Gauge transformation and Occam's razor

[blechman]

You're missing a term!

\Delta\mathcal{L}=g^2 A_\mu A^\mu \phi^{*}\phi

To see where this term comes from, better to use the covariant derivative:

\mathcal{L}_{\rm scalar QED}=|D_\mu\phi|^2-m^2|\phi|^2-\frac{1}{4}F^2

where

D_\mu\phi=(\partial_\mu+igA_\mu)\phi

The cross terms give you the current, but then there is that 2 photon - 2 scalar interaction that is not there in QED, and is vital for maintaining gauge invariance, as you so correctly just showed!

BTW: in case you're wondering, this is how the Higgs mechanism works: when \phi gets a vacuum expectation value (goes to a constant if you don't know what that means), then this term generates a mass term for the photon! This is how the W and Z bosons get masses in particle physics. Also (if you're interested) it's how to derive the Meisner effect in (Landau-Ginzberg) superconductivity, where you can think of the photon as acquiring a mass inside the superconducting material. This sometimes goes under the name of "Anderson-Higgs" mechanism. That's the problem with spontaneous symmetry breaking: there are too many big names that contributed to it!

 

[jostpuur]

Distant memories from the particle physics course are coming back. I didn't understand anything on that course, and wasn't following very carefully anyway.

Does this mean that the conserving current is also

<br /> \textrm{Im}(\phi^* D^{\mu} \phi),<br />

and not

<br /> \textrm{Im}(\phi^*\partial^{\mu}\phi)?<br />

 

[blechman]

Distant memories from the particle physics course are coming back. I didn't understand anything on that course, and wasn't following very carefully anyway.

Does this mean that the conserving current is also

<br /> \textrm{Im}(\phi^* D^{\mu} \phi),<br />

and not

<br /> \textrm{Im}(\phi^*\partial^{\mu}\phi)?<br />

That's right, since your first current is not even gauge invariant!

 

[jostpuur]

blechman, you claimed I was doing something wrong, when I said that total charge conserves trivially when the system consists of discrete particles, even without gauge symmetries of fields. What do you think about the mass conservation discussion https://www.physicsforums.com/showthread.php?t=209929? Can you tell what is the symmetry behind the mass conservation, or is it the mass instead something that is trivially conserved with discrete particles, even without the corresponding symmetry?

If you say, that charge cannot be conserved without corresponding symmetry, but mass can, then I'm curious to hear explanations for this. After all the only difference between

<br /> \sum_{k=1}^N q_k<br />

and

<br /> \sum_{k=1}^N m_k<br />

is the used symbol!

 

[blechman]

sigh... that thread makes my head spin!

First of all, mass is not TRULY conserved - or else there would be no nuclear bombs! Rather, mass is a form of energy, and it is the energy that is conserved when time is homogeneous, according to Noether's theorem. This is manifest directly in the relativistic formulation.

Noether's theorem specifically states that if there is a symmetry, then there is a conserved charge. The symmetry that enforces electromagnetic charge conservation is the gauge symmetry of E&M. The symmetry that enforces ENERGY conservation is homogeneity of time.

If a particle is losing mass, then time is not homogeneous: as you move forward or backward in time, the particle has changed (lost or gained mass). So the symmetry is broken and so is the conservation law!

As to formulating this with a Lagrangian, I don't know what those people on that thread were thinking! If mass is now a DYNAMICAL quantity, you must include its dynamics! That means including some Lagrangian density that describes the flow of mass. It's exactly the same as E&M! If you have a static field, you can just write the Lagrangian as a particle moving in a background field. But if your field is allowed to evolve in time as well, then you must include you \int d^4x (-1/4)F_{\mu\nu}F^{\mu\nu} term as well to take that into account! I'll post this on that thread as well.

 

[jdg812]

If a particle is losing mass, then time is not homogeneous

I would rather say: then Lagrangian depends explicitly on 't'

PS
I cannot believe that a crazy particle losing its mass can change nature of time around itself...

 

[jostpuur]

First of all, mass is not TRULY conserved

I understand that the mass is not conserved in reality in all circumstances, but for example if a system is described by the Lagrange's function

<br /> L(x_1,\ldots x_N,\dot{x}_1,\ldots \dot{x}_N) = \sum_{k=1}^N \frac{1}{2}m_k\dot{x}_k^2 \;-\; V(x_1,\ldots x_N)<br />

then the mass

<br /> \sum_{k=1}^N m_k<br />

is conserved. Similarly, if I describe the system by a Lagrange's function

<br /> L=-\int d^3x\; \frac{1}{2}\big(\partial_{\mu} A_{\nu}(x)\big)\big(\partial^{\mu} A^{\nu}(x)\big) - \sum_{k=1}^N \Big(q_k A^0(x_k) - q_k v_k\cdot A(x_k) + m_k \sqrt{1 - |v_k|^2}\Big).<br />

then the total charge

<br /> \sum_{k=1}^N q_k<br />

is conserved. Now you are trying to tell me, that the mass conservation is trivial, but the charge conservation would not hold without proper gauge symmetry?

 

[blechman]

I would rather say: then Lagrangian depends explicitly on 't'

PS
I cannot believe that a crazy particle losing its mass can change nature of time around itself...

You have to put these statements in context: putting a particle in a box breaks spatial homogeneity since now there is a wall! Therefore the momentum is not conserved (the particle bounces off the wall!). This is what physicists mean when they say things like "space is not homogeneous." On the cosmic scale, it still is, but on the scale of the particle in a box, no, it is not!

Same goes for time translation: anytime there is friction, for example, time translation invariance is LOST in the problem, and thus energy is not conserved (there is heat). Of course, in the COSMIC sense, energy is still conserved thanks to the first law of thermo (heat is energy!), but in the context of the particle experiencing the friction, we are perfectly in our right to say that it is sensitive to time! And that is what those statements mean.

This is common language in the physics community. I'm not pulling it out of my butt!

 

[blechman]

Now you are trying to tell me, that the mass conservation is trivial, but the charge conservation would not hold without proper gauge symmetry?

I challenge you, jostpuur, to find where I claimed that "mass conservation is trivial"! There is *NOTHING* trivial about it! What I claimed is that ENERGY conservation follows from time-translation symmetry, and that mass conservation is a form of energy conservation. Ergo, mass conservation also follows from time-translation symmetry.

I also pointed out, both in that thread you referenced and this one, that in order to understand where that symmetry comes in, you must include the dynamics of the mass flow! If the mass flow has no dynamics, then for G-d's sake, of COURSE it's trivially conserved! I mean, if the mass is just static, then how can it NOT be conserved? The only time mass conservation becomes interesting is when there is dynamics.

Perhaps this is the source of all your woes: if you turn off Maxwell's equations, then charge is conserved, since charge is charge, and there's no where for it to go! But if you allow for the charge to have its own dynamics (Maxwell) then it is not immediately obvious that it should remain conserved. The gauge symmetry forces it on you.

Same with mass. If you model your mass as a fluid and try to describe it with Navier-Stokes, for example, then you would find that mass conservation would be imposed on you as a result of a symmetry (and violations in the symmetry would imply violations in the conservation law!). Of course, if you just say that we set the mass to 1kg and that mass has no dynamics, then isn't it rather straightforward to conclude that mass is "conserved"??

Only when a quantity's dynamics is included does the question become "interesting".

Maybe this answers your questions better than my previous attempts.

 

[jdg812]

I'm not pulling it out of my butt!

... in order to understand where that symmetry comes in, you must include the dynamics of the mass flow! If the mass flow has no dynamics, then for G-d's sake, of COURSE it's trivially conserved! I mean, if the mass is just static, then how can it NOT be conserved? The only time mass conservation becomes interesting is when there is dynamics.
...
Of course, if you just say that we set the mass to 1kg and that mass has no dynamics, then isn't it rather straightforward to conclude that mass is "conserved"??

Only when a quantity's dynamics is included does the question become "interesting".

In another Thread I was trying to say practically the same thing ...

 

[sweetser]

The spins of the particles

Hello:

The Lagrangian in the first post of this thread makes no sense in terms of the symmetries that the field strength tensor needs to represent.

In EM, like charges repel. That means the particles that mediate this force must be spin 1. The antisymmetric tensor F^{\mu \nu} changes signs if the indexes are switched, the sign that it could represent a spin 1 particle.

The same cannot be said for the asymmetric tensor \partial_{\mu} A_{\nu}. Instead, this tensor is reducible into two irreducible tensors, the antisymmetric tensor F^{\mu \nu} and a symmetric tensor:

\partial_{\mu} A_{\nu} = \frac{1}{2}(\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}) + \frac{1}{2}(\partial_{\mu} A_{\nu} + \partial_{\nu} A_{\mu})

For the second rank 2 tensor, changing the signs does not alter any signs. If this tensor happened to have a trace of zero, then it could represent a spin two particle that travels at the speed of light, or the graviton.

Looks like you are stumbling into the briar patch of unified field theory my friends. There is a very long discussion going on this topic on the Independent Research boards, "Unifying Gravity and EM", https://www.physicsforums.com/forumdisplay.php?f=146. One of the fun challenges was to spot spin 1 and spin 2 symmetry in the phase space of the current coupling term (hint: analyze everything, not just the transverse current).

Doug

 

[jostpuur]

Hello:

The Lagrangian in the first post of this thread makes no sense in terms of the symmetries that the field strength tensor needs to represent.

In EM, like charges repel. That means the particles that mediate this force must be spin 1. The antisymmetric tensor F^{\mu \nu} changes signs if the indexes are switched, the sign that it could represent a spin 1 particle.

The same cannot be said for the asymmetric tensor \partial_{\mu} A_{\nu}. Instead, this tensor is reducible into two irreducible tensors, the antisymmetric tensor F^{\mu \nu} and a symmetric tensor:

\partial_{\mu} A_{\nu} = \frac{1}{2}(\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}) + \frac{1}{2}(\partial_{\mu} A_{\nu} + \partial_{\nu} A_{\mu})

For the second rank 2 tensor, changing the signs does not alter any signs. If this tensor happened to have a trace of zero, then it could represent a spin two particle that travels at the speed of light, or the graviton.

Looks like you are stumbling into the briar patch of unified field theory my friends. There is a very long discussion going on this topic on the Independent Research boards, "Unifying Gravity and EM", https://www.physicsforums.com/forumdisplay.php?f=146. One of the fun challenges was to spot spin 1 and spin 2 symmetry in the phase space of the current coupling term (hint: analyze everything, not just the transverse current).

Doug

I don't understand why these symmetries are relevant for the repelling interaction of the like charges.

If I define the system by the Lagrangian

<br /> L=-\int d^3x\; \frac{1}{2}\big(\partial_{\mu} A_{\nu}(x)\big)\big(\partial^{\mu} A^{\nu}(x)\big) - \sum_{k=1}^N \Big(q_k A^0(x_k) - q_k v_k\cdot A(x_k) + m_k \sqrt{1 - |v_k|^2}\Big).<br />

the equations of motion, given by the action extremizing demand, are

<br /> \partial_{\mu}\partial^{\mu} A^{\nu}(x) = \sum_{k=1}^N q_k\delta^3(x-x_k)(1,v_k)^{\nu}<br />

<br /> F_k = -q_k(\nabla A^0 + \partial_0 A) + q_k v_k\times(\nabla\times A)<br />

So if there's two charges at locations x1 and x2, the stationary field produced by the charge 1 is given by

<br /> -\nabla^2 A^0(x) = q_1 \delta^3(x-x_1)<br />

with solution

<br /> A^0(x) = \frac{q_1}{4\pi}\frac{1}{|x-x_1|}<br />

and A^i(x)=0. The force that the particle 2 experiences is given by

<br /> F=-q_2\nabla A^0(x_2) = \frac{q_1q_2}{4\pi} \frac{x_2-x_1}{|x_2-x_1|^3}<br />

We say that the like charges repel, because they accelerate away from each others! I thought that the mediating particles belong to the phenomenology of particle collisions. Whatever... I don't understand how these symmetry properties are relevant now. If they are relevant for something, at least not for the fact that like charges repel?

 

[sweetser]

Particle exchnages and spin symmetries

Hello jostpuur:

I learned this from Brian Hatfield's introduction to "Feynman lectures on Gravitation". So I don't mess up, I will quote extensively:

In order to produce a static force and not just scattering, the emission or absorption of a single graviton by either particle must leave both particles in the same internal state. This rules out the possibility that the graviton carries have-integer spin (for example, related to the fact that it takes a rotation of 720 degrees to return a spin-1/2 wavefunction back to itself). Therefore the graviton must have integer spin. Next, to decide which integer spins are possible, we examine the two cases where particle 2 is identical to particle 1 and where particle 2 is the antiparticle of 1, so when charged, the two particles will carry the same and opposite charge, respectively. When the potential is computed in both cases, and the appropriate limits are taken, we find that when the exchanged particle carries odd integer spin, like charges repel and opposite charges attract, just as in the example of electrodynamics. On the other hand, when the exchanged partilce carries even integer spin, the potential is universally attractive (like charges and opposite charges attract). Hence, the spin of the graviton must be 0, 2, 4,...

The laws of physics are consistent. This means I am not saying your derivation of the force law is wrong in any way (because your calculation is right!). The calculation Hatfield describes is different, dealing with quantum field theory. There he is taking into account issues like the exchange of particles, an issue that a classical force law does not touch. The spin of the particle for a force where like charges repel must be odd. Since gravity is a universally attractive force, the spin must be even. The field strength tensor for EM must have odd spin exclusively, which \partial_{\mu} A_{\nu} does not. Your classical argument will not persuade quantum field theorists!

Doug

 

[genneth]

Is all this a really simple issue which was mentioned, but then got lost in all the posts?

The proposed "simpler" Lagrangian is correct, but only if you have the gauge condition built into the variational principle via Lagrange multipliers. As already noted, this actually gives you back the conventional Lagrangian. So the problem has been a mis-application of maths --- we minimise action *with the constraints*. Doing otherwise is simply not solving the same problem.

 

[sweetser]

Hello Genneth:

I think it is valuable to have multiple views on why a theory such as the Maxwell equations is self-consistent. Looking at this issue as an action with the a gauge constraint makes sense to me. I liked your comment that "Doing otherwise is simply not solving the same problem". The question becomes what problem is that, since it looks darn similar? It would have to be a force that - based on the symmetry arguments above - has 'like and opposite charges attract.' That is why I bet my research dollar that \nabla_{\mu} A_{\nu} (not \partial_{\mu} A_{\nu}) might be the basis of a unified field theory for gravity and EM.

Doug