On Einstein's famous paper, in the section on the transformation of light energy between frames of reference, Einstein devises a thought experiment to determine it. He imagines two frames of reference, $K$ and $k$, in which $k$ moves relative to $K$ with velocity $v$ along the x-axis. The coordinates and time in frame $K$ are $x, y, z,$ and $t$. The coordinates in time for $k$ are $\xi, \eta, \zeta,$ and $\tau$. He imagines that a plane light wave travels at an angle from the axis with $l, m,$ and $n$ being the cosines between the wave-normal and the three axis, respectively. He then imagines an imaginary spherical surface which moves in the same direction as the light, such that as it travels, it always encloses the same 'bit' of light. The equation for this sphere is $$(x-clt)^2+(y-cmt)^2+(z-cnt)^2=R^2$$ To find the volume of this 'sphere' in the $k$ frame of reference, he applies the Lorentz Transformation to the equation of the sphere and obtaining an ellipsoid. Setting $\tau=0$: $$(\beta\xi-l\beta\frac{v}{c}\xi)^2+(\eta-m\beta\frac{v}{c}\xi)^2+(\zeta-n\beta\frac{v}{c}\xi)^2=R^2$$ So far so great, but then he states that "through a simple calculation" it can be shown that the ratio between the volume $S'$ of the ellipsoid and the volume $S$ of the sphere is $$\frac{S'}{S}=\frac{\sqrt{1-\frac{v^2}{c^2}}}{1-l\frac{v}{c}}$$ What is this 'Simple Calculation' through which we can determine the volume of the ellipsoid?

  • $\begingroup$ Should there be a $\zeta$ somewhere in the second formula? $\endgroup$ – Chappers Feb 25 '17 at 16:38
  • $\begingroup$ Ah, yes, my mistake. $\endgroup$ – user140323 Feb 25 '17 at 18:51
  • $\begingroup$ And no one ever answered. $\endgroup$ – user140323 Feb 27 '17 at 22:06

Okay, I think this is the simplest way to do it. The volume of the ellipsoid $$ \frac{X^2}{a^2}+\frac{Y^2}{b^2}+\frac{Z^2}{c^2} = 1 $$ is $4\pi abc/3$. Thus the ratio of this to the sphere $(X^2+Y^2+Z^2)/R^2 = 1$ is $abc/R^3$. Dividing your equation by a factor of $R^2$ means that the $R^3$ goes away. Hence we just have to find the coefficients of $X^2,Y^2,Z^2$.

To do this, the quickest way is to define new coordinates by $\xi' = \xi$, $\eta' = \eta-m\beta v\xi/c $, and $\zeta' = \zeta - n\beta v\xi/c$ (which are obviously linearly independent, so will span the space), which simplifies the ellipse equation to $$ \left(1-\frac{lv}{c}\right)^2\beta^2 \xi'^2 + \eta'^2 + \zeta'^2 = R^2. $$ In particular, this does not change the volume measure since the Jacobian is upper triangular with $1$s down the diagonal.

Now one can read off that the volume ratio is $$ abc = \frac{\beta}{1-lv/c}, $$ as required.


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