This is a summary of the second part of Appendix A in Albert Einstein’s book [1]. By second part, I mean the part immediately after the derivation of the Lorentz transformation. If there is no relative motion between the coordinate systems with respect to the $ y$ and $ z$ axis of $ K$, then the $ Y$ and $ Z$ coordinates of the light in $ K$ are equal to the coordinates of the light in $ K’$, since there is no need to modify these coordinates: $ Y’ = Y$ and $ Z’ = Z$.

Next, the propagation of light is investigated. As mentioned at the beginning of Derivation #1, suppose there is a flash of light at the origin of $ K$ at $ t=0$. This light travels away from the origin in every possible direction in three dimensions. The light travels with a speed of $ c$ in each direction. The distance from the origin that the light travels as a function of time $ t$ is $ R=ct$. This $ R$ can be thought of as a radius of a sphere with a center at the origin. From geometry, the equation for a sphere as a function of position coordinates is $ R^2 = X^2 + Y^2 + Z^2$, so $ R = \sqrt{X^2 + Y^2 + Z^2}$. The positive root is used because distances are positive by definition. The equation for a sphere in three dimensions as a function of position coordinates can be derived with the Pythagorean Theorem. Since $ R^2 = c^2t^2 = X^2 + Y^2 + Z^2$, it is clear that $ 0 = X^2 + Y^2 + Z^2 – c^2t^2$. This equation can be made more general with a variable $ \sigma$, by multiplying both sides of the equation by $ \sigma$: $ 0 = \sigma(X^2 + Y^2 + Z^2 – c^2t^2)$.

The same reasoning can be applied to light in the translating coordinate system $ K’$ with coordinates $ x’, y’, z’, t’$. This can be done because one of the assumptions of special relativity is that the speed of light is $ c$ regardless of whether the coordinate system is translating with a nonzero velocity or not. Hence $ R’^2 = c^2 t’^2$, $ (R’)^2 = X’^2 + Y’^2 + Z’^2$, and $ 0 = X’^2 + Y’^2 + Z’^2 – c^2 t’^2$.

Since $ 0=0$ suppose that it is valid to write $ X’^2 + Y’^2 + Z’^2 – c^2 t’^2 = \sigma(X^2 + Y^2 + Z^2 – c^2t^2)$.

Furthermore, suppose in $ K$ it is possible for light to only move along the $ x$-axis. If this is the case, then the equation of $ 0 = \sigma(X^2 + Y^2 + Z^2 – c^2t^2)$ reduces to $ 0 = \sigma(X^2 – c^2t^2)$ since the light’s position does not advance along the $ y$ and $ z$ axes. Similarly, for $ K’$, suppose it is possible for light to only move along the $ x’$-axis. If this is the case, then the equation of $ 0 = X’^2 + Y’^2 + Z’^2 – c^2 t’^2$ reduces to $ 0 = X’^2 – c^2 t’^2$ since the light’s position does not advance along the $ y’$ and $ z’$ axes.

Since $ 0=0$ suppose that it is valid to write $ X’^2 – c^2 t’^2 = \sigma(X^2 – c^2t^2)$. Einstein sets $ \sigma=1$ for a reason I have not been able to determine. More attention will be given to this point in the future. This is one question to which I do not have an answer.

This post brings the engaged reader up to equation (11) in Appendix A of Einstein’s book [1].

References

[1] Albert Einstein. Relativity; The Special and General Theory. Appendix A. Three Rivers Press, 1961.

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I am a research assistant in theoretical physics. This website, derive-it.com, serves to organize my ongoing learning and research as well as to provide a resource to other learners around the world.

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