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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.2012-08-08
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    @timur what is 1-form valued 2-form? Is it $\star P$ ?2012-08-08
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    No, but as I said it is just a guess. I am curious why do you think it be star P?2012-08-08
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    @timur I just don't know what is "1-form valued 2-form", I was guessing. Btw, see my update to the answer.2012-08-08
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    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.2012-08-08
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    Why it is convenient to write things in terms of forms is it does not require a measure to integrate, and it does not require an underlying metric, which makes the true structure transparent. But if you want you can express everything in terms of vectors and tensors. You seem to want to do something in between, which I think would require more prerequisites than vectors and tensors but would still require metrics and measure. So I would recommend choosing one of the extremes. I mean, why use forms if you stop halfway?2012-08-08
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    @timur I couldn't really understand your comment, but I'm going to study what are vector-valued forms and why do they exist. I just didn't know about them.2012-08-08
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    @timur I can understand/accept your suggestion about the surface force, but not about the stress tensor. It can't be a 2-form since, that would correspond to a certain surface, but there is no preferred surface in the bulk of liquid/body. Though given a surface, stress tensor should give a surface force.2012-08-09
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    You might consult PDE, Vol III, Ch. 17 by Michael Taylor.2012-08-09
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    Note that being a 1-fom valued 2-form does not give you any preferred surface. If you take a little surface element inside the body and integrate the stress over it, you get a force, which is 1-form.2012-08-09
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    @Neal: Ch 17 is Euler and Navier-Stokes equations. Can you be a bit more specific? I cannot find anything related to elasticity there.2012-08-09
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    @timur Actually Neal is quite close, I'm studying the chapter. Stress tensor is the same concept for elastic bodies and liquid and primary I'm interested in liquid (you see, I've never specified I need an elastic or viscous stress tensor). As for the "preferred surface" now I'm starting to understand you are right, viewing stress tensor as a molecular flux of momentum, which is the case for liquids.2012-08-10
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    I see stress tensor is mentioned in Section 4, as a symmetrized gradient of the velocity field. I suspect an engineer would say this is the strain tensor that is related to just the geometry of the situation rather than the mechanic aspect of it.2012-08-10
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    That *pseudo* has a precise mathematical meaning. It should simply mean an object that behaves like a non-pseudo object *under an orientation preserving coordinate transformation*, but change sign in non-preserving transformations. See [Pseudotensor](http://en.wikipedia.org/wiki/Pseudotensor).2012-08-10

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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?2012-08-21
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    @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 $i_Xp$.2012-08-23
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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.2012-08-09
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    Hendrik, I prefer coordinate-free presentation since coordinates seem "not very physical to me".2012-08-10
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    @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.2012-08-10
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    @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.2012-08-10
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    and then what is $\mathbf P$ in a coordinate-free view? What is the space it belongs to?2012-08-10
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    $\bf P$ belongs to $\Gamma(T^1_1(M)) = \Gamma(TM\otimes T^*M)$, where $\Gamma(T^*M) = \Omega^1(M)$.2012-08-10
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    wait, isn't $T^*M = \Omega^1(M)$?2012-08-10
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    No, $T^*M$ is the vector bundle containing all possible covectors at all points in $M$. $\Omega^1(M)$ contains all possible sections of that vector bundle, that is smooth functions assigning to every point a covector at that point.2012-08-10
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    I got your argument with @timur --- it is just like you saying density $\rho$ is 0-form, and timur saying it is 3-form (your zero from multiplied by volume form), right?2012-08-11