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This doesn't seem to be working? Relativity

217
0
Evaluate the derivative of [tex]\vec{F} = \frac {d} {dt} \frac {m \vec{v}} {\sqrt{1 - v^2/c^2}}[/tex] to find the acceleration [tex] a = \frac {dv}{dt} [/tex] of the particle.

So, basically, I just tried to use the quotient rule and treat m, and the whole bottom of the fraction as constants. I didn't end up getting the right answer and I can't figure out why. As reference, here are two answers I've tried, both wrong:

[tex]\frac {F}{m} \sqrt {1 - \frac {v^2}{c^2}}[/tex]
[tex]\frac {F}{m} (1 + \frac {v^2}{3c^2})[/tex]
 
Last edited:

SammyS

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Evaluate the derivative of [tex]\vec{F} = \frac {d} {dt} \frac {m \vec{v}} {\sqrt{1 - v^2/c^2}}[/tex] to find the acceleration [tex] a = \frac {dv}{dt} [/tex] of the particle.

So, basically, I just tried to use the quotient rule and treat m, and the whole bottom of the fraction as constants. I didn't end up getting the right answer and I can't figure out why. As reference, here are two answers I've tried, both wrong:

[tex]\frac {F}{m} \sqrt {1 - \frac {v^2}{c^2}}[/tex]
[tex]\frac {F}{m} (1 + \frac {v^2}{3c^2})[/tex]
How can you treat v2 as a constant?
 
260
1
Remember the fact that:

[tex]\frac{dv}{dt}=\frac{\vec{a}\cdot \vec{v}}{v}[/tex]
 
Last edited:

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