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The Lorentz transformations can be derived from (a) Principle of Relativity and (2) group axioms. I was looking at the derivation here, and I have problem understanding one specific step. In the derivation one somehow argues that the determinant of the matrix should be $1$. That is done in the following steps, I quote directly,

Combining these two gives $\alpha=\gamma$ and the transformation matrix has simplified, $$ \left[\begin{array}{l} t^{\prime} \\ x^{\prime} \end{array}\right]=\left[\begin{array}{cc} \gamma & \delta \\ -v \gamma & \gamma \end{array}\right]\left[\begin{array}{l} t \\ x \end{array}\right] $$ Now consider the group postulate inverse element. There are two ways we can go from the $K$ coordinate system to the $K$ coordinate system. The first is to apply the inverse of the transform matrix to the $K$ coordinates: $$ \left[\begin{array}{l} t \\ x \end{array}\right]=\frac{1}{\gamma^{2}+v \delta \gamma}\left[\begin{array}{cc} \gamma & -\delta \\ v \gamma & \gamma \end{array}\right]\left[\begin{array}{l} t^{\prime} \\ x^{\prime} \end{array}\right] $$ The second is, considering that the $K$ coordinate system is moving at a velocity $v$ relative to the $K$ coordinate system, the $K$ coordinate system must be moving at a velocity $-v$ relative to the $K$ coordinate system. Replacing $v$ with $-v$ in the transformation matrix gives: $$ \left[\begin{array}{l} t \\ x \end{array}\right]=\left[\begin{array}{cc} \gamma(-v) & \delta(-v) \\ v \gamma(-v) & \gamma(-v) \end{array}\right]\left[\begin{array}{l} t^{\prime} \\ x^{\prime} \end{array}\right] $$ Now the function $\gamma$ can not depend upon the direction of $v$ because it is apparently the factor which defines the relativistic contraction and time dilation. These two (in an isotropic world of ours) cannot depend upon the direction of $v$. Thus, $\gamma(-v)=\gamma(v)$ and comparing the two matrices, we get $$ \gamma^{2}+v \delta \gamma=1 $$

But comparing last two matrices I get,

$$\delta(-v)=\frac{-\delta}{\gamma^2+v\delta\gamma}$$ Therefore the determinant of the transformation matrix is $1$ only if $-\delta=\delta(-v)$. But how to argue that? I don't see the argument being presented here. Can someone help?

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If two matrices are equal, then they are equal component-wise. If you compare the top left entries of the two matrices, then you find that $$\gamma(-v) = \frac{\gamma(v)}{\gamma^2 + v\delta \gamma}$$

Since you've argued that $\gamma(-v)=\gamma(v)$, the result follows directly. If you then compare the top right entries of the two matrices, you find that indeed $\delta(-v)=-\delta(v)$ as well.

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  • $\begingroup$ OK! I have been really stupid! It is obvious. $\endgroup$ Nov 3, 2020 at 14:51

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