Distance & Relativity: Does Light Experience Distance?

  • Thread starter Thread starter Cluelessluke
  • Start date Start date
  • Tags Tags
    Light
Cluelessluke
Messages
29
Reaction score
0
Hey all, I have a quick question about distance and relativity. L=L(proper)/(gamma) Now if you are dealing something that is actually going to speed of light (like light itself), gamma = 1/0. If we bring limits into the mess gamma = infinity. This would make L (the distance of the object to the moving observer) 0. Does this mean that light doesn't "expierience" distance? Such as, to light, is it the same distance from the US to China as it is from one end of the galaxy to the other?
 
Physics news on Phys.org
Cluelessluke said:
Hey all, I have a quick question about distance and relativity. L=L(proper)/(gamma) Now if you are dealing something that is actually going to speed of light (like light itself), gamma = 1/0. If we bring limits into the mess gamma = infinity. This would make L (the distance of the object to the moving observer) 0. Does this mean that light doesn't "expierience" distance? Such as, to light, is it the same distance from the US to China as it is from one end of the galaxy to the other?
This is true as a statement about what happens in the limit as your speed approaches c relative to the Galaxy, but it doesn't really make sense to talk about what light "experiences". Keep in mind that even to say that the distances in your inertial rest frame are what you "experience" is a bit of an oversimplification, what it really means is that if you use rulers and synchronized clocks ('synchronized' using the Einstein synchronization convention) at rest with respect to yourself to give your definition of "distance", then distance contracts according to the Lorentz formula (the synchronized clocks are necessary so that you can locally measure the time that the front and back of a moving object were at particular positions on your ruler, and subtract the position of the front at a given time from the position of the back at the same time). But there's no physical way to generalize this to a photon, since you can't have physical rulers and clocks which are at rest relative to the photon, so considering what the photon "experiences" is more of a pure mathematical game without much physical meaning.
 
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(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...

Similar threads

Replies
31
Views
2K
Replies
4
Views
1K
Replies
42
Views
682
Replies
11
Views
1K
Replies
5
Views
1K
Replies
20
Views
2K
Back
Top