Relativistic sum with magnitude c

Question

Pick any two vectors (in 3 dimensions) having magnitude equal to c and check whether the relativistic sum of them also has magnitude c. Is u v equal to v u?

Answer 1

Let $ ||u||_2 = ||v||_2 = c$. Then: \begin{align} ||u \oplus v||_2 &= \frac{1}{c^2 + u\cdot v}\left\vert\left\vert c^2(u+v) + \frac{u\times(u\times v)}{1 + \sqrt{1-(u\cdot u)/c^2}} \right\vert\right\vert_2 \\ &= \frac{1}{c^2 + u\cdot v} || \underbrace{c^2(u+v) }_a + \underbrace{ u\times(u\times v) }_b ||_2 \\ &= \frac{1}{c^2 + u\cdot v} \sqrt{ a\cdot a + b\cdot b + 2(a\cdot b) } \\ \end{align} Expanding these dot products: \begin{align} a\cdot a &= ||c^2(u+v)||_2^2 \\ &= c^4[(u\cdot u) + (v\cdot v) + 2(u\cdot v)] \\ &= 2c^4[c^2 + (u\cdot v)]\\ b\cdot b &= ||u\times(u \times v)||_2 \\ &= ||u||_2^2 ||u\times v||_2^2 - \underbrace{[u\cdot (u\times v)]^2}_0 \\ &= c^2[||u||_2^2 ||v||_2^2 - (u\cdot v)^2] \\ &= c^2[c^4 - (u\cdot v)^2] \\ a\cdot b &= c^2[(u+v)\cdot (u\times(u\times v))] \\ &= c^2[ \underbrace{u \cdot (u\times(u\times v))}_0 + v \cdot (u\times(u\times v)) ] \\ &= c^2[(u\times v)\cdot (v\times u)] \\ &= -||u\times v||_2^2 \\ &= -[c^4 - (u\cdot v)^2] \end{align} Plugging back in: \begin{align} ||u \oplus v||_2 &= \frac{1}{c^2 + u\cdot v} \sqrt{ 2c^4[c^2 + (u\cdot v)] + c^2[c^4 - (u\cdot v)^2] -2[c^4 - (u\cdot v)^2]c^2 } \\ &= \frac{c}{c^2 + u\cdot v} \sqrt{ c^4 + 2(u\cdot v) c^2 + (u\cdot v)^2 } \\ &= \frac{c}{c^2 + u\cdot v} [c^2 + (u\cdot v)] \\ &= c \end{align} as expected.

As for the second part, when the norms are $c$, $u\oplus v = v\oplus u$, since $u$ and $v$ were arbitrary.