# Prove invariance for a particular transformation

Consider inertial frames $$\Sigma$$ and $$\Sigma'$$ that are coincident at time $$t\boldsymbol{=}t'\boldsymbol{=}0$$. The relative velocity of $$\Sigma'$$ with respect to $$\Sigma$$ is $$\vec \upsilon$$, not necessarily aligned with one of the axes. The transformation from $$\left(t,\vec r\right)$$ to $$\left(t',\vec r'\right)$$ is $$\begin{equation} t'\boldsymbol{=}\gamma_{\upsilon}\left(t\boldsymbol{-}\dfrac{\vec \upsilon \cdot \vec r }{c^2}\right)\: ; \quad \vec r'\boldsymbol{=}\vec r\boldsymbol{+}\alpha_{\upsilon}\left(\vec \upsilon \cdot \vec r\right)\vec \upsilon\boldsymbol{-}\gamma_{\upsilon}\vec \upsilon t \nonumber \end{equation}$$ where $$\upsilon\boldsymbol{=}\vert\vec \upsilon \vert$$ and $$\alpha_{\upsilon}\boldsymbol{=}\dfrac{\gamma_{\upsilon}\boldsymbol{-}1}{\upsilon^2}\boldsymbol{=}\dfrac{\gamma^2_{\upsilon}/c^2}{\gamma_{\upsilon}\boldsymbol{+}1}$$.

Show that $$c^2t^2\boldsymbol{-}\vec r\cdot \vec r$$ is invariant under this transformation.

We are asked to prove the invariance of the equation for the transformation above. I started by subbing in the equations for t' and $$\vec{r'}$$ where t and $$\vec{r}$$ appear in the invariant equation several time at this point however I can't seem to get it to reduce back down to the original equation. Is there something needed beyond algebra?

One of the areas I'm worried I might be wrong is in calculating r'.r', which I do the way I would do any algebraic expressions. do some of the dot products give zero or something? I know that $$\vec{v}.\vec{v}$$ = $${v^2}$$ and that cancels in some instances with the expression for alpha but I can't seem to reduce it beyond that.

So my first line is: $${\gamma^2}{c^2}{t^2}-2{\gamma^2}(\vec{v}.\vec{r})t+{\gamma^2}(\vec{v}.\vec{r})/c^2-\vec{r}.\vec{r}-2{\alpha}{(\vec{v}.\vec{r})^2}+2{\gamma}(\vec{v}.\vec{r})t-{\alpha^2}(\vec{v}.\vec{r}){v^2}+2{\alpha}{\gamma}{v^2}t-{\gamma^2}{v^2}{t^2}$$

Light signals in vacuum are propagated rectilinearly, with the same speed $$c$$, at all times, in all directions, in all inertial frames.
For your question now, you are a step before the proof : $$\begin{equation} \underbrace{{\gamma^2}{c^2}{t^2}}_{\boxed{1}}\underbrace{-2{\gamma^2}(\vec{v}.\vec{r})t}_{\boxed{2}}\underbrace{+{\gamma^2}\overbrace{(\vec{v}.\vec{r})^2}^{missing\, 2}/c^2}_{\boxed{3}}\underbrace{-\vec{r}.\vec{r}\vphantom{/c^2}}_{\boxed{4}}\underbrace{-2{\alpha}{(\vec{v}.\vec{r})^2}}_{\boxed{5}}\underbrace{+2{\gamma}(\vec{v}.\vec{r})t}_{\boxed{6}}\underbrace{-{\alpha^2}\overbrace{(\vec{v}.\vec{r})^2}^{missing\, 2}{v^2}}_{\boxed{7}}\underbrace{+2{\alpha}{\gamma}\overbrace{(\vec{v}.\vec{r})}^{missing}{v^2}t}_{\boxed{8}}\underbrace{-{\gamma^2}{v^2}{t^2}}_{\boxed{9}} \nonumber \end{equation}$$ Since $$\begin{equation} \boxed{1}+\boxed{9}+\boxed{4}=c^2t^2-\vec{r}\cdot\vec{r} \nonumber \end{equation}$$ you must prove that $$\begin{equation} \underbrace{\boxed{2}+\boxed{6}+\boxed{8}}_{A\cdot t}+\underbrace{\boxed{3}+\boxed{5}+\boxed{7}}_{B}=A\cdot t+B=0 \nonumber \end{equation}$$