Need tips to understand Relativistic Energy in Special Relativity

[mcastillo356]

TL;DR

I don't understand how Tipler/Mosca famous book accomplishes Relativistic Energy

As in classic mechanics, we will define kinetic energy as the work done by the net force in accelerating a particle from rest to some final velocity $u_f$. Considering one dimension only, we have

$$K=\displaystyle\int_{u=0}^{u=u_f}F_{net}\,ds=\displaystyle\int_{u=0}^{u-u_f}\cfrac{dp}{dt}\,ds=\displaystyle\int_{u=0}^{u-u_f}u\,dp=\displaystyle\int_{u=0}^{u-u_f}u\,d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )\qquad{39-21}$$

where we have used $u=ds/dt$. It is left as a problem for you to show that

$$d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )=m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

(I needed a quick browsing to understand it)

$$K=\displaystyle\int_{u=0}^{u=u_f}u\,d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )=\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

$$=mc^2\Bigg (\cfrac{1}{\sqrt{1-(u^2/c^2)}}-1\Bigg )$$

This last step is what I don't understand.

Attempt

$$(1-u^2/c^2)=t\Rightarrow{dt=-2u/c^2}$$

But it doesn't work: I can't get rid of $u$. I neither know how to deal with $c$

$$K=\cfrac{mc^2}{\sqrt{1-(u^2/c^2)}}-mc^2\qquad{39-22}$$

REST ENERGY

$$E_0=mc^2\qquad{39-23}$$

RELATIVISTIC ENERGY

$$E=K+mc^2=\cfrac{mc^2}{\sqrt{1-(u^2/c^2)}}\qquad{39-24}$$

 

[Ibix]

$$\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

I think this is the integral you are attempting to compute. If so, I'd do it by inspection. For any integrand with a $1/f^n(x)$ it's worth asking if $1/f^{n+1}(x)$ might be the solution and seeing if the rest of the integrand turns out to be $f'(x)$.

 

[SiennaTheGr8]

You can also derive $d \gamma = \gamma^3 (\vec \beta \cdot d \vec \beta)$ fairly easily and use that to do a change of variables (where $\vec \beta = \vec u / c$ and $\gamma = (1 - \beta^2)^{-1/2}$): https://www.physicsforums.com/threa...special-relativity.938269/page-3#post-5939051

An equivalent but arguably more elegant approach uses hyperbolic functions: https://www.physicsforums.com/threa...of-consts-of-integration.950967/#post-6022512

 

[Herman Trivilino]

Attempt

$(1−u^2/c^2)=t⇒dt=−2u/c^2$

Should be $dt=−2u/c^2\ du$

 

[PeroK]

$$K=\displaystyle\int_{u=0}^{u=u_f}u\,d\Bigg (\cfrac{mu}{\sqrt{1-(u^2/c^2)}}\Bigg )=\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

$$=mc^2\Bigg (\cfrac{1}{\sqrt{1-(u^2/c^2)}}-1\Bigg )$$

This last step is what I don't understand.

An integral is an antiderivative. The only thing you need to check is that:
$$mc^2\frac d {du} \bigg (\cfrac{1}{\sqrt{1-(u^2/c^2)}}\Bigg ) = m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u$$You should never be confused by an integral like that. All you need to do is check by differentiation.

That said, you might ask how you find that antiderivative? I think substitution is a clumsy way to do it, because you should be able to see that:
$$\frac d {du}\bigg((1 - au^2)^{-\frac 1 2}\bigg) = -\frac 1 2(1 - au^2)^{-\frac 3 2}(-2au) = au(1 - au^2)^{-\frac 3 2} $$And, that derivative is something that you should be familiar with by now. You should be able to write that down with minimal calculation.

Note that the author probably assumes you have the prerequisite calculus knowledge, and are not going to be stumped by something that you should already know.

 

[mcastillo356]

$$\displaystyle\int_0^{u_f}m\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du=m\displaystyle\int_0^{u_f}\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,u\,du$$

$$1-\cfrac{u^2}{c^2}=t\Rightarrow{-\cfrac{2u}{c^2}}\,du=dt\Rightarrow{u\,du=-\cfrac{c^2}{2}\,dt}$$

$$m\Big (-\frac{c^2}{2}\Big )\displaystyle\int_0^{u_f}\Bigg (1-\cfrac{u^2}{c^2}\Bigg )^{-3/2}\,dt$$

$$-m\cfrac{c^2}{2}\displaystyle\int_0^{u_f}t^{-3/2}\,dt=-m\frac{c^2}{2}\cdot\cfrac{t^{-1/2}}{-1/2}\Bigg |_0^{u_f}$$

$$mc^2\Bigg (\cfrac{1}{\sqrt{1-\cfrac{u_f^2}{c^2}}}-1\Bigg )$$