/sci/ - Science & Math » Thread #11798122

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Anonymous Sun 14 Jun 18:55:21 2020 No.11798122 View Reply Original Report

Quoted By: >>11798591 >>11798617 >>11799878

IS this true at all? got it from the electrostatic flow fomula

Anonymous

Anonymous Sun 14 Jun 2020 18:57:33 No.11798128 Report

Quoted By: >>11798135

how are you taking the divergence of a scalar quantity (the triple integral)

Anonymous

Anonymous Sun 14 Jun 2020 19:00:10 No.11798135 Report

Quoted By: >>11798148

>>11798128
not a scalar quantiy. r0-r is a vector

Anonymous

Anonymous Sun 14 Jun 2020 19:03:22 No.11798148 Report

Quoted By: >>11798164

>>11798135
it is being integrated over a region which returns a scalar

Anonymous

Anonymous Sun 14 Jun 2020 19:07:59 No.11798164 Report

Quoted By:

>>11798148
what? it is tree different triple integrals that give you the coordinates.

Anonymous

Anonymous Sun 14 Jun 2020 21:10:50 No.11798591 Report

Quoted By:

>>11798122
how did you possibly arrive at this from the "electrostatic flow formula". you mean something like gauss's law?

Anonymous

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Anonymous Sun 14 Jun 2020 21:20:54 No.11798612 Report

Quoted By: >>11798882 >>11798971

It is interesting to me that the charge density term in the triple integral is missing. I feel the intuition here was to set the divergence of the electric field equal to charge density per unit permitivity, than cancel out

Ah i think i know what you attempted here, you substituted the triple integral into gauses law expressed as the divergence of the electric field equal to charge density per unit permitivity..

he subbed it in literally canceled out rho and rho(r_vector), the permittivity constant on both sides, and shuffled 4pi to the end.

I would need further clarification on how the charge density function, one of r0-0 pops out as equivalent to the charge density so as to cancel out this term

Anonymous

Anonymous Sun 14 Jun 2020 21:21:50 No.11798617 Report

Quoted By:

>>11798122
No the density of charge is missing.

Anonymous

Anonymous Sun 14 Jun 2020 22:48:17 No.11798882 Report

Quoted By: >>11798971 >>11798993

>>11798612
yeah, you have a high IQ, wich is rare here.
I did exactly that, but I asumed ro as a constant cause otherwise the formula wouldnt apply otherwise, right?

Anonymous

Anonymous Sun 14 Jun 2020 23:10:48 No.11798971 Report

Quoted By: >>11798984 >>11799646

>>11798612
>>11798882
You can't cancel out rho. On one side of the equation rho is a function of unprimed r which is a variable and on the other side it is a function of r' which is being integrated over.

To give you an example of why what you did was wrong, it is true that

$\tan x \cos x=\int^x_{0} \cos x'dx'$

but it's not true that

$\tan x =\int^x_{0} dx'$

Anonymous

Anonymous Sun 14 Jun 2020 23:16:36 No.11798984 Report

Quoted By: >>11799123

>>11798971
To continue and tell you something useful, what you really can extrapolate from that formula is
$\nabla\cdot\left(\frac {1}{4\pi} \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}\right)=\delta^{(3)}(\vec{r}-\vec{r}')$

Anonymous

Anonymous Sun 14 Jun 2020 23:19:50 No.11798993 Report

Quoted By: >>11799762

>>11798882
Yes, this seems mostly correct, as long as r0 is inside the body.
First, this corresponds to rho constant in the body, and zero outside it.
Since the integrand is smooth, just move the divergence inside the integral. The integrand has zero divergence whenever r0 isnt r. As a result, we only have to integrate over a ball around r0.
So if r0 is outside the body, the integral is obviously 0. If r0 is inside the body, it is some fixed constant independant of the choice of body and r0 -- probably 4pi but i can't be assed to check.

Anonymous

Anonymous Mon 15 Jun 2020 00:16:07 No.11799123 Report

Quoted By:

>>11798984
\thread
Look up greens functions

Anonymous

Anonymous Mon 15 Jun 2020 03:39:14 No.11799646 Report

Quoted By: >>11799832

>>11798971
do you kno the difference betwen a constant and a function?
in the case i said ro(R) is constant, so it IS true that
Rf=integral(RF)
therefor
f=integral(F)

Anonymous

Anonymous Mon 15 Jun 2020 04:16:07 No.11799762 Report

Quoted By:

>>11798993
>The integrand has zero divergence whenever r0 isnt r. As a result, we only have to integrate over a ball around r0.
wtf with this two rules??

Anonymous

Anonymous Mon 15 Jun 2020 04:38:27 No.11799832 Report

Quoted By: >>11799851 >>11800990

>>11799646
Cool it with the attitude. I know more than you and was trying to teach you something. You aren't going to learn anything like that

Anonymous

Anonymous Mon 15 Jun 2020 04:45:14 No.11799851 Report

Quoted By:

>>11799832
Not the guy you were answering to but you should be the one cooling it. You made a mistake, he called it out in good faith. He told you rho was constant right at the beginning.

Anonymous

Anonymous Mon 15 Jun 2020 04:58:58 No.11799878 Report

Quoted By:

>>11798122
$<br />\triangledown \cdot \iiint \dfrac{ \bar{r}_0- \bar{r}}{ \left | \bar{r}_0- \bar{r} \right |^3} \, \text{d}x \, \text{d}y \, \text{d}z = 4 \pi <br />$

Anonymous

Anonymous Mon 15 Jun 2020 14:16:18 No.11800990 Report

Quoted By:

>>11799832
How would I learn from you if you are replying to a post were i said rho was constant because that is a hipothesis for guass law, and act if you didnt even read it? You just changed topic to "demonstrate" all that you know without any real interest in the topic.

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