# Would magnetic flux be necessary for analogous systems?

I learned about the analogousness between mechanical and electrical systems a few months back (with the help of Feynman). Yesterday, my professor was lecturing about this topic, when she told that electrical systems can be written in two ways - either by using the voltage source, or by using the current source (drawing the equivalent analogs, and writing differential equations).

I've always thought current sources to be abstract constructs for the electrical engineers to solve problems in their boring way, node analysis and stuff, since it's voltage that drives the free electrons (hence, current).

Anyways, my confusion was about something called "Force-Current" analogous system, where the situation has gone crazy. Here's my (rather ugly) drawing of her example. We were asked to draw the equivalent electrical analogs. (It's a convention that whenever we make use of current source, we analyze them using nodes) I can understand the $F\to V$ system alright. Because, it makes use of the analog:

• $x\to q$
• $m\to L$
• $c\to R$
• $k\to 1/C$

But, it becomes hard to perceive, when current source $F\to I$ comes into play. Now, the analogs are crazy... (Note: This analog wasn't discussed by Feynman)

• $x\to \phi$
• $m\to C$
• $c\to 1/R$
• $k\to 1/L$

Is it right, actually? I'm skeptical about this kind of mapping. In this case, $G=1/R$ is conductance (which is the thing analogous to the damping factor $c$). Would this mean that conductance is the actual resistance here? And, more weirdo - capacitance is analogous to mass.

The DEs for this system are somewhat long. I'll choose an RLC circuit to express my point.

In terms of $F\to V$, (using $x\to q$) $$V_R=R \frac{dq}{dt},\ \ V_L=L\frac{d^2 q}{dt^2},\ \ V_C=\frac{q}C$$

Now, using $F\to I$, (using $x\to \phi$) $$I_R=\frac{1}R\frac{d\phi}{dt},\ \ I_L=\frac{\phi}{L},\ \ I_C=C\frac{d^2 \phi}{dt^2}$$

The $x\to q$ was fine, because motion of charge is current, and that's the basis for all electrical systems. But, the $x\to \phi$ is inconceivable. The magnetic flux, how $\phi$ changes, induced emf, etc. are needed for inductors. Yeah, but how can they be applied to resistors or capacitors?

With this framework for writing the differential equations, it seems as if magnetic flux is the clockwork behind the working of such a system.

This horror made me think that the $F\to I$ analog (using $\phi$) is just another abstract mathematical construct for writing differential equations. Am I right? Is it reasonable at all?

• Of course that it's just a dictionary that shows the mathematical isomorphism between differential equations that appear on both sides. Some of your entries disagree with Feynman - why didn't you copy his table accurately? – Luboš Motl Jul 31 '14 at 18:39
• Dear @Luboš: Well, I'm able to understand the equivalence for the $F\to V$ system quite well. Can you specify which of those entries disagree? In his lecture, Feynman addressed only the voltage-analogous system (which I understand just fine). Only the latter entries (for the current-analog) are confusing me (which wasn't addressed by Feynman) – Waffle's Crazy Peanut Jul 31 '14 at 18:45
• Have you tried writing down the equations that govern the motion for the F-I system? That ought to clear this up for you. – Floris Jul 31 '14 at 18:56
• @Floris: I've revised my question with a sample of the differential equations which obviously show my confusion :) – Waffle's Crazy Peanut Jul 31 '14 at 19:08
• @Waffle'sCrazyPeanut, sorry, they don't disagree, you used the $c$ notation while I call it $m\gamma$. – Luboš Motl Aug 1 '14 at 3:36

I think there is something wrong with your mapping.

Looking at http://lpsa.swarthmore.edu/Analogs/ElectricalMechanicalAnalogs.html , I see the following table: This is inconsistent with the mapping you are showing.

I can understand this table - I can't understand yours. I think an error crept in - which would reasonably explain your confusion.

• Hmmm... That's an useful link. Looks like you've given the inverse relations $V\to F$ and $I\to F$. Based on your link (and Phonon's argument), I might guess that the current-source equivalent is (probably) mathematical, and so I don't think I should expect any intuitive explanation for that. Anyways, thanks for your answer :) – Waffle's Crazy Peanut Aug 1 '14 at 13:59

To add to Floris's reply, just to elaborate in a very basic manner in what way the flux linkage (analogous of displacement x in Force Current analogy) and magnetic flux differ.

Consider the scenario where a magnetic field is present, and we have an open circuit that has been closed using a metalic rod. As the picture shows: We know that the induced EMF is $$\epsilon = \frac{B A}{t}$$ ($$A$$: surface enclosed by the circuit) or better expressed in terms of rate of change of magnetic flux $$\frac{d\Phi}{dt}$$. Now in order to increase the induced EMF, other than increasing $$v$$ (hence increasing $$A$$) and increasing simply the field strength $$B$$, there is a further possibility, which is to increase the number of turns of wire N in the circuit. Doing so, the flux has not altered but the flux linkage $$N \Phi$$ will have increased. To complete Faraday's law of induction, Lenz's law is taken into account, finally the rate of change of flux linkage: $$\epsilon = - N \frac{d\Phi}{dt}$$

Finally intuitively speaking, it does make sense to replace the displacement x with the magnetic flux linkage $$N \Phi$$ as we know that in the circuit the springs are replaced by inductors.

(stretching the spring $$\leftrightarrow$$ increasing $$N$$ in inductors).

• Yep, I agree with your argument that voltage can be thought of as change in magnetic flux. Thanks :) – Waffle's Crazy Peanut Aug 1 '14 at 13:58