f=1r2
with proper definition of the source and distance. What is Coulomb's law and Newton's law look like in high dimensions?
To answer this question, we have to make assumptions. We assume the Lagrangian stays the same form in d+1 dimension. It means, the Maxwell equations hold; or equivalently, Poisson equation holds ∇2φ(x)=0.
Solving φ in free space will produce the potential hence the force. By doing Fourier transform, φ(x)=∫ddk(2π)deik⋅xk2.
Solving φ(r)
φ=1(2π)d∫dkkd−3dd−1Ωeikrcosθ1, where dd−1Ω is the (d−1)-D angular element. Now this integral involves the integral of one azimuthal angle θ. We can parametrize the coordinate in d-D spherical coordinate as: x1=rcosθ1;x2=rsinθ1cosθ2;⋯;xd=rsinθ1sinθ2⋯sinθd−2cosϕ;
Then the surface element becomes:
dd−1Ω=sind−2θ1sind−3θ2⋯sinθd−2dθ1dθ2⋯dθd−2dϕ. The integral over angles except θ1 is just the surface area of a (d−2)-D hypersphere (See appendix for a derivation) Sd−2=2πd−12Γ(d−12)
So φ=Sd−2(2π)drd−2Id, where Id=∫∞0dξ∫π0dθξd−3sind−2θexp[iξcosθ].
It's tempted to do the ξ integral first, because it gives gamma function and leaves the rest a integral over tanθ: ∫π/20dθtand−2θ+(−1)d−2∫π/2πdθtand−2θ. The problem is that integral of tan function at π/2 is singular . We can do the θ integral first, which yields (using mathematica):Id=∫∞0dξ√πΓ(d−12)0F1(d2,−ξ24)Γ(d2)ξd−3=2d−3√πΓ(d−22)Γ(d−12).
Therefore, φ=Sd−2(2π)drd−22d−3√πΓ(d−22)Γ(d−12)=Γ(d−22)4πd21rd−2
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| Coulomb potential in higher dimensions |
Gauss Law
There is a much easier method to solve this problem. We note that Gauss theorem (in mathematics) hence Gauss law (in physics) still holds. E(r)⋅Sd−1=1 where Sd−1 is the area of a d−1 D hypersphere.
So we get Coulomb's law in d+1 dimension as:f=Γ(d2)2πd21rd−1.
It can be checked, −∂∂rφ(r)=E(r), Just as we expected. Of course, the direct integration can be used in where Gauss law does not hold.
Coulomb's law for massive boson exchange
Another interesting result is the Coulomb's law for classical theories with massive intermediate boson in higher dimentions. The Poisson equation becomes (∇2−m2)φ(x)=0.By doing Fourier transform, φ(x)=∫ddk(2π)deik⋅xk2+m2.
Apply the same technique again (except the Gauss Law), φ(r)=(mr)d2−1Kd2−1(mr)(2π)d21rd−2
where Kn(x) is the Bessel function of the second kind.
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| Comparison of field potential of massless and massive boson exchange in higher dimensions |
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| Comparison of field potential of massless and massive boson exchange in higher dimensions at large r |
Appendix: the surface area of a (d−1)-D hypersphere
Consider the following Gaussian integeral: ∫dnxexp(−x2)=(∫dxexp[−x2])n=πn2The left hand side can be written as ∫drrn−1exp[−r2]Sn−1. So 12Γ(n2)Sn−1=πn2. Sn−1=2πn2Γ(n2).
See also:
http://naturalunits.blogspot.com/2013/01/a-remark-on-high-dimension-propagators.htmlupdate, March 21, 2014:
- I was solving the Green's function with free space boundary condition. The Poisson equation should have been
∇2φ(x)=δ(x)
- This is not a the only generalization. It may not even be the natural generalization, from the point view the Newtonian approximation in general relativity. In GR, one should write down the Einstein equation in d+1 D and do the linearization there, as our commentator explained. (That means one fix G, which is not always appreciated.)
