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Is there something wrong with the following proof (see below)? To me, it seems like the third line should show $$\frac{dP_{ab}}{dt}=-\int_a^b \frac{\partial}{\partial t}J(x,t)dx$$ Am I missing something obvious?

enter image description here

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Yes in the very first equation after the horizontal line, the right absolute value line, |, is missing. – Mew Mar 11 '13 at 4:54
@Chris: That's just a typo. My main concern is the swapping of a $\partial x$ for a $\partial t$ in the third line. – Joebevo Mar 11 '13 at 4:56
Ok. I'd be more inclined to think the last equation on line 2 is wrong. I think it should be the partial with respect to x, not t. – Mew Mar 11 '13 at 4:59
Griffith's 2nd Ed problem 1.14. – CHM Jan 30 '14 at 6:42
up vote 1 down vote accepted

If we look at line 2, we have an integral set equal to:

enter image description here

We then must note that $J(x,t)$ is defined as the negative of:

enter image description here

So the last equation on line 2 should be: $-\frac{\partial J(x,t)}{\partial x}$ as opposed to $-\frac{\partial J(x,t)}{\partial t}$.

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