Relativistic Uniform Accelerated Motion

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SUMMARY

The discussion focuses on calculating the distance traveled by a particle undergoing uniform acceleration equal to Earth's gravity (9.81 m/s²) over proper time intervals of 1, 10, and 100 years. The derived equations for position are t(s) = (c/a)sinh(as/c) and x(s) = (c²/a)cosh(as/c). Initial calculations yielded a distance of 0.6 lightyears for 1 year, but results for 10 and 100 years exceeded 10 and 100 lightyears, respectively, raising concerns about exceeding the speed of light. The participant seeks clarification on whether to first calculate proper time using the first equation before substituting it into the second equation.

PREREQUISITES
  • Understanding of relativistic physics concepts, particularly uniform acceleration.
  • Familiarity with hyperbolic functions, specifically sinh and cosh.
  • Knowledge of the speed of light as a universal constant.
  • Basic grasp of time dilation effects in special relativity.
NEXT STEPS
  • Review the derivation of the four-vector equations for relativistic motion.
  • Learn how to calculate proper time in relativistic contexts.
  • Explore the implications of time dilation on distance measurements in special relativity.
  • Investigate the relationship between acceleration, velocity, and distance in relativistic frameworks.
USEFUL FOR

Students and enthusiasts of physics, particularly those studying special relativity and its applications in calculating relativistic motion and distances. This discussion is beneficial for anyone looking to deepen their understanding of uniform acceleration in a relativistic context.

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Homework Statement


Hi I am supposed to calculate the distance traveled by a particle that is uniformly accelerated with acceleration equal to the Earth's gravity (i.e. a=9.81m/s^2) after 1, 10 and 100 years proper time.

Homework Equations


We derived in class the four vector which describes the position of the particle:
t(s)=\frac{c}{a}sinh\left(\frac{as}{c}\right)
x(s)=\frac{c^2}{a}cosh\left(\frac{as}{c}\right)

The Attempt at a Solution


So far all I did was to substitute in the second equation the values of 1,10,100years. The first result is reasonably since it yields 0.6 lightyears. However the other results yield values which are way above 10 lightyears and 100 lightyears respectively. This can't be possible, because the particle can never exceed the speed of light. So I should get a value which is below 10 and 100 light years.
I am wondering whether I have to do something else. Do I have to find the proper time s through the first equation and then substitute that value in the second equation?
 
Physics news on Phys.org
It does not have to exceed the speed of light to travel over 10 lightyears in 10 years of proper time. If you have something moving close to the speed of light, it will move essentially 10 lightyears in 10 of your years. However, time dilation means that essentially no time will have elapsed for the traveller.

Edit: Of course, these 10 light years are in the frame of an external observer in an inertial frame.
 

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