How Does the Reynolds Transport Theorem Apply to Electromagnetics?

[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.

 

[Clausius2]

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.

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 traveling with the flow velocity (and thus cancelling the convective transport terms) or with another different velocity (enhancing the convective transport).

For instance, I am going to 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 realize 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).

 

[charng]

I completely agree with your assessment of the Reynolds Transport Theorem. It is indeed a powerful tool in the study of fluid dynamics, and it has many applications in various fields of science and engineering.

In terms of its applicability in electromagnetics, you are correct in noting that the surface integral term resembles Gauss' law. This is because the Reynolds Transport Theorem is a generalization of the continuity equation, which is a fundamental principle in many areas of physics, including electromagnetics.

In fact, the Reynolds Transport Theorem can be applied to any system where there is a transfer of a property (such as mass, energy, or momentum) through a control volume or control surface. This makes it a versatile tool in many fields, including electromagnetics.

As for the specific interpretation of the terms in the equation, the first term on the right represents the change in the property B within the control volume due to internal processes, while the second term represents the change in B due to flux across the control surface. In electromagnetics, this could be applied to the flow of electric charge or magnetic flux through a control volume or surface.

Overall, the Reynolds Transport Theorem is a valuable tool in understanding the dynamics of various systems, and its applications extend beyond just fluid dynamics.