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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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    Please describe the hamiltonian and give a reference. "At some point there's a derivative of the Hamiltonian of the system with respect to time Tau." – my2cts Jul 18 '19 at 11:40
  • @my2cts H = - (1-2M/r)^-1 p_{t}^2/2 + ( 1-2M/r) p_{r}^2/2 + (p_{ \theta}^2/(2r^2) +p_{\phi}^2/(2r^2*sin^2(\theta))) in Latex : 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). The hamiltonian is the potential and kinetic energies i think – poissonrouge Jul 18 '19 at 11:44
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    "At some point there's a derivative of the Hamiltonian..." This sounds like it's referring to some paper that you're basing your paper on. Tell us what it is. –  Jul 18 '19 at 13:07
  • In your comment you need to use single dollar signs to wrap around Mathjax expressions so they display properly. – StephenG - Help Ukraine Jul 18 '19 at 13:31
  • Answered here: https://physics.stackexchange.com/q/465136/ – safesphere Jul 18 '19 at 15:20
  • @BenCrowell https://www.physik.uzh.ch/~psaha/magicenv/magicenv.pdf pages 51-55 – poissonrouge Jul 19 '19 at 08:06

2 Answers2

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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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