0
$\begingroup$

Photons are massless particles.

Time dilation is caused between two observers either by the relative speed of the observers or by the observers being in different gravitational zones (stress-energy difference).

According to the theory of relativity, time dilation is a difference in the elapsed time measured by two observers, either due to a velocity difference relative to each other, or by being differently situated relative to a gravitational field. Special relativity indicates that, for an observer in an inertial frame of reference, a clock that is moving relative to him will be measured to tick slower than a clock that is at rest in his frame of reference. This case is sometimes called special relativistic time dilation. The faster the relative velocity, the greater the time dilation between one another, with the rate of time reaching zero as one approaches the speed of light (299,792,458 m/s). This causes massless particles that travel at the speed of light to be unaffected by the passage of time.

https://en.wikipedia.org/wiki/Time_dilation

This states that, massless particles are unaffected by time dilation.

What is the time component of velocity of a light ray?

where Izzhov says:

Four-velocity actually isn't well-defined for light. This is because four-velocity is the derivative of the position four-vector with respect to the proper time, i.e. the time in the moving object's rest frame. Since you can't pick an inertial rest frame for a beam of light, you can't take the derivative with respect to the proper time. Mathematically, the proper time of light is zero, so to take the derivative of light's position four-vector with respect to proper time, you would be dividing by zero, returning infinity/undefined (for all components, including the time component).

So this says that the four velocity of light is not well defined.

Now we have to use an affine parameter λ instead of proper time for massless particles.

What this does not explain is whether for a photon, we can talk about time dilation or not, because according to SR, photons should not be affected by time dilation.

Now if we can define a four velocity for photons with an affine parameter λ, then we should be able to calculate the time dilation too.

It does not make sense to talk about the view of the photons, because photons do not have a frame of reference as per SR.

Photon time dilation paradox

where S. Mcgrew says:

That is not how it works. To us, time appears to pass much more slowly in a system that is moving rapidly with respect to us. If * you * were part of that system, you would observe the opposite: that we were moving rapidly and that our clocks appear to run more slowly than yours. Time is not frozen in either frame, regardless of their relative speed.

Now one says that massless particles are not affected by time dilation, the other one says it is not defined.

Question:

  1. Which one is right, are photons affected by time dilation or not(If we can calculate the four velocity of photons with an affine parameter λ, does this mean that photons are affected by SR (velocity) time dilation)?
|cite|improve this question|||||
$\endgroup$
  • $\begingroup$ One thing about proper time is that it is the the unique parameter that normalizes all time-like paths $x(s)$ so that $g(\dot{x}, \dot{x}) = g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} = 1$. This gives us a universal way to define time for massive objects. When it comes to photons, you don't have that ability to define such a parameter. The closest analog to four-velocity for photons is four-frequency, which ties in to the deBroglie relation $p^{\mu} = \hbar k^{\mu}$ ($k$ is four-wavevector), and this is subject to Lorentz trans b/c of doppler shift. $\endgroup$ – Maximal Ideal Jul 13 '19 at 18:45
  • $\begingroup$ However, I'm having a difficult time finding a reason why you can't define a new parameter that acts similar to time. There would need to be an arbitrary decision about its scale. It's hard to get a contradiction without using limits, so maybe this is a possible hypothetical piece of physics. The parameter would definitely act very differently than proper time, of course. $\endgroup$ – Maximal Ideal Jul 13 '19 at 18:50
  • $\begingroup$ Everything moves with the speed of time $c$, including light. The difference is the direction. Yoy move with the speed of time in time; light moves with the speed of time in space. The quote that 4-velocity "is not well defined for light" is incorrect. The correct statement is that the 4-velocity of light cannot be viewed in the rest frame of light, because light doesn't rest. This is a frame issue that has absolutely nothing to do with 4-velocity. 4-velocity is defined for light perfectly well as a tangent vector to the worldline. Same as for everything else. $\endgroup$ – safesphere Jul 13 '19 at 19:17
  • $\begingroup$ @safesphere You are 100% right saying that light has a tangent vector to its worldline. The four-frequency is an example of this. However, four-velocity for time-like curves is defined to be $u^{\mu} = \frac{dx^{\mu}}{d\tau}$ where $\tau$ is the unique parameter such that the norm $g_{\mu\nu} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau}$ is set to $1$ (assuming the $(+1, -1, -1, -1)$-metric and theoretical units with $c=1$). This is impossible to define for light-like trajectories as I have explained in my first comment. If you say otherwise, can you give an explicit definition? $\endgroup$ – Maximal Ideal Jul 13 '19 at 19:52
  • $\begingroup$ @SpiralRain Would this answer work? - physics.stackexchange.com/questions/66422/… $\endgroup$ – safesphere Jul 13 '19 at 22:11
1
$\begingroup$

The infinitesimal proper time $d\tau$ elapsed along any segment of a worldline in flat spacetime is

$$c^2 d\tau^2=c^2 dt^2-dx^2-dy^2-dz^2.$$

This is zero along the worldline of a photon traveling at speed $c$, so no proper time elapses along the entire worldline. This means that photons experience infinite time dilation relative to all inertial observers, as would be expected since the Lorentz factor that determines time dilation,

$$\gamma=\frac{1}{\sqrt{1-v^2/c^2}},$$

becomes infinite as $v\to c$.

By the way, you wrongly interpreted Wikipedia’s phrase “unaffected by the passage of time” to mean “unaffected by time dilation”. “Unaffected by the passage of time” means “experiences zero elapsed proper time” which implies infinite time dilation.

|cite|improve this answer|||||
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.