# A problem in deriving Lorentz transformation from homogeneity and isotropy of spacetime and the principle of relativity

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 $$$$\begin{pmatrix}t'\\x'\\y'\\z'\end{pmatrix}=A(v)\begin{pmatrix}t\\x\\y\\z\end{pmatrix} \ ,$$$$ where $$A(v)$$ is a $$4\times4$$ matrix. Using the principle of relativity we find that directions perpendicular to the motion don't change, so $$$$A(v)=\begin{pmatrix}A_1(v)&A_2(v)\\\mathbf{0}&\mathbf{1}\end{pmatrix} \ ,$$$$ where $$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?

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

## 1 Answer

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 $$A_2(v) = \mathbf{0}$$.

• So if the $A(v)$ is a boost in a general direction, space is no longer isotropic? Jan 22, 2020 at 21:14
• 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 $\hat{n}$... Jan 22, 2020 at 21:25
• Thank you! Very clear! Jan 23, 2020 at 7:41
• @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. Jan 23, 2020 at 22:00
• @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. Jan 23, 2020 at 22:54