# Analogies between equations [closed]

What properties of fields and matter are related to the analogy of the Schrödinger equation and the Navier-Stokes equation, between the equation of general relativity and the Navier-Stokes equation? I have only assumptions in this regard. As I think, this is due to the general structure of fields and matter, from the same elements. But I have a description of this connection only in Russian. So you have to fantasize.

• It's very odd. I have never seen an analogy as you suggest. As far as I can tell, the Schrodinger and the Navier-Stokes equations are not really all that analogous, nor are the GR field equations. Can you elaborate? – puppetsock Aug 29 '19 at 19:38
• The solution of the Navier-Stokes and Schrödinger equations is related by the relation $V_l=-i\frac{\hbar}{m}\frac{\partial ln\psi}{\partial x_l}$, where velocity is the solution of the Navier-Stokes equation, and the wave function is the solution of the Schrödinger equation. In the equation of the general theory of relativity, the relationship is determined by the implicit equation. – Evgeniy Yakubovskiy Aug 29 '19 at 20:10
• The relationship between gravity and the Navier-Stokes equations is described in 1. Bredberg, Irene. 2012. The Einstein and the Navier-Stokes Equations: Connecting the Two. Doctoral dissertation, Harvard University. 2. Rodrigues F. G., Rodrigues Jr W. A., da Rocha R. The Maxwell and Navier-Stokes equations that follow from Einstein equation in a spacetime containing a Killing vector field //AIP Conference Proceedings. – AIP, 2012. – Т. 1483. – №. 1. – С. 277-295. – Evgeniy Yakubovskiy Aug 29 '19 at 20:19
• Um... Nope. The time derivative in the Schrodinger equation isn't a velocity field. I don't see any reason to say that it's an analogy. The fact that they are both differential equations with space and time derivatives does not make them analogous. It may mean there are situations in which solution techniques are related. – puppetsock Aug 29 '19 at 20:28
• Can you possibly edit the info into your question? And maybe give complete citations? I mean, what department did Irene submit to? – puppetsock Aug 29 '19 at 20:33

In a very odd problem I had to do for a PDE class, for the free particle Schrodinger equation on $$\mathbb{R}\times\mathbb{R}^n$$: $$u_t-i\Delta u = 0$$ we had to calculate a symmetric rank 2 stress-energy tensor $$T_{ij}$$ for $$i,j=0,\cdots, n$$ with the following assumptions: \begin{align} & u \in \mathcal{S} \\ &T_{00} = \frac{1}{2}|u|^2 \\ &\partial_i T_{ij} = 0 \\ \end{align} In other words, we made an analogy to fluid flows where the probability density plays the role of the mass density of the "fluid" and the fluid was also under a divergence-free condition.
Surprisingly enough, the role of the momentum density of this probability fluid was played by $$\frac{i}{2}(u\nabla \bar{u} - \bar{u} \nabla u)$$, which, when you integrate over all of $$\mathbb{R}^n$$, is the expected value of the momentum operator!
• This is not surprising momentum equals $-iu\frac{\partial u}{\partial x_k}=-i|u|^2\frac{\partial lnu}{\partial x_k}=\rho V_k$ – Evgeniy Yakubovskiy Aug 30 '19 at 2:56
• @EvgeniyYakubovskiy apologies I had the wrong formula. I edited my answer accordingly. What is $V_k$ exactly? – Ninad Munshi Aug 30 '19 at 3:04
• '@Ninad Munshi' This is the velocity of the medium - "fluid". Substitute the expression instead of the wave function $u=exp[-i(Et-px)]$ and you get a dimensionless momentum, i.e. speed – Evgeniy Yakubovskiy Aug 30 '19 at 6:47
• '@Ninad Munshi' Substituting an expression for a plane wave in your first formula and the second leads to the same result $u=exp[-i(Et-px)]$ and you get a dimensionless momentum, i.e. speed – Evgeniy Yakubovskiy Aug 30 '19 at 7:11