# 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.