Region the light never reaches the 'uniformly-accelerated' observer?

  • Thread starter Thread starter johnahn
  • Start date Start date
  • Tags Tags
    Light Observer
johnahn
Messages
8
Reaction score
0
Is there a region that the light never reaches the 'uniformly accelerated' observer?

Of course, light travels in the same direction the observer moves.

It sounds weird for me...

I derive the parameterization of t and x, and gets hyperbola.

So I try to find with drawing that in the ST diagram, but I cannot see any clues for the reason why there exists such things...
 
Physics news on Phys.org
I am not at all clear on what "region" you are talking about. Ignoring expansion of space itself, no observer can move faster than light with respect to any star so light from any star will eventually reach any observer.
 
Oh, I find such region... It was very simple. If we draw the asymtotic line of the hyperbola that is the uniformly accelerated observer, then that line will meet with the x-axis at a point. And the light emitted by the source behind that point, never reaches the observer forever because the world line of the light and the observer do not meet at all.

I throw a question, and I answer it lol
 
HallsofIvy said:
... . Ignoring expansion of space itself, no observer can move faster than light with respect to any star so light from any star will eventually reach any observer.

Not true. Have a look at the Minkoski chart in this post https://www.physicsforums.com/showpost.php?p=2858281&postcount=37 depicting accelerating observers in flat Minkowski space. Light emitted from the region to left of the diagonal dotted line will never catch up with the accelerating Rindler observers. The boundary of this region is known as the Rindler horizon and is an flat space analogue of the curved space Schwarzschild event horizon.
 
johnahn said:
Is there a region that the light never reaches the 'uniformly accelerated' observer?

Of course, light travels in the same direction the observer moves.

It sounds weird for me...

I derive the parameterization of t and x, and gets hyperbola.

So I try to find with drawing that in the ST diagram, but I cannot see any clues for the reason why there exists such things...

Yes, there is a region of space-time a uniformly acceleratig observer cannot see. The boundary between the region he can see and can't see is called the Rindler horizon. It's rather similar to a black hole horizon in many respects.

The Rindler horizon appears to be a constant distance behind the accelerating observer in his accelerated frame of reference (using the defintion of simultaneity of a co-moving inertial observer to create said frame of reference).
 
Here's a spacetime diagram of the accelerating observer A. A and L have synchronised clocks and L sends a light beam to A at the time the acceleration starts. If A drops a device C, then C intercepts the light beam fairly quickly. It seems strange that the horizons existence depends on acceleration, not velocity. The moment the acceleration stops the horizon disappears.
 

Attachments

  • rind-h.png
    rind-h.png
    2.1 KB · Views: 492
Last edited:
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