# Intuitively arguing that the determinant of Lorentz boost matrices is unity

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?

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.