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- Thread starter quasi426
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PerennialII

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[tex]

\nabla \cdot \bfseries\sigma + \rho b = \rho \frac{D}{Dt}v

[/tex]

where [itex]\sigma[/itex] 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),

[tex]

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

[/tex]

(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, [itex]\rho[/itex] density, b body force vector, v velocity of a material point and I'm using [itex]D/Dt[/itex] for the material derivative operator. Substituting to the above the law of balance of momentum (take the [itex] r\times[/itex] off and got it), the Cauchy's equation of motion and somewhat lengthy manipulation the above reduces to

[tex]

\int e_{ijk}\sigma_{jk}i_{i}dV=0

[/tex]

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

[tex]

e_{ijk}\sigma_{jk}=0

[/tex]

and writing the permutation symbol open leads to

[tex]

\sigma=\sigma^{T}

[/tex]

.... 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.

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please check out this post

https://www.physicsforums.com/showthread.php?t=90636

https://www.physicsforums.com/showthread.php?t=90636

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tiny-tim

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[tex]

e_{ijk}\sigma_{jk}=0

[/tex]

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, K_{n}-> 1, e.g. Non-Newtonian fluid, which can lead to rotationally non-invariant fluids, such as polymers.

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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

- #6

samalkhaiat

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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.

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

[tex]

T_{ab} = \frac{\partial L}{\partial \partial_{a} \phi} \partial_{b} \phi - \eta_{ab}L

[/tex]

is conserved;

[tex] \partial^{a} T_{ab} = 0[/tex]

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

[tex]

\Theta_{ab} = T_{ab} + \partial^{c} X_{cab}

[/tex]

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

[tex]X_{cab} = - X_{acb}[/tex]

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

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

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

[tex]R_{ab} - \frac{1}{2} g_{ab} R = - k \Theta_{ab}[/tex]

Since the (geometrical) Ricci tensor [itex]R_{ab}[/itex] and the metric tensor [itex]g_{ab}[/itex] are both symmetric, so [itex]\Theta_{ab}[/itex] 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;

[tex]\mathcal{M}_{cab} = \Theta_{ca} x_{b} - \Theta_{cb} x_{a}[/tex]

is conserved! But

[tex]\partial^{c} \mathcal{M}_{cab} = \Theta_{ab} - \Theta_{ba}[/tex]

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

[tex]\Theta_{ab} = \Theta_{ba}[/tex]

regards

sam

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