A Question About Gauge Symmetry and the Interaction Term

Doug Sweetser

I spent the week seeing if I could give the good old college try and find a "way that worked" for the gauge games I have been playing. Nothing panned out. Bummer, I had a blog to write. While collecting crumpled paper, there was an old issue I wasn't able to figure out. The answer was probably known a hundred years ago. The blog will explain where I got stuck, and who knows, maybe someone will provide the answer.

In my previous blog, I focused on the fields, the derivatives of the potential. In the comments, CuriousReader pointed out that careful thought had to be devoted to the interaction term, the contraction of the current with the potential. The potential A will change for both the interaction and the fields due to the gauge. Having never thought about the interaction term in this context, I was the proverbial deer in the headlights, not doing too much but worrying about shotguns.

Here is how gauge symmetry is presented in the wiki way:

$\\ \phi \rightarrow \phi' = \phi - \frac{\partial f}{\partial t} \\ \\ A \rightarrow A' = A + \nabla f$

Drop those into the definition of the electric field E and the magnetic field B [corrected sign errors]:

$\begin{align*} E \rightarrow E' &= -\frac{\partial A}{\partial t} - \frac{\partial \nabla f}{\partial t} - \nabla \phi + \nabla \frac{\partial f}{\partial t} = E \\ \\ B \rightarrow B' &= \nabla \times A + \nabla \times \nabla f = B \end{align*}$

Fini!

In my limited reading, that was presented as the full story. Yet something is being added to the potential. Shouldn't that mess up the interaction term?

$J_{\mu} A^{\mu} \rightarrow J_{\mu} A'^{\mu} \ne J_{\mu} A^{\mu}$

Look at the interaction with the gauge field close up:

$J_{\mu} A'^{\mu} = \rho \phi - \rho \frac{\partial f}{\partial t} - J \cdot A - J \cdot \nabla f$

This caused me to scratch my head.The derivatives of the scalar function f are free to wander in spacetime. Don't read anything into the plus and minus signs because the derivative of f with respect to the four spacetime variables can be positive in one location, and negative in another.

The way out is to realize how the interaction term is used to generate the field equations. One takes the derivative of the current coupling with respect to the potential with or without the gauge field:

$\frac{\partial J_{\mu} A^{\mu}}{\partial A^{\nu}} = \frac{\partial J_{\mu} A'^{\mu}}{\partial A'^{\nu}} = J_{\nu}$

No matter what odd things the gauge field does, the derivative with respect to the new potential will still be the 4-current J. One can find solutions to the field equations that are independent of the choice of the gauge field because it doesn't matter for the field term nor the interaction term. Those solutions when put into the standard force equation are consistent with experiments. So far, so good.

Here is something that still doesn't quite make sense to me. I don't see how the derivation of the equations of motion remains unaltered with a gauge field. To be concrete, suppose the function and gauge field were defined like so:

$\begin{align*} f &= x^2 + y^2 \\ \\ f' &= (0, 2 x, 2 y, 0) \end{align*}$

[ERROR? Looking into this in detail, I think f must be a differential function of t, x, y, and z for this to work out.]
By the preceding analysis, the derivation of the field equations are unaltered because the derivatives are with respect to the potential A and changes in the potential. The same solutions to the field equations will be found. No mystery.

To generate the force equation however, one takes the derivative of the action with respect to position and velocity. There will be an extra factor of 2x and 2y for this gauge field. It looks like the form of the force equation is altered by this gauge choice.

Hopefully my misunderstanding can be corrected in the comments.

A tentative simple road out.

Both David and CuriousReader said I should keep real close track of what the coordinates are. This is the world of calculus of variations, not integrate the area under the curve. The goal is to find extremums of functions are variables that are varied. I will need to spend a few days doodling to see if I can track what is going on.

I was wondering if there might be a more direct solution to the force derivation + gauge transformation riddle, which is only to me, not a mystery to physics.

Do the force derivation the "standard way" as I did a number of blogs ago. Notice that the results can be expressed in terms of the fields E and B. I showed above how those two fields are invariant under a gauge transformation. Ego the force will be invariant under the same gauge transformation.

Doug

Next Monday/Tuesday: A Tour of the Private Pop Science Art Collection

Comments

If you take a Lagrangian and add a total derivative, you have not changed anything because this will only add an overall constant to the action. So unless there is some strange "boundary" term to your action calculation, the Lagrangian represents the same physics, so they are equivalent Lagrangians.

You found that after the gauge transformation you got
L' = L - (rho (partial f / partial t) + J_x (partial f / partial x) + J_y (partial f / partial y) + J_z (partial f / partial z) )

Is this addition a total derivative?
I already gave the answer last week, and the question itself is very leading, so of course the answer is yes.

To make it more clear, write it in tensor notation:
L' = L - J^b (partial_b f)
and as long as the charges 'q' on the particles in your theory are some fixed parameter, we have charge conservation
(partial_b J^b) = 0
and therefore
L' = L - J^b (partial_b f) - f (partial_b J^b) = L - partial_b (f J^b)

So the Lagrangian differs only by a total derivative, and is therefore equivalent. The Lagrangian is invariant to the gauge transformation.

If the charges 'q' are not some fixed parameter, this will not work. This is another way in which gravity will throw a wrench in your ideas. Unlike electromagnetism, gravity is not sourced by a scalar. It is sourced by a tensor -- the stress energy tensor -- which is not some fixed quantity for an object. For example take a particle at rest, then give it some kinetic energy and the trace of the stress energy tensor changes.

There are many issues with your ideas that have been enumerated by others, none of which were complaints about gauge symmetries. I'm not sure why you think the key to your theory lies in adding a gauge symmetry to your Lagrangian. It would be worthwhile to go back and learn from the actual problems if you did not understand them. For example maybe clear up your potential <--> curved spacetime "equivalence" misconceptions, or maybe learn why a scalar gravity charge will not work, etc. As you can hopefully see by now, it would be useful to really learn a subject before you try proposing changes to it.

CuriousReader (not verified) | 02/07/12 | 00:58 AM

I hope you don't mind, but I am going to quote your comment completely since I cannot parse LaTeX, only the output of LaTeX.

<<Begin full quote.>>
If you take a Lagrangian and add a total derivative, you have not changed anything because this will only add an overall constant to the action. So unless there is some strange "boundary" term to your action calculation, the Lagrangian represents the same physics, so they are equivalent Lagrangians.

You found that after the gauge transformation you got

$L' = L - (\rho ~\frac{\partial f }{ \partial t} + J_x ~\frac{\partial f }{ \partial x} + J_y ~\frac{\partial f}{ \partial y} + J_z ~\frac{\partial f}{ \partial z} )$

Is this addition a total derivative?
I already gave the answer last week, and the question itself is very leading, so of course the answer is yes.

To make it more clear, write it in tensor notation:

$L' = L - J^b ~\partial_b f$

and as long as the charges 'q' on the particles in your theory are some fixed parameter, we have charge conservation

$\partial_b \,J^b = 0$

and therefore

$L' = L - J^b \,\partial_b f - f \,\partial_b J^b = L - \partial_b \, f J^b$

So the Lagrangian differs only by a total derivative, and is therefore equivalent. The Lagrangian is invariant to the gauge transformation.

If the charges 'q' are not some fixed parameter, this will not work. This is another way in which gravity will throw a wrench in your ideas. Unlike electromagnetism, gravity is not sourced by a scalar. It is sourced by a tensor -- the stress energy tensor -- which is not some fixed quantity for an object. For example take a particle at rest, then give it some kinetic energy and the trace of the stress energy tensor changes.

There are many issues with your ideas that have been enumerated by others, none of which were complaints about gauge symmetries. I'm not sure why you think the key to your theory lies in adding a gauge symmetry to your Lagrangian. It would be worthwhile to go back and learn from the actual problems if you did not understand them. For example maybe clear up your potential <--> curved spacetime "equivalence" misconceptions, or maybe learn why a scalar gravity charge will not work, etc. As you can hopefully see by now, it would be useful to really learn a subject before you try proposing changes to it.
<<end full quote>>

That is a pretty argument when I can see it as math, not LaTeX. I don't quite get it yet, but I can see it now :-) Given your difficulty with Internet connection, I hope you don't mind if I translate these from time to time since it greatly increases the efficiency of communication for me.
This was nice:

$L' = L - J^b \,\partial_b f - f \,\partial_b J^b = L - \partial_b \, f J^b$

I guess I missed this part of my training:

So the Lagrangian differs only by a total derivative, and is therefore equivalent.

I am having trouble connecting the dots. No one sees a Lagrangian. Instead, we take different derivatives of Lagrange densities and get different results. Let me add a detail, to show that f is a function of position:

$L' = L - \partial_b \, f(x) J^b$

Take the derivative of L' with respect to A and the derivatives of A, and one gets the field equations. Take derivatives of L' with respect to position x and one gets the force equations. The derivative of L' with respect to x is different from the derivative of L with respect to x. Symbolically:
$\\ \frac{\partial \mathcal{L}}{\partial A} = \frac{\partial \mathcal{L'}}{\partial A'} \\ \\ \frac{\partial \mathcal{L}}{\partial x} ?\ne? \frac{\partial \mathcal{L'}}{\partial x'} \\$

Writing these equations pointed out a detail I don't quite get. The derivative with the A's is simple. With respect to the x's, what I had said and thought in words was this:

$\frac{\partial \mathcal{L}}{\partial x} ?\ne? \frac{\partial \mathcal{L'}}{\partial x} \\$

Notice the lack of a prime on the x. That was the way I was thinking about it. It didn't seem right to not have a prime on the x as I wrote it out, so I added it, but I don't know what it is :-( This may be a clue.

This blog is suppose to be entirely about EM and gauge theory. I will limit myself to three paragraphs

Unlike electromagnetism, gravity is not sourced by a scalar. It is sourced by a tensor

That is true and logically consistent for rank 2 field theories for gravity. It is the accepted dogma. People have plenty of thought experiments to prove it. I recall one in a GR intro article by Price. He had two boxes that had 6 particles in each. Both boxes attracted each other due to gravity. Then he imagined one of the particles was destroyed, but all the energy turned into an equivalent amount of kinetic motion. It makes sense to say the attraction wouldn't change between the boxes.

The problem with such though experiments is that if the one particle destroyed happened to have an electric charge, then there would be no way to do the experiment. One cannot destroy electric charge, that would be very bad. That is what I do when I hear about thought experiments and gravity: sprinkle electric charges around and see if they are still OK. I cannot accept the conclusions of a thought experiment that requires electric charge is no longer conserved.

I can only solve one problem at a time. There are lots of Catch-22 issues here. I tried to work with solutions to the field equations but then someone said I couldn't choose the gauge so I didn't have field equations. If I had stated the issue of fixing the gauge precisely, I could have proceeded (but I a skeptical of that). Solving these problems is hard, and no, I haven't solved them yet. But I will proceed to try, I have been consistent in that way.

Doug Sweetser | 02/07/12 | 10:00 AM

"Given your difficulty with Internet connection, I hope you don't mind if I translate these from time to time..."

That's fine by me; it does look nicer.

"Notice the lack of a prime on the x. That was the way I was thinking about it. It didn't seem right to not have a prime on the x as I wrote it out, so I added it, but I don't know what it is :-( This may be a clue."

I think I see what is going on now. I'm running late for a class, so I don't have time to type the details up. It'll just have to be a hint or exercise for you to work on. The issue is the sloppiness in which you and I and most textbooks, expect it to be understood from context what the position coordinates are referring to. Try writing the Lagrangian and Action out, being very explicit what each term is a function of and also distinguishing between the particle position (let's say (x,y,z,t) ) and just the spacetime coordinates used to label the fields at each point (let's say x', y', z', t' ) which will be integrated over at some point.

If it still isn't clear, then go ahead real quick and reproduce your particle and field equation of motion derivations. The key is that you choose a Lagrangian and the generalized coordinates. That second choice is crucial, for the Lagrangian itself doesn't specify the model (depending on what you choose to vary, the same Lagangian can be used for different models, which I've seen in some gravity theories that all start with the Einstein-Hilbert action). You then vary the action with respect to these coordinates to get the equations of motion for those generalized coordinates. Note for example that you don't vary the action with respect to just the spacetime coordinate labels that you integrate the Lagrangian with.

Now I am late, so I'll have to end here.

CuriousReader (not verified) | 02/07/12 | 11:23 AM

Doug:

Remember that the Lagrangian density (or even just a plain old ordinary Lagrangian) is actually within an integral over all spacetime (or, at least, the region of spacetime of interest). Also, remember what the integral of a total derivative is.

This also means that the derivative of the Lagrangian density with respect to position (x) is not meaningful, since spacetime coordinates are all integrated "away". (Yes, I know that when one uses a Lagrangian integrated over time, or proper time—as a path integral—one still has "paths" [like x(t)] that can be varied. Then one can, meaningfully, take the "derivative" of the Lagrangian with respect to these "paths", and this often looks—symbolically—like we are taking a derivative of the Lagrangian with respect to x, position. However, one should not get fooled by appearances—let alone symbology. ;) )

David

David Halliday | 02/07/12 | 11:42 AM

CuriousReader:

I think you may have been noticing that I have been setting out some "carrots" for Doug. (I've noticed that many of them have piqued your curiosity even more than Doug's, or at least more than what Doug has shown.) My hope has been that he would pursue them enough to answer them (mostly) on his own, so he would more fully gain the deeper understanding he lacks.

Now you've gone and practically given him this one. ;)

David

David Halliday | 02/07/12 | 11:52 AM

Doug:

Glad to see you have taken care of the sign error in, at least, your transformation from E to E'.

David

David Halliday | 02/07/12 | 10:46 AM

Apart from gauge and Lorentz invariance of $jA$ , it would be nice to get equations with physical solutions and it is not the case, unfortunately. To that interaction one have to add the counter-terms and consider all that as an interaction. It is so at least in QED and QFT. So the physics of interaction is more tricky than just $jA$ .

Vladimir Kalitv... | 02/08/12 | 07:03 AM

A very serious note to Doug and all readers:
Vladimir has some severe misunderstandings of Lagrangians and in particular electrodynamics. He also has a personal pet theory that he has been pushing for years, that I've seen the issues explained to him in detail on multiple sites to no effect. It is not worth going into here, the point is that anyone struggling to understand this material, just ignore Vladimir.

_Sam_ (not verified) | 02/08/12 | 11:25 AM

Sam, whatever pets I have, here I just cited the famous Sidney Coleman lecture about what is the interaction Lagrangian:

$L_{CT}$ here is the counter-term Lagrangian density that must be added to $L_{int}$ . Shame on you.

Vladimir Kalitv... | 02/08/12 | 11:52 AM

Sorry, Vladimir, this is quite off topic for this (classical) discussion: We are not talking about quantum fields, we are not talking about perturbation theories of any kind, and we are not having to deal with the vagaries of any "renormalization".

It will be up to Doug as to whether to delete these off topic messages.

David

David Halliday | 02/08/12 | 12:28 PM

For Doug and other readers:

[EDITED BY DOUG, originally by _Sam_] Solid technical objection to Vladimir's work has been removed, sorry Sam! I don't like to do that, but there is no need to battle this battle here, since as you point out, odds of communication are low. I did want to keep in the gem below though, so thanks for that observation.]

While we are on the subject, Lagrangians that are equivalent in classical mechanics may not be equivalent in quantum mechanics. This may sound very strange if you had not heard it before, so it is an important physics tidbit to point out. Therefore David and CuriousReader's comments should be read in the context of classical mechanics.

Doug Sweetser | 02/08/12 | 13:51 PM

Thanks, Doug, for editing _Sam_'s comment so I can respond to this tidbit without worrying (so much) that I was polluting the discussion. (Sorry _Sam_.)

To quote _Sam_:

While we are on the subject, Lagrangians that are equivalent in classical mechanics may not be equivalent in quantum mechanics. This may sound very strange if you had not heard it before, so it is an important physics tidbit to point out. Therefore David and CuriousReader's comments should be read in the context of classical mechanics.

Absolutely. Our discussion has been focused upon classical mechanics in this article and blog. When we do deviate into quantum mechanical aspects (as we have done in other articles and blog discussions) we (or, at least, I) try to make that explicit.

Now, while there are different ways one may interpret _Sam_'s remarks, above, one interesting example of what I believe he is alluding to is the Aharonov–Bohm effect: Wherein, even though, classically, the fields (as opposed to the potentials) experienced by the electrons never change, and even though, classically, the vector potentials are free to have any arbitrary gauge applied without any (classical) affect; the electron interference pattern is clearly modified by the changes wrought in the vector potential "seen" by the electrons.

Of course, from a Quantum Field Theoretic (QFT) perspective, the "classical" action (spacetime integral of the Lagrangian density) involves the (Quantum Mechanical [QM]) Dirac (wave) equation for the electron, and a local change in the gauge of the electromagnetic field is "matched" by a local change in the "phase" of the (Dirac) electron's wave-function, in order to preserve gauge invariance. So what can be meant by "classical" and "equivalent" Lagrangian densities (or actions, or "Lagrangians"), and even "quantum mechanics", can get a little fuzzy.

However, I think _Sam_'s observation—that one must be careful to distinguish between classical vs. (various levels of) Quantum Mechanical situations/descriptions/expectations/etc.—is well worth keeping in mind.

David

David Halliday | 02/08/12 | 14:57 PM

I do let people post a bit about their own concerns, but I also cap such discussions (a form of strikeout rule). This blog is really supposed to be about plain-Jane classical not quantized EM. I limited a discussion of my own worked to three paragraphs in response to a specific question , and fortunately the dialog stopped there (thanks CuriousReader, there are serious questions, but this is not the blog to hash the issues out).

I don't think it is ever necessary to warn readers about other commenters since the readers are bright enough a) to read this stuff and b) figure out whose words are of most value. It does break the spirit of the Golden Rule - Vladimir had to reply to Sam's comment, and back and forth it can go. There will be a diversity of opinions about folks here. For example, I am pretty sure David has a better rep than I do ;-)

Just to be clear about my line of action: any further comments to this subthread will be deleted because they are too far off topic.

Doug Sweetser | 02/08/12 | 13:30 PM

So with that complaint about a J . A interaction term set aside, where does that leave everything?
Since the purpose of this article was to ask some questions about gauge symmetry in the classical electrodynamics Lagrangian, I'm curious if after the discussion you still have questions on this.

David is a great teacher, and I guess he feels some lessons can really only be learned by doing. I'm sorry I got impatient and ruined some of it. I think everything was laid out clearly, but in case some of it wasn't, I hope you were able to work out the final details by doing the math yourself.

CuriousReader (not verified) | 02/09/12 | 01:25 AM

I spent my limited free time focused on the quotient group. I altered the main blog to put up one possible way to resolve my issue. The force equation is expressed in terms of E and B, and since E and B are invariant under a gauge transformation, the case is closed.
No need to apologize for jumping in, I know I wouldn't have found the total derivative business. Here is the action with the gauge transformation:

$S = \int d^4 x (\mathcal{L} - \partial_b f J^b)$

We want to work with variations with respect to postion and velocity.

$\delta S(x, x') = \int d^4 x ( \frac{\partial \mathcal{L} }{\partial x} + \frac{\partial \mathcal{L} }{\partial x'} - \frac{ \partial \; \partial_b f J^b }{\partial x} - \frac{ \partial \;\partial_b f J^b }{\partial x'})$

<<Begin correction>>
BAD error: no variation terms on the right hand side...try again
NOTE: x does too many things! I have now changed the position and velocity variations with a capital X to distinguish it from the lower case x over the volume of the intergration. The primes here are for derivatives with respect to position, in other words velocities.

NOTE: The variation is with respect to position, capital X. I am further presuming the Lagrangian is a function of changes in position, but not higher order derivatives.

$\delta S(X) = \int d^4 x \left( \frac{\partial \mathcal{L}}{\partial x} \delta X + \frac{\partial \mathcal{L}}{\partial x'} \delta X' - \frac{\partial \frac{d \, f J^b}{d x^b}}{\partial x} \delta X - \frac{\partial\frac{\partial f J^b}{\partial x'^b}}{\partial x'} \delta X' \right)$

I think this is the point were we can say the total derivatives of the gauge terms will merely redefine the constants of integration.
As penance, continue on to the Euler-Lagrange equation.

$\delta S(X) = \int d^4 x \left( \frac{\partial \mathcal{L}}{\partial x} \delta X + \frac{\partial \mathcal{L}}{\partial x'} \delta X' \right)$

By the chain rule:

$\left(\frac{\partial \mathcal{L}}{\partial x'} \delta X \right)' = \left(\frac{\partial \mathcal{L}}{\partial x'} \right)' \delta X + \frac{\partial \mathcal{L}}{\partial x'} \delta X'$

This is the step I don't get at this time, but there is a theorem that says the above evaluates to zero. That means on the right, one term equals the negative of the other. Eliminate the variation by x':

$\delta S(X) = \int d^4 x \left( \frac{\partial \mathcal{L}}{\partial x} - \left(\frac{\partial \mathcal{L}}{\partial x'} \right)' \right)\delta X$

This will reach an extremum if the integrand is zero.
<<End correction>>

I am familiar with the road that leads from here to the Euler-Lagrange equation using the first two terms in the integral. I don't see why the total derivative terms can be ignored, probably because I have a garbled expression or am missing a basic message of the fundamental theorem of calculus.

Doug Sweetser | 02/11/12 | 12:34 PM

Doug:

I don't know what you think you are doing with your second equation. In fact I'm reasonably certain it displays your lack of understanding of the relationship between the variational principle and "the Euler-Lagrange equation".

On the other hand, "the total derivative business" involves nothing but the second term on the right-hand side of the first equation.

What is the integral, from x=a to x=b, of df/dx with respect to x, as in $\int_{a}^{b}{\frac{df}{dx}dx}$ ?

Now, can you generalize that to more than a single dimension?

David

David Halliday | 02/10/12 | 17:57 PM

Integral of a full derivative is just a constant C, and a constant does not vary in the variation principle. In other words, A and A+C give the same equations independent of C (A being the action).

Vladimir Kalitv... | 02/11/12 | 06:16 AM

Vladimir:

Actually, you are partially to mostly correct. However, you are most certainly not totally and completely correct.

The issue is that the "constant" term, "C", is actually a function on the boundary. So, depending on the boundary conditions placed upon the functions within the action, "A"—over which the action is a functional—"C" may or may not play a part in the variational principle.

In "the usual cases", where the functions are restricted to be "set" on the boundary, so they aren't allowed to vary on the boundary when one varies the functions in the variational procedure, "C" is completely constant, and you are correct.

On the other hand, this subtle fact has some very interesting things to "say" about appropriate boundary conditions, and other things, about the quantum mechanical wave equations, such as the Dirac equation.

David

David Halliday | 02/11/12 | 09:01 AM

Yes, I implied fixed boundary conditions.

Vladimir Kalitv... | 02/11/12 | 09:05 AM

Yep. It totally works under fixed boundary conditions.

David Halliday | 02/11/12 | 09:20 AM

DON"T bother with this comment. I wandered too far away from the Euler-Lagrange sort of variation principle and got lost, my bad.

An effort was mad to correct my errant comment. I am starting to see it finally.
$S =\int_a^b \int_c^d \frac{df}{dx ~dy} ~dx ~dy = f(b) + f(d) -f(a) -f(c) = k$
I think I need a brain transplant. I know a little voice mumbled "Mixed derivative? WTF?". I have to pick up on those critical voices.
$\delta S[f] =\int_a^b \int_c^d ~dx ~dy \left( \frac{\partial f}{\partial x } + \frac{\partial f}{\partial y }\right) \delta x ~\delta y= f(b) + f(d) -f(a) -f(c)$

This looks wrong to me, too many x's doing jobs that are different. I also am aware of a common blindness I have, the trouble recognizing the difference between total and partial derivatives. Sure, the partials are more curvy than the total derivatives. It is not now it is written, it is what it implies what is going on.

I had a minor debate with my inner critic about using partials above. The partial means I took the derivative with respect to x while holding y constant, then the derivative with respect to y holding x constant. If f is only a function of x and y, then this is the total derivative of the function f, all the derivatives that can be taken. I am not coming up with a way to write the gradient so that it looks like the total derivative.

One minor change is to make the things that get varied look a little different:
$\delta S[f] =\int_a^b \int_c^d ~dx ~dy \left( \frac{\partial f}{\partial x } + \frac{\partial f}{\partial y }\right) \delta X ~\delta Y= f(b) + f(d) -f(a) -f(c)$
ADDED comment...
Perhaps the best thing to note about this garbled comment is why the above variation is dead dull. The variation of S is not a ratio like a standard derivative. Do a small enough variation, and nothing will be seen. What is needed is another layer deeper of derivatives of f.

$\delta S[f]=\int_a^b \int_c^d d x ~d y\left( \frac{\partial f}{\partial x} ~\delta X + \frac{\partial f}{\partial x'} ~\delta X' + \frac{\partial f}{\partial y} ~\delta Y + \frac{\partial f}{\partial y'} ~\delta Y' \right)$

While better, I don't think this is quite there, but also is not worth the bother.
END added comment

Varying something is different from the volume integration.

The "tricky" part will be to see that the remaining integration is over the boundary of the space over which the original (multi-dimensional "volume") integral was taken.

For this to work, I must presume f is only a function of x and y, not x, y, and z.

Doug Sweetser | 02/12/12 | 00:01 AM

Doug:

You need to do some serious work on this equation. It is quite incorrect, as it stands. :(

Besides, the appropriate integrand is not something like a second derivative, but more like a divergence of a vector function. So, it will retain all but one level of integration.

The "tricky" part will be to see that the remaining integration is over the boundary of the space over which the original (multi-dimensional "volume") integral was taken.

David

David Halliday | 02/13/12 | 22:55 PM

Doug:

First, some notes on notation:

The functional S, of functions X and Y, should be expressed as S[X, Y].
The variation is of S[X, Y], not the functional of a variation of X and a variation of Y: The variation of S[X, Y] can be considered to be the limit, as the variations (δX and δY) become very small, of S[X + δX, Y + δY] - S[X, Y].

Then, f should be a function of X and Y (these functions, X and Y, are functions of both x and y), and may also depend explicitly on the parameters x and y, but that's not particularly important.

So, since the variation is of the functions X(x,y) and Y(x,y), the limit leads to our being able to express the integrand in terms of derivatives of f with respect to the functions X and Y.

Furthermore, notice that the limit in item 2, above, looks very much like the definition of a single derivative (except that we haven't divided by the "size" of the thing we are taking the limit with respect to). So, the right-hand side should only have terms to first order in δX and δY, so even the product of these two should be gone.

Therefore, the integrand should look like the sum of products of variations in the functions (δX and δY) and the derivatives of f with respect to the functions (X and Y).

Now, since you only have f depending on the functions, and not on derivatives of the functions, you have no total derivatives, and no cause for the use of the integration by parts. So your elimination of the integral makes absolutely no sense.

To make this more like the case with typical actions (Lagrangian densities) you will need f (or, rather, the Langrangian density) to be a function of both these functions, and their derivatives (with respect to the parameters x and y).

By the way, it may help you to begin with a functional of only a single function. It may further help to use some name for this function that is completely different from your parameter names.

May I suggest having S be a functional of the function f, S[f], where the functional is in the form of the integral (over x and y, as you have it) of a Lagrangian density (cursive uppercase L) that is a function of f(x,y) and the partial derivatives of f(x,y) with respect to the parameters/arguments x and y. (So, if you spell out the explicit dependance of the Lagrangian density you will have at least the three arguments f, partial of f with respect to x, and partial of f with respect to y.)

Then try the variational principle.

David

David Halliday | 02/11/12 | 19:19 PM

Doug:

In order to see where the Euler-Lagrange equation comes from—in other words, how it is derived—you will need to get "down and dirty" with the calculus of variations. Since you appear to be rather comfortable with the fundamental theorem of calculus—that only applies to a single dimension—I suggest you begin by investigating the calculus of variations in one dimension: The usual Lagrangians, where the action is a single integral of the Lagrangian over some interval (such as a time interval).

$S\left[f \right]=\int_{a}^{b}{L\left(f\left(t \right),\dot{f}\left(t \right) \right) dt}$

(The dot notation is being used for time derivatives.)

Now take the variation of S[f], as f varies:

$\delta S\left[f \right]=S\left[f+\delta f \right]-S\left[f \right]$

$\delta S\left[f \right]=\int_{a}^{b}{L\left(f\left(t \right)+\delta f\left(t \right),\dot{f}\left(t \right)+\delta \dot{f}\left(t \right) \right) dt}-\int_{a}^{b}{L\left(f\left(t \right),\dot{f}\left(t \right) \right) dt}$

For δf (and its time derivative) very small, we have (via Taylor series expansion)

$\begin{align*} L\left(f\left(t \right)+\delta f\left(t \right),\dot{f}\left(t \right)+\delta \dot{f}\left(t \right) \right) = L\left(f\left(t \right),\dot{f}\left(t \right) \right)+\frac{\partial L}{\partial f}\delta f\left(t \right) &+\frac{\partial L}{\partial \dot{f}}\delta \dot{f}\left(t \right)\\ &+O^{2} \end{align*}$

So, to first order,

$\delta S\left[f \right]=\int_{a}^{b}{\left( \frac{\partial L}{\partial f}\delta f\left(t \right)+\frac{\partial L}{\partial \dot{f}}\delta \dot{f}\left(t \right)\right) dt}$

Now, apply integration by parts to the second term within the integral in order to obtain something proportional only to δf (along with "boundary" terms).

I know you can do it, Doug.

David

David Halliday | 02/13/12 | 22:32 PM

Doug:

Well, you're getting closer.

I now know more about what you were and are trying to say with what was your second equation.

I now see that you are using x (and x') to be the functions over which the action, S, is a "functional". I also see that you are using the prime notation to imply derivatives, rather than the gauge transformation, used further above.

OK. So now I think I see what's tripping you up.

Just as the gauge transformation creates a part of the integrand (of the action integral) that is a total derivative, the variational principle creates a part (the left-hand side of your "chain rule" portion, when all is done correctly) that is similarly a total derivative.

The exact same theorem is used in both of these cases. ;)

David

P.S. You cannot, properly, have the action, S, be a "functional" of "functions" (or variables) over which you are integrating. Furthermore, the derivatives of the "functions" (what you were symbolizing with x') must be over all the variables over which you are integrating in the formation of your action "functional".

So try something more on the lines of S[y(x), partial y(x)/partial x] (where the square brackets are typically used to distinguish "functionals" from simply functions over their arguments, and must not be confused with the way Mathematica uses the square brackets in place of parentheses), where the partial derivatives represent all the partial derivatives with respect to the entire space (x) over which you are integrating to form your action. (Similarly, your Lagrangian densities are functions [using plain old ordinary parentheses to set off their arguments] of y(x) and partial y(x)/partial x.)

P.P.S Technically speaking, the action, S, is strictly a functional over the functions, and not also over the derivatives of those functions. So the action should be written as S[y], rather than as shown above. However, the Lagrangian densities are still to be shown as above, and can even involve higher order derivatives, that must be handled (and go beyond, or generalize, the Euler-Lagrange equation).

David Halliday | 02/11/12 | 10:07 AM

The exact same theorem is used in both of these cases. ;)

At least I am a consistent airhead. Another kind of blindness is not appreciating the functional. I know it is "different", but don't have a good grasp of it, surprise. Not. The idea is at the core of the calculus of variations, something I only tried to teach myself.

Doug Sweetser | 02/11/12 | 12:30 PM

Doug:

I'm not sure how much I would say the concept of the functional "is at the core of the calculus of variations". Yes, it's kind of basic, for the calculus of variations, but no more so than the concept of the function "is at the core of" optimization (minimization/maximization) of a function by choosing the parameters of the function.

Just as a function is a map from some space of numbers (which may be a product space: Multiple function arguments means a product space of the number spaces of the arguments) to another space of numbers, a functional is a map from a space of functions (which may also be a product space) to a space of numbers (usually real numbers).

The key is that instead of varying some set of parameters, we vary some set of functions.

David

David Halliday | 02/11/12 | 18:32 PM

The exact same theorem is used in both of these cases. ;)

Could you provide the name of this theorem? It looks vaguely like the fundamental theorem of calculus in a calculus of variations context.

Doug Sweetser | 02/11/12 | 21:41 PM

Doug:

The theorem of which I speak has nothing particular to do with "a calculus of variations context." It is nothing but the fundamental theorem of calculus generalized to arbitrary dimensions, and arbitrary orientable manifolds: It is Stokes' Theorem.

It's usually written in the language of differential geometry, in the language of "forms". However, it does not have to be used in that form. One only needs to understand how the language of forms maps into the "usual" language of multidimensional integration. (However, I have seen many people, even scientists, make mistakes using the "usual" language that would be easily avoided in the language of forms.)

Anyway, I hope that helps. Unfortunately, I fear it may be more of a distraction. But what do I know. ;)

David

David Halliday | 02/12/12 | 01:38 AM

Doug:

After you've completed the integration by parts exercise, perhaps you should try a two dimensional integral of a two dimensional divergence:

$I\left(f, \vec{J}; x_0, x_1, y_0, y_1 \right)=\int_{x_0}^{x_1}{\int_{y_0}^{y_1}{\partial_bfJ^b\,dy}dx}=\int_{x_0}^{x_1}{\int_{y_0}^{y_1}{\left( \frac{\partial fJ^x}{\partial x}+\frac{\partial fJ^y}{\partial y}\right)dy}dx}$

Once you have both of these exercises completed successfully, I believe you will be ready for a two (or more) dimensional calculus of variations problem.

Are you up for the challenge?

David

David Halliday | 02/13/12 | 22:29 PM

CuriousReader:

While, based upon my experience, "doing" is the best way to learn (with guidance, as appropriate), please don't feel like your "impatience" "ruined some [or any] of it". That's why I used the winking smiley when I said "Now you've gone and practically given him this one. ;)"

Especially since he hadn't gone for this particular "carrot" before your comments, I think you probably helped. Sometimes people need a little encouragement, or prodding before they undertake the unfamiliar, in order to learn something new.

David

David Halliday | 02/13/12 | 22:48 PM

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