it will be the same length as all the other meter sticks in the vicinity of the telescope.

You know this, because you follow up immediately with:

(Conversely, I think we both agree that if an observer climbs up and down the towers, physically checking the length of each meter stick with one he carries, he will find that all of the meter-sticks are the same length, it's just when he makes the comparison by sight. i.e. nonlocal observation of the metric, where the difference in lengths can be observed.)

I agree with this; as I said, I believe we are in agreement on all of the actual physical observables. The only thing I object to is your contention that, because the meter sticks look shorter from far away if they are closer to the planet, they somehow *are* shorter, in some sense. They aren't, because, as you agree, you can take a meter stick anywhere you like and it will be exactly the same length as all the other meter sticks in its vicinity. So the fact that they "look" shorter from far away is an optical illusion, so to speak; it is not caused by any "real" change in length, it is caused by the way gravity bends light rays.

 

[JDoolin]

What does "the coordinates remaining in place" even mean? Coordinates are numbers used to label events. If you mean "an event labeled by a given 4-tuple of coordinates remains in place", what does *that* even mean? The only physical interpretation I can give this is that the "location" of an event labeled by a given 4-tuple is "the same" whether a gravitating body is there or not, and since "location" is determined by the metric, this is obviously false; or perhaps a better way to say it would be that any interpretation of the term "location" that has the "location" of an event labeled by a given 4-tuple being "the same" when the metric changes is not a good interpretation for doing physics. You could salvage it somewhat by steadfastly refusing to draw *any* deductions that you would normally draw from the phrase "the location is the same"--for example, you could refuse to deduce that two "locations" which remain the same will be the same physical distance apart (since they won't if the metric changes). But I don't see the point; the usefulness of the term "location" is completely thrown away, so why use it at all?

Is this another major point of disagreement, then?

In my thinking, the metric (being ∫ds or ∫dτ along a specific spacetime path) is an observer independent quantity which determines how many meter-sticks fit between two specific events along a specific path, or how many ticks on a clock following a specific path between two events.

But just knowing the length of a path does not convey a sense of its location. The location of the path is represented by its coordinates; not it's length.

I can easily see how changing the mass changes the metric, but I can't see how you can claim that changing the mass changes the coordinates. And of course, you can't see how I can claim that changing the mass WOULDN'T change the coordinates. So this issue may be a stalemate.

Okay, so what DO we agree on here. The Schwarzschild metric gives us a RELATIONSHIP between the metric and the coordinates. My central theme is that both the coordinates and the metric are concretely definable constructs. The metric is the actual measurement, while the coordinates are "what the measurements of distance and time would be (referenced from the location of the central mass) if the central mass were zero."

Another idea would be to say that the metric is concretely definable, but that the coordinates are NOT concretely definable. But that begs the question, if the coordinates are meaningless, or not concretely definable, and the Schwarzschild metric gives you the RELATIONSHIP between the neaningFUL metric and the meaningLESS coordinates, wouldn't that mean that the Schwarzschild metric itself is meaningless?

There was a *long* thread about this recently. The thread started with a different question, but the "meaning" of the Schwarzschild metric as regards radial measures of length came up, starting more or less with this post:

https://www.physicsforums.com/showpost.php?p=3557020&postcount=138No, they will agree that the meter sticks *look* shorter from far away. But locally, each meter stick still measures one meter; if I am far away from the planet, and I have a meter stick with me, which checks out as being exactly the same length as all other meter sticks, and I take it down to the surface of the planet, it will be the same length as the meter sticks there; I can put it side by side with any meter stick and they will exactly overlap. Or, I can take a meter stick from the surface of the planet, that "looks shorter" as seen from far away, put it in my rocket and bring it far away to the location of the telescope through which that meter stick "looked shorter", and it will be the same length as all the other meter sticks in the vicinity of the telescope.

You know this, because you follow up immediately with:
I agree with this; as I said, I believe we are in agreement on all of the actual physical observables. The only thing I object to is your contention that, because the meter sticks look shorter from far away if they are closer to the planet, they somehow *are* shorter, in some sense. They aren't, because, as you agree, you can take a meter stick anywhere you like and it will be exactly the same length as all the other meter sticks in its vicinity. So the fact that they "look" shorter from far away is an optical illusion, so to speak; it is not caused by any "real" change in length, it is caused by the way gravity bends light rays.

You're saying that the meter-sticks "look" shorter, but that is an optical illusion caused by the way gravity bends light rays. I agree that there is going to be an additional optical effect caused by the bending of light rays, but the Schwarzschild metric itself does not include this optical effect.

I generated an image to give further explanation to what I am saying.

The innermost circle has radius r=1.1

The circles beyond that are drawn so that \int ds = \int_{1.1}^{r_f}\frac{1}{\sqrt{1-\frac{1}{r}}}dr evaluates to 0.1, 0.2, 0.3, 0.4, ... 6.1, 6.2, 6.3. i.e. the circles are drawn so that the radial distance between each has the same "physical observable" distance.

The axes give values for r=2, r=4, r=6, where r=1 represents 1 schwarzschild radius. The figure is drawn so that the innermost circle is drawn at 1.1 Schwarzschild radius, staying well clear of the singularity.

From the diagram, it is clear that length "AB" looks shorter than length "CD" But I feel it is not just a matter of some optical illusion. From a global point-of-view, the length "AB" really is shorter than length "CD"

Also, from a global point of view, the clocks at A would be going much slower than the clocks at D. The time-slowing effect may be a stronger argument that this is no illusion. A person may climb up to D, and verify that his pocketwatch is going at the same rate as the clock at D. Then he goes down to A and verifies that his pocketwatch is going at the same rate as the one at A. However, as he continues to do this, he will find that when he goes above, the time on his clock really gets ahead of the one below. And/or when he goes below, his clock really gets behind the one above. It's a slightly different phenomenon than with the rulers, which don't have a memory, but with the clocks, they will lose their synchronization, so you can measure locally that there is a difference in the rates of time "globally."

There should still be an additional distortion effect on top of this one, whereby the image is even further distorted, caused by the bending of light rays on their way toward the camera. This distortion is calculatable based on where the camera is positioned.

 

[JDoolin]

There are some great things you can do with that dτ; in particular, setting it equal to zero, to see what the path of a photon through your curved space is. But that path is through the t, r, θ, and φ coordinates. Those coordinates are not meaningless. They are the overlying GLOBAL LORENTZ FRAME COORDINATES IN SPHERICAL FORM, COMOVING WITH THE CENTRAL MASS.

If you really believe this, then please demonstrate how, in a curved spacetime with gravity present, two worldlines with the same constant θ and φ, and values of r that differ by a small amount dr, at a given instant of coordinate time t = 0, will continue to remain separated by that same amount dr for all times t. That is what is physically required for your global Lorentz frame coordinates to be valid. If this requirement is violated (which it is in the presence of gravity), your global Lorentz frame coordinates *will not work* the way you are claiming they do. You can *assign* such coordinates as arbitrary labels, but they will not support the claims about the physics that you appear to be making.

This is obvious. Surely you already know the answer I would give: Upon a non-rotating gravitating sphere, place one object on the floor at radius r. Place another object on a table directly above it at radius r+dr. The two objects will remain on the floor and on the table, maintaining their positions in r, theta, phi, but moving through time.

Ah, I see I left out a key word: "freely falling". Neither of the objects in question are freely falling.

You asked how, in a curved spacetime with gravity present, two worldlines with the same constant θ and φ, and values of r that differ by a small amount dr, at a given instant of coordinate time t = 0, will continue to remain separated by that same amount dr for all times t. I gave the correct answer, and apparently you knew the correct answer. But you are also right; you cannot have freely falling objects maintain the same constant dr for all times t. That's the essential flaw of using geodesics as a coordinate system.

No, that's the essential flaw in assuming that SR is globally valid, i.e., that spacetime is flat. It isn't flat. If we don't have agreement on this point, then any further discussion is useless unless you can give an actual *physical* observable that picks out your "flat" lines in spacetime instead of the "curved" ones. I emphasize that this isn't just about geodesics; there are no "flat" lines in your sense in a curved spacetime, geodesic or otherwise. None. If you think there are, show me one. It doesn't have to be a geodesic; any "flat" line will do.

I thought this particular thread of the argument was interesting. I have been claiming all along that geodesics are NOT straight lines.

But when I offered an example of an actual straight line, you rejected it because it wasn't a geodesic.

Do you have any better argument that the lines of constant r, theta, phi are not straight?

 

[PeterDonis]

In my thinking, the metric (being ∫ds or ∫dτ along a specific spacetime path) is an observer independent quantity which determines how many meter-sticks fit between two specific events along a specific path, or how many ticks on a clock following a specific path between two events.

No problem here.

But just knowing the length of a path does not convey a sense of its location. The location of the path is represented by its coordinates; not it's length.

If you choose to define "location" this way. "Location" is just a word. All the *physics* is conveyed by the statement about the integral of the metric along a specific path. Given the physics, I don't think the definition of "location" you appear to be using is a very useful one; but either way it doesn't change the physics.

I can easily see how changing the mass changes the metric, but I can't see how you can claim that changing the mass changes the coordinates. And of course, you can't see how I can claim that changing the mass WOULDN'T change the coordinates. So this issue may be a stalemate.

I don't say that changing the mass changes the coordinates, or that it doesn't. I say that the whole idea of "changing the coordinates" or not, when you're talking about two different spacetimes, is physically meaningless; it's just words.

The coordinates are just numbers used to label events. When you have a mass present, you have a different spacetime, a different geometry, and a different set of events. Saying that an event in the curved spacetime with a given (t, r, theta, phi) is "the same event" as an event in the flat spacetime with those coordinate values is just words; it has no physical meaning. Setting up such a correspondence may help someone (such as you) to visualize what is going on when the mass is present, but that's not the same as it having an actual physical meaning.

The metric is the actual measurement, while the coordinates are "what the measurements of distance and time would be (referenced from the location of the central mass) if the central mass were zero."

And that statement has no physical meaning. It may help you with visualization, but all the physics can be captured without taking it into account at all.

Another idea would be to say that the metric is concretely definable, but that the coordinates are NOT concretely definable. But that begs the question, if the coordinates are meaningless, or not concretely definable, and the Schwarzschild metric gives you the RELATIONSHIP between the neaningFUL metric and the meaningLESS coordinates, wouldn't that mean that the Schwarzschild metric itself is meaningless?

I didn't say the coordinates are meaningless. I said they are arbitrary numbers used to label events. The metric tells you how the coordinate labeling expresses the physics. Pick a different coordinate labeling, and the metric looks different. Look up Painleve coordinates, or Eddington-Finkelstein coordinates, or Kruskal coordinates; all of them describe the *same* geometry, the *same* physics, as Schwarzschild coordinates, they just label events differently, so the metric looks different. Does that mean they're "meaningless"? Does the fact that Cartesian and polar coordinates on a plane label points differently mean those are "meaningless"? The metric looks different for those two cases too, but both describe the same geometry (a flat Euclidean plane).

You're saying that the meter-sticks "look" shorter, but that is an optical illusion caused by the way gravity bends light rays. I agree that there is going to be an additional optical effect caused by the bending of light rays, but the Schwarzschild metric itself does not include this optical effect.

Yes, it does, because the metric describes *all* curves, not just timelike or spacelike ones; it describes null curves too. To find out how a meter stick appears to someone far away, you figure out the paths that light rays follow from the meter stick to the observer. How do you do that? By using the metric to figure out what paths null curves (curves with ds = 0) will follow.

I thought this particular thread of the argument was interesting. I have been claiming all along that geodesics are NOT straight lines.

But when I offered an example of an actual straight line, you rejected it because it wasn't a geodesic.

Do you have any better argument that the lines of constant r, theta, phi are not straight?

The word "straight" has different possible definitions. I was pointing out that according to the definition which is used in GR, geodesics are straight and accelerated worldlines are not. But I was also pointing out that, even according to a *Euclidean* (or "Minkowskian", since we're talking about spacetime) definition of "straight", lines of constant r, theta, phi are not straight in Schwarzschild coordinates. If those lines were straight by the Minkowskian definition of "straight", then objects following them would have to be in free fall, since the lines remain the same distance apart for all time. But objects following those lines are not in free fall. So the lines are not "straight" according to any definition of the word "straight" that is usual in relativity physics.

You are obviously using a different definition of "straight". That's a matter of words, not physics; the physics is the same regardless of which lines we call "straight", just as it's the same regardless of how we define a "location" (see above). I happen to think your choice of words is not a very useful one for studying the physics we're talking about, but that doesn't make your choice of words "wrong", just not very useful IMO. I don't really care about the words; I care about the physics. If any misunderstanding is suspected based on not using words the same way, we can just switch to math terminology, as you did in the first passage I quoted in this post (about the integral of the metric); that expresses the physics involved clearly and concisely, and that's all that's necessary.