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In the Mori-Zwanzig formalism, the following identification for the generalised shear viscosity $\eta(t)$ is given: $$ \eta(t) = \frac{V}{k_B T} \langle \sigma(t) \sigma(0) \rangle, $$ identified as such because it appears in what amounts to a viscoelastic constitutive relation: $$ \sigma(t) = -\int_0^t \mathrm{d} \tau \ \eta(t-\tau) \gamma(\tau). $$ Here $\sigma$ is the shear stress, $\gamma$ the shear strain rate, $V$ volume, $T$ temperature and $\langle \ldots \rangle$ the equilibrium phase average. This approach was done in the books by both Evans & Morriss and Hansen & McDonald on liquid theory. I'm happy with the general usage of memory functions coming from Liouvillian evolution, however here something strange seems to have happened. Substituting the expression for $\eta$ into the constitutive relation gives: $$ \sigma(t) = -\int_0^t \mathrm{d} \tau \ \frac{V}{k_B T} \langle \sigma(t-\tau) \sigma(0) \rangle \gamma(\tau), $$ which seems to just be an expression for the shear stress in terms of itself. The viscosity, which describes the relationship between stress and strain rate, is itself a function of the history of shear stress. This makes sense, but when you put it together you seem to get a sort of recursive equation. It doesn't seem like much use as a constitutive equation given that it contains the quantity (the stress) which is being solved for. How do I interpret this?

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  • $\begingroup$ Can you provide a reference in either Evans & Morriss or Hansen & MacDonald for the relation $\sigma(t) = -\int_0^t \eta(t-\tau)\gamma(\tau)\mathrm{d}\tau$? That is not the way I am used to seeing memory functions appear in constitutive relations. $\endgroup$ – Endulum Jan 10 at 21:19
  • $\begingroup$ In the "Non-Markovian constitutive relations: viscoelasticity" section (2.4) in Evans, equation 2.76 in the second edition. $\endgroup$ – Senrade Jan 10 at 21:25
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The equations are correct, but some confusion appears to have arisen from the notation and the nomenclature.

Firstly, in your last equation $$ \sigma(t) = -\frac{V}{k_BT} \int_0^t d\tau \, \langle \sigma(t-\tau)\sigma(0)\rangle \, \gamma(\tau) $$ the quantity inside $\langle\cdots\rangle$ is an equilibrium time correlation function. It does not reflect the history of the shear stress in a particular nonequilibrium experiment involving a time-dependent strain rate. It is just a function of time. Secondly, a better notation on the left of the equation would be $\langle \sigma(t)\rangle_{\text{ne}}$ to emphasize that it is not an instantaneous dynamical variable, but is also an average, taken in the nonequilibrium ensemble produced by the applied strain rate.

So it should be clear that there is no recursive element to the equation: the same quantities do not appear on both sides. It simply links a nonequilibrium average response $\langle \sigma(t)\rangle_{\text{ne}}$ to an applied perturbation $\gamma(t)$. The response is governed by the equilibrium time correlation function.

Remember that dynamical linear response equations take the same form, in general. If the perturbation can be represented as a term $-\alpha(t) A$ in the hamiltonian, where $A$ is a dynamical variable and $\alpha(t)$ a time-dependent field, the response in any other dynamical variable $B$ can be written $$ \langle B(t)\rangle_{\text{ne}} = \frac{1}{k_B T} \int_{-\infty}^t \alpha(t') \langle \dot{A}(0) B(t-t')\rangle \, dt' , $$ where $\dot{A}$ is the time derivative of $A$. The presence of $B$ on both sides of this equation is not a cause for concern (and indeed, we can choose $B$ to be $A$ if we wish). The memory function approach is useful, especially when discussing some of the subtleties of transport coefficients and time scale separation, but this issue is not specific to memory functions. Cases like the shear viscosity need more careful handling, because the perturbation is not a hamiltonian one, but again this aspect does not seem to be crucial to your question.

Secondly, although it is useful to have a symbol to denote that response function, it can be confusing to call it $\eta(t)$ and to refer to it as a generalized viscosity. (I'm not actually sure that Evans and Morriss, and Hansen and McDonald, precisely refer to $\eta(t)$ as a generalized viscosity, but maybe you have seen a place where they do). It has the wrong dimensions for a viscosity. Remember the Green-Kubo relation $$ \eta = \frac{V}{k_BT} \int_0^\infty dt \, \langle \sigma(0)\sigma(t)\rangle $$ so clearly the function $$ \eta(t) = \frac{V}{k_BT} \langle \sigma(0)\sigma(t)\rangle $$ is missing a factor with dimensions of time. E & M introduce this firstly through its Fourier transform, in eqn (2.73), which they term a "frequency dependent Maxwell viscosity" $$ \tilde{\eta}_M(\omega) = \frac{\eta}{1+i\omega\tau_M} $$ which makes clear that, when inverse transformed, it will not have the same dimensions as $\eta$. Then when they do the inverse transform to get the time-dependent function $\eta_M(t)$ which appears in eqn (2.74) they refer to it as the "Maxwell memory function" (again, it does not have viscosity dimensions) and then in eqn (2.76), which is your viscoelastic constitutive equation, they drop the restriction to the Maxwell model, and the subscript $M$, so it looks like a time dependent viscosity, but still has the wrong dimensions. Call it a minor hobby horse of mine, but I wish people would use a different symbol.

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  • $\begingroup$ As I understand it, this relation was derived in E&M (in sec 4.3) just from the Generalised Langevin equation for the transverse current density. $$ \dot{J} = \int d \tau \langle \sigma(0) \sigma(t) \rangle J(t - \tau) $$ in which $\dot{J}$ was replaced by $i k \sigma$, which is where the $\sigma$ in the integral come from too. They're time derivatives of the phase variable $J$ so they're phase variables too, right (not averages)? (the strain rate was squeezed out of this equation by some hand-waving) $\endgroup$ – Senrade Jan 11 at 12:22
  • $\begingroup$ This problem seems to be more general. Following the formalism in E&M, for a single variable, $F(t) = \mathrm{exp}(i Q L t) \dot{A(0)}$, such that $$ \dot{A(t)} = -\int d \tau \frac{\langle \mathrm{exp(i L Q t} \dot{A(0)} \dot{A(0)} \rangle}{\langle A A \rangle} A(t-\tau) + \mathrm{exp(i L Q t} \dot{A(0)} $$ And then the claim that in the low $k$ limit the anomalous propagator becomes $\mathrm{exp(i L t)}$ gives us $\dot{A(t)}$ on both sides. $\endgroup$ – Senrade Jan 11 at 13:22
  • $\begingroup$ It is true that the Mori formulation produces a "transport equation" which is written in terms of dynamical variables, not non-equilibrium averages. But then one gets into semantics: this is only correct if the dynamical variables are all "slow" (i.e. all the fast ones have been projected away). Remember, just before eqn (4.45) E&M say "ignoring the random force". Obviously, the ultimate aim of the Mori approach is to produce the familiar deterministic transport equations. I prefer to regard the slow variables as coarse-grained. You may prefer to see them as genuine microscopic variables. $\endgroup$ – LonelyProf Jan 11 at 13:24
  • $\begingroup$ The low-$k$ limit is crucial to justifying the time scale separation, which is why the Mori approach works for conserved variables (whose rate of change goes to zero as $k\rightarrow 0$). As for the final term on the right of your last comment, it is supposed to be the random force term, and is defined to be orthogonal to $A$. Not $\dot{A}$ but $Q\dot{A}$. I suggest looking more closely at Hansen & McDonald's treatment, which I think is clearer. Anyway, I don't particularly want to argue about this aspect. My main point was the one concerning the equilibrium time correlation function. $\endgroup$ – LonelyProf Jan 11 at 13:35
  • $\begingroup$ Oh yes, I've been utterly dense - I can see that I looked like I was arguing over unimportant terms. I was reading the word equilibrium without processing it in context - thank you! $\endgroup$ – Senrade Jan 11 at 14:39

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