Laplacian = tr(H)

1. Feb 19, 2010

icurays1

Stupid thing I noticed today:

$$\nabla^2 U=tr(H(U))$$

Or, in other words, the Laplacian of a function is just the trace of its Hessian matrix. Whoop-de-frickin do, right? Is this useful knowledge or should I forget it immediately?

N!

2. Feb 19, 2010

Ben Niehoff

This is not true in curvilinear coordinates.

3. Feb 19, 2010

icurays1

But then isn't the 'Hessian' a tensor with co/contravariant components? is the trace even defined for tensors like that? And how is a differential operator like $$\nabla^2$$ even defined in a nonlinear coordinate system? (Genuine questions, I'm just starting to learn tensors and differential geometry and whatnot...)

4. Feb 19, 2010

Ben Niehoff

5. Feb 19, 2010

icurays1

okay, cool. thanks!!

would the trace of an arbitrary tensor be $$\sum_{i}{A_{ii...i}}$$ i.e. summing over the elements of the tensor with identical indicies? For tensors rank>2 'diagonal' is sort of vague, or does 'diagonal' always mean 'elements with the same index'? I guess this doesn't have anything to do with the original question...thanks again Ben!

6. Feb 19, 2010

Ben Niehoff

Tensor traces are taken by contracting any two indices with the metric tensor. The result will be a tensor whose rank has been reduced by 2. For example, the Ricci tensor is the trace of the Riemann tensor on the 1st and 3rd indices:

$$R_{bd} = g^{ac} R_{abcd}$$

and then the scalar curvature is the trace of the Ricci tensor, or the double trace of the Riemann tensor:

$$R = g^{bd} R_{bd} = g^{ac} g^{bd} R_{abcd}$$.

In flat space, the metric tensor is just the identity matrix, so this reduces to the more familiar definition of the trace.