Why is the Stress Tensor Symmetrical?

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The symmetry of the stress tensor arises from the balance of moment of momentum, which is crucial for maintaining equilibrium in continuum mechanics. This relationship is established through the Cauchy's equation of motion and leads to the conclusion that the stress tensor must satisfy the condition σij = σji. In general relativity, the requirement for a symmetric stress tensor is reinforced by its coupling to gravity through the Einstein equations, where both the Ricci tensor and the metric tensor are symmetric. Additionally, Lorentz symmetry necessitates that the angular momentum tensor is conserved, further implying the symmetry of the stress tensor. Overall, the symmetry of the stress tensor is essential for the physical consistency of the governing equations in mechanics and relativity.
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Can someone explain to me why the stress tensor is symmetrical. I understand that Sij=Sji , but can someone give me the assumption or the physical reason why this is true. Thanks.
 
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Continuum mechanics is based essentially on laws of conservation of mass, balance of momentum and balance of moment of momentum, the two latter based on Newton's 2nd. Balance of momentum leads to the Cauchy's equation of motion :

<br /> \nabla \cdot \bfseries\sigma + \rho b = \rho \frac{D}{Dt}v<br />

where \sigma is the Cauchy (true) stress tensor. The symmetricity of the Cauchy stress tensor arises from the law of balance of moment of momentum (complete presentations, and best IMO, are typically in thermomechanics books & papers),

<br /> \oint (r\times T) dA + \int (r \times \rho b) dV = \frac{D}{Dt} \int (r \times \rho v) dV<br />

(if the presentation looks unfamiliar you can 'tie' it to for example 'typical' presentations in relation to Newton's 2nd in dynamics books)

where r is a vector from an arbitrary point to a material point, V is an arbitrary subsystem volume, A its area, T traction vector, \rho density, b body force vector, v velocity of a material point and I'm using D/Dt for the material derivative operator. Substituting to the above the law of balance of momentum (take the r\times off and got it), the Cauchy's equation of motion and somewhat lengthy manipulation the above reduces to

<br /> \int e_{ijk}\sigma_{jk}i_{i}dV=0<br />

where e_{ijk} is the permutation symbol, and since the integrand of the above has to vanish everywhere within the system one arrives at

<br /> e_{ijk}\sigma_{jk}=0<br />

and writing the permutation symbol open leads to

<br /> \sigma=\sigma^{T}<br />

... so all in all it results from balance of moment of momentum, in a sense it's understandable that it requires the stress tensor to be symmetric considering its role in equilibrium equations.
 
Hi quasi426! :smile:
PerennialII said:
<br /> e_{ijk}\sigma_{jk}=0<br />

There's a derivation of this at http://en.wikipedia.org/wiki/Stress_Tensor#Equilibrium_equations_and_symmetry_of_the_stress_tensor, followed by:
However, in the presence of couple-stresses, i.e. moments per unit volume, the stress tensor is non-symmetric. This also is the case when the Knudsen number is close to one, Kn -> 1, e.g. Non-Newtonian fluid, which can lead to rotationally non-invariant fluids, such as polymers.
 
please beware
if you read the pdf file i attached in my post you can understand it better, I'm not talking about magnetic fields or non-Newtonian fluids. i want to say cauchy principles are not the exact and the most general forms of relation between stress and direction.
please check it again.
https://www.physicsforums.com/showthread.php?t=90636
 
quasi426 said:
Can someone explain to me why the stress tensor is symmetrical. I understand that Sij=Sji , but can someone give me the assumption or the physical reason why this is true.

Symmetry under spacetime translations implies (by Noether theorem) that the canonical energy-momentum (or stress) tensor

<br /> T_{ab} = \frac{\partial L}{\partial \partial_{a} \phi} \partial_{b} \phi - \eta_{ab}L<br />

is conserved;

\partial^{a} T_{ab} = 0

But it is not, in general, symmetric! Well, it is not unique either, for you could define a new tensor

<br /> \Theta_{ab} = T_{ab} + \partial^{c} X_{cab}<br />

which is also conserved, \partial^{a}\Theta_{ab} = 0, provided that

X_{cab} = - X_{acb}

In a Lorentz invariant theories, we may choose X_{cab} to make [ the new stress tensor] \Theta_{ab} symmetric.

So, your question should have been: Why do we want the stress tensor to be symmetric?

There are two reasons for this:
1) In general relativity, the matter fields couple to gravity via the stress tensor and this is given by the Einstein equations

R_{ab} - \frac{1}{2} g_{ab} R = - k \Theta_{ab}

Since the (geometrical) Ricci tensor R_{ab} and the metric tensor g_{ab} are both symmetric, so \Theta_{ab} must be also.

2) The second reason for requiring a symmetric stress tensor comes from Lorentz symmetry:
Lorentz invariance implies that the ungular momentum tensor;

\mathcal{M}_{cab} = \Theta_{ca} x_{b} - \Theta_{cb} x_{a}

is conserved! But

\partial^{c} \mathcal{M}_{cab} = \Theta_{ab} - \Theta_{ba}

Thus, conservation of ungular momentum requires the stress tensor to be symmetric;

\Theta_{ab} = \Theta_{ba}

regards

sam
 
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