# Stress-energy tensor as a 2-form

1. Oct 9, 2006

### pervect

Staff Emeritus
I've been looking off and on recently at how the stress-energy tensor can be interpreted as a 2-form. I can't quite see how one manages to convert a symmetric rank 2 tensor into a 2-form, though, given that a 2-form is by defintion anti-symmetric, in spite of some reading. I'm hoping that someone here can help give me a "jump start".

The text that does this is MTW's "Gravitation", BTW.

2. Oct 9, 2006

### masudr

Heh, upon seeing the thread title, I immediately came to the same conclusion as you did, but clicked on it hoping to see how the extra components in a symmetric tensor will be dealt with in an antisymmetric one. But it seems you have the same question too.

Anyone?

3. Oct 10, 2006

### George Jones

Staff Emeritus
Could you give a page number?

General tensors can be handled as forms by using not just a single form, but by using a set of forms.

Also, there are types of forms other than the "standard" type, e.g, vector-valued forms.

4. Oct 10, 2006

### pervect

Staff Emeritus
What I`m trying to understand is on pg 371, where the conservation of the stress-energy tensor is written as d *T = 0, d being the exterior derivative, and *T being the dual of the stress-energy tensor.

This section refers to exercise 14.18 on pg 362.

Your remark about vector-valued forms sounds promising. My motivation for thinking that *T must be a two-form is that it was an argument for the exterior derivative operator (which as I understand it acts on differential forms) - perhaps this is where I was going wrong.

5. Oct 10, 2006

### George Jones

Staff Emeritus
It looks to me like *T is a vector-valued 3-form.

Feeding 3 vectors into a "standard" 3-form produces a scalar.

From the last 2 lines of (b) on 363, it looks as if feeding 3 vectors into *T produces a vector, i.e, each w gobbles a vector to produce a scalar, leaving scalars times the e_mu's.

6. Oct 10, 2006

### pervect

Staff Emeritus
Let's see if I have this straight now:

T is being interpreted as a vector-valued one form (NOT a two form). And the duality operator applies only to the one-form, making *T a vector-valued 3-form.

When we take the exterior derivative, d *T, we apply the exterior derivative operator to each element of the vector. The exterior derivative maps a p form into a p+1 form. In this specific case, we start with a vector-valued 3-form, so we get a vector-valued 4-form (pseudoscalar) when we take the extrior derivative. That seems to match the final result on pg 363.

7. Oct 12, 2006

### George Jones

Staff Emeritus
Yes, this all looks nice and consistent.

8. Oct 12, 2006

### coalquay404

If T is to somehow be regarded as a two-form then *T must be a two-form also since the Hodge dual over an m-dimensional space is defined as

$$\star:\Lambda^p(M)\to\Lambda^{m-p}(M)$$

Thus, if $$T\in\Lambda^2(M)$$ and $$\dim M=2$$, then $$d\star T\in\Lambda^3(M)$$.

Last edited: Oct 12, 2006
9. Oct 12, 2006

### pervect

Staff Emeritus
The defintion is a little odd. To quote the exercise:

"Let the duality operator *, as defined for exterior differential forms, act on the forms but not on the contravariant vectors, which appear when the stress-energy tensor T or the Einstein tensor G is writtten as a mixed (1,1) tensor."

This whole setup had me scratching my head for quite a while, fortunately George put me back on the right track.

10. Oct 13, 2006

### George Jones

Staff Emeritus
I fail to see how your remark is relevant.

Neither $T$ nor $\star T$ are 2-forms - as I said, $\star T$ is a vector-valued 3-form, and, as pervect noted in post #6, $T$ is a vector-valued 1-form.

For example, in an abuse of nomenclature, since, for simplicity I won't consider fields, if $V$ and $W$ are vector space, then a vector-valued 3-form $\alpha$ is an anti-symmetric multilinear map

$$\alpha :V\times V\times V\rightarrow W.$$

Defining

$$\tilde{\alpha} :W* \times V\times V\times V\rightarrow \mathbb{R}$$

by

$$\tilde{\alpha} \left( f,v_{1},v_{2},v_{3}\right) =f\left( \alpha \left( v_{1},v_{2},v_{3}\right) \right)$$

shows that there is a natural way for the space of vector-valued 3-forms to be taken as $W\otimes \Lambda ^{3}\left( V* \right)$.

If $W = \mathbb{R}$, then the "standard'' forms result.

Vector-valued forms are useful for connections on bundles, and are used in gauge theory.

Last edited: Oct 13, 2006