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I'm a bit stuck in the part of special relativity, as all the books I've read assume previous knowledge of the topic.

I would like to know how can I show that a quantity is invariant under Lorentz rotations and boosts. From my understanding with Lorentz boosts, I just have to prove that the quantity remains constant after applying the transformation.

For example, for the invariant:

$(ct)^2-x^2-y^2-z^2=(ct')^2-x'^2-y'^2-z'^2$

If I apply, for example, a Lorentz boost of the form:

$t'=\gamma(t-vx/c^2) \ \ , \ \ x'=x(\gamma-vt) \ \ , \ \ y'=y \ \ , \ \ z'=z$

It's just a matter of substituting and developing terms, so it's straightforward.

However, for Lorentz rotations I'm not entirely sure how to proceed as I don't know how to build the rotation that is still Lorentz invariant (all the articles only show the case for Lorentz boosts).

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Rotations only change the spatial coordinates $(x^i)$; the time coordinate $(x^0)$ stays unchanged.

Now suppose you're rotating around the $z$ axis. Then the rotation matrix $(R)$ for this is:

\begin{equation} R = \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0\\ \sin{\theta} & \cos{\theta} & 0\\ 0 & 0 & 1 \end{bmatrix} \end{equation}

This induces a rotation of coordinates $(x \rightarrow x')$ in component form as: \begin{equation} x'^i = R^i_{\ j}x^j \end{equation}

Note that the $3 \times 3$ matrix above is the spatial part of the $4 \times 4$ Lorentz transformation matrix $\Lambda$.

Now to show Lorentz invariance under this special case of rotation around $z$ axis, we just need to show that $(x')^2 + (y')^2 = (x)^2 + (y)^2$, which is trivial.

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  • $\begingroup$ Thank you very much, so I only had to apply a normal rotation (for some reason I thought I had to include gamma factors in it). I understand now, this really helped me. $\endgroup$ – Charlie Aug 21 '18 at 0:04
  • $\begingroup$ @Charlie $\gamma$, or more usually, rapidity $\eta = v/c$ is the parameter for boosts. For rotations, the parameter is the angle $\theta$. $\endgroup$ – Avantgarde Aug 21 '18 at 0:15
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Why a Rotation matrix $R$ don't "destroy" the Lorentz invariant

Prove what Mr. Avantgarde wrote

\begin{align*} & \text{General Lorentz Transformation matrix}\\\\ \Lambda&=\begin{bmatrix} \gamma & \gamma \,\vec{v}^T \\ \gamma \,\vec{v}^T & I+\frac{\gamma-1}{\vec{v}^2}\,\vec{v}\,\vec{v}^T \\ \end{bmatrix}\\\\ &\text{with:}\\ \gamma&=\frac{1}{\sqrt{1-\frac{\vec{v}^2}{c^2}}}\\\\ \vec{v}&=\begin{bmatrix} v_x \\ v_y \\ v_z \\ \end{bmatrix}\,,\quad\text{velocity boost vector}\\ I&=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end{bmatrix}\,,\quad\text{$3\times3$ identity matrix}\\\\ &\text{Line}\\ s^2&=x_\mu\,x^\mu=\eta_{\mu\nu}x^\nu\,x^\mu\\ s^2&=\vec{x}^T\,\eta\,\vec{x}=(x^0)^2-(x^1)^2-(x^2)-(x^3)\,,\quad \text{with}\\ \vec{x}&=\begin{bmatrix} x^0 \\ \vec{x}^i \\ \end{bmatrix}\,,\quad \text{and}\\ \eta&=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix} \end{align*}

I) Lorentz transformation: Pure boost

\begin{align*} &\text{line}\\ s'^2&=x'_\mu\,x'^\mu\,,\quad\text{with}\quad x'^\mu=\Lambda\,\vec{x}\quad\text{and}\quad x'_\mu=\eta\,\vec{x}\,\quad\Rightarrow \\ s'^2&=\vec{x}^T\,\underbrace{\Lambda^T\,\eta\,\Lambda}_{=\eta}\vec{x}= \vec{x}^T\,\eta\,\vec{x}=s^2\,,\quad\text{Lorentz invariant} \end{align*}

II) Lorentz transformation: Boost plus rotation matrix $R$

\begin{align*} \vec{x}&\mapsto \begin{bmatrix} x^0 \\ R\,\vec{x}^i \\ \end{bmatrix}\\ \quad \Rightarrow\\ &\text{line}\\ s'^2&=\vec{x}^T\,\underbrace{\Lambda^T\,\eta\,\Lambda}_{=\eta},\vec{x}= \vec{x}^T\,\eta\,\vec{x}= \begin{bmatrix} x^0 & \left(R\,\vec{x}^i\right)^T \\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix} \begin{bmatrix} x^0 \\ R\,\vec{x}^i \\ \end{bmatrix}\\ &=\begin{bmatrix} x^0 & \vec{x^i}^T R^T \\ \end{bmatrix} \begin{bmatrix} x^0 \\ -R\,\vec{x}^i \\ \end{bmatrix}=(x^0)^2-(x^1)^2-(x^2)-(x^3)\,, \quad\text{again Lorentz invariant}\\ &\text{with}\quad R^T\,R=I \end{align*}

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