18
$\begingroup$

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.

$\endgroup$
14
  • 1
    $\begingroup$ 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. $\endgroup$
    – timur
    Commented Aug 8, 2012 at 0:18
  • $\begingroup$ @timur what is 1-form valued 2-form? Is it $\star P$ ? $\endgroup$
    – Yrogirg
    Commented Aug 8, 2012 at 4:00
  • $\begingroup$ No, but as I said it is just a guess. I am curious why do you think it be star P? $\endgroup$
    – timur
    Commented Aug 8, 2012 at 5:34
  • $\begingroup$ @timur I just don't know what is "1-form valued 2-form", I was guessing. Btw, see my update to the answer. $\endgroup$
    – Yrogirg
    Commented Aug 8, 2012 at 7:26
  • 2
    $\begingroup$ 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. $\endgroup$
    – user14972
    Commented Aug 8, 2012 at 7:29

3 Answers 3

6
+50
$\begingroup$

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.

$\endgroup$
9
  • 3
    $\begingroup$ 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. $\endgroup$
    – timur
    Commented Aug 9, 2012 at 23:55
  • $\begingroup$ Hendrik, I prefer coordinate-free presentation since coordinates seem "not very physical to me". $\endgroup$
    – Yrogirg
    Commented Aug 10, 2012 at 4:21
  • $\begingroup$ @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. $\endgroup$ Commented Aug 10, 2012 at 11:32
  • $\begingroup$ @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. $\endgroup$ Commented Aug 10, 2012 at 11:38
  • $\begingroup$ and then what is $\mathbf P$ in a coordinate-free view? What is the space it belongs to? $\endgroup$
    – Yrogirg
    Commented Aug 10, 2012 at 11:54
4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Given an inviscid fluid with a 0-form $p$ for preassure, how would you make a stress tensor for it? $\endgroup$
    – Yrogirg
    Commented Aug 21, 2012 at 12:47
  • $\begingroup$ @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 $i_Xp$. $\endgroup$
    – timur
    Commented Aug 23, 2012 at 1:39
0
$\begingroup$

They're components of a tensor density, as the names already indicate, e.g. "mass density" for $ρ$. The same is true of the Lagrangian density $𝔏$ that comprises the integrand of an action integral $\int 𝔏 d^4x$, where $$d^4x = dx^0 ∧ dx^1 ∧ dx^2 ∧ dx^3,$$ assuming, here, we're speaking of 4D manifolds, like space-time.

The cardinal example is the canonical stress tensor associated with $𝔏$, which I will denote here $𝔓$, which may be written as: $$𝔓^ρ_ν = \sum_{0≤A<n} \frac{∂𝔏}{∂v^A_ρ} v^A_ν - δ^ρ_ν 𝔏, \quad v^A_ρ = ∂_ρ q^A,$$ for a field given in component form by $q(x) = \left(q^A(x): A = 0≤A<n\right)$, where $∂_ρ = ∂/∂x^ρ$. In here, and the following, I will use the summation convention (sum over repeated index pairs in monomials), so that the above sum may be written as: $$\sum_{0≤A<n} \frac{∂𝔏}{∂v^A_ρ} v^A_ν ⇔ \frac{∂𝔏}{∂v^A_ρ} v^A_ν.$$

The Lagrangian may be written as a 4-form, while the components of the stress tensor density may be written as a 3-form as follows: $$L = 𝔏 d^4x, \quad P_ν = 𝔓^ρ_ν ∂_ρ ˩ d^4x,$$ where the (bi-linear) contraction operator $˩$ is defined recursively on differential forms by $$∂_ρ ˩ (a ∧ α) = a_ρ α - a ∧ (∂_ρ ˩ α), \quad ∂_ρ ˩ b = 0,$$ where $a = a_νdx^ν$ is a 1-form and $b$ a 0-form (i.e. scalar). Thus, for instance, $$ ∂_0 ˩ d^4x = +dx^1∧dx^2∧dx^3, \quad ∂_1 ˩ d^4x = -dx^0∧dx^2∧dx^3,\\ ∂_2 ˩ d^4x = +dx^0∧dx^1∧dx^3, \quad ∂_3 ˩ d^4x = -dx^0∧dx^1∧dx^2. $$ Thus, we could also write $$\begin{align} P_ν &= \left(\frac{∂𝔏}{∂v^A_ρ} v^A_ν - δ^ρ_ν 𝔏\right) ∂_ρ ˩ d^4x\\ &= v^A_ν \frac{∂𝔏}{∂v^A_ρ} ∂_ρ ˩ d^4x - δ^ρ_ν ∂_ρ ˩ 𝔏 d^4x\\ &= ∂_νq^A p_A - ∂_ν ˩ L, \end{align}$$ where we define the 3-form $p_A$ conjugate to $q^A$ by: $$p_A = \frac{∂𝔏}{∂v^A_ρ} ∂_ρ ˩ d^4x.$$

In turn, this is associated with the flow of a vector field $X = X^ν ∂_ν$ by $$P_X ≡ X^ν P_ν = (Xq^A) p_A - X ˩ L.$$ You can think of this as the response of a deformation given locally in coordinate form by $X = Δx$, with $Δq^A$ given by the chain rule as $Δxq^A = Δx^ν ∂_νq^A$, and $$P_{Δx} = Δq^A p_A - Δx ˩ L.$$

Generically, for a stress tensor density with components $𝔗^ρ_ν$, one can define: $$T_ν = 𝔗^ρ_ν ∂_ρ ˩ d^4x, \quad T_X = X^ν T_ν = X^ν 𝔗^ρ_ν ∂_ρ ˩ d^4x.$$ The stress tensor which arises in contexts, such as general relativity or fluid dynamics, would then be expressed in terms of the canonical stress tensor as $𝔗^ρ_ν = 𝔓^ρ_ν + ∂_μ 𝔭^{ρμ}_ν$, where the components $𝔭^{ρμ}_ν = -𝔭^{μρ}_ν$ would be associated with a 2-form as: $$p_ν = \frac{1}{2}𝔭^{ρμ}_ν ∂_μ ˩ ∂_ρ ˩ d^4x,$$ hence $T_ν = P_ν + dp_ν$, where the adjustment $dp_ν$ is given by the Belinfante correction or a generalization of it.

Expressing the stress tensor this way as a 3-form allows one to express the divergence of the stress tensor density as an exterior differential: $$\begin{align} dT_ν &= d\left(𝔗^ρ_ν ∂_ρ ˩ d^4x\right)\\ &= ∂_μ𝔗^ρ_ν dx^μ ∧ \left(∂_ρ ˩ d^4x\right)\\ &= ∂_μ𝔗^ρ_ν δ^μ_ρ d^4x\\ &= ∂_ρ𝔗^ρ_ν d^4x. \end{align}$$

Adopting coordinates $x^0 = t$ and $\left(x^1, x^2, x^3\right) = (x, y, z)$, and denoting the dimension of length, time duration and mass, respectively by $(L,T,M)$, of action by $H = ML^2/T$ and of $d^4x$ by $Ω = TL^3$, the Lagrangian 4-form, coordinate deformation, stress tensor 3-form, and their components will have the dimensions: $$ [L] = H, \quad [𝔏] = \frac{H}{Ω} = \frac{M}{LT^2},\\ \left[Δx^ν\right] = [ν] ≡ \left[x^ν\right], \quad \left[∂_ν\right] = \frac{1}{[ν]}, \quad [Δx] = 1,\\ \left[T_{Δx}\right] = \frac{H}{Ω}, \quad \left[T_ν\right] = \frac{H}{Ω}\frac{1}{[ν]}, \quad \left[𝔗^ρ_ν\right] = \frac{H}{Ω}\frac{[ρ]}{[ν]}, $$ therefore, using $[0] = T$ and $[1] = [2] = [3] = L$, we have the following: $$\begin{align} \left[𝔗^0_0\right] &= \frac{H}{Ω}\frac{T}{T} = \frac{1}{L^3}\frac{ML^2}{T^2},\\ \left[𝔗^i_0\right] &= \frac{H}{Ω}\frac{L}{T} = \frac{1}{L^3}\frac{ML^2}{T^2}\frac{L}{T}\quad(i=1,2,3),\\ \left[𝔗^0_j\right] &= \frac{H}{Ω}\frac{T}{L} = \frac{1}{L^3}\frac{ML}{T}\quad(j=1,2,3),\\ \left[𝔗^i_j\right] &= \frac{H}{Ω}\frac{L}{L} = \frac{1}{L^3}\frac{ML}{T}\frac{L}{T}\quad(i,j=1,2,3), \end{align}$$ which are, respectively, the dimensions for energy density, energy flux density, momentum density and momentum flux density (and pressure and stress).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .