To deduce the expression of force in relativistic mechanics we can use :

$$\vec{F}=\frac{d\vec{p}}{dt}=\frac{d(\gamma m\vec{v})}{dt}=m\left(\frac{d\gamma}{dt}\vec{v}+\gamma\frac{d\vec{v}}{dt}\right) = m\left(\frac{|\vec{v}||\vec{a}|\gamma^3}{c^2}\vec{v} +\gamma\vec{a}\right)= m\gamma\left(\frac{|\vec{v}||\vec{a}|}{c^2-v^2}\vec{v} +\vec{a}\right)$$

But in one book i found :

$$\vec{F}=m\gamma\left(\frac{\vec{v}\cdot\vec{a}}{c^2-v^2}\vec{v} +\vec{a}\right)$$

How could one justify the exchange of $|\vec{v}||\vec{a}|$ for $\vec{v}\cdot\vec{a}= |\vec{v}||\vec{a}|cos\theta $ ?

  • $\begingroup$ If $\theta=0$ ...which you would know already? $\endgroup$
    – JMLCarter
    Nov 24 '17 at 2:48
  • $\begingroup$ but in general, an acceleration is not in the same direction than the velocity, and i don't think i have made any imposition of that beeing so in my derivation $\endgroup$ Nov 24 '17 at 2:52
  • $\begingroup$ And is there any suggestion of it "in one book" where you found the equation? $\endgroup$
    – JMLCarter
    Nov 24 '17 at 2:54
  • $\begingroup$ No it just uses this in a derivation, without actually proving it $\endgroup$ Nov 24 '17 at 2:59

First, notice that:

  • $\displaystyle \frac{d |\vec{v}|}{dt}$ stands for the time derivative of the magnitude of the velocity vector $\vec{v}$.

  • $\displaystyle \left|\frac{d\vec{v}}{dt}\right|$ stands for the magnitude of the time derivative of the velocity vector.

Generally, these quantities are different. Only the second is to be identified with the magnitude of the acceleration vector, $|\vec{a}|=|d\vec{v}/dt|$.

Because of that, you have to correct your first formula using

$$\frac{d\gamma}{dt} = \frac{\gamma^3}{c^2} |\vec{v}| \frac{d|\vec{v}|}{dt} \neq \frac{\gamma^3}{c^2} |\vec{v}| |\vec{a}|.$$

Finally, you get your last (and correct) formula noticing that

$$ |\vec{v}| \frac{d|\vec{v}|}{dt} = \frac{1}{2} \frac{d}{dt}|\vec{v}|^2 = \frac{1}{2} \frac{d}{dt}(\vec{v}\cdot\vec{v}) = \vec{v} \cdot \frac{d\vec{v}}{dt} = \vec{v} \cdot \vec{a}.$$


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