# I Lorentz Transformations in the context of tensor analysis

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1. Aug 20, 2016

### Diego Berdeja

Hello everyone,

There is something that has been bugging me for a long time about the meaning of Lorentz Transformations when looked at in the context of tensor analysis. I will try to be as clear as possible while at the same time remaining faithful to the train of thought that brought me here.

Most, if not all, derivations of Lorentz Transformations start by requiring a set of matrices (linear transformations) which leave the components of the Minkowski metric $\eta$ invariant. These would be a set of matrices $\Lambda^\alpha_\beta$ such that
$$\Lambda^\alpha_\beta\Lambda^\mu_\nu\eta _{\alpha\mu}=\eta_{\beta\nu}.$$
Such derivations claim that this is a direct consequence of requiring the constancy of the speed of light, i.e., the constancy of the space-time interval.

Nevertheless, if one looks at it from the point of view of tensor analysis, this appears to be false (at least to me). Let us consider flat space-time. Suppose we find a reference system $x^\mu$ such that the components of the metric tensor, $g$, do take the values of Minkowski space-time. This means that $g_{\alpha\beta}=\eta_{\alpha\beta}$. Let us consider the four-vector $P$, which describes the infinitesimal displacement of a ray of light. Then
$$g(P,P)=(g_{\alpha\beta} dx^\alpha\otimes dx^\beta)(P^\rho\partial_\rho)(P^\lambda\partial_\lambda)=g_{\alpha\beta}P^\rho P^\lambda\delta^\alpha_\rho\delta^\beta_\lambda=\eta_{\alpha\beta}P^\alpha P^\beta.$$

Let us now look for the most general transformation $x\rightarrow x'(x)$ such that the interval remains invariant. In the new reference system, we have
$$g(P,P)=(g'_{ab} dx'^a\otimes dx'^b)(P'^r\partial'_r)(P'^l\partial'_l)=g'_{ab}P'^a P'^b.$$

Since the interval is invariant, we must have
$$g'_{ab}P'^a P'^b=\eta_{\alpha\beta}P^\alpha P^\beta,$$
but we know how vectors transform, so
$$g'_{ab}P'^a P'^b=g'_{ab}\frac{\partial x^\alpha}{\partial x'^a}\frac{\partial x^\beta}{\partial x'^b}P^\alpha P^\beta,$$
$$g'_{ab}\frac{\partial x^\alpha}{\partial x'^a}\frac{\partial x^\beta}{\partial x'^b}=\eta_{\alpha\beta}.$$
This equation is the transformation law for the components of a tensor of type $(0,2)$, and imposes no
restrictions on the allowed transformations. In other words, as far as tensor analysis is concerned, if one defines Lorentz Transformations to be those coordinate transformations which leave the speed of light invariant, then all coordinate transformations are Lorentz Transformations. This has to be true, since tensors are geometric objects that do not depend on the reference system.

It is my understanding that the actual restrictions on the transformations come from requiring that they be a symmetry of Maxwell's Equations. Indeed, this is how they were first derived.

Another thing that really gets to me is that the treatment of Lorentz Transformations is usually restricted to the use of cartesian systems. I think this looses generality.

What do you guys think? Can Lorentz Transformations be derived from the constancy of the speed of light, without reference to Maxwell's Equations, and in the language of tensor analysis?

Cheers!

2. Aug 20, 2016

### samalkhaiat

Constancy of the speed of light is nothing but the condition $\bar{g}_{\mu\nu} = \eta_{\mu\nu}$. This leads to the defining relation of Lorentz transformations in Minkowski space-time: $$\eta_{\mu\nu} \frac{\partial \bar{x}^{\mu}}{\partial x^{\rho}} \frac{\partial \bar{x}^{\nu}}{\partial x^{\sigma}} = \eta_{\rho \sigma} .$$

3. Aug 20, 2016

### Orodruin

Staff Emeritus
Note that the definition of Minkowski coordinates, which is what a Lorentz transformation relate, is exactly equivalent to defining Cartesian coordinates in a Euclidean space.

This is the same as requiring the frame invariance of the speed of light with the assumption that the coordinates are times and lengths measured by an inertial observer.

4. Aug 21, 2016

### stevendaryl

Staff Emeritus
Yes, every coordinate transformation leaves the interval unchanged, since the interval is a scalar. But only a combination of Lorentz transformations, translations and rotations leave $g_{\mu \nu} = \eta_{\mu \nu}$.

5. Aug 21, 2016

### Diego Berdeja

What about transformations between non-Cartesian inertial coordinates? For example, a transformation from an inertial, Cartesian coordinate system to the same system in spherical coordinates? After all, any coordinate transformation results in a linear map (Carroll, p. 129-133).
If I am correct, these transformations are excluded because they are not symmetries of Maxwell's equations, i.e. $A_\phi,$ $A_\theta,$ and $A_r$ do not obey the same differential equations as do $A_x,$ $A_y,$ $A_z$.

Isn't this condition too restricting according to the computation above?

6. Aug 21, 2016

### Diego Berdeja

What is the basis of the requirement $g_{\mu\nu}=\eta_{\mu\nu}$? It is my understanding that this requirement comes from the necessity that transformations between inertial frames be symmetries of Maxwell's equations (cf. Springer), but I have never seen it stated in any paper or book (probably out of my own ignorance).

7. Aug 21, 2016

### stevendaryl

Staff Emeritus
No, I don't think the requirement has anything to do with Maxwell's equations. The requirement that $g_{\mu \nu} = \eta_{\mu \nu}$ is just what you would get if you create an operationally defined coordinate system based on clocks and measuring sticks. The Lorentz transformation (plus rotations and translations) relate any two such operationally defined coordinate systems.

8. Aug 21, 2016

### Orodruin

Staff Emeritus
No, these are not Minkowski coordinates. Just as little as they are Cartesian coordinates in Euclidean space.

Of course you can use whatever coordinates you like - but that is not how we define Lorentz transformations. They are defined as the coordinate transformation from one set of Minkowski coordinates to another. Just as rotations are defined as transformations from one Cartesian coordinate system to another. You can use whatever coordinate system you want in Euclidean space, but not all coordinate transformations are rotations.

9. Aug 21, 2016

### samalkhaiat

The constancy of the speed of light: $$c = \sqrt{(\frac{dx^{1}}{dx^{0}})^{2} + (\frac{dx^{2}}{dx^{0}})^{2} + (\frac{dx^{3}}{dx^{0}})^{2}} = \sqrt{(\frac{d \bar{x}^{1}}{d\bar{x}^{0}})^{2} + (\frac{d\bar{x}^{2}}{d\bar{x}^{0}})^{2} + (\frac{d\bar{x}^{3}}{d\bar{x}^{0}})^{2}} .$$ If you square both sides you get
$$\eta_{\mu\nu}dx^{\mu}dx^{\nu} = \eta_{\rho \sigma}d\bar{x}^{\rho}d\bar{x}^{\sigma} = 0 .$$

10. Aug 22, 2016

### stevendaryl

Staff Emeritus
That raises the question as to exactly what is the meaning of the common way of summarizing Special Relativity: "The laws of physics are invariant under Lorentz transformations". You can write the laws using tensors so that they are invariant (or covariant) under all coordinate transformations, so what, exactly, is special about the Lorentz transformations, when it comes to SR?

It might have to do with active versus passive transformations. All laws are invariant (or can be written so that they are invariant) under passive coordinate transformations, but SR is also invariant under active Lorentz transformations.

11. Aug 22, 2016

### vanhees71

No, you are not restricted to pseudo-cartesian bases. You can work with any basis you like or even with completely "generalized" coordinates ("curvilinear" coordinates). Then of course the pseudometric coefficients become a general function $g_{\mu \nu}(q)$. All this is completely analogous to the Euclidean vector calculus known from 3D space.

Nevertheless, the space-time symmetry is the (proper orthochronous) Poincare group not the general covariance of GR.