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## The Lorentz Transformations and a Few Concerns

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.

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Blog Entries: 1 				 				 				 				         Recognitions:        Science  Advisor  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.

Quote by 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.
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!

## The Lorentz Transformations and a Few Concerns

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]

Quote by 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.
These are the priceless words of the century--from Bill_K

Mentor  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.

Mentor

Quote by 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.
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).

Blog Entries: 47 				 				 				 				         Recognitions:      Gold Member      Homework  Help     Science  Advisor  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).

Blog Entries: 4 				 				 				 				         Recognitions:      Gold Member  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.

Blog Entries: 4 				 				 				 				         Recognitions:      Gold Member  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.

Quote by JDoolin 					 				 				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.
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.

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?

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

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Quote by 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?
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 Thumbnails

Mentor

Quote by JDoolin 					 				 				(sorry, I can't find the parsing error in the LaTeX.)
There's a ^ followed by a space.

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Quote by 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?
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.

Quote by CarrollLecturesPage8 					 				 				"But inspiration aside, it is important to think of these vectors as being located at a single point, rather than stretching from one point to another.  (Although this won't stop us from drawing them as arrows on spacetime diagrams.)"
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:

A vector is a perfectly well-dened geometric object, as is a vector field, defined as a set of vectors with exactly one at each point in spacetime.
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.

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Quote by 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.
A Galilean Transformation "warps" space-time like this:

where the events move along lines of constant t.

whereas a Lorentz Transformation "warps" space-time like this:

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.

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