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/sci/ - Science & Math


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3978598 No.3978598 [Reply] [Original]

/sci/, suppose you're in a 4 dimensional space.
You want to calculate the electric field of a point charge.You build an hypersphere that includes the charge. You apply Gauss Theorem.
E* 2*pi^2*R^3 = Q/ε0. E is proportional to 1/R^3.
WTF? What's wrong?

>> No.3978608

(i forgot to sai that 2*pi^2*R^3 is hypersphere surface)

>> No.3978613

why should that be wrong?

>> No.3978617

that's what's supposed to happen

>> No.3978624

The dimensions of the lefthandside and righthandside are different.

>> No.3978647

>>3978613
>>3978617
I don't know, it's quite strange
>>3978624
lefthandside?
righthandside?

>> No.3978680
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3978680

You could even do this with a circle in the plane and then wonder why E is proportional to 1/r. However Electrodynamics is inherently three dimensional (three space, one time), i.e. there is a reason why the number of field components E_x,E_y,E_z,B_x,B_y,B_z is six (there are six possible "angles" between x, y, z and t). The Maxwell equations are fundamental, not Gauss Theorem applied to 3 dimensional \vec{E} in 3 dimensions, i.e. the conservation of the total flux of the point charge field.

For a more general study of this subject (more general regarding dimensions) read
http://en.wikipedia.org/wiki/Harmonic_analysis

>> No.3978696

>>3978680
okay I just saw that the harmonic analysis link is not very helpful. My point was to consider the Poisson equation for a point charge in d dimensions, then you get an analog situation as in 3+1-dimensional electrodynamics. (not by postucalting E·A \propto Q)
This will soon lead to harmonic analysis, sperical harmonics in d dimesnions and alike.

>> No.3978791

<div class="math">\mathbf{1.: 3D.}</div>
Let <span class="math">E=E_i\mathrm du^i[/spoiler] and <span class="math">\mathcal M^3[/spoiler] be a ball with radius <span class="math">R[/spoiler], then by applying Stokes' theorem (using <span class="math">\varepsilon_0=1[/spoiler]),
<div class="math">\int_{\partial\mathcal M^3}*E = \int_{\mathcal M^3}
\mathrm d*E</div>Using Maxwell's equation <span class="math">d*E
=\varrho[/spoiler] (classical: <span class="math">\vec\nabla\vec E=\varrho[/spoiler]), this yields<div class="math">\int_{\mathcal M^3}\varrho~.</div><span class="math">\varrho[/spoiler] is a 3-form with the coordinate expansion <span class="math">\sum_{i<j<k}\varrho_{ijk}\mathrm du^i\wedge\mathrm du^j\wedge\mathrm du^k:=\sum_{i<j<k}\varrho_{ijk}
\mathrm du^{ijk} = \varrho_{123}\mathrm du^{123}[/spoiler] (since there is only one index combination satisfying <span class="math">i<j<k[/spoiler] in 3-dimensional space), therefore<div class="math">\int_{\mathcal M^3}\varrho
= \int_{\mathcal M^3}\varrho_{123}\mathrm du^1\mathrm du^2\mathrm du^3 = Q</div>
Assuming <span class="math">E=const[/spoiler], <div class="math">\int_{\partial\mathcal M^3}*E=E\int_{\partial\mathcal M^3}\mathrm{vol}^3 = E V_{\partial\mathcal M^3}</div>where <span class="math">V_{\partial\mathcal M^3}[/spoiler] is the surface area of your sphere. In D dimensions, it is given by <span class="math">DR^{D-1}\frac{\pi^{D/2}}{\Gamma(D/2+1)}[/spoiler].

<div class="math">\mathbf{2.: n\!\!-\!\!D.}</div>
Replace <span class="math">\varrho_{123}\rightarrow\varrho_{12\cdots n}[/spoiler], done. Taking all the stuff from above together, the D-dimensional result is
<div class="math">E\;DR^{D-1}\frac{\pi^{D/2}}{\Gamma(D/2+1)}=Q</div>
<div class="math">\Leftrightarrow E=Q \frac{\Gamma(D/2+1)}{DR^{D-1}\pi^{D/2}}</div>

>> No.3978811

Here's the thing, OP. Electrodynamics in arbitrary dimensions is different. Another anon already pointed it out: we have 3 different E-Fields and B-Fields in 3 spatial dimensions .. but that's just a nice coincidence. By going to arbitrary dimensions, you'll see that the number of E-Fields increases with D^2 and the number of B-Fields with D (or the other way around? I forgot).

The book on the introduction to String Theory by Barton Zwiebach treats this problem I believe, you might wanna look in there.

>> No.3978820

>>3978811 E-Fields increases with D^2
Oh, I didn't know that. I assumed E is always a 1-form. Do you know where the exponent comes from? I don't see where the field strengh tensor should pick up additional degrees of freedom. Is U(1) only valid for 3+1D, but runs into problems for higher spatial dimensions?

>> No.3978910
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3978910

>>3978820
>I assumed E is always a 1-form.
Since we are talking hypotetical stuff here, the meaning of E is pretty much undefined. If you consider a Yang-Mills theory in arbitrary dimensions, then F is certainly the relevant object and it makes sense to define E via i_{∂_t}F and so it will be a 1-form. Then the components will be D. The remaining components, i.e. the B-Field can't be D:
http://en.wikipedia.org/wiki/Hodge_star#Dimensions_and_algebra
For 2-Forms, E and B will only have equal number of components if D=4.
>I don't see where the field strengh tensor should pick up additional degrees of freedom.
Even if the components of B will go quadratically in D, the "degrees of freedom" will certainly go with D, since they dont depend on the number of components, but on A.
>Is U(1) only valid for 3+1D?
Unrelated. Why should the bundle be affected by the dimension of the base space?

>> No.3978917

>>3978910
I mean
"Then the components will be D-1"
if D=d+1

>> No.3979466

>>3978910 Unrelated. Why should the bundle be affected by the dimension of the base space?
No idea, maybe it has unphysical consequences, non-renormalizable etc. It was the first thing that came to my mind in order to justify the <span class="math">D^2[/spoiler] thing.

>> No.3980659

>>3979466
This is interesting. Could I get more information about electrodynamics in more than 3-D space?

>> No.3980666

>>3980659
My first book on that was Frankel - the geometry of physics. It's not about the EM itself, but it uses it numerous times as physical examples. The book itself is aimed as a practical introduction to geometry for physicists; it's more rigorous than physics books, but not as strict as math books.