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I'm writing a paper about the motion of photons near a Schwarzschild black hole. At some point there's a derivative of the Hamiltonian of the system with respect to time $\tau$. I need to explain what the proper time is $\tau$, but it's quite odd because photons don't have any proper time.

The Hamiltonian that I have is

$$H = - \left( 1-\frac{2M}{r} \right)^{-1} \frac{p_{t}^2}{2}+\left( 1-\frac{2M}{r} \right) \frac{p_{r}^2}{2}+\left( \frac{p_{ \theta}^2}{2r^2}+\frac{p_{\phi}^2}{2r^2sin^2\theta} \right).$$

  • So what would be the definition in this case?
  • "the proper time is the time for the photon although he doesn't have one?"

Does anyone know?

Qmechanic
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poissonrouge
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2 Answers2

5

I actually asked a question about the four velocity of the photon.

Four-velocity vector of light

where Chiral Anomaly says:

Short answer: - If the term "four-velocity" is used in the strict sense of $d x^\mu/d\tau$ where $\tau$ is the object's proper time, then four-velocity is undefined for light because the elapsed proper time is always zero ($d\tau=0$) along a lightlike worldline. - If the term "four-velocity" is used in the generalized sense of $dx^\mu/d\lambda$ where $\lambda$ is an affine parameter that increases monotonically along the lightlike worldline, then the four-velocity is perfectly well-defined for light.

So photons do not have a proper time, but you can use a $\lambda$ affine parameter that increases monotonically along the lightlike worldline.

Árpád Szendrei
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Please note that no reference frame can move at $ c $ speed of light in vacuum, so no measure of proper time can be done for a photon traveling at this speed.

Special relativity states that the elementary interval of proper time for a photon with speed $ c $ is zero.

A possible demonstration is as follows:

if $ ds $ is the elementary interval of length, the elementary interval of proper time $ d\tau $ can be defined as $$ d\tau=\frac{ds}{c}\ \ \ \ [1] $$ Considering the elementary interval of length squared as $ ds^2=c^2dt^2-dx^2-dy^2-dz^2=c^2dt^2-\vec{dr}^2 $,

for a photon you have $ |\frac{dr}{dt}|=c $, which leads to $ ds^2=0 $, and then with $ [1] \Rightarrow d\tau=0 $.

Hoping to have answered the question,

Best regards.

Cornelius Fyla
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