# Non symmetry transformation of lagrangian

Gold Member

## Homework Statement

Show that if a transformation ##\Phi \rightarrow \Phi + \alpha \partial \Phi/ \partial \alpha## is not a symmetry of the Lagrangian, then the Noether current is no longer conserved, but rather ##\partial_{\mu}J^{\mu} = \partial L/ \partial \alpha##. Use this result to show that for a massive Dirac fermion the conservation of the chiral current is softly broken by the mass term: ##\partial_{\mu} j^{5,\mu} = -2m\bar{\psi} \gamma^5 \psi##

## Homework Equations

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$$\delta L = \frac{\partial L}{\partial \Phi} \delta \Phi + \frac{\partial L}{\partial(\partial_{\mu} \Phi)} \delta (\partial_{\mu}\Phi)$$

## The Attempt at a Solution

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I understand ##\Phi \rightarrow \Phi + \alpha \partial \Phi/ \partial \alpha## to not be a symmetry of the Lagrangian to mean that when this transformation is imposed on the fields the lagrangian is not invariant.

Can rewrite the equation in relevant equations as $$\frac{\partial L}{\partial \Phi} \delta \Phi + \frac{\partial L}{\partial(\partial_{\mu} \Phi)} \partial_{\mu} (\delta \Phi) = \alpha \left( \frac{\partial L}{\partial \Phi} + \frac{\partial L}{\partial( \partial_{\mu}\Phi)} \partial_{\mu} \right) \frac{\partial \Phi}{\partial \alpha} = \alpha \left( \frac{\partial L}{\partial \alpha} + \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \partial_{\mu} \frac{\partial \Phi}{\partial \alpha}\right)$$

The question doesn't state whether if the piece of the lagrangian after the transformation not making it invariant is a total derivative. If it was, then I could use the equations of motion since the action wouldn't change.

Many thanks for any tips to proceed!

Hi CAF,
I'm not too sure but how about considering a variation of the action resulting from non vanishing endpoint variations while the equations of motion hold? Like:
$$\delta S=\int d^4x\left[\alpha\left(\frac{\partial L}{\partial \Phi}-\partial_{\mu}\frac{\partial L}{\partial(\partial_{\mu} \Phi)}\right)\frac{\partial \Phi}{\partial \alpha} +\alpha\partial_{\mu}\left(\frac{\partial L}{\partial(\partial_{\mu} \Phi)}\frac{\partial \Phi}{\partial \alpha}\right)\right]$$
with
$$J^{\mu}=\left(\frac{\partial L}{\partial(\partial_{\mu} \Phi)}\frac{\partial \Phi}{\partial \alpha}\right)$$
Then if we are looking for non-conserved currents along solution trajectories, we keep only the boundary term:
$$\delta S = \int d^4x \,\alpha \,\partial_{\mu}\left(\frac{\partial L}{\partial(\partial_{\mu} \Phi)}\frac{\partial \Phi}{\partial \alpha}\right)=\int d^4x \,\alpha \,\partial_{\mu}J^{\mu}$$
The Lagrangian is
$$L=\partial_{\mu}S=\alpha\partial_{\mu}J^{\mu}$$
and so
$$\partial_{\mu}J^{\mu}= \frac{\partial L}{\partial\alpha}$$

How does that sound?

Gold Member
Hi muscaria,
I'm not too sure but how about considering a variation of the action resulting from non vanishing endpoint variations while the equations of motion hold? Like:
$$\delta S=\int d^4x\left[\alpha\left(\frac{\partial L}{\partial \Phi}-\partial_{\mu}\frac{\partial L}{\partial(\partial_{\mu} \Phi)}\right)\frac{\partial \Phi}{\partial \alpha} +\alpha\partial_{\mu}\left(\frac{\partial L}{\partial(\partial_{\mu} \Phi)}\frac{\partial \Phi}{\partial \alpha}\right)\right]$$
with
$$J^{\mu}=\left(\frac{\partial L}{\partial(\partial_{\mu} \Phi)}\frac{\partial \Phi}{\partial \alpha}\right)$$
Then if we are looking for non-conserved currents along solution trajectories, we keep only the boundary term:
$$\delta S = \int d^4x \,\alpha \,\partial_{\mu}\left(\frac{\partial L}{\partial(\partial_{\mu} \Phi)}\frac{\partial \Phi}{\partial \alpha}\right)=\int d^4x \,\alpha \,\partial_{\mu}J^{\mu}$$
Hmm, I guess this is the part I don't understand. It is said in the question that the lagrangian is not a symmetry under the field transformation, so this meant to me that the lagrangian changes in an arbitrary fashion so as to not permit the use of the classical equations of motion. Of course, if the transformation caused the lagrangian to change by a total derivative or a trivial rescaling or additive constant then it would be ok but this is not said.

I actually found an answer I think. In the last equality in my OP there, the last line should also include a piece due to the fact the derivatives would also depend on alpha. So $$\frac{\partial L}{\partial \alpha} = \frac{\partial L}{\partial \Phi}\frac{\partial \Phi}{\partial \alpha} + \frac{\partial L}{\partial (\partial_{\mu} \Phi)}\frac{\partial (\partial_{\mu}\Phi)}{\partial \alpha}.$$
Input into second last equality in OP and rearrange gives ##\delta L = \partial L/ \partial \alpha##. If transformation had been symmetry then ##\delta L = 0## and thus can infer ##\partial_{\mu}J^{\mu} = 0##. Here have ##\partial_{\mu}J^{\mu} = \partial L/\partial \alpha##.

What do you think about this? Incidentally, the solution the professor proposes does use the equations of motion so what you write maybe does make sense and I'm misinterpreting something, although I also asked a tutor for a second opinion and he agreed with me.

Thanks!

Hey CAF,
Sorry for the late reply, I've been a wee bit busy of late!
Hmm, I guess this is the part I don't understand. It is said in the question that the lagrangian is not a symmetry under the field transformation, so this meant to me that the lagrangian changes in an arbitrary fashion so as to not permit the use of the classical equations of motion. Of course, if the transformation caused the lagrangian to change by a total derivative or a trivial rescaling or additive constant then it would be ok but this is not said.
I would have thought that a change in the form of the Lagrangian, still gives rise to certain equations of motion, just not the the same class of solutions. Continuous symmetry transformations just mean that the generator of the transformation commutes with the time evolution of the system. Taking the example of a central potential which has the feature of conservation of angular momentum and whose solutions are some form of elliptic function, time evolving would carry the particle around the ellipse and a rotation transformation rotates the ellipse. Operating in inverse order, we can rotate about the origin so as to change the initial position and then time evolve and we would end up at the same point as the previous case.

A symmetry transformation is therefore a special type of transformation which "flows" in a manner which doesn't change the form of the equations of motion. However, this doesn't seem to imply that there are no solutions to the equations of motion if our transformation is not a symmetry, just that the form of the solutions will change.

That was where I was coming from. I had in mind that although there is no conserved steady current flow (we'll have diverging flow), we can express this non-steadiness in the form of some kind of injection of current at the boundary. Given that the variation of the action $$\delta S =\int d^4x\left(\frac{\partial L}{\partial\phi}\delta\phi-\partial_{\mu}\left(\frac{\partial L}{\partial(\partial_{\mu}\phi)}\right)\delta\phi +\partial_{\mu}\left(\frac{\partial L}{\partial(\partial_{\mu}\phi)}\delta\phi\right)\right)$$
is at an extremum infinitesimally throughout the motion, we can consider non steady current as arising continuously from the boundary term when the equations of motion hold, given that we want the current along solution trajectories, albeit non-conservative. I like this way of doing things as it gives a bit more of a feel of what's going on.
Incidentally, the solution the professor proposes does use the equations of motion
I guess this is maybe where he was coming from also? I have a feeling this may be connected nicely to infinitesimal canonical transformations ;).
Input into second last equality in OP and rearrange gives δL=∂L/∂α
I'm guessing you mean ##\delta L =\alpha\frac{\partial L}{\partial\alpha}##, right?
If transformation had been symmetry then ....Here have....
Do you not have to carry out an integral to be able to say the last statement? Effectively carrying a similar procedure to what I tried?
Anyway, let me know what you think. I haven't really studied this stuff properly before, so there may well be flaws in my reasoning! I enjoyed delving into it though, thanks for that :). Cheers and all the best.

Gold Member
Hi muscaria, thanks! Here are my thoughts,
I would have thought that a change in the form of the Lagrangian, still gives rise to certain equations of motion, just not the the same class of solutions.
Ok, so the argument is that given $$S = \int d^4 x \mathcal L \Rightarrow \delta S = \int d^4 x \delta \mathcal L = \int d^4 x \left(\frac{\partial \mathcal L}{\partial \Phi} d\Phi + \frac{\partial \mathcal L}{\partial (\partial_{\mu} \Phi)} \delta (\partial_{\mu}\Phi)\right).$$ Imposing ##\delta S = 0## gives rise to the Euler Lagrange equations, which are equations satisfied for a particular configuration of field solutions. So I see that these equations are derived assuming nothing about how ##\mathcal L## changes? We just give the generic change in ##\mathcal L## as a change due to the field configuration and its derivatives but say nothing about what this change actually is. Is that why you can still impose these equations of motion independent of whether the transformation was a symmetry transformation or not?

A symmetry transformation is therefore a special type of transformation which "flows" in a manner which doesn't change the form of the equations of motion.
And such a transformation is coined to be a symmetry transformation if it leaves the lagrangian invariant up to a total derivative?

However, this doesn't seem to imply that there are no solutions to the equations of motion if our transformation is not a symmetry, just that the form of the solutions will change.
The form of the solutions will change because under a non symmetry transformation, we will have that $$\partial_{\mu} \frac{\partial \mathcal L}{\partial (\partial_{\mu} \Phi)} - \frac{\partial \mathcal L}{\partial \Phi} = \partial_{\mu} \frac{\partial \mathcal L'}{\partial (\partial_{\mu} \Phi)} - \frac{\partial \mathcal L'}{\partial \Phi} + K(\Phi, \partial \Phi) = 0$$ So this extra piece at the end here changes the form of the solutions and is zero in the case of a symmetry transformation, ie leaves the form of the equations invariant?

I'm guessing you mean ##\delta L =\alpha\frac{\partial L}{\partial\alpha}##, right?
Yup
Do you not have to carry out an integral to be able to say the last statement? Effectively carrying a similar procedure to what I tried?
Yea I suppose I do. When you are left with just a total derivative of the quantity you called ##J_{\mu}## why doesn't ##\int d^4 x \partial_{\mu} J^{\mu} = 0## identically since it is just a surface term? Do we not put it to zero to reflect the fact this is a non conserved current we are talking about so leakage out of the volume is allowed, hence the flux through the surface is non zero and therefore the surface integral does not vanish?
Anyway, let me know what you think. I haven't really studied this stuff properly before, so there may well be flaws in my reasoning! I enjoyed delving into it though, thanks for that :). Cheers and all the best.
Thanks :) I hope my thoughts above make a bit of sense, I just tried to disentangle everything you wrote to see if I understood it

Hey CAF,
Just saw your post.. I'm getting ready for travel just now, so I'll get back to you with a proper reply tomorrow most probably and will edit this post. Had a quick read through your post, and yeah first impression seems like you disentangled it all nicely :p. Will be in touch! All the best,
Muscaria

Gold Member
Hey CAF,
Just saw your post.. I'm getting ready for travel just now, so I'll get back to you with a proper reply tomorrow most probably and will edit this post. Had a quick read through your post, and yeah first impression seems like you disentangled it all nicely :p. Will be in touch! All the best,
Muscaria
Ok thanks :)

Good morning from the French alps :),

Just to make sure, the E-L equations of motion are satisfied for any "reasonable" initial field configuration, in the same way that the harmonic oscillator problem gives rise to solutions for any initial position and velocity of a particle. It is the form of the Lagrangian which dictates the form of the solutions, for the HO the form of the Lagrangian has a potential term which is quadratic in position.. If we changed from quadratic to quartic, we would still have solutions to the equations of motion but the form of these solutions would change so that it would not be a symmetry transformation: e.g. for the 1-d HO we would get phase space solutions corresponding to ellipses and if we changed to a quartic potential, the ellipse would be flattened and become more oval and in the limit where our potential was ##lim_{n\rightarrow\infty}V = \alpha x^{2n}## our solution trajectory in phase space becomes a rectangle.. The initial ellipse gets more and more flattened as we go higher up in even powers of position and tends to a rectangle in the limit ##n\rightarrow\infty##.

All this to say that changing the form of the Lagrangian doesn't dictate whether or not the equations of motion will be satisfied, you can always find trajectories that will solve them (in these simple instances), it in fact dictates the form these solution trajectories must take, if that makes sense?
In the case of a field it's essentially the same story, by changing the form of the Lagrangian, you are generally changing the form of the interaction between your field degrees of freedom, and the solutions which describe how the field values change over time will as a consequence, also change in form. But we still have solution trajectories which solve the equations of motion! A symmetry transformation is then a change in the Lagrangian which does not change the form of solution trajectories, corresponding to very special types of transformations.

So when you say:
We just give the generic change in L as a change due to the field configuration and its derivatives but say nothing about what this change actually is. Is that why you can still impose these equations of motion independent of whether the transformation was a symmetry transformation or not?
You will in fact always have equations of motion no matter what transformation of L you carry out (within reason :p), just that the form of the solutions to these equations will change generally, unless it is a symmetry transformation. Again thinking of planetary motion for example: the central potential means that if I carry out a rotation I will still observe an ellipse as solution trajectory which has the same form as if I hadn't rotated, both solutions are just rotated with respect to each other. If however instead of rotating, I translate the planet away from the sun, I would still observe an ellipse being traced out over time, but the form of the ellipse would be different to the form of the ellipse without this translation transformation, meaning this translation is not a symmetry transformation but still has a trajectory which solves the equations of motion. Does this make sense? Sorry if you had already got all this and if I'm being overly garrulous, but from what you were saying it seemed like there may be confusion around this point and a lengthy explanation with examples may have helped clear this. Anyway let me know if this helps and we can discuss the other points after ;)

CAF123
Just read through the rest of your post and it seems good, to me anyway!
And such a transformation is coined to be a symmetry transformation if it leaves the lagrangian invariant up to a total derivative?
Yep.
The form of the solutions will change because under a non symmetry transformation, we will have that $$\partial_{\mu} \frac{\partial \mathcal L}{\partial (\partial_{\mu} \Phi)} - \frac{\partial \mathcal L}{\partial \Phi} = \partial_{\mu} \frac{\partial \mathcal L'}{\partial (\partial_{\mu} \Phi)} - \frac{\partial \mathcal L'}{\partial \Phi} + K(\Phi, \partial \Phi) = 0$$ So this extra piece at the end here changes the form of the solutions and is zero in the case of a symmetry transformation, ie leaves the form of the equations invariant?
That seems good to me.
Yea I suppose I do. When you are left with just a total derivative of the quantity you called ##J_{\mu}## why doesn't ##\int d^4 x \partial_{\mu} J^{\mu} = 0## identically since it is just a surface term? Do we not put it to zero to reflect the fact this is a non conserved current we are talking about so leakage out of the volume is allowed, hence the flux through the surface is non zero and therefore the surface integral does not vanish?
Yeah precisely, that's what I meant by current injection. A non-steady current (diverging current) can be obtained by injecting current at the boundary points and given that the action is extremised infinitesimally throughout the motion, we can inject the appropriate current (leakage to or from) where needed throughout the motion.
As a side note, the use of these boundary terms became apparent in analytical mechanics, where allowing a variation of the generalised coordinates at the boundary (in the form of a endpoint variation corresponding to the actual motion undergone by the system: i.e. ##\delta q =\dot{q} dt \equiv dq##), enables one to express the action as a function (not functional) along solution trajectories, i.e. ##S(q,t)##. One then readily sees that the canonical momenta correspond to ##p_i = \frac{\partial S}{\partial q_i}## or the slope of the action w/r to ##q_i##, describing the analogue of the current we've been talking about. It also paved the way for understanding infinitesimal canonical transformations and Hamilton-Jacobi theory. If you're interested in all this I would suggest Lanczos's textbook: "The variational principles of mechanics", specifically chapter V section 3, chapter VI section 8 and chapter VIII. Also a terrific place for getting science textbooks to check whether a book is good or not before purchasing is http://gen.lib.rus.ec/ ! All the best,
Muscaria

CAF123