High School How does this derivative work?; and why the speed of light remains?

[mcastillo356]

TL;DR

There is a step I don't understand.

Relativistic Momentum, Mass, and Energy

Momentum and mass

(...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...)

The momentum of a particle moving with velocity $v$ is given by

$$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$

ENERGY

In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional motion only, we have

$$F_{net}=\cfrac{dp}{dt}\qquad{R-11}$$

We wish to find an expression for the kinetic energy. To do this, we multiply both sides of equation $R-11$ by the displacement $dl$. This gives

$$F_{net}\,dl=\cfrac{dp}{dt}\,dl\qquad{R-12}$$

where we identify the term on the left as the work and the term on the right as the change in kinetic energy. Substituting $v\,dt$ for $dl$ in the term on the right we obtain

$$dK=\cfrac{dp}{dt}\,v\,dt=v\,dp$$

Integrating both sides gives

$$K=\displaystyle\int_0^{P_f}v\,dp\qquad{R.13}$$

To evaluate this integral, we first change the integration variable from $p$ to $v$. Using Equation $R-10$ and the quotient rule, we have

$$dp=d\Bigg (\cfrac{v}{\sqrt{1-(v^2/c^2)}}\Bigg )$$

Why in the step above $m$ disappears?

Attempt

$dp=m\cdot{dv}=d\Bigg (\cfrac{m\cdot{v}}{\sqrt{1-(v^2/c^2)}}\Bigg )$, so they cancel.

(...)

$\cfrac{[1-(v^2/c^2)]^{1/2}\,dv-v\frac{1}{2}[1-(v^2/c^2)]^{-\frac{1}{2}}\bigg (-\cfrac{2vdv}{c^2}\bigg )}{1-(v^2/c^2)^{3/2}}=\cfrac{dv}{[1-(v^2/c^2)]^{3/2}}$

Substituting for $dp$ in Equation $R-13$ gives

$$K=\displaystyle\int_0^{P_f}vdp=m\displaystyle\int_0^{v_f}\cfrac{vdv}{[1-(v^2/c^2)]^{3/2}}=mc^2\Bigg (\cfrac{1}{\sqrt{1-(v_f^2/c^2)}}-1\Bigg )$$

So

$$K=\cfrac{mc^2}{\sqrt{1-(v^2/c^2)}}-mc^2\qquad{R-14}$$

(In this expression, because the only speed is $v_f$, the subscript $f$ is not needed)

Defining $mc^2/\sqrt{1-(v^2/c^2)}$ as the total relativistic energy $E$, Equation $R-14$ can be written

$$E=K+mc^2=\cfrac{mc^2}{\sqrt{1-(v^2/c^2)}}\qquad{R-15}$$

where $mc^2$, called the rest energy $E_0$, is energy the particle has when it is at rest.

By multiplying both sides of Equation $R-10$ by $c$ and then dividing the resulting equation by Equation $R-15$, we obtain

$$\cfrac{v}{c}=\cfrac{pc}{E}\qquad{R-16}$$

which can be useful when trying to solve for the speed $v$. Eliminating $v$ from Equations $R-10$ and $R-16$, and solving for $E^2$, gives

$$E^2=p^2c^2+m^2c^4$$

 

[GlitchedGluon]

What is the question exactly?

 

[mcastillo356]

Why in the change of variable from $p$ to $v$ vanishes $m$. And which is the behavior of $c$: is this last an invariant of the derivative?. These are my doubts.

 

[GlitchedGluon]

Substituting for $dp$ in Equation $R-13$ gives

$$K=\displaystyle\int_0^{P_f}vdp=m\displaystyle\int_0^{v_f}\cfrac{vdv}{[1-(v^2/c^2)]^{3/2}}=mc^2\Bigg (\cfrac{1}{\sqrt{1-(v_f^2/c^2)}}-1\Bigg )$$

But the mass does not dissapear, you have it right there in your expression for K.
c is the speed of light, it is a constant.

 

[mcastillo356]

Let me think it over again. Your reasoning is perfect. I must be wrong somewhere, but still the change of variable in $R-10$ keeps me wondering.

 

[GlitchedGluon]

$$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$

If you take the derivative of that, m just "gets out" because it is constant:

$$dp =m \, \, d \left(\cfrac{v}{\sqrt{1-(v^2/c^2)}}\right) = m \, \, {dv \over [1-(v^2/c^2)]^{3/2}}$$

That is the change of variables you do in your equation R-13.

 

[PeroK]

Note that you can avoid the differentiation, by expressing $v$ as a function of $p$:
$$p = \gamma mv \implies v = \frac p {\sqrt{m^2 + \frac {p^2}{c^2}}}$$integrating gives:
$$K = c^2\sqrt{m^2 + \frac{p^2}{c^2}} - mc^2$$$$= \frac{pc^2}{v} -mc^2$$$$ = \gamma mc^2 - mc^2$$

 

[mcastillo356]

Understood inmediatly. Thanks Forum.