Time Contraction in Lorentz Transforms?

cynopolis
Messages
8
Reaction score
0
The question I have is to do with the Relativity of Simultaneity of the type described by Einsten; whereby two light emitters are placed pulsing once every millisecond on board a spaceship traveling at the speed of light. One of these lights faces forward and the other faces aft. If time dilates in order for the laws of physics to remain the same in all reference frames, shouldn't time in the aft part of the ship contract to deal with the fact that the light has arrived at the rear of the ship faster than intended (so to speak)?

Also if the spaceship has undergone a length contraction, while the angle of the light remains constant in space-time, shouldn't the the light ray - that makes contact with the rear wall of the ship early - continue to move in time to account for the disparagement in the spatial dimension, resulting in it appearing to be frozen to the wall?

If you draw this out you will notice that the light will remain on the rear wall until the forward light reaches its destination, reinstating simultaneity, if only for a brief and, admittedly, lopsided time. Your thoughts on this?
 
Physics news on Phys.org
Does anyone follow me?
 
cynopolis said:
The question I have is to do with the Relativity of Simultaneity of the type described by Einsten; whereby two light emitters are placed pulsing once every millisecond on board a spaceship traveling at the speed of light. One of these lights faces forward and the other faces aft. If time dilates in order for the laws of physics to remain the same in all reference frames, shouldn't time in the aft part of the ship contract to deal with the fact that the light has arrived at the rear of the ship faster than intended (so to speak)?
No. Clocks anywhere in the moving ship will be seen as running slow by the same factor.

Also if the spaceship has undergone a length contraction, while the angle of the light remains constant in space-time, shouldn't the the light ray - that makes contact with the rear wall of the ship early - continue to move in time to account for the disparagement in the spatial dimension, resulting in it appearing to be frozen to the wall?
No.

If you draw this out you will notice that the light will remain on the rear wall until the forward light reaches its destination, reinstating simultaneity, if only for a brief and, admittedly, lopsided time. Your thoughts on this?
Not sure what you are talking about.
 
Ok thanks for those answers.

With regard to the first one though. Time dilation doesn't effect the speed of light (which is a constant) and therefore could not effect the amount of time it takes to the aft light to reach the rear wall. As the rear wall is accelerating towards the beam of light, causing that beam to collide sooner, and the speed of light is supposed to dictate order of events doesn't that mean that time should speed up at the back of the space ship? Please explain your answer.
 
Oh no wait, I get it.
 
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