A Photon speed for an observer at the photon sphere

  • A
  • Thread starter Thread starter alexriemann
  • Start date Start date
  • Tags Tags
    Orbital period
alexriemann
Messages
3
Reaction score
0
TL;DR Summary
Photon speed for an observer at the photon sphere
I am asked to compute the orbital period of a photon, in the Scwarzschild spacetime, at the photon sphere for an observer at the same radius, ##r^\star=3M##. I have computed the result, ##\Delta T=6\pi M## where ##c=G=1## ,comparing with the proper time of an observer at infinity. However, as the result gives directly ##\Delta T=2\pi r^\star##, I wonder if I can skip making the calculation by inferring that, for the observer sitting at the photon sphere, the speed of the photon is exactly ##c=1##. If that is the case, how can I argue that this is true?
 
Physics news on Phys.org
Please show your computation.

Note that you cannot have a time coordinate difference of ##\Delta T = 6\pi M## at the same time as you are saying that a local hovering observer at ##r^\star## observes light to pass by at ##c = 1## as the hovering observer is gravitationally time dilated relative to the observer at infinity.
 
##\Delta T## is not the coordinate time difference, it is the orbital period of the photon for the observer sitting at the photon sphere. Actually, for the observer at infinity, the orbital period of the photon would be ##6\sqrt3\pi M##. Maybe ny notation wasn't clear.

My guess is that I can argue that locally the speed of light is always c, and for the observer at the same radius as of the photon, this remains true because the photon radius coordinate remains the same throughout its motion.
 
alexriemann said:
ΔT is not the coordinate time difference
I’m not sure how else to read this:
alexriemann said:
comparing with the proper time of an observer at infinity
But anyway …

Yes, the definition of the ##r## coordinate relates to the area of the sphere*. As such, a great circle on such a sphere will have circumference ##2\pi r## and it will take ##2\pi r/c## for a light singnal to go around for the local observer.

* Consequently note that ##r## does not correspond to a radial distance of any sort.
 
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...

Similar threads

Replies
3
Views
1K
Replies
3
Views
1K
Replies
5
Views
2K
Replies
55
Views
3K
Replies
8
Views
1K
Replies
15
Views
2K
Replies
13
Views
3K
Replies
10
Views
1K
Back
Top