# How to introduce stress tensor on manifolds?

I want to understand the type of stress tensor $\mathbf{P}$ in classical physics.

Usually in physics it is said that the force $\text d \boldsymbol F$ (vector) acting on an infinitesimal area $\text d \boldsymbol s$ (vector) equals

$\text d \boldsymbol F = \mathbf{P} \cdot \text d \boldsymbol s$

where $\cdot$ is a "scalar product".

How can it be rigourised? I guess directed area can be $\star s$ where $s$ is a 2-form, but can I avoid using $\star$ by employing the volume form for example? The force should be 1-form.

How is the power of surface forces is written? Usually it is given by

$$\frac{dA}{dt} = \int_S \boldsymbol v \cdot \text d \boldsymbol F$$

$\boldsymbol v$ being the speed of the surface of the deformed body.

What would be the corresponding local form, that is the power density of surface forces?

UPDATE 1

If it helps, I found a whole appendix "The Classical Cauchy Stress Tensor and Equations of Motion" in the book "The Geometry of Physics: An Introduction" by Theodore Frankel. Particularly it says

The Cauchy stress should be a vector-valued pseudo-$(n - 1)$-form.

However currently I don't know what does it mean. Further development in the book is rather obscure and I'm afraid of that "pseudo". If a thing called "pseudo-something" I would prefer it stated as "actual another thing".

UPDATE 2

Stress tensor can also be viewed as a (molecular) flux of momentum. Then the equation for balance of momentum would be the Newton's second law. Probably this approach would be more fruitful, analogues can be made with the flux of density.

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My guess: $P$ is a $1$-form valued $2$-form. Surface force $f$ is also a $1$-form valued $2$-form, and power density is the $2$-form that results from contracting $f$ with the surface velocity. –  timur Aug 8 '12 at 0:18
@timur what is 1-form valued 2-form? Is it $\star P$ ? –  Yrogirg Aug 8 '12 at 4:00
No, but as I said it is just a guess. I am curious why do you think it be star P? –  timur Aug 8 '12 at 5:34
@timur I just don't know what is "1-form valued 2-form", I was guessing. Btw, see my update to the answer. –  Yrogirg Aug 8 '12 at 7:26
A "foo valued" 2-form is, roughly speaking, something that, when combined with a bivector (or an ordered pair of tangent vectors), yields a "foo". The kind of n-form you're used to is a "scalar-valued" n-form. –  Hurkyl Aug 8 '12 at 7:29

I found a paper supporting my comment that $P$ is a 1-form valued 2-form, that surface force f is also a 1-form valued 2-form, and power density is the 2-form that results from contracting f with the surface velocity. The paper is

R. Segev and L. Falach. Velocities, stresses and vector bundle valued chains. J. Elast. 105:187-206, 2011.

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Given an inviscid fluid with a 0-form $p$ for preassure, how would you make a stress tensor for it? –  Yrogirg Aug 21 '12 at 12:47
@Yrogirg: Pressure is a 3-form (If you have a 0-form then just take its Hodge dual). The stress tensor $s$ corresponding to the 3-form $p$ is the following: Given a vector field $X$, the contraction $s(X)$, which is supposed to be a 2-form, is given by $i_Xp$. –  timur Aug 23 '12 at 1:39

It is important to distinguish between covariant and contravariant indices of a tensor. Differential forms are totally antisymmetric covariant tensor fields. So a 2-form has 2 covariant indices, and when you swap them, the sign changes. Contravariant indices are written as upper indices and covariant indices as lower indices. You can raise and lower indices by use of a metric. Now, the stress tensor has one covariant index and one contravariant index. When you lower the contravariant index, you get a symmetric tensor field, not a differential form. In local coordinates, you simply have a matrix associated to every point, say ${\bf P}(\vec x)$.

The easiest way to understand what the stress tensor does is to imagine the effect of infinitesimal deformations inside the body, described by a vector field, say $\vec v(\vec x)$. The actual displacement at $\vec x$ could be written as $\vec v(\vec x) dr$. Now, the Energy density released by this displacement is $dE = P^j_iv^i_{;j}\ dr$, or, if you take $\vec v(\vec x)$ as a velocity, $P^j_iv^i_{;j}$ will simply be the power density. The semicolon indicates the covariant derivative. You can compute it by taking local coordiantes such that at $\vec x$ the metric is the Euclidean metric and all derivates of the metric are zero. In such local coordinates, $P^j_iv^i_{;j} = {\rm tr}({\bf PJ}_{\vec v})$, with ${\bf J}_{\vec v}$ the Jacobi matrix of $\vec v$.

Edit: What I am saying is that you cannot use differential forms alone. They are special tensors, but you need more general tensors. The stress tensor is a vector-valued 1-form (which, in 3 dimensions, is equivalent to a vector-valued 2-form, by Hodge duality, which gives a little more weight to the surface interpretation you formulated above). A vector is a contravariant 1-tensor, a 1-form is a covariant 1-tensor. Using the metric, you can transform one into the other, so you could even write the stress tensor as a 1-form-valued 1-form (or $(n-1)$-form, in $n$ dimensions), but that seems not very physical to me.

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You can go from one to another using the metric, but I think the ideal formulation should be (or at least should try to be) independent of metric. An example I have in mind is Maxwell's electrodynamics, where the electric and magnetic fields are naturally 1- and 2-forms, and the metric enters only through the laws. –  timur Aug 9 '12 at 23:55
Hendrik, I prefer coordinate-free presentation since coordinates seem "not very physical to me". –  Yrogirg Aug 10 '12 at 4:21
@timur: What we need at any rate is a connection. We can make the stress tensor independent of the metric by multiplying it with the volume form: $P^j_i\omega_{klm}$ – we get a (vector-valued 1-form)-valued 3-form. We cannot make it independent of the connection though. –  Hendrik Lönngren Aug 10 '12 at 11:32
@Yrogirg: The covariant derivative is coordinate-free, I just gave a method to compute it. You could write ${\rm tr}({\bf P\nabla}v)$ instead for the power density, which does not depend on coordinates. –  Hendrik Lönngren Aug 10 '12 at 11:38
and then what is $\mathbf P$ in a coordinate-free view? What is the space it belongs to? –  Yrogirg Aug 10 '12 at 11:54