How do you solve this system of equations ($c$ and $v$ are constants):
$\beta\left(\beta+v\gamma\right)=1$
$c^2\gamma(\beta+v\gamma)=-v$
I somehow seem to be stuck and just can't figure out how to solve them in an effective way.
0
$\begingroup$
$\endgroup$
-
$\begingroup$ If $\beta$ and $\gamma$ are your unknowns, then your system ain't linear. $\endgroup$ – Ivan Neretin Sep 29 '16 at 20:20
-
$\begingroup$ @IvanNeretin Oh yeah, I inadvertently put "linear" in the title. I fixed it. $\endgroup$ – Jannik Pitt Sep 29 '16 at 20:22
-
$\begingroup$ Have you tried solving the first equation for $\gamma$ and then substituting into the second? $\endgroup$ – cdwe Sep 29 '16 at 20:24
1
$\begingroup$
$\endgroup$
Hint:
From the first equation you have $$ \gamma=\frac{1-\beta^2}{v\beta} $$
so the second equation becomes: $$ c^2(1-\beta^2)=-v^2\beta^2 $$
answered Sep 29 '16 at 20:31
-
$\begingroup$ For $\nu=0$ we have $\beta^2=1$ and $c\gamma=0$. Of course $\beta$ can't be zero. $\endgroup$ – Dietrich Burde Sep 29 '16 at 21:03
