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I'm trying to understand a step in a derivation of the Lorentz transformation, that my professor gave in class. We start assuming the homogeneity and the isotropy of the 4 dimensional spacetime and then we consider two inertial frames of reference $S$ and $S'$, with $S'$ moving at speed $v$ along the $x$-axis of $S$. We also assume that $S$ and $S'$ have parallel axes and their origins coincide at time $t=0$ in $S$. So a general transformation between the coordinates of $S$ and $S'$ respectively is, \begin{align} t\qquad &\to\qquad t'=T(t,x,y,z,v)\\ x\qquad &\to\qquad x'=X(t,x,y,z,v)\\ y\qquad &\to\qquad y'=Y(t,x,y,z,v)\\ z\qquad &\to\qquad z'=Z(t,x,y,z,v) \ . \end{align} Then, applying homogeneity we find that the transformation must be linear, so \begin{equation} \begin{pmatrix}t'\\x'\\y'\\z'\end{pmatrix}=A(v)\begin{pmatrix}t\\x\\y\\z\end{pmatrix} \ , \end{equation} where $A(v)$ is a $4\times4$ matrix. Using the principle of relativity we find that directions perpendicular to the motion don't change, so \begin{equation} A(v)=\begin{pmatrix}A_1(v)&A_2(v)\\\mathbf{0}&\mathbf{1}\end{pmatrix} \ , \end{equation} where $A_1(v)$ and $A_2(v)$ are $2\times 2$ matrices, and $\mathbf{0}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$, $\mathbf{1}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$. Now comes the step that I don't understand, it says that:

From the isotropy of space, follows that $A_2(v)=\mathbf{0}$

Can please someone help me to understand how the isotropy of space have this implication?

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  • $\begingroup$ $A(v)$ is a boost in the $x$ direction. $\endgroup$ Jan 22, 2020 at 20:53

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I have a feeling there should be a physical reason for the off-diagonal elements both having to be zero, but I can't think of one off-hand. Here's another way to show it, though:

Consider, for example, a point particle whose rest frame is $S^\prime$. To you, the observer sitting in $S$, this particle would be moving away along your $x$ axis. Now, what about the $y$ and $z$ axes? Well, they shouldn't be important here, as these directions are orthogonal to the motion of the particle. In other words, once you have chosen your $x$ axis to be along the direction of motion of the particle, you have an infinite number of $y$ and $z$ axes that can be chosen -- all related by simple rotations around the $x$-axis -- which must all give the same $A(v)$ matrix. This is one of the assumptions of isotropy.

Suppose, instead of the $(t,x,y,z)$ you used $(t,x,Y,Z)$, where $Y$ and $Z$ are two different mutually perpendicular directions that are also perpendicular to $x$. Since space is isotropic, your definition of $y$ and $z$ should not affect your transformation matrix, and so

$$\begin{pmatrix}t^\prime\\x^\prime\end{pmatrix} = A_1(v) \begin{pmatrix}t\\x\end{pmatrix} + A_2(v) \begin{pmatrix}y\\z\end{pmatrix}$$

$$\begin{pmatrix}t^\prime\\x^\prime\end{pmatrix} = A_1(v)\begin{pmatrix}t\\x\end{pmatrix} + A_2(v) \begin{pmatrix}Y\\Z\end{pmatrix}$$

Or $$A_2(v) \begin{pmatrix}y\\z\end{pmatrix} = A_2(v) \begin{pmatrix}Y\\Z\end{pmatrix}$$

It should be intuitively clear that since $Y$ and $Z$ could be any possible orthogonal set (also orthogonal to $x$), this must mean that $A_2(v)=\mathbf{0}$, but if you wish to be a little more rigorous, these new $Y,Z$ axes can be obtained from $y,z$ by rotation of some angle $\theta$ around the $x$ axis, and so

$$\begin{pmatrix}Y\\Z\end{pmatrix} = R(\theta) \begin{pmatrix}y\\z\end{pmatrix},$$ where $R(\theta)$ is the usual rotation matrix. The above equality then means that for any arbitrary value of $\theta$,

$$A_2(v) \left(\mathbf{1} - R(\theta) \right) = \mathbf{0}.$$

Since $\theta$ and $v$ are both arbitrary, it must be that $A_2(v) = \mathbf{0}$.

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  • $\begingroup$ So if the $A(v)$ is a boost in a general direction, space is no longer isotropic? $\endgroup$ Jan 22, 2020 at 21:14
  • $\begingroup$ I don't quite understand, the Lorentz transformations do distinguish between the direction parallel to the boost and those perpendicular to it. If the boost were in a general direction, then one could always rotate one's coordinate system to be along that direction and call it $x$, and the same argument I made would hold. I think the exact argument could also be made for a general transformation, by defining a direction $\hat{n}$ and two orthogonal basis vectors $Y,Z$ that are perpendicular to $\hat{n}$... $\endgroup$
    – Philip
    Jan 22, 2020 at 21:25
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    $\begingroup$ Thank you! Very clear! $\endgroup$ Jan 23, 2020 at 7:41
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    $\begingroup$ @Philip: you just need to careful when considering more then one boost since two boost create a rotation. And I don't believe the OP can prove space is isotropic mathematically. Locally, it can inferred from conservation of linear and angular momentum. But in general relativity, it isn't that simple - homogeneity and the isotropy of space are typically assumed and conservation of linear and angular momentum demonstrated. $\endgroup$ Jan 23, 2020 at 22:00
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    $\begingroup$ @Philip - actually, I just realized I misread the post - the OP assumes the isotropy of space - my bad. In which case, I presume the OP just didn't realize when the boost is in the $x$ direction, the boost plane is $t-x$ - and it doesn't include the other coordinates. You were right. $\endgroup$ Jan 23, 2020 at 22:54

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