Deriving Lorentz transformations

[facenian]

1. The 2D Lorentz transformation can be derived from the following mathematical assumptions, which all have physical motivations.

I think it's a good warm up to attack the real important case in 4D.
In that respect I think that strangerep and Fredrik gave the answers

 

[Erland]

I think it's a good warm up to attack the real important case in 4D.

The main diificulty in going to 4D is my point 8, about isotropy. How to formultate this mathematically in a sufficiently simple manner?

 

[samalkhaiat]

Very interesting post. However I would modify the demonstration because it has one flaw. The problem I see is the assertion that since straight lines must transform into straight lines, the transformation must be linear. I think this is a common mistake(for an example see, for instance, "The Special Theory of Relativity" by Aharoni)

The term “linear” in that thread is not the same as linearity in the algebraic sense: $f(ax + by) = af(x) + bf(y)$. Linearity there means polynomial of degree one in the coordinates, i.e., solutions to the system of 2-order PDE’s
$$\frac{\partial^{2}F^{\sigma}}{\partial x^{\mu}\partial x^{\nu}} = 0 .$$
This should have been clear to you because the inhomogeneous relation $F(x)=Ax + b$ is not linear in the algebraic sense, but it is a linear relation in the sense used in analytic geometry. Having said this, one can still speak of linear Poincare’ transformations: with every element $(\Lambda , a)$ of the Poincare’ group, we can associate a $5 \times 5$ matrix $\Gamma$ defined by
$$\Gamma = \begin{pmatrix} \Lambda^{\mu}{}_{\nu} & a^{\mu} \\ 0_{4\times 4} & 1 \end{pmatrix} . \ \ \ \ \ \ \ \ (1)$$
Then the multiplication law in the Poincare’ group $(\Lambda_{2},a_{2})\cdot (\Lambda_{1},a_{1}) = (\Lambda_{2}\Lambda_{1} , a_{2} + \Lambda_{2}a_{1})$ shows that the correspondence $(\Lambda , a) \to \Gamma$ is an isomorphism of the Poincare’ group on the subgroup of $GL(5,\mathbb{R})$ consisting of matrices of the form (1), where $\Lambda$ satisfies $\Lambda^{T}\eta \Lambda = \eta$ and $a$ is an arbitrary 4-vector. The Poincare group can, therefore, be identified with this matrix Lie group. And Minkowski spacetime $M^{4}$ can be identified with the hyperplane $x^{4}=1$ in $\mathbb{R}^{5}$ with coordinates $(x^{0}, x^{1}, … , x^{4})$. Then the linear operator (1) acts on this hyperplane as the corresponding Poincare transformation $(\Lambda , a)$.

The principle of Relativity only implies the conformal group and this means

This is incorrect. The conformal group does not even act on Minkowski space $M^{4}$. It acts on the (conformally) compactified version of Minkowski space $\bar{M}^{4}\cong S^{3}\times S^{1} / Z_{2}$.

An argument like the one given in Landau and Lifshitz Volume 2 now proves $\alpha=1$

Did you read the whole post? I used argument similar to Landau’s argument to show $\alpha = 1$.

Finally it can be shown(see, for instance, "Gravitation and Cosmology" by S. Weingberg) that the only transformations that leave $ds^2$ invariant are linear tranformations.

The theorem you are talking about is the following
“The coordinate transformation from one Minkowski chart to another is a Poincare’ transformation” , or equivalently stated “The Poincare’ group is the maximal symmetry group of Minkowski spacetime”.
Again, you should read my posts in that thread, because I proved the infinitesimal version of this theorem in there.

 

[strangerep]

The main diificulty in going to 4D is my point 8, about isotropy. How to formultate this mathematically in a sufficiently simple manner?

Well, here is a sketch to get you started...

Pick an arbitrary (fixed) direction in 3-space, denoted by the unit vector $\widehat {\bf v}$. We consider a transformation to a frame with 3-velocity ${\bf v} \equiv v \widehat {\bf v}$, where the nonbold $v$ is a real number.

Write the linear transformations in the form: $$t' ~=~ a(v)t + {\bf b}({\bf v})\cdot {\bf x} ~,~~~~ {\bf x'} ~=~ {\bf d}({\bf v})t + {\bf E}({\bf v}) {\bf x} ~,~~~~~~~ (1)$$ where ${\bf b}, {\bf d}$ are 3-vector-valued functions and ${\bf E}$ is a $3\times 3$ matrix-valued function.

Since ${\bf b}({\bf v})$ is 3-vector-valued and depends only on ${\bf v}$, it is necessarily of the form $b(v){\bf v}$, where the nonbold $b$ is a new function, now scalar-valued. A similar argument applies to ${\bf d}({\bf v})$. So we can rewrite the transformation equations (1) as: $$t' ~=~ a(v)t + b(v) {\bf v}\cdot {\bf x} ~,~~~~ {\bf x'} ~=~ d(v){\bf v}t + {\bf E}({\bf v}) {\bf x} ~.~~~~~~~ (2)$$
Now decompose ${\bf x}$ wrt $\widehat {\bf v}$ as $${\bf x} ~=~ {\bf x}_\| + {\bf x}_\perp$$ into parts parallel and perpendicular to $\widehat {\bf v}$. I.e., ${\bf x}_\| = x_\| \widehat {\bf v}$ and $\widehat {\bf v} \cdot {\bf x}_\perp = 0$. Also decompose ${\bf x}'$ similarly.

I invite you (or any other readers of this thread) to continue the above via the following exercises.

Exercise 1 (Easy): Using $\widehat {\bf v} \cdot {\bf x} = v x_\|$, what form do the transformation equations now take?

Exercise 2 (Harder): Contracting both sides of the ${\bf x'}$ equations with $\widehat {\bf v}$, and using the definition of spatial isotropy I gave earlier, deduce 2 distinct equations governing the transformations for $x'_\|$ and ${\bf x}'_\perp$ separately.

[Edit: Heh, who am I kidding? No one's going to do those.]

 

[facenian]

The term “linear” in that thread is not the same as linearity in the algebraic sense: f(ax+by)=af(x)+bf(y) f(ax + by) = af(x) + bf(y). Linearity there means polynomial of degree one in the coordinates, i.e., solutions to the system of 2-order PDE’s

∂ 2 F σ ∂x μ ∂x ν =0.​

Yes, and this is precisely the kind of linearity I was reffering to. However I reaffirm my only objection to your demonstration, i.e. that straight lines transforming into straight lines requires that the transformation be linear.(this is the only argument you should attack)

This is incorrect. The conformal group does not even act on Minkowski space M 4 M^{4}. It acts on the (conformally) compactified version of Minkowski space M ¯ 4 ≅S 3 ×S 1 /Z 2 \bar{M}^{4}\cong S^{3}\times S^{1} / Z_{2}.

You are right, I should have said that the light principle(LP) only implies that the transformation is conformal

Did you read the whole post? I used argument similar to Landau’s argument to show α=1 \alpha = 1.

As for the rest of your comments ,I was only sketching a demonstration and citing known authors because some other people may read the post. Yes I read the post and I noticed you used a similar agument.

 

[samalkhaiat]

However I reaffirm my only objection to your demonstration, i.e. that straight lines transforming into straight lines requires that the transformation be linear.(this is the only argument you should attack)
.

How about you give me a counter-example?
Okay, I give you a transition functions $F_{21}: x \to \bar{x}$ between two Minkowski charts. It ia assumed that $F_{21}$ is a $\mathscr{C}^{3}$ homeomorphism, i.e., continuosly thirce differentiable (smooth and regular). Furthermore, we demand that $F_{21}$ (or its inverse) maps straight (world) lines onto straight lines. Now, you show me one such homeomorphism that does not correspond to linear tranformation?

 

[facenian]

How about you give me a counter-example?
Okay, I give you a transition functions F21:x→x¯F_{21}: x \to \bar{x} between two Minkowski charts. It ia assumed that F21F_{21} is a C3\mathscr{C}^{3} homeomorphism, i.e., continuosly thirce differentiable (smooth and regular). Furthermore, we demand that F21F_{21} (or its inverse) maps straight (world) lines onto straight lines. Now, you show me one such homeomorphsm that does not correspond to linear tranformation?

Well, the discussion, as I understood it, is not at the mathematical level you are using. I' m sorry if I misunderstood that. At the level I took it there is no place for homeomorphisims, diffeomorphisms, bijective maps, etc. May be I will look like an ignorant fool to you but I'm not the only one, for instance, "Einstein Gravity in a Nutshell" by Zee was written at the mathematical level I'm using in this discussion. Please don't take this the wrong way, I respect the mathematics and I'm not saying it in a demeaning way, on the contrary, I think it's way over my head.
Having clarified this point, what I meant is that many authors working at the mathematical level to which I am referring (Einstein included), attribute the linearity either to the homogeneity and isotropy of space and time or to thecondition that straight lines must be transformed in straight lines, without further explanation, which I find inappropriate even for this level of rigor. At this level of rigor, a fractional linear transformation can work as a counter example, of course even I Know that this is not an homeormorphism for R4

 

[PeterDonis]

the discussion, as I understood it, is not at the mathematical level you are using.

This thread is at an "I" level, which means undergraduate level. However, the subthread you are participating in is probably on the borderline between "I" and "A" (which is graduate level). That's probably unavoidable given the nature of the topic; to talk about "derivation" of something you have to have enough rigor to be able to precisely specify the starting point and the conclusion.

 

[facenian]

his thread is at an "I" level, which means undergraduate level. However, the subthread you are participating in is probably on the borderline between "I" and "A" (which is graduate level). That's probably unavoidable given the nature of the topic; to talk about "derivation" of something you have to have enough rigor to be able to precisely specify the starting point and the conclusion.

This thread started with an innocent question and I believe by now it went too far, however I must confess I really enjoyed it and learned a lot from it.