Solving Integration by Parts for Relativistic Kinetic Energy

[Robaj]

Hi,
I've been following a derivation of relativistic kinetic energy. I've seen other ways to get the end result but I'm interested in finding out where I've gone wrong here: I'm struggling with integrating by parts.
The author goes from $$\frac{\textrm{d}T}{\textrm{d}t}=\frac{m}{\sqrt{(1-u^2/c^2)^3}}u\frac{\textrm{d}u}{\textrm{d}t}$$to$$\frac{\textrm{d}T}{dt}=\frac{\textrm{d}}{\textrm{d}t}\left[\frac{mc^2}{\sqrt{(1-u^2/c^2)}}\right].$$

Are they are using a chain rule like$$
\frac{\textrm{d}T}{\textrm{d}t}=\frac{\textrm{d}T}{\textrm{d}u}\frac{\textrm{d}u}{\textrm{d}t}?$$
If so then I need to find $\frac{\textrm{d}T}{\textrm{d}u}.$

From the first equation I have$$\textrm{d}T = \frac{m}{\sqrt{(1-u^2/c^2)^3}}u\textrm{d}u.$$ If integrating by parts is $$\int A\frac{\textrm{d}B}{\textrm{d}u}\textrm{d}u = AB-\int\frac{\textrm{d}A}{\textrm{d}u}B\textrm{d}u$$ then I choose $A = \sqrt{(1-u^2/c^2)^3}$ and $\frac{\textrm{d}B}{\textrm{d}u}= u.$ So $B = u^2/2$ and from the chain rule$$\frac{\textrm{d}A}{\textrm{d}u}=\frac{3u}{c^2\sqrt{\left(1-\frac{u^2}{c^2}\right)^5}},$$ but this gets me right back to the start with an integral of the form $$\int u f(u) \textrm{d}u.$$ Choosing A and B' the other way round in the integration by parts doesn't give me an integral I can solve by inspection to find B.

Where have I gone wrong?
Thanks

 

[Charles Link]

It's the chain rule: Take the second equation you have and differentiate it: It's rather straightforward. $d [v^{-1/2}]=-(\frac{1}{2}) v^{-3/2} \, dv$. In this case $v=1-\frac{u^2}{c^2}$ so that $dv=-\frac{2u}{c^2} \, du$.

 

[Robaj]

It's the chain rule: Take the second equation you have and differentiate it: It's rather straightforward. $d [v^{-1/2}]=-(\frac{1}{2}) v^{-3/2} \, dv$. In this case $v=1-\frac{u^2}{c^2}$ so that $dv=-\frac{2u}{c^2} \, du$.

Thanks for your reply. I understand the differentiation and substitution you've used, but not how they help me get from the first equation to the second equation in the top post. Could you clarify? I may have misunderstood how to get from $uf(u)\frac{du}{dt}$ to $\frac{d f(u)}{dt}.$

 

[Charles Link]

It's $f'(u) \frac{du}{dt} =\frac{d f(u)}{dt}$. Only it gets confusing, because the function is $v=1-\frac{u^2}{c^2}$. Let's change the equation to $f(v)$: $f'(v) \frac{dv}{dt}=\frac{d f(v)}{dt}$. [Note: $\frac{d f(v)}{dt}=(\frac{d f(v)}{dv})(\frac{dv}{dt})=f'(v) \, (\frac{dv}{dt})$]. $\\$ Then $f(v)=v^{-1/2}$, and $f'(v)=-(\frac{1}{2}) v^{-3/2 }$. Meanwhile $\frac{dv}{dt}=-(\frac{2u}{c^2})( \frac{du}{dt})$. $\\$ And you go from the second equation in the top post to the first one. The order is reversed.

 

[Robaj]

Thanks for your patience - I'm struggling to join the dots. I want to get from the first equation to the second but your substitution doesn't match the denominator of that first equation. Just to be clear, I'm looking at $$\begin{equation} \frac{1}{\sqrt{(1-u^2/c^2)^3}}u\frac{\textrm{d}u}{\textrm{d}t}\end{equation}$$ and hoping to end up at $$\begin{equation}\frac{\textrm{d}}{\textrm{d}t}\left[\frac{c^2}{\sqrt{(1-u^2/c^2)}}\right].\end{equation}$$

If $f'(u)\frac{\textrm{d}u}{\textrm{d}t} = \frac{\textrm{d}}{\textrm{d}t}f(u)$, how come I don't need to integrate to 'undo' the differentiation?

 

[Charles Link]

Differentiate equation (2). That's all that is necessary. $\\$ For this case, write it as $f'(v)\frac{dv}{dt}=\frac{d f(v)}{dt }$. Otherwise, the $u$ makes it confusing.

 

[Robaj]

Differentiate equation (2). That's all that is necessary. For this case, write it as $f'(v)dv/dt=d(f(v))/dt$. Otherwise, the $u$ makes it confusing.

I see what you're saying. But surely the derivation goes in one direction only, so we should be able to go forward at each point based only on what we've derived previously. It doesn't make sense to me to go backwards!

 

[Charles Link]

I see what you're saying. But surely the derivation goes in one direction only, so we should be able to go forward at each point based only on what we've derived previously. It doesn't make sense to me to go backwards!

This one really works best by starting with the second equation. Someone very skilled in calculus, (like the author), goes easily from (1) to (2), but it takes practice to see how the two are quite equivalent. Going from (2) to (1) is much easier for someone who is relatively new to calculus.

 

[Robaj]

Ah, I see. Understanding how the author got from (1) to (2) is really what I'm after! But I appreciate your help. Your second post has cleared up some confusion I had about differentials.

 

[PeroK]

Thanks for your patience - I'm struggling to join the dots. I want to get from the first equation to the second but your substitution doesn't match the denominator of that first equation. Just to be clear, I'm looking at $$\begin{equation} \frac{1}{\sqrt{(1-u^2/c^2)^3}}u\frac{\textrm{d}u}{\textrm{d}t}\end{equation}$$ and hoping to end up at $$\begin{equation}\frac{\textrm{d}}{\textrm{d}t}\left[\frac{c^2}{\sqrt{(1-u^2/c^2)}}\right].\end{equation}$$

If $f'(u)\frac{\textrm{d}u}{\textrm{d}t} = \frac{\textrm{d}}{\textrm{d}t}f(u)$, how come I don't need to integrate to 'undo' the differentiation?

To take your last question. There's an integration technique called integration by guessing. That's effectively what the author has done here.

You guess something that looks like it might be the integral, differentiate it and then adjust it if it wasn't quite right.

You really need to accept that it's valid to work backwards in such cases. Physicists do this sort of thing a lot. So, you will waste a lot of time worrying. in this case, you should simply have differentiated the second expression and checked you got the first.

That said, what you need to do here is:

You have something of the form:

$\frac{dT}{dt} = f(u)\frac{du}{dt}$

Idea: try to express the right hand side as an exact derivative. So, let:

$\frac{dg(u)}{dt} = f(u)\frac{du}{dt}$

Now integrate wrt $t$ to give:

$g(u) = \int f(u) du$

To integrate the function $f(u)$ that you have here use substitution. Not parts.

Note that $T$ doesn't come into this whole calculation. It could be anything on the left hand side of the original equation.

 

[Robaj]

You really need to accept that it's valid to work backwards in such cases. Physicists do this sort of thing a lot. So, you will waste a lot of time worrying. in this case, you should simply have differentiated the second expression and checked you got the first.

Ah I understand. This is very interesting! Thanks for both explanations.