Divergence theorem/ Surface integral

[vaibhavtewari]

I am not able to find any good reference to answer my question, so I will post here

how does divergence theorem translates to 4 dimensional curved spacetime. I understood how volume integral changes but I am not able to understand how surface integral changes.

I will be glad if some one can help.

Thanks

 

[Sam Gralla]

This is a bit too big of a question to tackle in a thread, at least for me. But Sean Carroll's book has a pretty clear treatment, in the appendices if I remember correctly.

 

[pervect]

Wald, "General Relativity" talks about this some, in the apendix on Stokes theorem, in the language of differential forms.

A n-form is an n-dimensional anti-symmetric tensor, usually described as a map from n vectors to a scalar, a map that's anti-symmetric so if you interchange any two vectors, the sign inverts.

Stokes theorem says that the intergal over a surface of a 2-form, A, is the intergal over a volume of the three-form dA, where dA is the "exterior derivative" of A.

If we apply this to two-forms, we say that the intergal of a 2-form A over a surface is the intergal of a 3-form dA over a volume.

This formulation is actual the Hodges dual of the probably more familiar one which uses the normal to the surface. The integral of the two-form described above is the integral of the hodges dual (usually described by the * operator) that doesn't use the normal vector at all. There's a natural association between a small square of area on any surface and a two-form, which describes an oriented area. So if you integrate the surface area of a 2-d surface, you break the surface up into many trapezoidal pieces. Each small trapezoid has a natural representation as the two vectors that form the sides of the trapezoid (strictly speaking,you use the exponential map to go from the sides of the trapzoid to the vectors). So you feed these two vectors that are the sides of the trapezoidal surface element into the two form that you're integrating, and you get a scalar quantity. You sum this scalar quantity over all the little trapezoids comprising the suface to and you get a scalar quantity. That's the intergal of the two-form over the surface.

The more traditional integral would be the integral of the hodges dual vector of the two form above, which gets fed the normal vector to the surface and the unit time vector instead of the two vectors that form the "sides" of a small parallelogram shaped surface element.

Likewise, to integrate the three-form, you divide the volume element into parallel piped (like cubes, but while the faces are parallel they aren't generally at right angles), and the sides of the parallelpiped define three vectors that you feed to the three form to get a scalar , which gets summed up. You can also integrate the hodges dual of the three-form, which is a one-form. You've probably seen volume elements represtned as 1-forms before, though it may not have been explained in these terms exactly. You recover the scalar volume by multiplying the dual of the three form by a unit time vector.

Hope this helps some, it's a bit rambly and not teribly rigorous.

 

[vaibhavtewari]

Thankyou very much for the reply. Inspired by your explanation I did some reading. I did get some doubts..
in context of
\int_Md\omega=\int_{\partial M}\omega

exterior derivative of a (n-1)-form gives a n-form. While I am not sure what is exterior derivative of a vector, will it still be a vector ?
i.e.
dx^{\mu}

looking at this equation
ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}
is is still a vector..

Please clarify

Thanks.

 

[vaibhavtewari]

Also, I will be glad if some one can reply what will be normal vector to a constant time surface.

Thanks

 

[Bill_K]

vaibhavtewari, It's a lot easier than all the Hodge operator stuff! The divergence theorem is no different in four dimensions than it is in three. If you understand what the surface element is in ordinary spherical coordinates, say, then you've got it made.

The theorem starts with ∫√-g V^μ_;μ d⁴x. There's a vector identity by which the covariant divergence of a vector can be expressed in terms of the ordinary divergence: V^μ_;μ ≡ (1/√-g) (√-g V^μ)_,μ. So the integral is ∫(√-g V^μ)_,μ d⁴x. Integrate by parts and you have ∫√-g V^μ dS_μ. The theorem is

∫√-g V^μ_;μ d⁴x = ∫√-g V^μ dS_μ

Now your question revolves around dS_μ. It's the surface element, and indeed it can be written out in general using differential forms. However you almost always adapt your coordinates to the surface, in which case it's quite simple. For example if the surface is r = const, and r is the proper distance normal to the surface, then you get dS_μ just by striking out dr from the volume element. So taking spherical coordinates in Minkowski space as an example, √-g d⁴x = r² sin Θ dr dΘ dφ dt and the surface element is just r² sin Θ dΘ dφ dt times the unit normal.

Also, I will be glad if some one can reply what will be normal vector to a constant time surface.

This is where it gets interesting! dS_μ is the outward pointing normal, and "outward" must be defined continuously over the surface. Take a surface r = const capped off by time slices t = t₁ and t = t₂. If dS_μ points outward on the r = const surface, then it must point inward on the t = const surfaces! This fact is sometimes difficult to swallow. Try it with a smooth surface, which makes a smooth transition from timelike to spacelike. Draw an outward pointing arrow at one point and follow it around the surface, keeping it orthogonal to the surface. (But remember what orthogonal looks like on a Minkowski diagram! It does not look like 90 degrees!) Where the surface goes null, the normal vector also goes null and is at the same time a tangent. Continuing on, where the surface becomes spacelike you will find the normal pointing inward.

 

[pervect]

Going back a bit to the differential forms approach, the exterior derivative of a vector is a 2 form (sometimes called a bi-vector) i.e. a rank 2 anti-symmetric tensor, and hence not a vector.

Clifford algebras help a lot (IMO) if you choose to pursue differential forms. You can view the dot product of two vectors as the symmetric part of a "geometric product", and the wedge product as the anti-symmetric part.

http://www.av8n.com/physics/clifford-intro.htm

 

[vaibhavtewari]

Thank You very much for reply. I did learn a lot of things. I will keep approach using differential forms for future, I think it is a good tool.

Thanks

 

[vaibhavtewari]

Hey Bill,
another doubt popped in my head when I was using the stokes equation you gave. You used \sqrt{-g} on both sides. Well for the surface side won't there be some sort of reduced metric, I believe this is called induced metric. Please clarify at your leisure.

Thanks

 

[Bill_K]

vaibhavtewari, This step gets you the induced metric:

For example if the surface is r = const, and r is the proper distance normal to the surface, then you get dSμ just by striking out dr from the volume element.

This suffices 99 times out of 100, I have never encountered a surface integral where the coordinates couldn't be adapted to the surface like this. However, if it's a sphere you're integrating over, and you insist on doing it in Cartesian coordinates..!

Well, it's not that hard, just messier. For example let x^μ, μ = 1...3 be the exterior coordinates, and yⁱ, i = 1...2 be the interior coordinates on the surface. The surface is given parametrically by a set of equations

x¹ = F₁(y¹, y²)
x² = F₂(y¹, y²)
x³ = F₃(y¹, y²)

If the exterior metric is g_μν, the induced metric will be h_ij = g_μν ∂x^μ/∂yⁱ ∂x^ν/∂y^j and the surface element will be √h.

 

[vaibhavtewari]

looking at your definition for induced metric, don't we also have to strike out rows and columns of metric containing g_{rr}, in that case \sqrt{-g} will be different for volume and surface integral ?

Please clarify.