Consistent Trajectory for a non-zero rest mass particle?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 1K views
DeldotB
Messages
117
Reaction score
8

Homework Statement


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

Homework Equations


None

The Attempt at a Solution



Im not sure what "consistent" means. Does it mean, [itex]\frac {dx}{dt} < c[/itex] 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:
Physics news on Phys.org
PeroK: Not sure what you mean...
I get "c" as the limit. Maybe my work is wrong? [tex]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[/tex]. So as t approaches infinity, the velocity approaches c.
 
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.