A Stress Tensor: Definition, Ideas & Discussion

[wrobel]

Inspired by the closed thread about pressure:) Here is some of my fantasies about a definition of the stress tensor. Nothing here claims to be a correct theory but just as a matter for discussion.

 

[Frabjous]

I am not comfortable with coordinate free notation, so my comment might be inappropriate.

I have not been exposed to arbitrary additions to the stress tensor. Would you expand on what this means? It seems incompatible with equations of state.

 

[wrobel]

for example, the Navier-Stokes equation is not changed if you replace p with p+f(t)

 

[Frabjous]

I am thinking of something like the ideal gas law, PV=nRT.

 

[wrobel]

then in my definition $\tilde\omega$ should be chosen such that this equality holds

 

[Frabjous]

Is setting the pressure of a vacuum equal to zero sufficient? Is there a more elegant (probably thermodynamic) criterion?

 

[Chestermiller]

The divergence of the stress tensor comes in when the divergence theorem is use to convert from a surface integral to a volume integral.

 

[vanhees71]

The stress tensor is symmetric rather than antisymmetric. It corresponds to a surface force of adjacent parts of the medium. This force is given by
$$F_l=\int_A \mathrm{d}^2 f_k \sigma_{kl},$$
where $\mathrm{d}^2 f_k$ is the surface-normal vector and $\sigma_{kl}=\sigma_{lk}$ the stress tensor.

In relativistic physics it's the space-space part of the energy-momentum tensor (up to a sign).

The pressure of a fluid is a special case, here $\sigma_{kl}=-p \delta_{kl}$.

 

[etotheipi]

This Landau did show nicely in volume seven. Here's basically just a repeat of his argumentation; the force exerted on any portion of the body by the rest of the body is purely a surface force due to the short range of intermolecular forces. Thus the body force density $F_i$ must be a divergence $\partial \sigma_{ik} / \partial x_k$, because when integrated over a volume $\Omega$ this is transformed into an integral of $\sigma_{ik} df_k$ over $\partial \Omega$.

The moment acting on the portion $\Omega$ due to the internal stresses is\begin{align*}

M_{i} = \int_{\Omega} (\mathbf{r} \times \mathbf{F})_i dV &= \int_{\Omega} \epsilon_{ijk} x_j \frac{\partial \sigma_{kl}}{\partial x_l} dV \\

&= \epsilon_{ijk}\int_{\Omega} \frac{\partial}{\partial x_l} (x_j \sigma_{kl}) dV - \int_{\Omega} \epsilon_{ijk} \delta_{jl} \sigma_{kl} dV \\

&= \epsilon_{ijk} \oint_{\partial \Omega} x_j \sigma_{kl} df_l - \int_{\Omega} \epsilon_{ijk} \sigma_{kj} dV

\end{align*}so if $\mathbf{M}$ is to be purely a surface integral then the second integrand has to vanish identically i.e. $\epsilon_{ijk} \sigma_{kj} = 0$, which happens if $\sigma_{ij} = \sigma_{ji}$ (i.e. then $\epsilon_{ijk} \sigma_{kj} = \epsilon_{ikj} \sigma_{jk} = - \epsilon_{ijk} \sigma_{jk} = - \epsilon_{ijk} \sigma_{kj}$ and thus as with any contraction of a symmmetric and antisymmetric part $\epsilon_{ijk} \sigma_{kj} = 0$). So the stress tensor must be symmetric.

 

[wrobel]

The stress tensor is symmetric rather than anti symmetric

I did not say that the stress tensor is anti symmetric

 

[vanhees71]

Then I don't understand your notation. For me the wedge product is antisymmetric (as in Cartan's calculus of alternating (differential) forms).

 

[wrobel]

I did not say that $$\omega_{ij}^k$$ is a stress tensor. I said about the correspondence. In the standard Euclidean coordinates this correspondence looks like
$$\omega^i_{12}=\sigma ^i_3,\quad \omega_{23}^i= \sigma_1^i,\quad \omega_{31}^i= \sigma_2^i$$

 

[vanhees71]

That doesn't make sense either. Again a stress tensor is a symmetric 2nd-rank tensor (field). It's meaning together with the standard proof that it's a symmetric tensor is nicely explained in @etotheipi 's posting #9.

 

[wrobel]

I corrected my previous post little bit

 

[vanhees71]

Now it makes sense, but your $\omega_{ab}^c$ are not called "stress" usually.

 

[wrobel]

Now it makes sense, but your $\omega_{ab}^c$ are not called "stress" usually.

sure

 

[etotheipi]

Ohhh, the $n$ in $\omega^n_{ij}$ is an index too! That would explain why you said it had $9$ instead of $3$ independent components, haha. I was very confused

 

[wrobel]

the correspondence $\omega_{ij}^k\leftrightarrow\sigma_{sr}$ resembles the Hodge star

 

[vanhees71]

But "Hodging" makes only sense for alternating forms too.

It's very unusual to introduce your $\omega_{ij}^k$'s. In some way you seem to want to incorporate the $\mathrm{d}^2 \vec{f}$'s into the definition of the stress tensor, but usually what one wants are tensor (densities) to be integrated, and that's for a good reason: It's the way to express all physical laws in terms of local field equations, which is the most natural language of relativistic physics.

 

[wrobel]

It's very unusual to introduce your 's

Sure! And I do not propose to use $\omega$ instead of $\sigma$ I just try to formalize the mathematical part of the topic. We want to have the force as an integral over boundary. Ok, but we can integrate differential forms only. Take the form $\omega$. Due to the Riemann metric we have a canonic isomorphism between the space of tensors of type $\omega$ and the space of tensors of type $\sigma$. Etc