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Post by Admin on Dec 8, 2019 15:01:41 GMT
I have posted my answers to exercises 15-19 inclusive. The solutions can be found in the usual place.
Vasco/Admin
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Post by telemeter on Apr 18, 2020 9:56:00 GMT
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Post by Admin on Apr 18, 2020 10:22:18 GMT
Hi telemeter
I don't see any attachment
Vasco
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Post by telemeter on Apr 18, 2020 10:40:42 GMT
Updated now
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Post by Admin on Apr 18, 2020 15:40:51 GMT
telemeter
The definition of gradient is:
The gradient can be interpreted as the "direction and rate of fastest increase". At a point p the direction of the gradient is the direction of fastest increase of the function at p, and its magnitude is the rate of increase in that direction.
In this exercise $n$ is the distance in the direction of $N$ and since $\mathcal{G}$ increases in the opposite direction then we must have the minus sign.
When we say that a vector field is sourceless we mean locally sourceless. It can have point sources at a finite number of points. Under this definition $\mathcal{G}_a$ is sourceless everywhere except at $a$ and $T$ is also sourceless.
On the infinitesimal circle $k$ the value of $\mathcal{Q}_a$ is constant and equal to $1/\rho$ and so it is fine to use GMVT at the bottom of page 2 of my answer and on page 3.
Vasco
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Post by telemeter on Apr 18, 2020 17:21:56 GMT
No, I'm still going to argue! Here's my case...If G gets smaller in the direction n, dG/dn will be inherently negative. Grad G.n will also be negative. After all, this means what is the projection of grad G in the direction n...answer than same as dG/dn...negative.
You assume dG/dn increases along n. It does not. If you want gradG.n=-dG/dn you are saying they vary "inversely". They do not.
In general, grad (phi).a = d(phi)/da....a fact of vectors, not particular circumstances. To say otherwise is like saying Div grad phi sometimes = - del squared phi.
On GMVT, thanks, I was unclear on this...I still am! If $G_a$ (gosh I used LaTex!) is not defined at a how can you say the integral = G(a)? Surely one needs a Cauchy integral at this point?
telemeter
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Post by telemeter on Apr 18, 2020 17:33:54 GMT
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Post by Admin on Apr 18, 2020 17:43:07 GMT
No, I'm still going to argue! Here's my case...If G gets smaller in the direction n, dG/dn will be inherently negative. Grad G.n will also be negative. After all, this means what is the projection of grad G in the direction n...answer than same as dG/dn...negative. You assume dG/dn increases along n. It does not. If you want gradG.n=-dG/dn you are saying they vary "inversely". They do not. In general, grad (phi).a = d(phi)/da....a fact of vectors, not particular circumstances. To say otherwise is like saying Div grad phi sometimes = - del squared phi. On GMVT, thanks, I was unclear on this...I still am! If $G_a$ (gosh I used LaTex!) is not defined at a how can you say the integral = G(a)? Surely one needs a Cauchy integral at this point? telemeter telemeter I don't say the integral is $\mathcal{G}_a$ do I? Where exactly? And which integral? Vasco |
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Post by telemeter on Apr 20, 2020 14:13:20 GMT
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Post by telemeter on Apr 20, 2020 14:18:44 GMT
Re Ex 15 ii)
Ah re $G_a$ , apologies, my confusion (as ever). I need to go away and have a deep thought on GMVT....especially as I have just argued everything is well-defined at a in my 15 iii) attempt!
telemeter
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