Consistent Trajectory for a non-zero rest mass particle?

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Homework Help Overview

The discussion revolves around the trajectory of a non-zero rest mass particle, specifically examining the expression \(\frac{dx}{dt}=\frac{cgt}{\sqrt{1+g^2t^2}}\) and its consistency with the principles of physics, particularly the speed of light as a limit.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster questions the meaning of "consistent" in the context of the trajectory and considers whether it implies that \(\frac{dx}{dt} < c\) for all \(t\). They express confusion regarding the limit of the velocity as \(t\) approaches infinity. Other participants suggest checking the limit of the expression for large \(t\) and discuss the implications of reaching the speed of light.

Discussion Status

Participants are actively exploring the implications of the trajectory's limit as \(t\) increases, with some noting discrepancies in the original poster's calculations. There is recognition of a potential error in the original post, but the overall consistency of the trajectory remains under scrutiny.

Contextual Notes

The discussion highlights a lack of clarity regarding the definition of "consistent" in the problem statement and the implications of the trajectory approaching the speed of light, which raises questions about the assumptions involved.

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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