Ngô Quốc Anh

June 19, 2013

The Bochner formula for vector fields

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 2:50

The Bochner formula for a gradient vector field is a important tool in geometric analysis which basically says that

\displaystyle\frac{1}{2}\Delta (|\nabla u{|^2}) = |{\nabla ^2}u{|^2} + \langle \nabla \Delta u,\nabla u\rangle + {\rm Ric}(\nabla u,\nabla u)

for any function f. Whenever |\nabla u| \ne 0, we have a  formula for |\nabla u| as follows

\displaystyle\frac{1}{2}\Delta (|\nabla u|) = \frac{1}{|\nabla u|} \left( \nabla u \cdot \nabla u ( \Delta u) + {\rm Ric}(\nabla u,\nabla u) + |\nabla \nabla u|^2 - \bigg| \Big\langle \nabla \nabla u, \frac{\nabla u}{|\nabla u|}\Big\rangle\bigg|\right).

Today, we discuss a variant of it called the Bochner formula for vector fields. I found this identity in a recent preprint of Li Ma [here].

Theorem. Let (M, g) be a Riemannian manifold of dimension n. Let X be a smooth vector field on M. Then we have

\displaystyle\frac{1}{2}\Delta (|X|^2) = |\nabla X|^2 + {\rm div}({\mathbb L_X}g)(X) - {\nabla _X}{\rm div}X - {\rm Ric}(X,X)

where {\mathbb L_X}g is the Lie derivative of the vector field X with respect to underlying metric g.

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June 14, 2013

MuPAD: Drawing asymptotically flat spacetime

Filed under: Linh Tinh — Tags: — Ngô Quốc Anh @ 23:06

Today, I will show how to draw an asymptotically flat spacetime with two ends using MuPad. To draw ends, we use hyperbolic functions.

First, we talk about the end. The function that I am going to use is h(t)=\frac{1-4t}{t-2}+12. It is clear that h blow up at t=2 and approaches -4 at infinity. To use h, we use the following

h := proc(t)
begin
  (1-4*t)/(t-2)+12
end_proc

I have added 12 to the function h so that h approaches 8 at infinity. Then to draw the (upper) end, I use the following function

f := proc(x, y)
begin
  if 
    x^2 + y^2 > 3.4 
  then
    h(x^2+y^2)
  else
  end_if
end_proc

I have used the number 3.4 because I do not want my end is too tall, keep in mind that 2 is the blow-up number. We can calculate

h(3.4)

to see that this number is nothing but 3 which is close to 1 as I need. We are now able to draw the first end using the following

plot(
     plot::Function3d(f, x = -6 .. 6, y = -6 .. 6, Submesh = [3, 3]),
     ViewingBox = [Automatic, Automatic, 3 .. 8], 
     Scaling = Constrained)

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