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by jsprogrammer
3744 days ago
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At the end you say: >Let me remind you that these are not the real equations of motion for the photon. The real equations of motion would have been achieved by calculating all of the geodesic equations from the Lagrangian. There are four of them instead of the three equations (in vector form) above. However, the three spatial equations will generate the exact same spatial curve as the real geodesic equations would, and they were relatively easily achieved. What do you mean that these are not the real equations of motion for the photon, but that they give the exact same spatial curve as the real geodesic equations? Are there cases where these equations would not give the exact same curve as the real geodesic equations? |
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In the process of deriving the said equations, an equation for the radial coordinate of the photon was achieved. This was identified with a classical, Newtonian system of one particle with unity mass. As the real, massless photon lives in four-dimensional spacetime and the said massive "test particle" lives in three-dimensional space, these systems just can't be dynamically the same (in spacetime, the massive particle would take a timelike curve).
To reword the statement, the derived "equation of motion" will yield the same trajectory in the spacelike components (x, y, z), but possibly with a different parametrization - in the classical system, we're integrating the equations of motion with respect to the time. However, this has nothing to do with the "coordinate time" of the four-vectors nor the proper time of the particle (for the photon, proper time doesn't even make sense).
Hope this helps! You could also see [0] for an alternative take on this derivation. I will try and clarify my article a bit as well.
[0] http://rantonels.github.io/starless/