-1
$\begingroup$

The work energy theorem is: $$\int_{s_1}^{s_2}\frac{dp}{dt}ds=\frac{1}{2}m(v_2^2-v_1^2)$$

Here $\frac{1}{2}mv^2$ is non-relativistic kinetic energy. Now, if we use $\frac{dp}{ds}$ instead of $\frac{dp}{dt}$, then it is easy to show that: $$\int_{t_1}^{t_2}\frac{dp}{ds}dt=m(\ln{(v_2)}-\ln{(v_1)})$$

This is similar to the work-energy theorem, so I thought $m\ln{(v)}$ must be a quantity analogous to kinetic energy, i.e. $\frac{1}{2}mv^2$ and $\frac{dp}{ds}$ analogous to force.

So, I tried to get a relativistically correct expression of $\int_{t_1}^{t_2}\frac{dp}{ds}dt$. But I got an absurd result:

$$\int_0^t\frac{dp}{ds}dt=\int_0^v\frac{d\left(\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\right)}{v}$$ $$=\left|\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}\right|_0^v-\int_0^v\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}d\left(\frac{1}{v}\right)$$ $$=\left|\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}\right|_0^v-\int_0^v\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\frac{-1}{v^2}dv$$ $$=\left|\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}\right|_0^v+\int_0^v\frac{m}{v\sqrt{1-\frac{v^2}{c^2}}}dv$$ Now, the second term is: $$I=\int_0^v\frac{m}{v\sqrt{1-\frac{v^2}{c^2}}}dv$$ $$=mc\int_0^v\frac{dv}{v\sqrt{c^2-v^2}}$$ Putting $v=c\sin{x}$ and $dv=c\cos{x}dx$, we get: $$I=mc\int_0^v\frac{c\cos{x}dx}{c^2\sin{x}\cos{x}}$$ $$=m\int_0^v\csc{x}dx$$ $$=m|\log{|\csc{x}-\cot{x}|}|_0^v$$ $$=m\left|\log{\left|\csc{sin^{-1}\left(\frac{v}{c}\right)}-\cot{sin^{-1}\left(\frac{v}{c}\right)}\right|}\right|_0^v$$ $$=m\left|\log{\left|\frac{c-c \sqrt{1-\frac{v^2}{c^2}}}{v}\right|}\right|_0^v$$ But, this expression is not defined at the lower limit $v=0$. Have I done something wrong in the maths or does the relativistically correct expression of $\int_0^t\frac{dp}{ds}dt$ not exist?

$\endgroup$
0
$\begingroup$

This is similar to the work-energy theorem, so I thought $m\,ln(v)$ must be a quantity analogous to kinetic energy, i.e. $\frac 1 2 mv^2$

The mass energy equivalence in (special) relativity means you get :

$$E = \frac {m_0 c^2} { \sqrt { 1 - \frac {v^2} {c^2} } }$$

But this, for $\frac v c$ small, is :

$$E \approx m_0 c^2 + \frac 1 2 m_0 v^2$$

So you already have an analogue of energy-mass equivalence in Newtonian mechanics, as the kinetic energy in classical mechanics already tells you the change in energy due to motion and you can, if you wish, equate that with the mass :

$$m \approx m_0 ( 1 + \frac 1 2 \frac {v^2} {c^2} )$$

You're simply going around the houses trying to find an "analogue" for this as you already have the actual expression.

$\endgroup$
  • $\begingroup$ The question is simply this: Get a relativistically correct expression for $\int_0^t\frac{dp}{ds}dt$. Forget about analogues. $\endgroup$ – Dove Apr 5 '17 at 11:41
  • $\begingroup$ It sound to me like you want the maths on this page. $\endgroup$ – StephenG Apr 5 '17 at 11:49
0
$\begingroup$

The thing you're taking the logarithm of at the end simplifies, I think, to $$\frac{c-c \sqrt{1-\frac{v^2}{c^2}}}{v}.$$ Expand the sqrt to second order in $v$ and I think you'll see that it (the above expression) goes to zero as v goes to zero. Still not nice when you have to take the log, but perhaps not surprising bearing in mind your starting point. Note that your earlier newtonian expression also blows up if you insist on putting $v_1=0$.

$\endgroup$
  • $\begingroup$ Thanks. But it still reduces to $\frac{0}{0}$. BTW, I'm just saying that in a world where $\frac{dp}{ds}$ is the definition of force, $m\ln{v}$ is the kinetic energy. $\endgroup$ – Dove Apr 5 '17 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.