Skip to the Main Content

Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser.

January 4, 2010

Coenergy

Posted by John Baez

It’s Newton’s birthday! Google has a nice homepage today, which shows an apple dropping

In honor of Sir Isaac, I thought I’d ask a question that’s related to classical mechanics: what’s coenergy?

I’m doing a lot of reading on electrical circuits these days — and most recently, the Hamiltonian and Lagrangian approaches to electrical circuits. And the word ‘coenergy’ keeps cropping up, but I’ve never seen it defined.

For example:

The function E^(q˙)\hat{E} (\dot{q}) is the sum of the electric coenergy of the capacitors in the tree Γ 1\Gamma_1 and the magnetic coenergy of the inductors in the cotree Λ 2\Lambda_2

from Dissipative Systems: Analysis and Control by Bernard Brogliato, Rogelio Lozano, Bernhard Maschke and Olav Egeland. Or:

CO-ENERGY (AGAIN)

In the linear case, energy and co-energy are numerically equal — the value of distinguishing between them may not be obvious.

Why bother with co-energy at all?

from Coenergy: Again, which seems to be notes from a course at MIT. Some other MIT course notes are more illuminating, but still awfully terse:

The constitutive equation for kinetic energy storage (inertia) is:

p=Mvp = M v

MM diagonal matrix of inertial parameters (masses, moments of inertia, e.g. about mass centers)

Kinetic co-energy is the dual of kinetic energy:

E k *=p tdv=12v tMv=E k *(v)E_k^* = \int p^t d v = \frac{1}{2} v^t M v = E_k^*(v)

Thus, by definition:

p=E k *vp = \frac{\partial E_k^*}{\partial v}

These are from Neville Hogan’s notes on Inertial Mechanics.

Now, I know that we can often think of kinetic energy either as a function of velocity or of momentum. In the first case, it’s a function on the tangent bundle TQT Q of configuration space QQ (some manifold or other). In the second case, it’s a function on the cotangent bundle T *QT^* Q. These two formulations are related by the ‘Legendre transform’.

Am I right in guessing from the above equations that ‘kinetic coenergy’ simply means kinetic energy regarded as a function of velocity? And maybe more generally ‘coenergy’ is energy regarded as a function on the tangent bundle? That seems perversely backwards to me — I’d say coenergy should be energy regarded as a function on the cotangent bundle. But who am I to say?

And I’m just guessing, anyway. I find it a bit frustrating that people would prefer to write E k *=E k *(v)E_k^* = E_k^*(v) rather than say ‘kinetic coenergy is just kinetic energy regarded as a function of velocity’ — if that’s what they actually mean. But I’m used to it.

Posted at January 4, 2010 6:14 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2147

13 Comments & 0 Trackbacks

Re: Coenergy

Principles of analytical system dynamics by Richard A. Layton tells us that “energy varies with momentum; coenergy varies with flow” and that kinetic coenergy is “the proper form to use in the classical Lagrangian” but kinetic energy is right for the classical Hamiltonian (p. 14).

There’s a nice table on the following page contrasting energy and coenergy of ideal kinetic stores.

Posted by: David Corfield on January 4, 2010 8:51 PM | Permalink | Reply to this

Re: Coenergy

Coenergy is (the quadratic form defined by) a Riemannian metric?

Posted by: Eugene Lerman on January 4, 2010 9:30 PM | Permalink | Reply to this

Re: Coenergy

Thanks, David! I seem to have made the mistake of searching the book under ‘coenergy’ and thereby reaching my page viewing limit before I even got to look at that “nice table”. But you’ve convinced me I was right.

‘Flow’ is the word for ‘velocity’ (a tangent vector) in general system theory, while ‘momentum’ is the word for ‘momentum’ (a cotangent vector). So, it sounds like ‘energy’ is energy considered as a function on the cotangent bundle of configuration space, while ‘coenergy’ is energy considered as a function on the tangent bundle. Just to keep life from getting too dull!

Eugene wrote:

Coenergy is (the quadratic form defined by) a Riemannian metric?

Yes: in the special case where all the energy is kinetic, and the kinetic energy varies quadratically with velocity. Then ‘coenergy’ is just

E(q,q˙)=12g(q˙,q˙)E(q,\dot{q}) = \frac{1}{2} g(\dot{q}, \dot{q})

where gg is the Riemannian metric.

Posted by: John Baez on January 4, 2010 11:27 PM | Permalink | Reply to this

Re: Coenergy

It seems that there’s a big analogy going on (as you describe in the last TWF):

  • velocity : momentum : mass
  • angular velocity : angular momentum : moment of inertia
  • current : flux linkage : inductance
  • inductor volume flow : inductor pressure momentum : fluid inertance (or inductance)

Coenergy and energy are defined in terms of the first and second quantities. Then there’s coenergy and energy for potential stores (rather than kinetic ones). E.g., for the linear electrical capacitor

coenergy=Ce 2/2;energy=q 2/2C. coenergy = C e^2/2; energy = q^2/2 C.

Posted by: David Corfield on January 5, 2010 10:07 AM | Permalink | Reply to this

Re: Coenergy

David: yes, you’re right!

A lot of this stuff is known — but not enough of it is known by enough people. Anyone who gets an undergraduate degree in physics is able to learn these analogies — and should. But the best explanations I’ve seen are in books that physics undergrads are unlikely to stumble upon, like this:

  • Dean C. Karnopp, Donald L. Margolis and Ronald C. Rosenberg, System Dynamics: a Unified Approach, Wiley, New York, 1990.

And even these books don’t seem to spell out everything as fully as one could.

Posted by: John Baez on January 6, 2010 3:01 AM | Permalink | Reply to this

Re: Coenergy

David wrote:

It seems that there’s a big analogy going on…

As a physicist I would say the big surprise is that linear ordinary differential equations are applicable to so many systems, and that’s it. It’s an example to that Einstein quote that it’s surprising that our simple models work at all, or that reductionism can be successful at all. But I’m ready to get mystified again by a topic that de-mystified itself to me a couple of years ago…

(In the class about differential equations that I attended the professor solved the equation for the linear approximation of the pendulum with initial condition identical to zero, which is the solution that is identical to zero for all time, stated that the theory tells us “that if nothing happens, nothing continues to happen” and called that a great triumph for the theory).

Posted by: Tim vB on January 6, 2010 10:52 AM | Permalink | Reply to this

Re: Coenergy

I feel like making a perverse remark about co-conservation, which may be equivalent to nservation, though perhaps not canonically…

Posted by: some guy on the street on January 4, 2010 10:51 PM | Permalink | Reply to this

Re: Coenergy

Go ahead — do it!

Posted by: John Baez on January 4, 2010 11:27 PM | Permalink | Reply to this

Co-Power Set; Re: Coenergy

So, is there a difference between Power and Copower? And is it problematic that the latter sounds close to “Coal Power”?

Posted by: Jonathan Vos Post on January 5, 2010 12:37 AM | Permalink | Reply to this

Re: Coenergy

It’s Newton’s birthday!

No it’s not.

Newton was born in England before 1752 so before Britain adopted the Julian calender. So his birthday was 10 days ago (well, 11 since I’m reading this a day after you posted it).

If you really want to be pedantic and say that it’s X years to the day since Newton was born then you ought to check that it really would be int(X*365.2425)\operatorname{int}(X*365.2425) first.

You’ll be telling me next that it’s Boxing Day today.

Bah, humbug.

Posted by: Andrew Stacey on January 5, 2010 11:47 AM | Permalink | Reply to this

Re: Coenergy

You mean while Britain was using the Julian calendar, before they adopted the Gregorian one.

Posted by: Mike Stay on January 5, 2010 4:34 PM | Permalink | Reply to this

Re: Coenergy

Whooops!

I blame the vikings.

Posted by: Andrew Stacey on January 6, 2010 8:03 AM | Permalink | Reply to this

Re: Boxing Day

Andrew wrote:

You’ll be telling me next that it’s Boxing Day today.

You mean that weird pugilistic holiday?

Posted by: John Baez on January 5, 2010 9:30 PM | Permalink | Reply to this

Post a New Comment