Find Time t_0 for a Beam to Not Catch Up with Earth

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Homework Help Overview

The problem involves a scenario where an observer accelerates away from Earth with uniform acceleration and seeks to determine a time \( t_0 \) such that a beam of light emitted from Earth after this time will not catch up to the observer. The context is rooted in concepts from relativity.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of accelerating away from Earth and the nature of light travel. There are attempts to derive expressions for time and conditions under which the light beam would not catch up. Questions arise regarding the interpretation of negative time values and the physical implications of reaching light speed.

Discussion Status

The discussion is active, with various interpretations being explored. Some participants provide insights into the mathematical behavior of the equations involved, while others question the assumptions about acceleration and light speed. There is no explicit consensus, but several productive lines of reasoning are being examined.

Contextual Notes

Participants are navigating assumptions about uniform acceleration and the limits imposed by the speed of light. The original poster's reference to a specific problem from a textbook suggests a structured approach, but the implications of the problem's conditions are under scrutiny.

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Homework Statement


You start at t=0 at rest on Earth and accelerate with uniform acceleration a away form earth.
Find a point in time [itex]t_0[/itex] such that when a beam emitted from Earth at [itex]t>t_0[/itex]won't catch up.


Homework Equations


[itex]x(t)=c^2/a(\sqrt{1+\frac{a^2}{c^2}t^2}-1)[/itex]


The Attempt at a Solution


I think that light travel with velocity c. So if the beam is emitted at [itex]t=t_1[/itex] then at time t, the beam have traveled [itex]c(t-t_1)[/itex]. So I try to find the solution for [itex]x(t)=c(t-t_1)[/itex], and I end up with the following expression for [itex]t[/itex]:
[itex]t=\frac{a}{2c}\frac{t_1(2-a/c t_1)}{(a/c - a^2/c^2 t_1)}[/itex]

According to this the time would be negatic in the intervall [itex]t_1=c/a[/itex] and [itex]t_1=2c/a[/itex] So I think in this intevall the beam won't be able to catch up, but after [itex]t_1=2c/a[/itex] the time becomes positive again, which I don't know how to interpret.
Am I approaching this problem the wrong way?
 
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Think about it this way: will you ever accelerate to a speed greater than that of light? If not, how can you possibly outrun light?
 
Well it is problem 3.9 in D'inverno Introducing Einsteins relativity. I agree that it seem impossible but the problem statement is that if you get a large enough headstart the light won't catch up.
 
What assumptions are you supposed to make?
 
negatic?
 
You should be able to show that your world line is a hyperbola. Find its asymptotes.
 
zardiac said:
[itex]t=\frac{a}{2c}\frac{t_1(2-a/c t_1)}{(a/c - a^2/c^2 t_1)}[/itex]

Note t approaches ∞ as the denominator on the right approaches 0.
 
Let's say you keep uniform acceleration a, relative to the stationary observer. After c/a time you will be moving at the speed of light. To keep uniform acceleration you need infinite amount of energy. I think the answer is c/a, just the problem is that you can't keep uniform acceleration.
 
  • #10
Myslius said:
Let's say you keep uniform acceleration a, relative to the stationary observer. After c/a time you will be moving at the speed of light. To keep uniform acceleration you need infinite amount of energy. I think the answer is c/a, just the problem is that you can't keep uniform acceleration.
You've misinterpreted the problem. The acceleration is uniform relative to the moving observer. As you noted, you can't have a uniform acceleration relative to the stationary observer indefinitely.
 

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