Force on a relativistic particle

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1. Dec 24, 2016

omoplata

1. The problem statement, all variables and given/known data

A particle of rest mass $m_0$ is is caused to move along a line in such a way that its postion is $$x = \sqrt{b^2 + c^2 t^2} -b$$What force must be applied to the particle to produce this motion?

2. Relevant equations

The velocity of the particle as seen from the rest frame is $v = \frac{dx}{dt}$ and the acceleration is $a = \frac{dv}{dt}$.

The mass of the particle as seen in the rest frame is $m = \gamma m_0$, where $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$.

Then the force should be $F = m a$.

3. The attempt at a solution

Differentiating $x$, I get $v = \frac{c^2 t}{\sqrt{b^2 + c^2 t^2}}$.

So, $\gamma = \frac{\sqrt{b^2 + c^2 t^2}}{b}$.

Differentiating $v$, I get $a = \frac{b^2 c^2}{(b^2 + c^2 t^2)^{\frac{3}{2}}}$.

So, $F = \frac{m_0 b c^2}{b^2 + c^2 t^2}$.

The answer given in the book is $F = \frac{m_0 c^2}{b}$. What did I do wrong?

2. Dec 24, 2016

TSny

Force in special relativity is generally defined as rate of change of relativistic momentum: $\vec{F} = \frac{d\vec{p}}{dt}$, where $\vec{p} = \gamma m_0 \vec{v}$. This is not equivalent to $\vec{F} = m \vec{a}$, where $m = \gamma m_0$ and $\vec{a} = \frac{d\vec{v}}{dt}$.

See

3. Dec 24, 2016

omoplata

OK, I get the book answer now. Thank you!