Understanding Relativity's F=y^3ma: How Does it Apply and What is Gamma?

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Pengwuino
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The question I'm given is:

Newton's second law is given by [tex]\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over F} = \frac{{d\vec p}}{{dt}}[/tex]. If the force is always parallel to the velocity, show that [tex]\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over F} = \gamma ^3 m\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over a}[/tex].

Now, how do I get started on this thing?

Also, what I'm really wondering is how this is actually applied. When they say [tex]\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over F} = \gamma ^3 m\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over a}[/tex]… what gamma is being used? Also, are we looking at the change in momentum from wrt to the K frame?
 
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Since [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex] are always parallel, you can remove the arrows, and just work with magnitudes. Write [itex]p[/itex] in terms of [itex]v[/itex], and differentiate using the product rule and the chain rule. After differentiating, find a common denominator

Regards,
George
 
What is the 'v'. Is v the speed of the moving frame?
 
Pengwuino said:
What is the 'v'. Is v the speed of the moving frame?

[itex]v[/itex] is the speed of the particle with respect to the lab frame, so that [itex]p = \gamma m v[/itex].

Regards,
George