The Lorentz Transformations and a Few Concerns


by Anamitra
Tags: concerns, lorentz, transformations
Anamitra
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Oct28-11, 11:55 AM
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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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Oct28-11, 01:17 PM
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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.
Anamitra
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Oct28-11, 01:54 PM
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Quote Quote by Bill_K View Post
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!

Anamitra
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Oct28-11, 02:08 PM
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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]
Anamitra
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Quote Quote by Bill_K View Post
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
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Oct30-11, 05:00 AM
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One way to find the Lorentz transformations is to consider a family of global coordinate systems [itex]\{x_i:\mathbb R^4\rightarrow\mathbb R^4|i\in I\}[/itex] such that each "coordinate change" function [itex]x_i\circ x_j^{-1}[/itex] takes straight lines to straight lines, and the set [itex]\big\{x_i\circ x_j^{-1}|i,j\in I\big\}[/itex] 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.
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Quote Quote by Bill_K View Post
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 [itex]\mathbb R^4[/itex] that satisfy a few additional conditions. The presence of matter in the actual universe obviously has no effect on [itex]\mathbb R^4[/itex], 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).
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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).
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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:
[tex]\begin{align*} x &= r \cos(\phi) \sin(\theta)\\ y &= r \sin(\phi)\sin(\theta)\\ z &= r \cos(\theta) \end{align*}[/tex]
(b) A rotation
[tex]\begin{align*} x' &= x \cos(\theta)+y \sin(\theta) \\ y' &= -x \sin(\theta)+y \cos(\theta) \\ z' &= z \end{align*}[/tex]
(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.
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Oct30-11, 02:58 PM
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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:


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

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

[tex]\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} \square & \square \\ \square & \square \end{pmatrix}\begin{pmatrix} r\\ \theta \end{pmatrix}[/tex]

However, if we take the derivatives,

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

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


[tex]\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}[/tex]

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.

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

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:

[tex]\begin{pmatrix} x'\\ y' \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}[/tex]

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

[tex]\begin{pmatrix} dx'\\ dy' \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}\begin{pmatrix} dx\\ dy \end{pmatrix}[/tex]

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
[tex]\begin{align*} ct'&=ct \cosh(\varphi)-x \sinh(\varphi)\\ x'&=-ct \sinh(\varphi) + x\cosh(\varphi) \end{align*}[/tex]
and
[tex]\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}[/tex]

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.
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Quote Quote by JDoolin View Post
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.
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Nov1-11, 02:31 AM
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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?
Anamitra
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Nov1-11, 03:13 AM
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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 Quote by Anamitra View Post
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:
[tex]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)[/tex]
(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.
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Fredrik
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Quote Quote by JDoolin View Post
(sorry, I can't find the parsing error in the LaTeX.)
There's a ^ followed by a space.
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Quote Quote by JDoolin View Post
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:

[tex]\left (\frac{\partial \tau}{\partial t} \right )_{\theta,\phi,r \, \mathrm{const}}[/tex]
[tex]\left (\frac{\partial s}{\partial r} \right )_{\theta,\phi,t \, \mathrm{const}}[/tex]
[tex]\left (\frac{\partial s}{\partial \phi} \right )_{\theta,r ,t \, \mathrm{const}}[/tex]
[tex]\left (\frac{\partial s}{\partial \theta} \right )_{\phi,r ,t \, \mathrm{const}}[/tex]

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 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

[tex]\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}[/tex]

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 Quote by PhilDSP View Post
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 [itex]c t^2 - x^2[/itex]

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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Quote Quote by JDoolin View Post
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.
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.

Quote Quote by JDoolin View Post
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.
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.

Quote Quote by JDoolin View Post
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?
Yes, exactly. Gravity is everywhere, therefore curvature is everywhere.

Quote Quote by JDoolin View Post
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.
And the answer is, in a universe with gravity, no, they don't.

Quote Quote by JDoolin View Post
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.
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.)

Quote Quote by JDoolin View Post
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).
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.

Quote Quote by JDoolin View Post
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.


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