- #1
bobinthebox
- 29
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During lecture today, we were given the constitutive equation for the Newtonian fluids, i.e. ##T= - \pi I + 2 \mu D## where ##D=\frac{L + L^T}{2}## is the symmetric part of the velocity gradient ##L##. Dimensionally speaking, this makes sense to me: indeed the units are the one of a pressure. Also, since the velocity gradient is what we use to model the relative motion between fluid particles, it makes sense for ##T## to have such a form.
However, my professor said that we need the symmetric part of ##L##, and not only ##L##, because otherwise we loose frame indifference. Unfortunately, he didn't prove this. I'd like to get some feedback on the following argument which should prove the bold sentence.
Assume we have a rotation tensor ##Q(t)## and let ##B_t## and ##B_t^{*}## two configurations related by a rigid motion ##x^{*}=Q(t)x + c(t)##. Then we have, for ##L^{*}=\nabla_{x^*}v## the following relation ##L^{*}=QLQ^T + \dot{Q}Q^T##. Let's focus on the second addendum of ##T##, i.e. ##2 \mu D##. If we were only considering ##S=2 \mu L## instead of the symmetric part of ##L##, we would obtain ##S^{*}=2 \mu L^*=(2 \mu) (QLQ^T + \dot{Q}Q^T) = QSQ^T + 2 \mu \dot{Q}Q^T## and from here we can see that we are not satisfying the requirement for frame indifference.
As another matter of fact, it's clear that if ##S=2 \mu L##, then we don't have symmetry anymore, because of that ##\dot{Q}Q^T## term. However, this is only an algebraic point of view, which has nothing to do with the frame-indifference requirement.
Do you think that this may be what the professor had in mind? Or maybe is there some more physical interpretation?
However, my professor said that we need the symmetric part of ##L##, and not only ##L##, because otherwise we loose frame indifference. Unfortunately, he didn't prove this. I'd like to get some feedback on the following argument which should prove the bold sentence.
Assume we have a rotation tensor ##Q(t)## and let ##B_t## and ##B_t^{*}## two configurations related by a rigid motion ##x^{*}=Q(t)x + c(t)##. Then we have, for ##L^{*}=\nabla_{x^*}v## the following relation ##L^{*}=QLQ^T + \dot{Q}Q^T##. Let's focus on the second addendum of ##T##, i.e. ##2 \mu D##. If we were only considering ##S=2 \mu L## instead of the symmetric part of ##L##, we would obtain ##S^{*}=2 \mu L^*=(2 \mu) (QLQ^T + \dot{Q}Q^T) = QSQ^T + 2 \mu \dot{Q}Q^T## and from here we can see that we are not satisfying the requirement for frame indifference.
As another matter of fact, it's clear that if ##S=2 \mu L##, then we don't have symmetry anymore, because of that ##\dot{Q}Q^T## term. However, this is only an algebraic point of view, which has nothing to do with the frame-indifference requirement.
Do you think that this may be what the professor had in mind? Or maybe is there some more physical interpretation?