3
$\begingroup$

Morse and Feshbach state on p. 313 without elaboration that the diffusion equation for temperature or concentration $\psi$ and its "conjugate" $\psi^*$ (quotation marks theirs) has the Lagrangian density:

$$L=-\nabla\psi\cdot\nabla\psi^* -\frac{1}{2}a^2(\psi^*\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^*}{\partial t}). $$

I don't understand what the conjugate field, $\psi^*$, is. Since the classical (non-Schrödinger) field should be real, I suspect the conjugation symbol * refers to something other than complex conjugation. With a real field, $\psi^*=\psi$, and only $-\nabla\psi\cdot\nabla\psi$ remains, which would be the Lagrangian for the Laplace equation (steady state diffusion).

$\endgroup$
3

1 Answer 1

2
$\begingroup$

The conjugate field ψ∗ is but the complex conjugate of ψ, so an extra degree of freedom to expedite derivation of the diffusion equation, $$ \nabla^2 \psi = a^2 \partial_t \psi , $$ analogous to the Lagrangian of the free Schroedinger equation, real in that case--only.

  • But, since this equation does not mix real and imaginary parts, take its imaginary part to be zero at the very end, and safely interpret ψ as a concentration, etc.

It is just computational expedience, namely extending to the complex plane, as, e.g. in electromagnetism, to avoid the quandary you observed if you take the imaginary part of the Lagrangian, (proportional to a2) to vanish.

Alternatively, integrating by parts in the action and discarding the surface terms nets a Lagrangian density $$ L= \psi^* (\nabla^2 -a^2 \partial_t )\psi, $$ so ψ∗ may be thought of as an extraneous Lagrange multiplier gimmick to brutally enforce the diffusion equation as is, and concentrate on its real solutions.

Note $\int dx \psi $ is a constant in time, as physically required for your diffusing quantity.

A central solution of this equation underlying its propagator is $$ \psi({\mathbf x},t)= \frac{a^3}{8 (\pi t )^{3/2}} ~ e^{-a^2 {\mathbf x}^2 /4t} $$ which starts out at t =0 as a Dirac $\delta ({\mathbf x})=\psi({\mathbf x},0)$. As a result, any initial concentration profile f can be written as a linear superposition of such δs, $$ \tilde \psi({\mathbf x},0)= \int d^3 y ~ f({\mathbf y} ) \delta ({\mathbf x} -{\mathbf y}) , $$ and propagated through each component thereof, by the above solution, $$ \tilde \psi({\mathbf x},t)= \frac{a^3}{8 (\pi t )^{3/2}} ~\int d^3 y ~ \tilde \psi({\mathbf y},0 ) ~ e^{-a^2 ({\mathbf x}-{\mathbf y})^2 /4t} . $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.