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Knight
6:02 AM
@JohnRennie Sir, why $$ \oint dU= 0$$ for any U ?
John Rennie
The integral of $dU$ is just the function $U$.
Knight
@JohnRennie Yes I agree
John Rennie
If you integrate round a loop you start and end at the same point $\mathbf x$, so your integral is $U(\mathbf x) - U(\mathbf x)$ and of course that's zero.
This assumes your function isn't multivalued.
Knight
$U$ is a scalar
@JohnRennie Is the result $$\oint dU =0$$ implies $U$ a scalar?
John Rennie
I think it would apply to any function.
Knight
6:10 AM
@JohnRennie I think I need to study Vector Calculus once more.
:)
Guru Vishnu
@JohnRennie: Hi sir. Good morning :-)
John Rennie
@GuruVishnu hi :-)
Guru Vishnu
@JohnRennie After Aladdin's discussion, if possible, could you come to the other room, sir?
John Rennie
@GuruVishnu ok
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