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While reading Feynman lectures on Physics, volume I, Chapter 13-4, I found following assumption, which I don't understand:

Then, since $r^2 = \rho^2 + a^2$, $\rho\,d\rho = r\,dr$. Therefore ...

leading to equation 13.17.

Could somebody please explain why is it correct? However I think that it relates to derivation, but I don´t know how.

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$$r^2=\rho^2+a^2 \tag{1}$$

We know that $r$ and $\rho$ are variables while $a$ is a constant.

Differentiate both sides of $(1)$ wrt some arbitrary variable $u$ (because we knew that both $r$ and $\rho$ are variables,their change will depend on some other variables. Thus we can treat them as $r(u)$ and $\rho(u)$) and use the chain rule

$$2r\frac{\rm dr}{\rm du}=2\rho\frac{\rm d\rho}{\rm du}+0$$ $$r\;\frac{\rm dr}{\rm du}=\rho\;\frac{\rm d\rho}{\rm du}\tag{2}$$

Now recall the differential of a function $f(x,y)$ is defined as $$\rm df=\frac{\partial f}{\partial x}\;\rm dx+\frac{\partial f}{\partial y}\;\rm dy$$

Therefore for $r$ its differential is $$\rm dr=\frac{\rm dr}{\rm du}\;\rm du$$ Now multiply both sides by $r$ to get $$r\;\rm dr=r\frac{\rm dr}{\rm du}\;\rm du$$ Now substitute $(2)$ to obtain $$r\;\rm dr=\rho\frac{\rm d\rho}{du}\;\rm du$$ Now use the differential definition again to collapse the above into the required result $$r\rm \;dr=\rho \rm\; d\rho$$

The chain rule is the reason why in physics we can use the short-cut in $(2)$ by formally "cancelling the $\rm du$ both sides to obtain the result directly"

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  • $\begingroup$ Thanks a lot for your and others explanation! I think I understand it now. $\endgroup$ – Michal Vaclavek Jan 30 '16 at 16:32
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Maybe a change of notation can help: if $f^2(x)=x^2+c$, then (differentiating both sides) $f(x)f'(x)=x$.

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This is a mathematical question but still physically relevant.

The point here is that you need to take the differential on both sides with respect to the different variables.

You have the equation $r^2=\rho^2+a^2$ where $r$ and $\rho$ are both variables but $a$ is a constant (Assuming this from the result, since you have not provided enough context.)

You then take the differential of $r$ on the left side and the differential of $\rho$ on the right side and get the required result.

$$2r\rm dr=2\rho \rm d\rho \Rightarrow r\: \rm dr=\rho \: \rm d\rho$$

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The result that the gravity emerging from a infinitly extended sheet of matter is independent of the distance ( it´s a uniform gravityfield) from the sheet can be easily seen if we consider the lines of gravity. They are all parallel, so there is no de- or increase with distance from the sheet. A problem arises if we consider (in the general relativity theory) the pace of time.

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