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


Hints :

For your information the particular transformation is the 3+1-Lorentz Transformation of Special Relativity (3-space + 1-time coordinates). The invariance of the expression we want to prove is identical to the 2nd Einstein's axiom :

Light signals in vacuum are propagated rectilinearly, with the same speed $c$, at all times, in all directions, in all inertial frames.

Usually, we follow the inverse path : we try to find the transformation from the demand of invariance of this expression. We derive first the 1+1- and later on the 3+1-Lorentz Transformation.

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}


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