# Reynolds Transport Theorem

1. Jun 22, 2006

### Cyrus

I have come across the Reynolds Transport Theorm in my study of Fluid Dynamics, and it's a very powerful tool.

$$\frac {DB_{sys}}{Dt} = \frac{ \delta}{\delta t} \int_{cv} \rho b dV + \int_{cs} \rho b \vec {V} \cdot \hat {n} dA$$

Where B is any extensive property of the system, and b is any intensive propery of the system. The term on the left is the material derivative of the system, the first term on the right is the rate of change of the property B in the control volume and the second term on the left is the rate of change of B through the control surface.

This seems like something that might be useful in many other areas. Usually the same equations are found in nearly all areas of science. Does this have any applicability in say, E&M? Look's like it should. The surface integral term looks like Gauss' law, though I am not sure what the other terms would possibly represent.

Last edited: Jun 22, 2006
2. Jun 25, 2006

### Clausius2

For instance, take the first Maxwell equation in absence of charges:

$$\nabla \cdot \overline{E}=0$$

and the equation of continuity for an incompressible flow:

$$\nabla\cdot\overline{v}=0$$

Both fields are solenidal, then

$$\oint_S \overline{E}\cdot \overline{dS}=\oint_S \overline{v}\cdot \overline{dS}=0$$

The first integral is the conservation of flux of electric field over any closed boundary, whereas the second integral is the conservation of velocity flux (mass flux) over any closed boundary.

The Reynolds transport gives you the definition of a material derivative. In this case, even though a parcel which travels with the flow velocity cannot have a mass variation, then the right hand side of the RTT has to cancel because the material derivative is a a derivative viewed from the laboratory frame. The RTT is esentially a change of frame of reference when computing derivatives. In my opinion, it does not make sense to do the same at least in EM. Why? Well, the Navier Stokes equations have Transport terms (the convective terms: $$u\cdot\nabla u$$) whereas the Maxwell equations don't allow transport of electric quantities by the background. That means that in the case of a fluid you can have different variations if you are travelling with the flow velocity (and thus cancelling the convective transport terms) or with another different velocity (enhancing the convective transport).

For instance, I am gonna put an example of the importance of the RTT which cannot be achieved in EMs. Imagine a turbulent shear flow. I have a mean shear profile in y direction (vertical) in a water tunnel and the flow is turbulent. The Turbulent Kinetic Energy equation (derived from RANS equations) says to me:

$$\frac{\partial K}{\partial t}+<U>\frac{\partial K}{\partial x}=P-\epsilon$$

where P is the production, $$\epsilon$$ is the turbulent dissipation rate and $$<U>$$ is the mean velocity (the shear). Imagine I change the frame of reference, such that I analyze the turbulent flow from a frame moving with the mean velocity. Thus, I cancel the convective term and there is no transport from my new frame but only a local variation:

$$\frac{ dK}{dt}=P-\epsilon$$

Moreover this equation can be integrated using a $$K-\epsilon$$ model or knowing some experimental data, and one will realise that the turbulent kinetic energy increases exponentially with time in this reference frame because of the shear flow which feeds the production term.

Hence, one can simplify a lot the analysis of a fluid field because of the fact that the physics of the fluid mechanics is being transported with the flow velocity in the general case, whereas it does not happen with Maxwell equations in general (except with EM waves).