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Is it possible to derive the matrix: $$\Lambda=\begin{pmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma \end{pmatrix}$$ From the condition: $$\Lambda g\Lambda^T=g \ \ \ \ g=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$ I couldn't find this way of deriving the Lorentz transformations online and also couldn't think of a way to derive them myself. I want to see them derived that way because this condition is deeply connected with preserving the canonical form of the wave equations arising from the Lorentz gauge of the electromagnetic potentials. Here is my shot:

We assume: $$\Lambda=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & c \\ b & d \end{pmatrix}=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -a & -c \\ b & d \end{pmatrix}=\begin{pmatrix} -a^2+b^2 & -ac+bd \\ -ac+bd & -c^2+d^2 \end{pmatrix}=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \\ \Rightarrow \begin{cases} a^2-b^2=1 \\ d^2-c^2=1 \\ ac-bd=0 \end{cases}$$ Where do I go from here?

Krum Kutsarov
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4 Answers4

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Let $$\Lambda = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ $$\Lambda g \Lambda^T = g$$ By taking the determinant of both sides we conclude $$\det{\Lambda}\det{\Lambda^T} =1 $$ $$\det{\Lambda} = \pm 1$$ We expect that the identity matrix is one valid form of a Lorentz transformation, and so $\det{\Lambda}=1.$ $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} a & c \\ b & d \end{pmatrix} =\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} $$ We also assume that the Lorentz transform is invertible: $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} d & -c \\ -b & a \end{pmatrix} $$

So $$\begin{pmatrix} -a & b \\ -c & d \end{pmatrix} = \begin{pmatrix} -d & c \\ -b & a \end{pmatrix}.$$ Implying $a= d,$ $b= c.$ Now the matrix looks like $$\begin{pmatrix} \gamma & b \\ b & \gamma \end{pmatrix}$$ together with the determinant condition $a^2 - b^2 =1$ lets us rewrite $b=\pm\sinh w=\gamma\beta,\,a=\cosh w=\gamma$ in terms of rapidity. So we have $$\begin{pmatrix} \gamma & \pm \gamma \beta \\ \pm \gamma \beta & \gamma \end{pmatrix}$$ The factor of $\pm$ here reflects the fact that both $\beta$ and $-\beta$ are valid, they are just velocities in different directions.

J.G.
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Jbag1212
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Is it possible to derive the matrix: $$\Lambda=\begin{pmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma \end{pmatrix}$$ From the condition: $$\Lambda g\Lambda^T=g \ \ \ \ g=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$$

No. It is not possible to derive the matrix $\Lambda$ from the above condition alone.

Depending on a choice of sign, you will either end up with: $$ \Lambda = \left(\begin{matrix}\gamma & -\beta\gamma \\ -\beta\gamma & \gamma\end{matrix}\right) $$ or $$ \Lambda = \left(\begin{matrix}\gamma & -\beta\gamma \\ +\beta\gamma & -\gamma\end{matrix}\right)\;. $$

If you also demand the condition $$ \det(\Lambda)=1 $$ then you can derive the form of $\Lambda$ that you seek, but note that at this point $\gamma$ is still an arbitrary real number, not necessarily greater than 1 (you need to impose further conditions on $\gamma$ to make physical sense).

Update:

As mentioned in the comments, an orthochronous condition should be imposed since, usually, we take $\gamma$ to mean: $$ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \ge 1 $$

Regarding the condition that $$ \det(\Lambda) = 1\;, $$ the transformations with $\det(\Lambda)=1$ are called "proper," by definition. These includes boosts and rotations. The "improper" transformations include parity and time reversal.

hft
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As I mentioned in one of the comments. You can solve for the fact that $\beta = \frac{v}{c}$ if you add the condition that, $$ \Lambda \begin{bmatrix}1 \\ 0 \end{bmatrix} = \frac{1}{\sqrt{1 - v^2}} \begin{bmatrix} 1 \\ v\end{bmatrix} $$ For simplicity, I am working in units where $c = 1$. This simply says that a boost does what we expect, it maps a four-velocity with three-velocity zero to a four-velocity with three-velocity $v$.

Josh Newey
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Here's a quick way to see that you can't do it, choose $\Lambda = g$. Then, $\Lambda^T=\Lambda=g$ and you can check that: $$ \Lambda g \Lambda^T = g g g = g $$

But $g$ is definitely not a Lorentz transformation.

Amit
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