(Physics) Is every field theory a gauge theory?

[CptBork]

But the fact that you can equate this connection with a physical particle is indeed amazing, and the more you study these things, the deeper a respect you gain for people like Yang and Mills who were thinking about these things in the 1950's.

Indeed, it really is quite a stroke of genius... Define a symmetry on your Lagrangian, find the resulting conserved current, and then there is pretty much only one simple way you can couple a gauge boson to that current so as to preserve the original symmetry. In this manner, 3 simple symmetries are all you need to describe all the nuclear forces. I'll need to look into this stuff again at some point though, as I remember when I learned this recipe that it seemed like there were potentially other, more messy mathematical ways in which such current couplings could be achieved without breaking gauge invariance.

Does Yang Mills really have to be the scheme by which all possible forces in nature are described? Is this the dominant approach theoretical physicists use when trying to describe quantum gravity?

 

[AlphaNumeric]

In field theories kinetic terms are usually completely symmetric, so if we make (infinitesimal) transformations which don't change potential (like changing phase), situation remains the same - they are gauge invariant.

What do you mean 'symmetric'? Besides, its not true for local symmetries where the phase shift is space-time location dependent.

For instance, consider the free scalar field Lagrangian $$\mathcal{L} = \partial_{\mu}\phi\partial^{\mu}\phi + m^{2}\phi^{2}$$ under the gauge transformation $$\phi \to e^{-i\alpha(x)}\phi = \phi - i\alpha\phi + \ldots$$. The mass term is invariant but the kinetic terms pick up contributions involving $$\partial_{\mu}\alpha$$. Without too much work it follows that in order for the gauge symmetry to exist you need to include a gauge field $$A_{\mu}$$, which also transforms under the gauge choice in a particular way. Furthermore you'll get more fields when you gauge fix.

It just so happens I wrote up precisely this method into a short pdf a few years ago, its here. It considers a massive field but you get the massless result by just setting m=0 (ie you're only considering the kinetic term) and its clear from the final expression that there's a ton of stuff you get from gauging the kinetic term (ie all those which don't involve the m parameter). Ben mentioned recently how gauge symmetry locks the photon mass to be zero but it goes further than that, you cannot have a phase shift gauge symmetry without a photon. It is the photon which keeps everything in check under the gauge transformation.

So higher dimensional filed theories usually have some gauge invariance

Gauge symmetry is something you put in by hand generally. The Lagrangian I just mentioned is perfectly valid as a Lagrangian, it just doesn't have a gauge symmetry. The kinetic terms are not automatically gauge invariant individually, they might just contribute things which cancel out. This is why the A in the pdf also transforms under the gauge transformation, its needed to counter things from the scalar field kinetic term.

That's the general way you get Lagrangians for particular field theories, you start with the simplest possible case of the fields you're interested in (ie just a kinetic term and perhaps a mass term) and then you have it transform under the symmetries you want it to. You then keep adding fields in particular ways until they cancel one another's changes out. It means you can consider as general as case as possible in a controlled manner, as symmetries are often very restricting in what they allow you to consider. In higher dimensional theories there might be too much freedom in your construction if you don't impose enough symmetries. It isn't because there's something special about kinetic terms in higher dimensions, its because in high dimensions its especially important to make things easier for yourself. For instance the entire motivation for looking at Calabi-Yaus in string theory is that they are equivalent to imposing N=1 or N=2 supersymmetry (which isn't a gauge symmetry though). An example from this thread would be how if you say $$\phi \to - \phi$$ is a symmetry then you can ignore terms of the form $$\phi^{3}$$ or anything else of the form $$O(\phi^{2n+1})$$.

 

[QuarkHead]

OK, I have abandoned the task of responding item-by-item to all the responses above. Instead I am going on a ramble, which I think goes part way the addressing them.

First, we have a connection $$A$$ which is a 1-form (Ben called it a vector, which it is in a certain sense, but I like to make the distinction).

This is by definition $$\mathfrak{G}$$-valued.

From this beastie we take the Maxwell tensor $$F_{\nu\mu}$$ and make the 2-form $$F:= F_{\nu\mu}\,\,dx^{\nu}\wedge dx^{\mu}$$ (since this tensor is skew-symmetric, I should perhaps have a factor of one-half in there?) such that

$$F = dA+A \wedge A$$, the curvature of the connection, which becomes $$D \equiv dF +F \wedge A + A\wedge F = dF +[F,A] = 0$$. This is, of course, the Bianchi identity, and thus $$[D_{\nu},D_{\mu}]A = [F_{\nu\mu},A] =0$$.

Now in U(1) all brackets vanish, so that in EM this identity reads $$dF =0$$ which is essentially Faraday's law and the absence of magnetic monopoles.

Now, as Ben and Alpha point out, any sensible physical theory must allow for "charge" in some loose sense, so I follow Ben in writing the covariant derivative as

$$D\psi = d \psi +ieA \psi$$, which from the above implies that $$e = 0$$. So we are dealing with "system" that has no mass and no charge

This is all fine and well, but this piece of extravagant and quite unnecessary showing off doesn't completely satisfy me.

So we have such a "system" and the fields $$A_{\rho}$$ are potential fields related by a gauge transformation $$A_{\rho} \to A_{\rho'} := A^g_{\rho}$$ where here $$ g \in U(1)$$.

I still don't see how this gives me a boson. Let me try this as an heuristic.

Suppose that $$P$$ is a principal bundle over the manifold $$M$$, with structure group $$G$$ and projections $$\pi: P \to M,\,\,\,\pi(p)= m\,\, \forall p \in P$$ and some $$m \in M$$. This is "highly surjective", in that $$\pi^{-1}(m) \sim G$$ which is called a fibre over $$M$$ given these definitions.

Now it is easily shown (I think) that $$P$$ is a manifold, and as such is entitled to its own tangent spaces, call them $$T_pP \,\, \forall p \in P$$. Now decompose $$T_pP$$ as $$T_pP = H_pP \oplus V_pP$$, these being, respectively, the horizontal and vertical subspaces. Note that $$V_pP$$ "runs parallel" to (is tangent to) each fibre over $$m$$.

OK. Since each fibre $$\pi^{-1}(m) \sim G$$, and since $$\mathfrak{G} \sim T_eG$$ (e is obviously the identity) then by transitivity $$ V_pP \sim \mathfrak{G}$$ and $$A(m) \in \mathfrak{G}$$, which is full circle.

I don't know where to go from here.

Sorry for my long-winded thickness!

 

[AlphaNumeric]

I still don't see how this gives me a boson. Let me try this as an heuristic.

It's nothing to do with its Lie algebra valued-ness but its space-time structure, ie the fact its a vector. Since its a connection you have $$D_{\mu} = \partial_{\mu} + ieA_{\mu}$$. If something is a vector field then by the spin statistics theorem its a boson.

 

[Green Destiny]

This is a good thread.

Now and again, i am learning something new, just by seeing it from anothers perspective too.

 

[Green Destiny]

It's nothing to do with its Lie algebra valued-ness but its space-time structure, ie the fact its a vector. Since its a connection you have $$D_{\mu} = \partial_{\mu} + ieA_{\mu}$$. If something is a vector field then by the spin statistics theorem its a boson.

See, this is where is breaks down for me. I haven't studied Lie Algebra.

Alphanumeric, do you know any good links?

 

[BenTheMan]

$$D\psi = d \psi +ieA \psi$$, which from the above implies that $$e = 0$$. So we are dealing with "system" that has no mass and no charge

Hmm. I with you up to here.

Why does e = 0 necessarily? If psi is a spinor (i.e. an electron), this shouldn't be true.

Regarding how we know A_{\mu} must be a boson: the indices here are Lorentz indices, which tell you about the object's transformation properties under the Lorentz group. The Lorentz group tells us the statistics that the particle must obey by the spin statistics theorem, as AN pointed out.

GD: Check here under ``particle physics'': http://www.sciforums.com/showthread.php?t=73777 The book by Cahn (Semi-Simple Lie Algebras and their Representations) is good. You should also be able to find ``Group Theory for Unified Model Building'' online by Slansky.

 

[Jarek Duda]

AlphaNumeric, to understand the real symmetry behind field theories, let's look at something more fundamental than Lagrangian - let's look at energy density (called Hamiltonian in QM) - it's what is really conserved. It usually looks like $$\sum_{i=0}^3 |\partial_i\phi|^2 + V(\phi)$$
where $$\phi$$ can be generally scalar, vector, tensor, ..., functional (QFT).
We see that it's completely 4D symmetric - doesn't emphasize any coordinate system and so any time direction - like '4D jello': each point is in equilibrium with its 4D neighborhood, such that tensions in all 4D directions are optimized.
It's completely deterministic picture (Einstein's block universe) - we move accordingly to entropy gradient in already created spacetime (caused probably by relatively low entropy after Big Bang).
We break this symmetry by choosing direction along which we want to calculate evolution (time) and transform it to Lagrangian density for given time direction - mathematical trick called 'Wick rotation' which make time 'imaginary'.

So these theories are really full of symmetries: 4D rotations (mainly boosts), translations ... or if potential don't mind: rotating all phases in the same way ...
But changing phases in locally varying way, changes their difference and so the whole theory - to allow for them we introduce new terms (like A).
By assuming that: 'ok, there are phases in all points ... but Physics doesn't really care about them' - is nice mathematical trick to motivate EM field (of extremely complicated: $$ E\cdot E + B \cdot B$$ energy density...).

So do you ask if for any symmetries we would require, we can modify the original theory by some artificial terms (transforming with symmetries) to make them fulfilled?
I think you should be more specific about the question ...

 

[AlphaNumeric]

let's look at something more fundamental than Lagrangian - let's look at energy density (called Hamiltonian in QM) - it's what is really conserved.

The Lagrangian and Hamiltonians of a system are Legendre transformations of one another and if the Hamiltonian is time invariant so is the Lagrangian. Which formalism is used is often a matter of preference or down to computational issues. The overwhelming majority of functional variation methods in particle physics are done on the Lagrangian of a theory.

where can be generally scalar, vector, tensor, ..., functional (QFT).
We see that it's completely 4D symmetric - doesn't emphasize any coordinate system and so any time direction - like '4D jello':

Its only 'symmetric' (as you define it, not emphasising any coordinate system) because you've suppressed the space-time indices. If $$\phi$$ were a vector field then the full notation would be $$\phi_{\mu}$$ but you haven't included that. You've written down what is formally a scalar field dependent expression, which by definition has no space-time indices, and then said "Ta-da, this tensor expression is symmetric!".

What you actually meant to say when you said '4 d symmetric' is that it doesn't transform under the Lorentz group, aka its a Lorentz scalar. A Lagrangian or Hamiltonian must be a Lorentz scalar else the theory wouldn't be Lorentz invariant and thus would violate special relativity. You haven't done that explicitly for a general setup, you've just written down a pretty much trivial case.

If $$\phi$$ were a vector field then you'd have to combine it with another vector, as in QED where $$A_{\mu}$$ is combined with $$\gamma^{\mu}$$ and the resultant matrix expression then turned into a function through a spinor inner product with the lepton field.

Your lack of familiarity with terminology and being unaware of how Lagrangians and Hamiltonians relate (and their equations of motion) has my spider sense tingling.... It warns me when cranks are nearby....

each point is in equilibrium with its 4D neighborhood, such that tensions in all 4D directions are optimized.

Equilibrium has nothing to do with it, its about how expressions transform under the Lorentz group. No indices means Lorentz scalar. Lorentz scalar Lagrangian (or equivalently Hamiltonian) means Lorentz invariant theory. Tensions has nothing to do with it. And 'optimised' seems like a buzzword you've thrown in for no reason.

It's completely deterministic picture (Einstein's block universe) - we move accordingly to entropy gradient in already created spacetime (caused probably by relatively low entropy after Big Bang).

Now you're into just making up your own stuff.

We break this symmetry by choosing direction along which we want to calculate evolution (time) and transform it to Lagrangian density for given time direction - mathematical trick called 'Wick rotation' which make time 'imaginary'.

A Wick rotation has nothing to do with turning a Hamiltonian into a Lagrangian. And there's no additional symmetry breaking by considering time separately, as the Lorentz group already considers it differently, hence why its SO(3,1). In fact, by doing a Wick rotation you create the symmetry as you make the symmetry group SO(4) with all 4 dimensions now treated precisely the same.

I suggest you go back to the book or website you are trying to parrot and read it again.

So do you ask if for any symmetries we would require, we can modify the original theory by some artificial terms (transforming with symmetries) to make them fulfilled?

No, I asked you to define a terminology you used in an unclear way and having now read your reply it is clear you were just throwing out terms and phrases for things you've read about but obviously not understood.

I think you should be more specific about the question ...

And I think you should understand something properly before trying to incorrectly correct someone on the subject. Particularly when that someone is not unfamiliar with the topic at hand.

 

[Jarek Duda]

I'm widely educated, but mathematician at heart - formalism is one thing, but what is really essential for me is to deeply understand the basics to create coherent picture.
So before going further, I see we should focus on the basics first: classical field theories.
Understanding the difference between Lagrangian and Hamiltonian is not only about notation, but it's much deeper - it wrongly leads to belief that 'time is imaginary'... it's why I've written energy density(Hamiltonian) using not suggesting notation, but pure math:
Kinetic term of energy density usually is: $$\sum_{i=0}^{3}|\partial_i\phi|^2$$ for scalar field, $$\sum_j\sum_{i=0}^{3}|\partial_i v_j|^2$$ for vector field, $$\sum_{jk}\sum_{i=0}^{3}|\partial_i M_{jk}|^2$$ for tensor field (like skyrmions)...
It's simple mathematical picture with full 4D symmetry: if we transform everything by 4D orthogonal matrix, energy density remains the same.
Now if we choose some direction along which we want to calculate evolution - like standard $$e_0$$ (but we could also choose any other versor) - for this chosen direction we transform energy density into Lagrangian density, what results in that it looks that this chosen time direction is imaginary now.

This energy density literally says about minimizing tension in all 4D direction, like in static '4D jello'.
Anyway, solving Lagrangian mechanics is about finding solution which optimize action (integral), like trajectory or valuation of the field in a part or the whole spacetime - about living in static 4D world in which we move along time direction.
Having such solution, we can introduce effective parameters to this picture, like thermodynamical ones: for example average over energy in a ball with center in given point gives us thermodynamical energy density in this point (smoothen original energy density) ... and so on we introduce the whole thermodynamics onto this picture - also entropy density for each point of spacetime.
We know that entropy grows while our time is passing - that mathematically means that our 'arrow of time' generally agrees with entropy 4D gradient.
Trying to extrapolate it back in time, entropy is lowering and finally we get into Big Bang - so it should be entropy minimum ... and could create its gradient - be the reason (?)...
From this direction runs our natural reason-result relation chain: though Earth creation, evolution, embryogenesis, our development - so we can only think this way ... but accepting Lagrangian theories as describing nature (GRT, EM, Klein-Gordon, QFT) says that there is some time (CPT) symmetry - not only past, but also the future is determined, what from our perspective is equivalent with that 'it's already there' in spacetime (what is clearly seen in general relativity).

Let's find a common language in these basics first before going to more abstract and hidden behind notation - where exactly do you disagree with this simple mathematical picture of our reality?

 

[AlphaNumeric]

I'm widely educated, but mathematician at heart

Be specific.

So before going further, I see we should focus on the basics first: classical field theories.

Its been irrelevant to anything we've discussed so far.

Understanding the difference between Lagrangian and Hamiltonian is not only about notation, but it's much deeper - it wrongly leads to belief that 'time is imaginary'

I said in my last post that Wick rotations have nothing to do with the relationship between Lagrangians and Hamiltonians. You are wrong. And the whole imaginary time thing is not what physicists think, its what laypersons who don't understand concepts like contour integrals and analytic continuation think. It is simply a tool to simply the algebra in the same way a change of variables simplifies an integral.

it's why I've written energy density(Hamiltonian) using not suggesting notation, but pure math:

When it is written in anything but 'pure math'. What's the difference between 'pure math' and the usual way? You sound like a recently banned member in how you don't grasp what the role of mathematics in physics is and your misuse of basic terminology. Or maybe its that hacks all sound the same.

It's simple mathematical picture with full 4D symmetry: if we transform everything by 4D orthogonal matrix, energy density remains the same.

That's after you've done the Wick rotation, not before. Before you do the Wick rotation you use Lorentz transformations. Only after a Wick rotation do they become 4d orthogonal matrices. You've gotten your information in the wrong order, a sign you don't really understand it.

Now if we choose some direction along which we want to calculate evolution - like standard (but we could also choose any other versor) - for this chosen direction we transform energy density into Lagrangian density, what results in that it looks that this chosen time direction is imaginary now.

There's no change in the kinetic term when you transform from Hamiltonians to Lagrangians or vice versa. Before a Wick rotation the time direction is already highlighted due to the metric signature and thus the Legendre transformation linking Hamiltonians and Lagrangians is clearly definable. Even if you've already done the Wick rotation and you have to pick a direction it doesn't give you anything different from doing the Legendre transformation then a Wick rotation. And if you're wanting to consider the $$t \to -i\tau$$ transformation you can do it for either Lagrangians or Hamiltonians.

You seem not to have any experience with this stuff, you're making elementary mistakes.

Anyway, solving Lagrangian mechanics is about finding solution which optimize action (integral), like trajectory or valuation of the field in a part or the whole spacetime - about living in static 4D world in which we move along time direction.

No, it finds extrema of the action, which is the Lagrangian density integrated over space-time. This is even more basic than quantum field theory.

From this direction runs our natural reason-result relation chain: though Earth creation, evolution, embryogenesis, our development - so we can only think this way ... but accepting Lagrangian theories as describing nature (GRT, EM, Klein-Gordon, QFT) says that there is some time (CPT) symmetry - not only past, but also the future is determined, what from our perspective is equivalent with that 'it's already there' in spacetime (what is clearly seen in general relativity).

Now you're just making up your own conclusions based on nothing but your flawed grasp of the physics. CPT symmetry doesn't mean that the future is determined, nothing of the sort. GR is, in principle, a deterministic model but quantum field theory kills that.

Let's find a common language in these basics first before going to more abstract and hidden behind notation - where exactly do you disagree with this simple mathematical picture of our reality?

Yes, you should actually learn what the models say and make an honest attempt to understand them, not just parrot back Wikipedia pages you don't understand. You've got to be pretty daft to try to lie and argue about bits of quantum field theory you don't understand in a thread whose subject matter attracts people who actually know some quantum field theory. Do you hate yourself or are you drinking your own koolaid?

 

[Jarek Duda]

Ok - piece by piece ... (my education: see tcs.uj.edu.pl/jarek )
Kinetic term in energy density is $$\sum_{i=0}^3 |\partial_i \phi|^2 $$, in Lagrangian density it transforms to $$|\partial_0\phi|^2-\sum_{i=1}^3 |\partial_i \phi|^2 $$
The same can be mathematically achieved for example by multiplying spatial coordinates by imaginary unit - so called 'Wick rotation' - in notation it's achieved by using pseudometric like diag(1,-1,-1,-1).
Am I wrong?

About 'quantum theory killing determinism' - but aren't Klein-Gordon and Quantum Field Theories Lagrangian mechanics?
Where in Lagrangian mechanics is a place for indeterminism????

 

[CptBork]

You still have to "quantize" the Lagrangian by introducing operators to represent the fields and their derivatives w.r.t. time.

 

[Jarek Duda]

'Quantization' procedure is mainly about writing evolution in eigenbase of evolution differential operator - in such coordinates it can be seen just as 'superposition of rotations' - like for normal modes of coupled pendulums.
We often use plane waves as these eigenfunctions - for linear filed theories (like EM waves) or after linearization (waves on water, gravitational waves).
But generally physics (potential term) isn't linear and so eigenfunctions are usually more complicated, like solitons - localized solutions ... as in classical-corpuscular picture of particles - but there still can be involved some periodic process like rotation of coordinate of this eigenfuction, giving them also 'wave' nature required for interference.

About indeterminism - please look at en.wikipedia.org/wiki/Wheeler%27s_delayed_choice_experiment

 

[QuarkHead]

Although I started this thread, I obviously have no propriety rights over it, so to speak. But lemme make a couple of general points here.

Jarek, well all here really.

If I (or anyone else, for that matter) asks a genuine question, moderately well-founded, essentially by definition they have no way to tell whether or not they are being fed soup by other members. In short, when responding to such questions please be sure you are sufficiently qualified to provide a mainstream answer. Otherwise this is a fraud on the questioner.

I, as should we all, feel fortunate that we have members who, between them, are able to answer almost any sincere question posed here and moreover are not afraid to be blunt when they they see soup on the menu. This may not always be the case however

Second general point: beware asking questions, however well-motivated, when you are not equipped to understand the answers.

It seems that is the case with me in his thread. In short I am totally lost. Shit!

I am sorry if I have wasted anyone's time, so in penance I just ordered a copy of Peskin & Schroeder, a snip at 30 quid from Albris.

 

[Jarek Duda]

I apology - I gave short concrete mathematical answer ... it occurred to be not so simple ... and so it ran away ... but such discussions generally have tendency to go offtopic ...
Anyway, if you really are mathematician, please don't get caught by suggesting formalism of physics, by mystical belief that e.g. there has to be some indeterminism there ... but just see pure math below, get its deep understanding (Evans, Rudin, Arnold are better) ... and the picture occurs really simple ...
Cheers, Jarek

 

[AlphaNumeric]

Ok - piece by piece ... (my education: see tcs.uj.edu.pl/jarek )
Kinetic term in energy density is $$\sum_{i=0}^3 |\partial_i \phi|^2 $$, in Lagrangian density it transforms to $$|\partial_0\phi|^2-\sum_{i=1}^3 |\partial_i \phi|^2 $$
The same can be mathematically achieved for example by multiplying spatial coordinates by imaginary unit - so called 'Wick rotation' - in notation it's achieved by using pseudometric like diag(1,-1,-1,-1).
Am I wrong?

To prove I'm not above saying so, I now see what you are referring to and I got it wrong in my comments in terms of signs. However, this doesn't mean the Hamiltonian is SO(4) invariant, it simply does not have manifest SO(3,1) Lorentz invariance and that's why people prefer to work with the Lagrangian. Energy conservation in the Hamiltonian is equivalent to time shift invariance in the Hamiltonian which is equivalent to the time shift invariance in the Lagrangian. You can get energy conservation directly from the Lagrangian. Furthermore you don't recover the difference by doing a Wick rotation as the potentials change sign. Else a Legendre transformation would only be a Wick rotation, which isn't the case. If it were Lattice QCD would start from a Hamiltonian formalism and not need to do a Euclideanisation.

About 'quantum theory killing determinism' - but aren't Klein-Gordon and Quantum Field Theories Lagrangian mechanics?
Where in Lagrangian mechanics is a place for indeterminism????

The quantities in the Lagrangian are fields, they are, in effect, the quantum field theory version of a wave function. The Schrodinger equation completely determines the time evolution of a wave function such that you can certainly tell what form it'll have in the future but that doesn't avoid the fact you still have to measure it and collapse it to a specific state. Same with quantum fields, you only end up getting probabilities and averages and differential cross sections. Differential cross sections tell you the probability a particular scattering process will have a particular outcome, while if you were bouncing tennis balls off one another Newtonian mechanics would not give you a probability distribution, it'd give you a specific outcome.

Various uniqueness and well-posed theorems tell you that if you have boundary conditions of the right kind then you'll be able to model the time evolution of a system precisely using Lagrangian mechanics but if that system is a probability distribution (or rather an amplitude distribution) all you get is a probabilistic outcome for the underlying physical processes. Uniqueness of PDE equations doesn't imply uniqueness of physical outcomes.

'Quantization' procedure is mainly about writing evolution in eigenbase of evolution differential operator - in such coordinates it can be seen just as 'superposition of rotations' - like for normal modes of coupled pendulums.

No, first quantisation is about upgrading conjugate variables (ie those related via a Lagrangian in the usual manner) to non-commuting operators, $$[x,p]=i$$ or in the case of fermionic fields anticommuting operators, $$\{ \psi,\Phi \} = i$$. Second quantisation is about applying path quantisation to the expression $$\exp \left(i\int_{\mathbb{R}^{4}}\mathcal{L} d^{4}x\right)$$. Coordinate choices are something completely different.

But generally physics (potential term) isn't linear

Non-linearity doesn't always reside in the potential term. The KdV Lagrangian doesn't split into such simple expressions and its the classic example of non-linear physics.

and so eigenfunctions are usually more complicated, like solitons - localized solutions ... as in classical-corpuscular picture of particles - but there still can be involved some periodic process like rotation of coordinate of this eigenfuction, giving them also 'wave' nature required for interference.

I fear you're wandering into 'Ramble country', though you might have crossed into 'BS-ville'. I fear this because the KdV equation is the defacto example of solitons, being motivated by the observation of a soliton in the real world long before mathematicians conceived of them, and its a counter example to your comment on non-linearity in a potential.

www.tcs.uj.edu.pl/jarek

Having seen that I'll be inclined to give you the benefit of the doubt on topics I'm unfamiliar with but I stand by my comments about things I am not unfamiliar with.

 

[Guest254]

Undeterred by history, let me say that I can certainly think of a field theory that doesn't have any local or global symmetries. For example, suppose I take a real field $$\phi$$ and add an interaction that looks like $$\phi^3$$. This action is not invariant under any discrete or continuous transformations that I can think of, but is a perfectly well-defined (albeit sick) field theory.

Firstly, it's clearly invariant with respect to translations: $$x^\mu\mapsto x^\mu+\epsilon$$, so the corresponding theory of physics will conserve energy and momentum. Secondly: that's one incredibly presumptuous statement to make! The fact that you can't think of one is hardly a reason to dismiss the existence!

To my knowledge, "field theory" is synonomous to a theory of physics which comes from extremising some Lagrangian. Suppose your Lagrangian density is:

$$\mathcal{L} = \partial^\mu\phi \partial_\mu \phi - V[\phi], \qquad [\phi] \sim (x,\phi,\partial\phi,\partial^2\phi, \ldots)$$

To find all the symmetries of such a Lagrangian, you can't seriously hope to just stare at it and find them out! And in general, it's not even enough to consider only symmetries that have some kind of geometric standing (called Lie-type symmetries). Such symmetries correspond to those with generators (i.e. the corresponding element of the associate Lie algebra) of the form

$$ V = \alpha^\mu(x) \frac{\partial}{\partial x^\mu} + \beta(x)\frac{\partial}{\partial\phi} \qquad (*)$$

and to find all the symmetries of the Lagrangian, you need to solve the (large) system of overdetermined PDEs

$$ \mathrm{pr}^N(V) \Delta[\phi] = 0 \quad \textrm{for all solutions to }\Delta[\phi]=0$$

where $$\Delta[\phi]$$ represents the E-L equations associated with your field theory, N is the order of these equations and $$\mathrm{pr}^N$$ is the prolongation of the vector field $$V\in TM$$ to the N-th jet bundle $$\mathcal{J}^NTM$$. This is non-trivial in itself (computers can solve the system if you specify polynomial $$\alpha^\mu(x)$$, $$\beta(x)$$ of some given order), but to actually answer the full question we need to consider more than just Lie-type symmetries, but those called "generalised" symmetries, or Lie-Bäcklund symmetries! These have generators of the form:

$$V = \alpha [\phi]\frac{\partial}{\partial \phi} $$

for which there is no (finite-dimensional) geometric interpretation. Note the function $$\alpha$$ is now a function of $$[\phi]\sim (x,\phi,\partial\phi, ....)$$.

A deep result of Noether tells us there is a one to one correspondence between generalised symmetries of the Lagrangian density and conservation laws for the system. So if you want to have a reasonable theories of physics (i.e. have some conservation laws!), then you'll need to have a few generalised symmetries at the very least. If you want your field theory have local symmetries (in the sense of gauge theories), then you want the Lie algebra spanned by the vector fields in (*) to be non-trivial, i.e. there exist non-commuting vector fields.