# How to define the proper time of a photon?

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?

• 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." Commented Jul 18, 2019 at 11:40
• @my2cts H = - (1-2M/r)^-1 *p_{t}^2/2 + ( 1-2M/r) *p_{r}^2/2 + (p_{ \theta}^2/(2*r^2) +p_{\phi}^2/(2*r^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 Commented Jul 18, 2019 at 11:44
• "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.
– user4552
Commented Jul 18, 2019 at 13:07
• In your comment you need to use single dollar signs to wrap around Mathjax expressions so they display properly. Commented Jul 18, 2019 at 13:31
• Answered here: physics.stackexchange.com/q/465136 Commented Jul 18, 2019 at 15:20

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.

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.