Show that m varies with v in theory of relativity

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Homework Help Overview

The discussion revolves around the relationship between mass and velocity in the context of the theory of relativity, specifically focusing on the derivative of mass with respect to velocity. Participants are attempting to understand the implications of their calculations and the definitions involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of taking the derivative of mass with respect to velocity and its implications, particularly regarding the constants involved. There are discussions about the expression for force and its relation to mass and acceleration, as well as confusion regarding the terms used in the equations.

Discussion Status

There is an ongoing exploration of the relationships between the variables involved, with some participants expressing uncertainty about their calculations and the definitions of terms like m' and a. Guidance has been offered regarding the interpretation of the equations, but no consensus has been reached on the correct approach or outcome.

Contextual Notes

Participants note the assumption that the force acts parallel to the velocity, which may affect the results. There is also mention of constants such as m0 and c, indicating potential constraints in the problem setup.

Feodalherren
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Am I supposed to take dm/dv? That means m0 is constant, correct?

If I do that I end up getting a m' on the Left hand side... What the heck does that mean?! Also where the crap is the "a" coming from?!
 
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Consider: F = d/dt (p) = m dv/dt + v dm/dt = m a + v dm/dt
Now introduce your idea of taking dm/dv in coming up with dm/dt as a function of a and dm/dv.

Warning: I have not succeeded myself in deriving this result. And I know it's correct, dadblame it!
Somebdy - help!
 
Feodalherren said:
image.jpg

image.png


Am I supposed to take dm/dv? That means m0 is constant, correct?

If I do that I end up getting a m' on the Left hand side... What the heck does that mean?! Also where the crap is the "a" coming from?!

a = dv/dt.
 
Feodalherren said:
That means m0 is constant, correct?
Correct.
If I do that I end up getting a m' on the Left hand side... What the heck does that mean?!
Since there is no [itex]m'[/itex] in the problem, the [itex]m'[/itex] means whatever you said it meant when you wrote it down.
Also where the crap is the "a" coming from?!
Read the problem again.

You'll also get a better response if you control your frustration and avoid even mild profanities here.
 
rude man said:
Consider: F = d/dt (p) = m dv/dt + v dm/dt = m a + v dm/dt
Now introduce your idea of taking dm/dv in coming up with dm/dt as a function of a and dm/dv.

Warning: I have not succeeded myself in deriving this result. And I know it's correct, dadblame it!
Somebdy - help!
It does work.
 
tms said:
It does work.

Yeah, I know it has to, I just could not get the result.
 
I think the problem should have stated that the force is assumed to act parallel to the velocity. Perhaps that was meant to be evident by the lack of vector notation. But, anyway, it's worth noting that if the force does not act parallel to the velocity, then you will get a different result for the relationship between the force and acceleration. (In fact, the acceleration is generally not even in the same direction as the force).
 
I'm still not getting it right. There must be something I'm missing.

The constants are m0 and c, correct?

when I do d/dv I end up with [ -2m0vc^(-2) ] / (1-(v^2/c^2))^3/2
 
F=d(mv)/dt.

dm/dt=dm/dv dv/dt = a dm/dv .

Your formula for dm/dv is not correct. The first "2" should be 1/2.

ehild
 

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