(gamma)ma=F-[F.(beta)]beta How derive this?

[vin300]

(gamma)ma=F-[F.(beta)]beta
How derive this?

 

[HallsofIvy]

Would you mind explaining what it is? Until you explain what "gamma", "m", "a", "F", and "beta" mean, it's just a string of symbols.

 

[vin300]

gamma is the lorentz factor, m is mass, a acceleration, F force; beta is v/c
Lev Okun, in his "concepts of mass" has derived it as follows:
E=ymc^2
p=ymv
dp/dt=F
and next he writes
a=[F-(F.beta)beta]/ym

 

[Tac-Tics]

Would you mind explaining what it is? Until you explain what "gamma", "m", "a", "F", and "beta" mean, it's just a string of symbols.

This is how I feel every time I try to read anything written by a physicist



But Halls is right. What is the context of the problem you're working on?

The left hand side, F-[F.(beta)]beta, looks a lot like you're trying to remove a component from the force vector (it's very close to the force perpendicular to the direction of motion... is this what you're going for?)

 

[Meir Achuz]

I think the equation you are looking for is
{\bf F}=\frac{d{\bf p}}{dt}
\frac{d}{dt}\frac{m{\bf v}}{\sqrt{1-v^2}}
=m{\bf a}\gamma+m{\bf v}({\bf v}\cdot{\bf a})\gamma^3.
This is the usual form.
To get your form, use the above equation to show
{\bf v}\cdot{\bf F}=m\gamma^3{\bf v}\cdot{\bf a}
in a few steps.
I use units with c=1.
I couldn't find my latex errors. I hope you can read it as is.

 

[Fredrik]

I couldn't find my latex errors. I hope you can read it as is.

Just add tex tags.

I think the equation you are looking for is
$${\bf F}=\frac{d{\bf p}}{dt}$$
$$\frac{d}{dt}\frac{m{\bf v}}{\sqrt{1-v^2}}<br /> =m{\bf a}\gamma+m{\bf v}({\bf v}\cdot{\bf a})\gamma^3$$.
This is the usual form.
To get your form, use the above equation to show
$${\bf v}\cdot{\bf F}=m\gamma^3{\bf v}\cdot{\bf a}$$
in a few steps.
I use units with c=1.

 

[Meir Achuz]

Thank you. I did use $$...$$, but it kept showing errors.
I guess it can read your handwriting.

 

[vin300]

I got it thanks but using the equation gives F=y^3ma when parallel and F=yma when transverse why is dt so?

 

[Meir Achuz]

You can't argue with algebra. There is no "why" in algebra.

 

[vin300]

Usually it is "y" times mass but here it increases to y^3 times as we move from transverse to longitudinal