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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
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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 at 6:42
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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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