Consistent Trajectory for a non-zero rest mass particle?

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SUMMARY

The discussion revolves around verifying the consistency of the trajectory defined by the equation \(\frac{dx}{dt}=\frac{cgt}{\sqrt{1+g^2t^2}}\) for a non-zero rest mass particle. The key point raised is whether the condition \(\frac{dx}{dt} < c\) holds for all time \(t\). As \(t\) approaches infinity, the velocity approaches the speed of light \(c\), indicating inconsistency with the expected behavior of such particles. The participants conclude that the trajectory does not meet the consistency requirement outlined in the assignment.

PREREQUISITES
  • Understanding of special relativity and the speed of light limit.
  • Familiarity with calculus, particularly limits and derivatives.
  • Knowledge of kinematic equations involving acceleration due to gravity.
  • Basic understanding of non-zero rest mass particle dynamics.
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  • Investigate the implications of relativistic velocity limits in particle physics.
  • Learn about the mathematical treatment of trajectories in special relativity.
  • Explore the concept of consistency in physical equations and its significance.
  • Review kinematic equations under gravitational influence and their limits.
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Students and educators in physics, particularly those studying special relativity and kinematics, as well as anyone involved in theoretical physics discussions regarding particle dynamics.

DeldotB
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Homework Statement


Good day all!
Quick question:
As part of a problem statement, I'm asked to verify if the trajectory: \frac{dx}{dt}=\frac{cgt}{\sqrt{1+g^2t^2}}
Is "consistent".

Homework Equations


None

The Attempt at a Solution



Im not sure what "consistent" means. Does it mean, \frac {dx}{dt} &lt; c for all t? If that's so, I run into a problem because in the limit as t approaches infinity, the velocity = the speed of light (the limit goes to c). Am I approaching this the wrong way? (The trajectory is supposed to be "consistent")
 
Last edited:
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You may want to check the limit of that expression for large ##t##.
 
PeroK: Not sure what you mean...
I get "c" as the limit. Maybe my work is wrong? Lim\, \, t\rightarrow \infty (\frac{cgt}{\sqrt{1+(9.8))^2t^2}})=cg(Lim\, \, t\rightarrow \infty (\frac{t}{\sqrt{1+(9.8))^2t^2}}))=cg(5/49)=c. So as t approaches infinity, the velocity approaches c.
 
That's correct, but inconsistent with the limit of ##cg## you gave in the original post!
 
Ah, I see. A miss-type. Well, nevertheless, this trajectory doesn't seem to be consistent even though my assignment is saying it should be.
 

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