# The Lorentz Transformations and a Few Concerns

1. Oct 28, 2011

### Anamitra

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.

Last edited: Oct 28, 2011
2. Oct 28, 2011

### Bill_K

Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime. Nor do they depend on the presence or absence of material objects or EM or gravitational fields. Lorentz invariance is a local property of vacuum. All the equations of physics are local equations and rely on Lorentz invariance at every point for their consistency.

3. Oct 28, 2011

### Anamitra

Could you provide me with a sample derivation of the Lorentz Transformations in the "LOCAL CONTEXT"?

Hope you don't have to go back to Resnick!

4. Oct 28, 2011

### Anamitra

A Basic Issue to be addressed:

Can the anisotropy of space itself affect clock rates[considered apart from gravity]?

[Even if you consider an infinitesimally small region of space[surrounded by matter] you can have milloins of directions emanating from a point in it,providing enough scope for anisotropy]

Last edited: Oct 28, 2011
5. Oct 29, 2011

### Anamitra

These are the priceless words of the century--from Bill_K

6. Oct 30, 2011

### Fredrik

Staff Emeritus
One way to find the Lorentz transformations is to consider a family of global coordinate systems $\{x_i:\mathbb R^4\rightarrow\mathbb R^4|i\in I\}$ such that each "coordinate change" function $x_i\circ x_j^{-1}$ takes straight lines to straight lines, and the set $\big\{x_i\circ x_j^{-1}|i,j\in I\big\}$ is a group, with a subgroup that's isomorphic to the translation group, and another that's isomorphic to the rotation group.

To drop the requirement of isotropy or homogeneity would be to drop the requirement that the group of coordinate change functions must have the appropriate subgroups. I don't know what we get if you do. I suppose that we don't want to completely drop them, but rather weaken them somewhat, so that the group almost has the translation and rotation groups as subgroups. I don't even know how to make that statement precise.

7. Oct 30, 2011

### Fredrik

Staff Emeritus
Lorentz transformations are permutations of $\mathbb R^4$ that satisfy a few additional conditions. The presence of matter in the actual universe obviously has no effect on $\mathbb R^4$, and therefore no effect on the Lorentz transformations. But you certainly do have to make assumptions of homogeneity and isotropy to be able to "derive" them from Einstein's postulates. (You can't actually derive them from Einstein's postulates. You derive them from mathematical statements that can be thought of as expressing aspects of Einstein's postulates mathematically. Are homogeneity and isotropy such aspects, or are they separate assumptions? I think that's actually a matter of taste. Einstein's postulates aren't very precise, so you can interpret them in more than one way).

Last edited: Oct 30, 2011
8. Oct 30, 2011

### robphy

In Special Relativity, the Lorentz Transformations appear in two places... in spacetime (transforming events or coordinates of events) and in the tangent-spaces associated with each event (transforming tensors or components of tensors).

9. Oct 30, 2011

### JDoolin

Some questions

(1) What is your concept of the universe?
(a) Space-time curvature embedded in an overall flat space-time, or
(b) tiny little differential sections of flat space-time that are approximately flat, embedded in an overall curved universe?​

(2) Where does Lorentz Transformations apply?
(a) Throughout any flat region of spacetime?
(b) Only along a differential path element of space-time.​

(3) Which is the Lorentz Transformation more similar to?
(a) A transformation from spherical to rectangular coordinates:
\begin{align*} x &= r \cos(\phi) \sin(\theta)\\ y &= r \sin(\phi)\sin(\theta)\\ z &= r \cos(\theta) \end{align*}
(b) A rotation
\begin{align*} x' &= x \cos(\theta)+y \sin(\theta) \\ y' &= -x \sin(\theta)+y \cos(\theta) \\ z' &= z \end{align*}

(4) What is the radial limitation on a rotation transformation?
(a) Universal: If I turn my reference frame to the right, then even galaxies billions of light years away will move counterclockwise around me.
(b) Local: Objects in my room will move counterclockwise around me, but galaxies billions of light years away will maintain their positions.​

Last edited: Oct 30, 2011
10. Oct 30, 2011

### JDoolin

I'd like to simplify question #3 to two dimensions, and delve into it more deeply.

Let's look at the polar to rectangular transformation:

\begin{align*} x &= r \cos(\theta)\\ y &= r \sin(\theta) \end{align*}

This is a global, but nonlinear expression. There is no way to express these equations in the form:

$$\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} \square & \square \\ \square & \square \end{pmatrix}\begin{pmatrix} r\\ \theta \end{pmatrix}$$

However, if we take the derivatives,

\begin{align*} dx &= \cos(\theta)dr - r \sin(\theta) d\theta \\ dy &= \sin(\theta)dr + r \cos(\theta) d\theta \end{align*}

this becomes an expression which can be expressed in a form which at least resembles a linear equation:

$$\begin{pmatrix} dx\\ dy \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -r \sin(\theta)\\ \sin(\theta) & r \cos(\theta) \end{pmatrix}\begin{pmatrix} dr\\ d\theta \end{pmatrix}$$

However, since r and θ are functions of x and y, there's not any global linearity. The function is only "almost" linear in a "sufficiently small" region of space. One could say that the r,θ coordinates are almost linear in a small enough region.

This should be contrasted with the rotation transformation, which has no such restrictions about small regions.

\begin{align*} x'&=x \cos(\theta)-y \sin(\theta)\\ y'&=x \sin(\theta) + y\cos(\theta) \end{align*}

Unlike the polar-to-rectangular transformation which is a global but NONlinear transformation, this is a global LINEAR transformation. So when we try to change the form, as follows:

$$\begin{pmatrix} x'\\ y' \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$$

There is no difficulty. And if we take the derivatives:

$$\begin{pmatrix} dx'\\ dy' \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}\begin{pmatrix} dx\\ dy \end{pmatrix}$$

There is also no difficulty.

The key difference here is that in the polar-to-rectangular case, the value of r and θ actually are functions of position. In the rotation case, however, the value of θ is a global parameter that does not vary.

So what I'm trying to convince you of here, is that the rotation transformation is NOT a local transformation but it applies to every particle in the universe. If I turn my head to the left, everything in the universe moves clockwise around my head. If I turn my head to the right, everything in the universe moves counterclockwise around my head. There is no situation where the stuff in my room moves counterclockwise, but the stuff outside my room remains stationary, unless I'm not really rotating but my room is.

There's no way to say that the rotation only applies locally. It is a linear global transformation that affects all particles in the universe. I know some people will laugh at me, saying that my turning my head has no effect on distant particles in the universe, but that is not my argument. I'm not saying that turning my head exerts a force on galaxies billions of light years away causing them to go around in a circle.

When you do a rotation transformation, you are changing the observers' reference frame. And thus everything in the universe moves according to the transformation with respect to the observer.

Finally, the Lorentz Transformation equations are NOT ANALOGOUS to the polar-to-cartesian coordinate transformations. They are analogous to the rotation transformation equations.

They are global linear equations that can be expressed as
\begin{align*} ct'&=ct \cosh(\varphi)-x \sinh(\varphi)\\ x'&=-ct \sinh(\varphi) + x\cosh(\varphi) \end{align*}
and
$$\begin{pmatrix} c t'\\ x' \end{pmatrix}= \begin{pmatrix} \cosh(\varphi) & -\sinh(\varphi)\\ -\sinh(\varphi) & \cosh(\varphi) \end{pmatrix}\begin{pmatrix} c t\\ x \end{pmatrix}$$

The parameter φ is not a function of t or x; it is an independent global parameter, just like θ in the rotation transformation.

I don't know for sure, but it seems to me, these two types of transformations; global but nonlinear, and GLOBALLY LINEAR transformations are confused. The transformation that makes working with spherical coordinates convenient is nonlinear, and on the scale where the coordinates appear linear, it is local.

The Lorentz transformations, on the other hand, have nothing to do with converting from spherical to polar. They have to do with changing velocities; changing rapidity. The Lorentz Transformation equations are a GLOBAL LINEAR TRANSFORMATION, just like rotation.

11. Oct 31, 2011

### PhilDSP

Yes, I have to agree with that. If you analyze what the LT does on the basis of its original intent by Voigt and H. A. Lorentz (Voigt's use of the transformation preceded that of Lorentz by several years) then what you say (and more) becomes even clearer. As applied to produce both a change in form of the wave equation and the proper change of initial conditions to accompany the wave equation, it results in initial conditions which can never be reconciled between adjacent points of time or space unless a stretch or shift of all other points is accomplished. But each time that is done the situation remains recursively the same as if you're chasing something that can never be caught. There is no way to make the operation of the LT local unless you freeze the transformation and step back into a Galilean frame of reference.

Last edited: Oct 31, 2011
12. Nov 1, 2011

### Anamitra

If we look from the reverse direction[instead of the deductive side] the issue is apparently becoming much simpler.
The manifold[described by the metric coefficients] may take care of all anisotropies and inhomogeneities.A metric can include charges[eg:Reissner-Nordstrom metric]. If a metric can include charges it may include strong charges ,spin and other properties of matter with their distributions.

One could think in the direction of a generalized unified metric incorporating all known/conceivable properties of matter.

These should have their independent effects on clock rates and change of other physical dimensions. In fact the temporal part of the Reissner-Nordstrom metric contains charge as well as mass.
[The same is true for the spatial part]

One could just imagine reducing the mass and keeping the charge same[in this metric]. The manner in which the clock rate would be affected by charge alone could be examined theoretically. One could try out similar exercises with other properties like spin etc in other types of metrics.

The distribution[redistribution] of mass will change the value of the metric coefficients. Distribution[redistribution] of chargees etc will also alter the manifold properties taking care of inhomogeneties and anisotropies.At the same time,one must note, that the clock rates and the spatial dimensions will depend on the distribution/redistribution of these properties in respect of space and time

So by writing a general type of a metric one should be able to explain refraction of light.
But these metrics will fail to explain properties like the slowing down of light in a medium.Perhaps inclusion of refractive index in the metric could improve the situation.But this[ie,RI] is not a fundamental property like charge or mass. I am very much confused on this issue.

Even if RI is included the metric will not interpret the change of the speed of light [in a medium]if it has a form:
ds^2=g(00)dt^2-g(11)dx1^2- g(22)dx^2-g(33)dx3^2

Again if two frames are in relative motion at a uniform rate in curved space-time full of anisotropies and inhomogenities, should one apply the usual lorentz Transformations as we know them?

Last edited: Nov 1, 2011
13. Nov 1, 2011

### Anamitra

The Local Context:
[On Local Inertial Frames]

We may consider a local region in strong curved spacetime. It may be surrounded by inhomgenities and anisotropies. How do we get the Lorentz Transformations here?

Actually a local transformation to flat spacetime is not a physical transformation.We are doing it on paper--in our imagination. So to that little bit of flat spacetime[we got by transformation] add the rest of it---in your imagination.

It is something remote from reality--a workspace in our imagination.A freely falling lift in the earths field can have sufficient inhomogenities and anisotropies around it and does not seem to match with the workspace.
That describes our "frictionless plane"

We can derive the Lorentz Transformations with sufficient conditions of homogenity of space [and time] and isotropy of space[in the "ideal workspace"] from the Postulates of Special Relativity.

[In this process[local transformation from curved space to flat spacetime] we are considering the transformation of a small region of spacetime in total alienation from its surroundings[in so far as anisotropy is concerned], hoping to get correct interpretations from the reverse transformations, when the surroundings are present]

In the transformed flat spacetime[local transformation] the un-transformed anisotropies[and heterogeneities] of the surroundings [of the original curved space] are neglected/ignored. Rather we ASSUME "a Global Transformation" for the rest of the flat space when we consider the lorentz Transformations

[The gravitational field itself, in general will have an anisotropic and a heterogeneous distribution of the metric coefficients wrt some arbitrary point---one may consider this point in the small transformed space in this context]

The Lorentz transformations are of course correct-----only in the proper context of their application

Last edited: Nov 1, 2011
14. Nov 1, 2011

### JDoolin

I would say the Lorentz Transformations should be applied in exact analogy to the Rotation Transformations. That is, if we have some inhomogeneity or anisotropy in spacetime, and we turn and look the other direction, then the whole thing moves, out of our sight.

What coordinates are used when I do this rotation?

Let's look for a moment at a simple example; the Schwarzschild metric:
$$c^2 d\tau^2 = \left( 1 - \frac{2 G M}{c^2 r} \right)c^2 dt^2 - \left( 1-\frac{2 G M}{c^2 r}\right)^{-1} dr^2 - r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$
(Thanks for the help finding the parsing error, Fredrik)

It describes a transformation from dt to dτ. And if written in its negative form, it makes a transformation from ds to a function of dr, dθ, and dφ.

My suggestion would be, instead of treating t, r, θ, and φ as "meaningless" variables, we should treat them as spherical coordinates which can easily map to Minkowski coordinates, and are in fact, the domain and range of the Lorentz Transformation equation.

Again, people will laugh at me, and say that there is no place where matter is not present, so, we have schwarzchild metric embedded in another schwarzschild metric, embedded in another schwarzschild metric. In effect, they will say there is no place in the universe where I can get rid of the curvature altogether. And therefore we might as well not try?

It's a fallacious argument. We should always be aware that there may be masses and motion that we have not accounted for, and so lines we think are straight may be curved after all. But the question is not whether we can know for certain that lines are straight. The question is whether the straight lines exist at all.

As long as we have a concept of straightness. Can we not compare one geodesic path to another geodesic path, and unambiguously declare which of those two paths is straighter? Yes, of course. The straight line is the direction that an object would travel if there were no forces acting on it at all. The geodesic is the direction that an object travels when a force is acting on it.

In real life, we know the difference. But once you start reading a lot of General Relativity books, you start calling geodesics straight lines. And for some reason, you ignore t, r, θ, and φ as "meaningless" variables, and you only look at τ, and say. The object in a geodesic only has τ changing, and is otherwise motionless (with respect to itself).

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 rotate the coordinate system, r, θ, and φ are affected. If you Lorentz Transform the coordinate system, t, r, θ, and φ are affected.

#### Attached Files:

• ###### rotate1.gif
File size:
178 KB
Views:
491
Last edited: Nov 1, 2011
15. Nov 1, 2011

### Fredrik

Staff Emeritus
There's a ^ followed by a space.

16. Nov 1, 2011

### JDoolin

If I'm not mistaken, the General Relativity "textbook" answer to this question is (b). But I want to make the argument that the actual answer is (a).

For the Schwarzschild metric, in particular, let's discuss the meanings of these partial derivatives:

$$\left (\frac{\partial \tau}{\partial t} \right )_{\theta,\phi,r \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial r} \right )_{\theta,\phi,t \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial \phi} \right )_{\theta,r ,t \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial \theta} \right )_{\phi,r ,t \, \mathrm{const}}$$

What are the textbook definitions of of t, r, θ, and φ? And what are the labels for

Typically, I find textbooks tend to slide over the subject with a great deal of ambiguity.

I don't know exactly what Carroll's point is here, but to contrast with Carroll, I will say it is important to think of these vectors as stretching between one event and another event, rather than located at a single point. If you are doing derivatives, with a dt, a dr, a dθ, and a dφ, you can't PUT that at a single point. You can't have a variation in positional variables t, r, θ, and φ "at a single point."

There is an s-component parallel to the r-direction (space-like interval between events), an s-component parallel to the θ direction (space-like interval between events), and an s-component parallel to the φ direction (space-like interval between events), and another component in the t-direction (time-like interval between events), called τ (tau).

There are two different coordinate systems here, and they overlap. You have the three unnamed s-components in the same direction as r, θ, and φ, and the named tau component, in the same direction as t.

The three unnamed s-components and tau make up the curved coordinate system. But there is a continuous bijection from these curved coordinates to the t, r, θ, and φ coordinates, which are flat. The curved coordinates are embedded in the flat coordinates. You can do things easily with the flat coordinates that perhaps you can't easily do with the curved coordinates. For instance, you can do the rotation transformation.

Carroll goes on to say:

I say there is not one, but two sets of vectors defined at each point in spacetime.

(1) There are Δt, Δθ, Δφ, and Δr which are the flat coordinates, quite easily mapped into Cartesian coordinates, which can then be readily rotated or Lorentz Transformed to whatever angle and rapidity you like,

(2) then there are the "curved" components

$$\begin{matrix} \Delta \tau_{\left (r,\theta,\phi\;\mathrm{const} \right )}\\ \Delta s_\left ({r,\theta, t\;\mathrm{const}} \right )\\ \Delta s_\left ({r,\phi, t\;\mathrm{const}} \right )\\ \Delta s_\left ({\theta,\phi, t\;\mathrm{const}} \right ) \end{matrix}$$

It is these curved components which control the local acceleration and behavior of matter. It is within these curved components where you can set dτ=0 and find the geodesic of a photon, for instance, and all of the neat cool stuff you can do with General Relativity.

Last edited: Nov 1, 2011
17. Nov 1, 2011

### JDoolin

A Galilean Transformation "warps" space-time like this:
[URL]http://www.spoonfedrelativity.com/web_images/galileantransform.gif[/URL]
where the events move along lines of constant t.

whereas a Lorentz Transformation "warps" space-time like this:
[URL]http://www.spoonfedrelativity.com/web_images/lorentztransform.gif[/URL]
and the events move along hyperbolic arcs of constant $c t^2 - x^2$

I suppose the Galilean Transformation could be said to be "local" if by "local" you mean that it does not affect events which are occurring "now." But it has a big effect on events which occur in the far future or the far past.

And the Lorentz Transformation could be said to be "local" if by "local" you mean that it does not effect THE event which occurs "here" and "now".

(Either way calling a transformation local based on the only events that it doesn't affect seems backward.)

Accelerating toward an event in the future causes it to lean toward you. But accelerating toward an event in the past causes it to lean away.

Last edited by a moderator: Apr 26, 2017
18. Nov 2, 2011

### Staff: Mentor

There are at least two conceptual errors here.

First, the metric, which you have written as a line element, is *not* a coordinate transformation equation. It is an equation that tells you how to calculate an actual physical interval along a differential segment of a curve in spacetime, if you know the coordinates and their differentials along that segment. If you integrate it along a worldline, it tells you the physical interval along that worldline; for example, along a timelike worldline, it tells you the proper time experienced by an observer following that worldline. But proper time is *not* a coordinate; it's a physical observable. The t, r, θ, and φ coordinates are *not* observables; they're arbitrary numbers that are used to label events.

Second, you can't split up the line element into a "positive" and "negative" part, as though one part calculates dτ and the other part calculates ds. The line element is a single expression for calculating a single interval. Putting dτ vs. ds on the left-hand side is just a naming convention, corresponding to a sign convention for the metric coefficients; typically, if the LHS is written as dτ it means you are using a timelike sign convention, where timelike squared intervals are positive, so the "time" metric coefficient will be positive and the "space" ones will be negative, whereas if the LHS is written as ds it means you are using a spacelike sign convention, where spacelike squared intervals are positive, so the "space" metric coefficients are positive and the "time" one is negative. But "time" and "space" here are in quotes because not all metrics are diagonal and not all metrics have one timelike and three spacelike coordinates. Anyway, the sign convention is just that, a convention; it doesn't affect the physics. Same for the convention of writing the LHS as dτ or ds.

You can try to treat them this way, but it won't work. Minkowski coordinates, interpreted in the usual way, require a flat spacetime; they simply won't work the way you want them to work if spacetime is curved. Same for Lorentz transformations.

Yes, exactly. Gravity is everywhere, therefore curvature is everywhere.

And the answer is, in a universe with gravity, no, they don't.

No, this is wrong; in fact, again, there are at least two conceptual errors here.

First, you are misusing the term "geodesic". In a curved manifold, a "geodesic" is the closest thing you can get to a straight line. There aren't any curves that are straighter. And all geodesics are equally straight, so it makes no sense to say one is straighter than another. I realize that lots of people don't like calling these curves "straight lines", because they aren't Euclidean straight lines; but that's why the term "geodesic" was invented, so we could have a name for curves in curved manifolds that are the closest analogues to Euclidean straight lines; since in a curved manifold there are *no* Euclidean straight lines at all, geodesics are the best we can do.

Second, you are misunderstanding the physics. A geodesic is the worldline of an object on which *no* forces are acting, when the term "force" is defined correctly, as something that is actually *felt* by an object (and which can be measured by an accelerometer). You define gravity as a force, but an object moving solely under the influence of gravity is weightless; it feels no force, and an accelerometer attached to it would read zero. You can call the path of this object "curved" if you like, but it's the straightest path there is in a spacetime with gravity present. If you disagree, then please explain, in detail, how we are to *physically* pick out the "real straight lines", the ones compared to which we are supposed to view the path of a freely falling object as curved. Oh, and by the way, saying that "straight lines" are defined by the paths of light rays will *not* work; light is bent by gravity as well. It isn't bent as much because it moves faster than ordinary falling objects, but its path is still bent. (If you ask why light and a falling rock travel on such different paths if they're both geodesics, that's because there are many geodesics through any given event, corresponding to all the possible velocities a freely falling object can have at that event, up to and including the speed of light.)

Well, of course. Are you trying to say an object is moving with respect to itself?

The reason we pick out τ is that it's a physical observable; it's the proper time experienced by an observer traveling on that worldline. You can't observe t, r, θ, and φ directly. Sometimes you can observe things that are very *close* to those coordinates, but that only works in regions where gravity is very weak. You can't generalize to everywhere in a curved spacetime.

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.

19. Nov 2, 2011

### Staff: Mentor

Yes.

See my last post. The metric is a single expression for a single interval; it is not split up into a piece that calculates dtau and a piece that calculates ds.

Depends on the spacetime and the coordinate chart you are using to describe it. There is no single "definition" of any set of coordinates that applies generally in GR. You have to specify the spacetime first, and *then* you have to define how the coordinates label events in the spacetime.

The Wikipedia page actually does a pretty good job of explaining the definitions of the exterior Schwarzschild coordinates, which I think is what you're asking about:

http://en.wikipedia.org/wiki/Schwarzschild_coordinates

Carroll's point is that vectors are not objects in spacetime itself; they are abstract objects that "live" in the tangent space at each event.

No, but you can have a *rate* of variation at a single point, which is what a derivative describes. More precisely, it describes the limit in the variation between two points as the separation between them goes to zero, divided by the separation--but that means that once the limit is taken, you are effectively "at a single point". Derivatives do not describe change over a finite length; they describe change at a single point.

Again, see my previous post. s and tau are not separate. The positive and negative terms in the metric are added *together* to get a single interval, which may be timelike, null, or spacelike.

No, this is wrong. See above.

Nope. See above. However, there is one other thing worth mentioning here. You *can* construct a continuous bijection between two different sets of coordinates; but if one set is "curved" and the other set is "flat", then the bijection will obviously not preserve distances (or in spacetime, intervals) between events. Since preserving intervals between events is a requirement of a coordinate transformation in physics, any such bijection as you describe will not qualify.

Similar comments to the above on the rest of this post: there are not two sets of coordinates, only one.

20. Nov 2, 2011

### PhilDSP

Without taking the time to search for a more precise definition, I'd have to say that a mathematical model is local if the cause of the change of all parameters is limited in displacement across space to a period of time that is greater than or equal to the minimum propagation time allowed in the model.

A Galilean transformation has no intrinsic rate limitation and would seem to support the instantaneous translation of coordinates. So it would be local so far as the rate change of each parameter is consistent with the particular designation of the limiting velocity of EM propagation, gravity, weak and strong forces separately, for example, which don't need to share a single propagation speed.

Under SR, the propagation or displacement of anything is limited to c. Yet the operation of the Lorentz transformation in performing coordinate translations apparently must occur instantaneously. Maybe that's a moot issue as in thinking about this further, the lack of locality is more related to a general mathematical discontinuity between adjacent points of time and space rather than speed of translation. Whereas adjacent points conforming to a Galilean transformation have a shared linear connection, with the LT they don't share that and rather sort of dance around each other depending on how close you wish to make them.

21. Nov 3, 2011

### JDoolin

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.
In the Schwarzschild coordinates, theta, and phi are "arbitrary" in the following sense: Because the schwarzchild metric is spherically symmetric, you can choose the directions for θ=0 and φ=0 in any direction you like. The r coordinate is zero at the center of mass of the central gravitating body, so in that sense, r is not arbitrary. You might suggest that the scale of r is arbitrary since we can measure this in feet, meters, miles, light-years, whatever. And the time coordinate t, is measured from some arbitrarily chosen event, and you can choose that scale arbitrarily, as well, whether it be seconds, hours, months, or years, whatever. Illustration of spherical coordinates. The red sphere shows the points with r = 2, the blue cone shows the points with inclination (or elevation) θ = 45°, and the yellow half-plane shows the points with azimuth φ = −60°. The zenith direction is vertical, and the zero-azimuth axis is highlighted in green. The spherical coordinates (2,45°,−60°) determine the point of space where those three surfaces intersect, shown as a black sphere.
Let $$ds^2 = -\left ( 1 - \frac{2 G M}{c^2 r} \right )c^2 dt^2 + \left ( 1-\frac{2 G M}{c^2 r} \right )^{-1} dr^2 + r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$. If the right-hand-side comes out to be a positive number, then the space-time interval is space-like; meaning there is no way for a clock to move between the two events (t,r,θ,φ) and (t+dt, r+dr, θ+dθ, φ+dφ). But it is possible to stretch a physical object such as a ruler between those two events. On the other hand, if the RHS is negative, then the space-time-interval is time-like, meaning it would be impossible to stretch a ruler between those two events, but it is possible for a clock to be at both of those events at different times. I don't understand what you mean "the sign convention...doesn't affect the physics" If you aren't distinguishing between time-like intervals, and space-like intervals, I can't imagine how you can get any of the other physics right.

The t, r, θ, and φ coordinates are a flat spacetime; (well, to be more clear, θ, and φ are obviously curved, but there is no difficulty converting t, r, θ, and φ. of any given event, and converting it to (t,x,y,z) using \begin{align*} x &= r cos(\phi) sin(\theta) \\ y &= r sin(\phi) sin(\theta) \\ z &= r cos(\theta) \end{align*} From there, there is no difficulty performing a lorentz transformation on the flat coordinates,

It is not difficult to establish where gravity is present, and where gravity is less present. If you take any given gravitating body, and move further and further away from it, the geodesics become straighter and straighter. Also, as you increase the velocity of a body the trajectory becomes straighter and straighter. There is, of course a limiting factor that even light is affected by gravity, However, from far away, it is clear that objects deep in a gravitational well do not move in straight lines. Objects that are further from a gravitational well move in much straighter lines. It may be appropriate to say that no body will ever follow a perfect straight line in the universe, because of gravity. It is not appropriate or correct to say straight lines do not exist. What you have is curvature of a SUBgeometry which is present everywhere, but it can easily be mapped one-to-one; event by event onto a flat geometry.

Try to understand that I am not drawing a line ON the curved manifold. Think of it more like drawing a line on the plane in a flat surface "above" the curved manifold. Unless you can find a situation where the curved manifold curves down under itself, there is a one-to-one mapping from the curved coordinate system to an overlying global flat coordinate system.
https://www.physicsforums.com/attachment.php?attachmentid=40656&d=1320321617

It doesn't matter if the underlying surfaces are curved, so long as none of the particles in the underlying surface go faster than the speed of light in the overlying geometry, in which case there would be causality problems.

But if I am 4 light years from Alpha Centauri, and say some black hole passed between Alpha Centauri and here. What would happen? The light from Alpha Centauri would probably be bent in every which direction, and I would know there were a problem. Let's say also, that somehow the black hole warps the space-time so that an extra light-year of rulers (physical observables) could fit between Alpha Centauri and me.

Here is where I think we differ. I would say what you have is a bunch of really short rulers, but the distance is still only four light years. You say you have to follow all of the curves, and so the distance is five light years. I'm saying there is a global flat coordinate system overlying all of your curved space; you do NOT have to follow all of the curves. You're saying there is no global coordinate system, and the only way to measure distance is to place rulers end to end, and look at clocks you are physically carrying, ignoring the presence of distant bodies.

For the purposes of a Lorentz Transformation, It doesn't matter whether a body "feels" a force. What matters is whether it undergoes a change in velocity. I fully agree that when an object's path curves under a gravitational field, there is little or nothing it experiences internally. But I don't think we should limit our description of physics to beings so obsessed with themselves that they never look anywhere except at the clock on their navel. The fact is, I do not "feel" the motion of my body around the earth as it revolves. I do not "feel" the motion of the earth around the sun. I do not "feel" the motion of the sun around the center of the Milky Way. Yet I am aware of all of these motions . I am aware that these motions are not straight lines. The accelerometer reads zero, but it is still obvious that I am changing velocity, and changing my reference frame.

To explain, in detail, how we are to *physically* pick out the "real straight lines", it's not a question of picking out THE real straight line, it is in distinguishing between lines which are obviously not straight and those which are straighter.

For each of these motions, I can draw a picture of the orbital path, and then I can at draw a tangent line that is MUCH straighter than the geodesic. How will I pick out the real straight lines? For one, by using those variables t, r, θ, and φ, converting them to cartesian coordinates by \begin{align*} x &= r cos(\phi) sin(\theta) \\ y &= r sin(\phi) sin(\theta) \\ z &= r cos(\theta) \end{align*} and then using the usual method for describing straight lines in Cartesian Coordinates.

No. What I'm saying is that putting a falling rock in a box that is also falling does not make it move in a straight line. You think that the closest you can get to a straight line is by throwing a rock. That rock is not going in a straight line. It goes up and falls. Ahh, but now you say "Put the rock in a box. See now, in the box the rock is going in a straight line, relative to the box. Ergo, Hence, Therefore, straight line. QED" or From inside the box, the rock appears stationary. Therefore, the rock is only traveling through time. Hence a geodesic arc is a straight line. So by putting the rock in a box, now you think the rock is going in a straight line? The rock is NOT going in a straight line. It is just now that you have both rock and box moving in an arc, and because it is (roughly) the same geodesic, it looks like a straight line.

All I'm saying is, don't put the rock in the stupid box. Don't ignore the presence of gravity. Don't put me in a box and throw the box, because regardless of how it "feels" like I must be floating in free space, that doesn't change the fact that I'm going to hit the ground hard, and frankly hitting the ground IS a physical observable.

This tyranny of "observables" is overrated. You limit your notion of distance and time to the actual readings of clocks, and the lengths of rulers that you can physically touch. I watch someone throw a rock and it follows a geodesic. To me that path looks curved. You claim that it is not curved, because if we put the rock in a box that were traveling along with it, also following a geodesic, and rulers in the box, then it would appear from inside the box that the rock was not moving; or might be following a straight line.
The point is that it does not matter whether t, r, θ, and φ can be measured directly. We don't actually need to put clocks and rulers out there to get a pretty good idea where the sun is, where the moon is, where jupiter is, where the center of the galaxy is. (or was when the light left it). The point is, any nonlocal measurement of distance (i.e. the estimation of distance by sight or parallax, or luminosity.)

My point is, it does not make sense to constrain your physics to only things you can touch. You need to think about far-away objects as well. When discussing far-away objects, the distances are not determined by following every curve of the gravitational manifold, but are instead determined by an overlying global flat coordinate system.

22. Nov 3, 2011

### JDoolin

Well, whether we call it local or global, let's ask what the rotation transformation does, first of all.

Let's say I turn a little to the right, in such a way that the objects 10 feet in front of me move 1 foot to the left. Objects 1000 miles in front of me move 100 miles to the left. Objects a billion light years in front of me move 100 million light years to the left. The point is, all I did was local; turning to the right; but the transformation affects things in direct proportion to their distance. The further away the thing is, the further it moves.

The same can be said for the Lorentz Transformations. Events which are furthest away from the observer undergoing a velocity change are affected the most by the transformation.

In either case, I'm only changing a single local variable, for rotation the local variable is which direction I'm facing. For Lorentz Transformation, that local variable is my rapidity, or my velocity. However, the transformation of coordinates is global.

23. Nov 3, 2011

### Staff: Mentor

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

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.

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.

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.

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?

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.

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.

Velocity relative to what?

Relative to what?

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:

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.

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.

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.

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.

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

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.

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.

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.

24. Nov 3, 2011

### Passionflower

JDoolin, the Earth goes around the Sun just as much as the Sun goes around the Earth, it is simply a matter of relativity.

25. Nov 4, 2011

### JDoolin

The question I am asking is what are the textbook definitions of t, r, θ, and φ. The article describes a "coordinate singularity" at t=t0, r=r0, θ=0, and φ undefined being the north pole, and t=t0, r=r0, θ=Pi, and φ undefined being the south pole. That's fine. That is precisely what I meant by them. But the variable r extends from 0 to infinity, the variable θ from 0 to Pi, and φ from 0 to 2 Pi. The domain of r goes from 0 to infinity, even though you have a coordinate singularity at r=r0. This is the geometry of a black hole, and there is a certain radius, called the schwarzschild radius or the event horizon where light will fall in. My point is that the (t,r,θ,φ) variables are not meaningless. They represent position and time of a global flat coordinate system in spherical coordinates centered at (t=t0, r=0).
In order to have a rate of change, you have to allow variables to change. A derivative is not a change at a single point. It is the "limit" as the distance approaches zero. If you actually try to describe the change at a single point, it is 0/0, which is undefined. A "limit" in math means playing the game of whatever number you can give, I can give one smaller, but LARGER than zero. Once you actually say the denominator is dx=zero, then it is a division by zero error.
I misspoke. Obviously, (t,r, θ, φ) are four parameters while $$s^2 = - c^2 d\tau^2$$ is but a single parameter. So let me clarify and correct so that you may understand what I meant: If the value of s is imaginary, then it represents the time between time-like-separated events. I was thinking of the time on the clock being one of your "physical observables" and thinking there was only one physical observable there. But if we have a clock going between these two events, then the velocity of the clock has a component in the r direction, the θ direction and the φ direction; (I'm not sure if you count these components of velocity in your "physical observables" or not, but I did neglect to think of them when I made my earlier post). If the value s is real, it represents the distance between two events. (The events may have different values of t, in which case they are simultaneous only in some non-comoving reference frame. This is a more complicated problem and we would have to discuss the orientation of the ruler as well as its velocity.) But if the t-values are the same, then the orientation of the ruler has a component in the r direction, the theta direction, and the phi direction. Since you've got a ruler there, and its length is one of your "physical observables", the different components of direction are also "physical observables"

What I am saying is that the (t,r, θ, φ) coordinates overlap with what you call "physical observables" coordinates. See below.

But the Schwarzschild metric does NOT preserve distances. It does NOT preserve space-time intervals. Are you arguing that the Schwarzschild metric is invalid? The Schwarzschild metric does NOT preserve the speed of light. It makes the light travel slower and slower as it goes down toward the Schwarzschild radius. What Schwarzschild metric DOES do is maintain the local measure of distance and the local measure of time, so that locally it looks like the speed of light is preserved. But globally, the speed of light is not preserved. From the global perspective, the schwarzschild metric causes the light to slow, and turn and bend, not unlike the way a transparent crystal causes light to slow and bend.But the Schwarzschild metric does NOT preserve distances. It does NOT preserve space-time intervals. Are you arguing that the Schwarzschild metric is invalid? The Schwarzschild metric does NOT preserve the speed of light. It makes the light travel slower and slower as it goes down toward the Schwarzschild radius. What Schwarzschild metric DOES do is maintain the local measure of distance and the local measure of time, so that locally it looks like the speed of light is preserved. But globally, the speed of light is not preserved. From the global perspective, the schwarzschild metric causes the light to slow, and turn and bend, not unlike the way a transparent crystal causes light to slow and bend.

Which of the two sets of coordinates are you referring to as the only set of coordinates? In the schwarzcshild metric there is a coordinate system composed of (t, r, θ, φ), but then you keep referring to "observables" being the lengths of local rulers, and the times on local clocks. one could construct a sort of pseudo-coordinate system where you take as your measure of time the times on local clocks, and take as your measure of distance the lengths marked off on local rulers. In this pseudo-coordinate system, the local speed of light is constant. This pseudo-coordinate system is curved, since the rulers at different locations are different lengths, and the clocks at different locations have different rates. If you want to construct a coordinate system strictly from "observables" defined as lengths of local rulers, and you get a curved coordinate system.

However, in the schwarzschild metric, we also have (t,r,θ, φ) which we both agree do not mark off the lengths of local rulers or the times on local clocks. I think we can both acknowledge, at least, that the coordinates (t,r, θ, φ) are NOT the physical "observables" that you've been describing, that may help our communication. So to help me understand your point here, explain which coordinate system you mean; the one made of (t,r,θ, φ) or the one made up of local rulers and clocks?

When using these parameters (t, r, θ, φ) it is helpful to discuss many different derivatives and partial derivatives, in sort of a similar way as to when in thermodynamics, functions of (pressure, volume, temperature, chemical potential, and total energy) etc. used. The idea is you can hold certain parameters constant, and allow other parameters to vary, and find helpful formulas for isothermal, isobaric, adiabatic reactions, and many more. In particular, you can say "hold r, θ, φ constant, and find dtau/dt as a function of r." You can set ds^2=0 and hold theta and phi constant, and find dr/dt as a function of r, to find the velocity of a photon falling into the Schwarzschild geometry. You can set ds^2=0 and hold theta =Pi/2, and find a relationship between radial velocity, dr/dt and angular velocity dphi/dt, and this is a good start to working out the arc that a photon will follow in the schwarzschild geometry. I have listed the questions below, along with my approach to getting the answers, so you can understand exactly what I mean.

Exercise 1. Assume a photon is moving around a gravitating mass in the equatorial plane. Use the Schwarzschild metric to find a relationship between radial velocity, $v_r=dr/dt$ and tangential velocity $v_t=r dφ/dt$ .

Answer: Sed ds=0, for a photon, and Set θ=Pi/2 for the equatorial plane, then Schwarzschild metric simplifies to $$0 =\left( 1-\frac{r_s}{r} \right )c^2 dt^2+\left( 1-\frac{r_s}{r} \right )^{-1}dr^2+r^2 d\phi^2$$ Divide by dt^2. \begin{align*} 0 &=-\left( 1-\frac{r_s}{r} \right )c^2 +\left( 1-\frac{r_s}{r} \right )^{-1}\left (\frac{dr}{dt} \right )^2+\left (r \frac{d\phi}{dt} \right )^2 \\ 0 &=-\left( 1-\frac{r_s}{r} \right )c^2 +\left( 1-\frac{r_s}{r} \right )^{-1} v_r^2+v_t^2 \\ v_t^2 &=\left( 1-\frac{r_s}{r} \right )c^2 -\left( 1-\frac{r_s}{r} \right )^{-1} v_r^2 \\ v_t &=\pm \sqrt{\left( 1-\frac{r_s}{r} \right )c^2 -\left( 1-\frac{r_s}{r} \right )^{-1} v_r^2} \end{align*}

Exercise 2. Find the tangential velocity of a photon as it makes its closest approach to a gravitating body.

Answer: Using the results from previous problem, but dr/dt=0; $$v_t =\pm c \sqrt{1-\frac{r_s}{r} }$$

Exercise 3. Find the rate of time tau of a clock sitting on a non-rotating surface at radius r, as compared to the global coordinate time t.

Answer: In this case, dr=dθ =dφ = 0, since the position of the clock is not changing. The Schwarzschild metric simplifies to \begin{align*} c^2 d\tau^2&=\left ( 1-\frac{r_s}{r} \right )c^2 dt^2 \\ \frac{d\tau}{dt}&= \pm c \sqrt{ 1-\frac{r_s}{r}} \end{align*}

Exercise 4; Assume a photon is moving directly toward a gravitating mass. Use the Schwarzschild metric to find dr/dt, the radial velocity of the photon as a function of r.

Answer: $$ds^2 =- \left ( 1 - \frac{2 G M}{c^2 r} \right )c^2 dt^2 + \left ( 1-\frac{2 G M}{c^2 r} \right )^{-1} dr^2 + r^2 (d\theta^2 + \sin^2(\theta)d\varphi^2)$$ ds=0 because it is a photon. dθ=dφ=0 because it is coming straight in. Let $$r_s=\frac{2 G M}{c^2}$$ Then the equation simplifies to \begin{align*} 0 &=-\left ( 1-\frac{r_s}{r} \right )c^2 dt^2+\left( 1-\frac{r_s}{r} \right )^{-1}dr^2 \\ \left ( 1-\frac{r_s}{r} \right )c^2 dt^2 &= \left( 1-\frac{r_s}{r} \right )^{-1}dr^2\\ \left (\frac{dr}{dt} \right )^2 &=\left( 1-\frac{r_s}{r} \right )^2 c^2 \\ \frac{dr}{dt}&= \pm \left ( 1-\frac{r_s}{r} \right )c \end{align*}