Prove that x 2+y2+z2 -c2 t 2 is invariant under Lorentz transformation.

Answer: substituting the Lorentz transformation into \(x'^{2}+y'^{2}+z'^{2}-c^{2}t'^{2}\) gives back exactly \(x^{2}+y^{2}+z^{2}-c^{2}t^{2}\), so the quantity is invariant.

Lorentz transformation (S′ moving with velocity \(v\) along x), with \(\gamma = \dfrac{1}{\sqrt{1-v^{2}/c^{2}}}\):

\[x' = \gamma(x - vt),\quad y' = y,\quad z' = z\]

\[t' = \gamma\left(t - \frac{vx}{c^{2}}\right)\]

Since \(y' = y\) and \(z' = z\), we only need to show \(x'^{2} - c^{2}t'^{2} = x^{2} - c^{2}t^{2}\).

\[x'^{2} = \gamma^{2}(x^{2} - 2xvt + v^{2}t^{2})\]

\[c^{2}t'^{2} = \gamma^{2}\left(c^{2}t^{2} - 2xvt + \frac{v^{2}x^{2}}{c^{2}}\right)\]

Subtracting, the \(2xvt\) terms cancel:

\[x'^{2} - c^{2}t'^{2} = \gamma^{2}\left[x^{2}\left(1 - \frac{v^{2}}{c^{2}}\right) - c^{2}t^{2}\left(1 - \frac{v^{2}}{c^{2}}\right)\right]\]

Since \(\gamma^{2}\left(1 - \dfrac{v^{2}}{c^{2}}\right) = 1\):

\[x'^{2} - c^{2}t'^{2} = x^{2} - c^{2}t^{2}\]

Adding \(y'^{2} + z'^{2} = y^{2} + z^{2}\):

\[x'^{2}+y'^{2}+z'^{2}-c^{2}t'^{2} = x^{2}+y^{2}+z^{2}-c^{2}t^{2}\]

Hence the quantity is invariant under Lorentz transformation. (Physically, this is why a light pulse that spreads as a sphere, \(x^{2}+y^{2}+z^{2} = c^{2}t^{2}\), in one frame also spreads as a sphere at speed \(c\) in the other frame.)