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When $E$ is a conserved quantity, what does

$$\frac{\partial }{\partial E}$$

mean? Does that make any sense?

The reference where I stumbled over this problem is this PDF at eq. (A19)

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closed as off-topic by Shailesh, Shaun, Nosrati, Claude Leibovici, kingW3 Feb 18 '17 at 13:47

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    $\begingroup$ If you are reading that paper, then you should not need to ask this question. $\endgroup$ – Mark Viola Feb 18 '17 at 4:20
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    $\begingroup$ Eqs. (A4) define $R$ and $\Theta$ as expressions in, among other things, $E$. It thus makes sense to calculate $\partial/\partial E$ of those expressions, and that is what's going on in (A19) where we have $\partial(R+\Delta\Theta)/\partial E$. I do not know whether it makes any physical sense; I don't really get ordinary Hamiltonian mechanics, let alone this relativistic version. $\endgroup$ – ForgotALot Feb 18 '17 at 4:34
  • $\begingroup$ But E is constant (in the PDF it is said to be a conserved quantity), so there should be no change in it! Wouldn't that lead to a division by zero? $\endgroup$ – Yukterez Feb 18 '17 at 5:27
  • $\begingroup$ I don't see why. Think of good old planetary orbits. Sure, there are constants of the motion, angular momentum $L$ say; but we can write expressions involving them like $L-m\omega r^2$ and take derivatives. Maybe taking the derivative is part of a derivation which lets us calculate the value of the constant of the motion. The interatomic distance $l$ in a diatomic molecule like H$_2$ is fixed (leaving aside vibrations and silly things like that), but we often calculate it by $dE/dl=0$ for a suitable energy function $E$. $\endgroup$ – ForgotALot Feb 18 '17 at 7:52
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I don't know the context, but $\frac{\partial}{\partial E}$ is a partial derivative with respect to the variable (or whatever it is) $E$. If you look back in the PDF, you will likely find the definition of what $E$ means. If you were reading this whole file you would know what $E$ is, unless $E$ wasn't defined, then it is the author's fault.

If you don't know what are partial derivatives, then I suggest learning about it (depending on what level in mathematics you are, and why you stumbled on this file anyways).

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  • $\begingroup$ E is defined, it is the conserved energy of the particle who's trajectory is differentiated with respect to time. Therefore I wonder, because I don't understand why one would differentiate a constant? $\endgroup$ – Yukterez Feb 18 '17 at 4:56
  • $\begingroup$ Well, differentiating a constant can be seen, not really often but still. For example, in basic physics, if you have a velocity graph with a constant function, the acceleration would be 0. $\endgroup$ – KKZiomek Feb 18 '17 at 5:08
  • $\begingroup$ Sure, but in the context of the pdf that would unescessarily overcomplicate the equation which is praised to be as simple as possible to carry a term which is 0 anyway, I don't really understand what to do with this. Set it to 0? $\endgroup$ – Yukterez Feb 18 '17 at 5:12
  • $\begingroup$ Are you sure that the expression inside the parenthesis after $\frac{\partial}{\partial E}$ is constant? If you are really, really sure then yes, set it to 0. $\endgroup$ – KKZiomek Feb 18 '17 at 5:18
  • $\begingroup$ But wouldn't that lead to a division by zero, since E is in the denominator? The PDF says on page 30: "three additional quantities are conserved along Kerr geodesics: the energy E and..." so the change in E should be 0, shouldn't it? $\endgroup$ – Yukterez Feb 18 '17 at 5:29

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