Are More Complicated Lagrangians Wrong ?

[maverick_starstrider]

Are More Complicated Lagrangians "Wrong"?

When deriving physics from a postulated Lagrangian (like in Landau's books) we demand the simplest (i.e. the one with the lowest order terms) that obeys some symmetries. Are more complicated Lagrangians "wrong"? Or are they actually better approximations? Theoretically why isn't a Lagrangian with terms of all orders (that obey some symmetry) the most general and the most correct (and of course the most difficult to deal with)?

 

[dextercioby]

On the contrary, I think in all cases we assume the most general ones compatible with our known assumptions and mathematical symmetries.

 

[maverick_starstrider]

On the contrary, I think in all cases we assume the most general ones compatible with our known assumptions and mathematical symmetries.

Right, but one of those "assumptions" is always that the Lagrangian be "as simple as possible". That's what I'm asking about.

 

[RedX]

I read somewhere that terms in the Lagrangian with more than two derivatives have problems involving causality. I don't understand how that can be the case, since so long as the Lagrangian is Lorentz invariant, causality should be preserved. But that's what I read from some book.

Then again I don't understand how the propagator can have advanced waves, but it does. I understand that for neutral particles the propagator is symmetric in time and space, so the advanced waves will just give you the result of the retarded waves, so having advanced waves just multiplies everything by 2. And that antiparticles are a good interpretation if the propagator is asymmetric.

Actually, come to think about it, it's not true that if you just take the advanced propagator, you'll get a result equal to half the value had you taken the Feynman propagator (which is both advanced and retarded), since:

$$<0|\phi(x)\phi(y)|0> \neq <0|\phi(y)\phi(x)|0> \neq <0|T{\phi(y)\phi(x)}|0>$$

Only this is true:

$$<0|T{\phi(x)\phi(y)}|0>=<0|T{\phi(y)\phi(x)}|0>$$

 

[Matterwave]

A lot of times we throw out some higher order terms, higher order being more than quadratic in the derivatives, because we want the resulting differential equations to be linear. Only linear diff eqs obey the principle of superposition, and superposition is obeyed in many cases (e.g. Maxwell's equations are linear and therefore E and B fields obey superposition principle).

As for higher order derivatives, I'm not entirely sure, I will have to think about that.

 

[maverick_starstrider]

A lot of times we throw out some higher order terms, higher order being more than quadratic in the derivatives, because we want the resulting differential equations to be linear. Only linear diff eqs obey the principle of superposition, and superposition is obeyed in many cases (e.g. Maxwell's equations are linear and therefore E and B fields obey superposition principle).

As for higher order derivatives, I'm not entirely sure, I will have to think about that.

But to what precision is superposition verified? Do we discard non-linear terms because we're experimentally sure they're not there or because they're probably negligible and they greatly complicate things?

 

[Fredrik]

The higher order terms can be ignored because their contribution to interactions at low energies is negligible. See e.g. this transcript from a talk given by Steven Weinberg. No, he doesn't prove it there, but at least you'll get to hear this claim from someone smarter than me. (See the second paragraph on page 9).

 

[RedX]

But to what precision is superposition verified? Do we discard non-linear terms because we're experimentally sure they're not there or because they're probably negligible and they greatly complicate things?

Light can scatter light, so technically speaking electromagnetism is nonlinear. Of course that's purely quantum, since it requires light to turn into electron-positron pairs. Nevertheless there are non-linear terms in the interactions, but usually for the free-field Lagrangian superposition is nice so that in the absence of interactions, everything superposes. Of course for things like gluon fields the interactions are very important, so while you can separate the Lagrangian into terms quadratic in the gluon field and higher order terms so that the free gluon fields superpose, the coupling is not necessarily small so interactions are non-negligible and the failure of superposition becomes very important unlike in electrodynamics.

The higher order terms can be ignored because they their contribution to interactions at low energies is negligible. See e.g. this transcript from a talk given by Steven Weinberg. No, he doesn't prove it there, but at least you'll get to hear this claim from someone smarter than me. (See the second paragraph on page 9).

I think that's the effective Lagrangian approach. It requires a cutoff I think. I recall you can write down a spin-2 Lagrangian in the typical way of quantum field theory, and then treat higher order (non-renormalizeable) interactions as an effective field theory, and you recover general relativity (at least the form using vielbeins/tetrads). But at high energies the effective Lagrangian fails, so you still don't get quantum gravity.

 

[homology]

I read somewhere that terms in the Lagrangian with more than two derivatives have problems involving causality. I don't understand how that can be the case, since so long as the Lagrangian is Lorentz invariant, causality should be preserved. But that's what I read from some book.

RedX, I've also read that (in Srednicki's QFT text). I've talked it over with friend and this is all we have at the moment. The function evaluated at a point tells you only what's happening exactly at that point. The first derivative however does give you some information about what's happening around that point. Higher derivatives continue to give you more information further from that point and the example we thought of was the power series for the function. If you want to 'recapture' the function you need arbitrarily high derivatives. So in that sense the equations of motion might require information from arbitrarily 'far away' whereas the differential equations should be local in terms of the information they give.

Feel free to demolish the above if its codswallop.

 

[RedX]

RedX, I've also read that (in Srednicki's QFT text). I've talked it over with friend and this is all we have at the moment. The function evaluated at a point tells you only what's happening exactly at that point. The first derivative however does give you some information about what's happening around that point. Higher derivatives continue to give you more information further from that point and the example we thought of was the power series for the function. If you want to 'recapture' the function you need arbitrarily high derivatives. So in that sense the equations of motion might require information from arbitrarily 'far away' whereas the differential equations should be local in terms of the information they give.

You're right, it's locality, not causality.

The analog in mechanics would be a quantity called "jerk", which is the derivative of the acceleration. It doesn't seem to appear in mechanics, and the Lagrangian is just a function of position and velocity, but not of acceleration. Although since there is an official name for it, maybe it does appear in mechanics? I have no clue.

But a higher order differential equation certainly implies more boundary conditions. Indeed, in QFT the only boundary conditions I'm aware of are the value of the field on a closed hypersurface (a box at the 2 infinities of each of the 4 dimensions) which is taken to be zero. So you don't even need first derivatives of the field for boundary conditions (this is similar to electrostatics where for Laplace's equation, in spite of it being a 2nd order differential equation, only requires the potential on the box surrounding the system, and no derivatives).

Anyways, I have no idea.

 

[maverick_starstrider]

Light can scatter light, so technically speaking electromagnetism is nonlinear. Of course that's purely quantum, since it requires light to turn into electron-positron pairs. Nevertheless there are non-linear terms in the interactions, but usually for the free-field Lagrangian superposition is nice so that in the absence of interactions, everything superposes. Of course for things like gluon fields the interactions are very important, so while you can separate the Lagrangian into terms quadratic in the gluon field and higher order terms so that the free gluon fields superpose, the coupling is not necessarily small so interactions are non-negligible and the failure of superposition becomes very important unlike in electrodynamics.

So is using a linear Lagrangian and then applying perturbations for interaction absolutely identical to working with a Lagrangian with non-linear terms from the get go?

 

[RedX]

So is using a linear Lagrangian and then applying perturbations for interaction absolutely identical to working with a Lagrangian with non-linear terms from the get go?

The short answer is no, since there are things you can't see in perturbation theory, no matter how many higher loops you include. In order to see these things you have to include the interaction terms from the get go. For example a $$\phi^3$$ theory suffers from an unbounded energy in the ground state, but you can't see that in perturbation theory, and in factyou can calculate amplitudes using perturbation theory and not suspect anything is wrong. This is a little over my head though, so hopefulyl someone else with more knowledge can answer.

Also, I just remembered that accelerating charges cause electromagnetic waves. So there should be a reaction force. So forces can depend on acceleration. Does this imply the Lagrangian can depend on acceleration? I have no clue.

 

[maverick_starstrider]

The short answer is no, since there are things you can't see in perturbation theory, no matter how many higher loops you include. In order to see these things you have to include the interaction terms from the get go. For example a $$\phi^3$$ theory suffers from an unbounded energy in the ground state, but you can't see that in perturbation theory, and in factyou can calculate amplitudes using perturbation theory and not suspect anything is wrong. This is a little over my head though, so hopefulyl someone else with more knowledge can answer.

Also, I just remembered that accelerating charges cause electromagnetic waves. So there should be a reaction force. So forces can depend on acceleration. Does this imply the Lagrangian can depend on acceleration? I have no clue.

Well then I'm afraid I don't understand how we can say that there's a problem with merging QFT and GR or some such. How do we know that quantizing a Lagrangian of infinite order (if it were possible) would no produce the correct physics?

 

[RedX]

Well then I'm afraid I don't understand how we can say that there's a problem with merging QFT and GR or some such. How do we know that quantizing a Lagrangian of infinite order (if it were possible) would no produce the correct physics?

I think the trouble is we don't even know how to quantize a Lagrangian of higher order - the infinities become too much to overcome by the renormalization procedure. However, so long as we only consider low energies, we can still calculate using such a Lagrangian, and when we do this, QFT can produce everything that GR can, which is kind of amazing, since GR treats gravity as curved spacetime, while QFT treats gravity as particles (graviton) in flat space time: the two couldn't be farther apart in concept. Anyways, as I mentioned, this is beyond my level right now, but here is an article (claimed to be an easy article!) that sort of discusses it on page 7:

http://www.frankwilczek.com/Wilczek_Easy_Pieces/416_Fundamental_Constants.pdf

 

[A. Neumaier]

When deriving physics from a postulated Lagrangian (like in Landau's books) we demand the simplest (i.e. the one with the lowest order terms) that obeys some symmetries. Are more complicated Lagrangians "wrong"? Or are they actually better approximations? Theoretically why isn't a Lagrangian with terms of all orders (that obey some symmetry) the most general and the most correct (and of course the most difficult to deal with)?

Simplest means easiest to study. For a textbook, this is always important; and it is good science to use the simplest model matching the data. (This is called Ockham's razor, or Occam's razor.) But the standard model Lagrangian already has very many terms - one needs that many to describe our world locally. A theory of everything will probably even more complicated.

Then there are restrictions by methodology: Lagrangians containing higher derivatives invalidate the canonical quantization framework and also cause severe difficulties in functional integration approaches; so they are almost completely avoided. Products of too many terms make computations much harder (one nested integral more for each factor), so they are avoided when possible. If one wants to have a theory that is renormalizable in the strict sense, only very few nonquadratic terms are permitted.
For example, renormalized coupling of a Dirac and a Maxwell field already forces the minimal interaction defining QED. (Dropping all nonquadratic terms makes a QFT theory free if there are no classical external fields.)

 

[maverick_starstrider]

Simplest means easiest to study. For a textbook, this is always important; and it is good science to use the simplest model matching the data. (This is called Ockham's razor, or Occam's razor.) But the standard model Lagrangian already has very many terms - one needs that many to describe our world locally. A theory of everything will probably even more complicated.

Then there are restrictions by methodology: Lagrangians containing higher derivatives invalidate the canonical quantization framework and also cause severe difficulties in functional integration approaches; so they are almost completely avoided. Products of too many terms make computations much harder (one nested integral more for each factor), so they are avoided when possible. If one wants to have a theory that is renormalizable in the strict sense, only very few nonquadratic terms are permitted.
For example, renormalized coupling of a Dirac and a Maxwell field already forces the minimal interaction defining QED. (Dropping all nonquadratic terms makes a QFT theory free if there are no classical external fields.)

Sorry to necropost on my own thread but I've been thinking about this again. I understand the extreme, probably insurmountable, calculational difficulties associated with higher terms, however, what I don't understand is why it is not considered the ultimate theory of reality to simply take the framework we know works amazingly well (say path integrals of lagrangians) and just extend them to infinite order (and thus include paths that could be any analytic function). Why are people not of the opinion that if 3 terms creates some of the most accurate predictions ever made then 4 terms must be better and 5 terms even more so? With things like String Theory and such it seems like people have it in their heads that higher terms are not the ultimate description. I just don't understand why. I understand why they're a mathematical pain but why aren't most of the people in quantum gravity working on extending the math to higher orders rather than abandoning it and looking for new perspectives?

 

[A. Neumaier]

Why are people not of the opinion that if 3 terms creates some of the most accurate predictions ever made then 4 terms must be better and 5 terms even more so? With things like String Theory and such it seems like people have it in their heads that higher terms are not the ultimate description. I just don't understand why. I understand why they're a mathematical pain but why aren't most of the people in quantum gravity working on extending the math to higher orders rather than abandoning it and looking for new perspectives?

Taking more terms means having more parameters to fit to the known data, and therefore leads to less predictivity. This is the case already for very simple things like polynomial approximation. Usually approximations get worse beyond a particular fitting order.

 

[maverick_starstrider]

Taking more terms means having more parameters to fit to the known data, and therefore leads to less predictivity. This is the case already for very simple things like polynomial approximation. Usually approximations get worse beyond a particular fitting order.

But it depends if each successive order of polynomial has an ever smaller coefficient then the fit can improve, imagine an almost quadratic function, the x^2 coefficient will be large but the higher order term will be there but be small. Ultimately the issue of coefficients is really due to using a series solution rather than a closed-form one is it not? Is there any a priori reason to discount an infinite order field theory as a true description of reality (including GR) based on current experimental evidence?

 

[Dale]

But it depends if each successive order of polynomial has an ever smaller coefficient then the fit can improve, imagine an almost quadratic function, the x^2 coefficient will be large but the higher order term will be there but be small.

You are correct that if you have more adjustible parameters then you will be better able to fit the data, but you will be less able to predict the data. Necessary parameters improve your prediction of new data, but unnecessary adjustible parameters actually degrade the ability to predict as they add more uncertainty to your model. This is well understood in the context of Bayesian model selection or Bayesian hypothesis testing.

 

[A. Neumaier]

But it depends if each successive order of polynomial has an ever smaller coefficient then the fit can improve, imagine an almost quadratic function, the x^2 coefficient will be large but the higher order term will be there but be small. Ultimately the issue of coefficients is really due to using a series solution rather than a closed-form one is it not? Is there any a priori reason to discount an infinite order field theory as a true description of reality (including GR) based on current experimental evidence?

To use a theory one needs to know the coefficients. It is no good to have a theory of which you only know (if one could know that) that it is able to represent some part of reality if you can't get the model explicit enough to do some computation. Thus it is essential that the model parameters can be obtained by matxching them to real data.

If the number of parameters it too large and the data are noisy (real evidence is always noisy, except in very simple cases) then the data-fitting problem is ill-posed, which means that the resulting parameters are so sensitive to oise that they are useless for prediction. This is called overfitting.

To avoid overfitting, it is important to keep the number of parameters as small as possible. This is the mathematical justification for Ockham's razor.

 

[bonker]

When deriving physics from a postulated Lagrangian (like in Landau's books) we demand the simplest (i.e. the one with the lowest order terms) that obeys some symmetries. Are more complicated Lagrangians "wrong"? Or are they actually better approximations? Theoretically why isn't a Lagrangian with terms of all orders (that obey some symmetry) the most general and the most correct (and of course the most difficult to deal with)?

Occam's razor.

 

[maverick_starstrider]

To use a theory one needs to know the coefficients. It is no good to have a theory of which you only know (if one could know that) that it is able to represent some part of reality if you can't get the model explicit enough to do some computation. Thus it is essential that the model parameters can be obtained by matxching them to real data.

If the number of parameters it too large and the data are noisy (real evidence is always noisy, except in very simple cases) then the data-fitting problem is ill-posed, which means that the resulting parameters are so sensitive to oise that they are useless for prediction. This is called overfitting.

To avoid overfitting, it is important to keep the number of parameters as small as possible. This is the mathematical justification for Ockham's razor.

Well I'm less concerned with the calculational hurdles I'm simply wondering why this line of thought has pretty much been abandoned. As for fitting to data if we're really going for a complete theory of reality than the coefficients would be calculable ab initio (from first principles) and fitting to data doesn't matter.

 

[A. Neumaier]

Well I'm less concerned with the calculational hurdles I'm simply wondering why this line of thought has pretty much been abandoned.

Because it is sterile. Physic isn't usually done for aestethical reasons but to be able to make predictions.

 

[maverick_starstrider]

Because it is sterile. Physic isn't usually done for aestethical reasons but to be able to make predictions.

But I mean QFT was pretty useless until RG but it wasn't really abandoned as a correct theory in the meantime. It just seems to me that a higher-order QFT gets the least attention and resources of the possible Quantum Gravity theories

 

[A. Neumaier]

But I mean QFT was pretty useless until RG but it wasn't really abandoned as a correct theory in the meantime.

It was very useful in QED even without the RG, hence it survived the time until it was found how to handle the general case.

Nonrenormalizable QFTs (as you want to have them) are not abandoned either; but they are called ''effective theories'' and are not believed to be fundamental. There is ongoing work on canonical gravity considered as an effective theory, with more and more terms added as the accuracy demands increase. See the entry ''Renormalization in quantum gravity'' in my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html#renQG