Light's Instantaneous Wordline: Is Time Really Zero?

  • Thread starter Thread starter OS Richert
  • Start date Start date
  • Tags Tags
    Time Zero
OS Richert
Messages
35
Reaction score
0
The proper time of a photon which always has a lightlike wordline is zero, correct? Does this mean from the perspective of light (which doesn't have any conscience to have a perspective) that it travels instantaneously in zero time. So while we see light from Alpha Centari and can watch it travel for four years to get to earth, it thinks it traveled from Alpha Centari to Earth instantaneously?
 
Physics news on Phys.org
Yes. Also this approach is the easiest way to understand the twin paradox. One twin goes to Alpha Centauri at almost the speed of light and to him it takes a very short time. After stopping to catch his breath, he returns to Earth at the same speed, so to him the round trip took a very short time. On the other hand, his twin on eath has aged around 9 years,
 
mathman said:
One twin goes to Alpha Centauri at almost the speed of light and to him it takes a very short time.
Not only does it take him a short time but the distance to Alpha Centauri is much shorter for him than for an observer on Earth.
 
OS Richert said:
The proper time of a photon which always has a lightlike wordline is zero, correct? Does this mean from the perspective of light (which doesn't have any conscience to have a perspective) that it travels instantaneously in zero time. So while we see light from Alpha Centari and can watch it travel for four years to get to earth, it thinks it traveled from Alpha Centari to Earth instantaneously?

Actually it is mathematically inconsistent to say that a photon has a perspective in terms of time and space. The problem isn't so much with "proper time" for the photon which goes to zero, but with any idea of assigning a frame or distance to the photon.

What one can safely say is that the limit as one approaches light speed, the time it takes to get to Alpha Centuari from Earth approaches zero. A seemingly minor, but potentially important, distinction.

The closest thing to giving a photon a perspective is to adopt a system of null coordinates, which are neither timelike nor spacelike, but null.

For instance, if x and t are cartesian coordinates, U= x-t and V = x+t are null coordinates. U gives a constant number for the worldline of a photon moving in one direction, V gives a constant number for the worldline of a photon moving in another direction.
 
For those interested in null coordinates, mathpages has an interesting http://www.mathpages.com/rr/s1-09/1-09.htm" on it.

Note that the anonymous author expresses a remarkable level of lucidity on the subject which not everybody might appreciate.
 
Last edited by a moderator:
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
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...
Back
Top