6
$\begingroup$

Physics theories are always constrained by symmetry principles (Strong equivalence principle, Galilean invariance principle, Gauge invariance, …).

This means that each model of a theory inherits (by construction) the symmetries enforced by the theory. For example, in classical mechanics, every model is (supposedly) Galilean invariant (no absolute frame of reference, invariance by rotations, translations, boosts, …).

Now, take the classical expression of the Lorentz force: it depends explicitly on v and and cannot therefore be Galilean invariant. The same is true for the harmonic oscillator $F = -kx$ or the central force $F = -\frac{1}{r^2}$. In these cases, an absolute frame of reference is defined because an absolute point is discriminated (the origin) from all others.

As these models are not Galilean invariant, we should not expect them to preserve energy, momentum, angular momentum or any other quantity related to these broken symmetries (say, by Noether theorem).

How to explain that in some case the conservation laws are preserved while they a priori have reason to be, say, for the central force? It seems to me that we cannot even know when it i the case, so why are we even using them to begin with?

Isn't this kind of practice extremely bad and confusing for students too?

As an example, notice that in the central force case the momentum is not preserved because you don't have invariance by translation in space as a point is discriminated and fixed.

EDIT: apparently my question wasn't clear, so here's a new formulation. Given a model $M$, some physics claims are sometimes preserved by doing a limit process or by renormalizing some quantity in $M$, and sometimes they are not. Why? How can we know which claims are preserved a priori and which are not?

Knowing what symmetry is preserved or not is just one type of physics claim (you have infinite many others you can make), and it's only trivial in models that admit a lagrangian formulation and continuous symmetries by Noether theorem. It's not even trivial otherwise (say, in stat mech models).

$\endgroup$
  • $\begingroup$ I am slightly confused - in the cases where we use models which break Galilean/Lorentz invariance, we don't expect the corresponding conservation laws to hold, which is why we are not surprised e.g. that a mass on a spring does not have a conserved momentum. We expect such quantities to be conserved only if the corresponding symmetry holds. $\endgroup$ – J. Murray Jun 4 at 22:33
  • $\begingroup$ it breaks some very strong constraint of your theory, which means that your model is not a model of classical mechanics. Those are elementary so you can realize it, but notice that the central force is a renormalization process of the two body problems: you divide by an infinite quantity. You do the same in QFT, and here it's much harder to be convinced that we're preserving the symmetries we want to preserve, so how can we trust these models? $\endgroup$ – sure Jun 4 at 22:41
  • $\begingroup$ I'm still lost. Are you saying that we should impose the full symmetry of the Galilei group on all of our models, and discard those which do not fit into that category? Also, that's not at all what renormalization is in QFT. $\endgroup$ – J. Murray Jun 4 at 22:45
  • $\begingroup$ of course we should, otherwise you do not have a model of your theory by very definition of what a model of your theory is. Yes, renormalization in QFT is even worse than that, I agree $\endgroup$ – sure Jun 4 at 22:49
  • 4
    $\begingroup$ That is ludicrously restrictive. You're saying that if I model a ball bouncing off the floor, then I'm not doing physics unless I account for the recoil of the Earth and the minute amount of energy lost to sound waves in my calculations. $\endgroup$ – J. Murray Jun 4 at 22:56
13
$\begingroup$

You have an extremely restrictive opinion on what should be called physics. I suspect you would find very few people to agree that one is only doing physics if one imposes full Galilean or Lorentz symmetry on all of their models; that would rule out almost everything in vast swaths of the physics community.

How to explain that in some case the conservation laws are preserved while they a priori have reason to be, say, for the central force?

If you are referring to the Kepler problem with $\mathbf F \propto -\frac{\mathbf r}{r^3}$, then it possesses rotational and temporal symmetry but not translational symmetry, and so angular momentum and energy are conserved while linear momentum is not.

If you prefer, you can consider the dynamics of two masses $m_1$ and $m_2$, under the influence of an attractive force with magnitude $F \propto \frac{\mathbf r_1 - \mathbf r_2}{|\mathbf r_1-\mathbf r_2|^3}$. This system possesses full Galilean symmetry, and conserves linear momentum, angular momentum, and energy.

It seems to me that we cannot even know when it i the case, so why are we even using them to begin with?

Conservation laws hold if the corresponding symmetries hold, as per Noether's theorem. We know which conservation laws to expect because we know which symmetries apply.

Isn't this kind of practice extremely bad and confusing for students too?

No.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In these simple problems you can obviously apply Noether theorem, but you have no guarantee to be able to apply them in other instances. For example, you can find examples of limit of models that are "preserving" the syntax/algebra of your former model, but the limit is not preserving the physics at all. This is especially true if you add electromagnetism (think of an electron orbitting a proton of finite mass. Now, take the limit of the proton mass to go to infinity, and you're not preserving energy) $\endgroup$ – sure Jun 4 at 23:15
  • $\begingroup$ @sure The value of the proton mass would have no bearing on the conservation (or lack thereof) of energy in such a system. $\endgroup$ – J. Murray Jun 4 at 23:18
  • $\begingroup$ that's still irrelevant. The process of taking a limit (say, the thermodynamic limit in stat mech) is non-trivial, and it can break a lot of things you a priori can't know/see $\endgroup$ – sure Jun 4 at 23:22
  • 3
    $\begingroup$ @sure Of course you have to be careful in taking thermodynamic limits, but I don't see how your question is related to that at all. You asked why we expect conservation laws to hold when the corresponding symmetries are broken, and the answer is that we don't. $\endgroup$ – J. Murray Jun 4 at 23:27
  • 7
    $\begingroup$ @sure That is not at all apparent from the body of your question. I would suggest dramatically rewording things if that is what you mean to ask. $\endgroup$ – J. Murray Jun 4 at 23:31
0
$\begingroup$

I'm not really sure I understand your question, but I'll try to answer anyway.

You start with classical mechanics. Here, I guess the main points are:

  1. Classical mechanics itself (i.e. the "physical laws") are invariant under translations in space and time and under rotations, hence energy, momentum and angular momentum are conserved.
  2. You can consider any system of masses with forces between them, and the conservation still holds. In other words, the physical system does not have to be invariant under the transformations. (Otherwise, physics would be pretty boring.)
  3. You often encounter systems which can be, to a very good approximation, described in much simpler terms by taking some limit, e.g. taking the sun as infinitely heavy for the Kepler problem, taking the earth as infinitely heavy and flat for stuff moving on the surface, taking one end of a spring to be fixed to a rigid point (which is again fixed to the earth) etc. In all these cases, you end up with a set of physical laws that have different (usually fewer) symmetries than the orginal, and thus less conserved quantities (ball bouncing off the earth etc.). In all of these cases, the procedure is not that hard, and you can go through the details to see what the symmetries are, and, of course, for which limits the approximation holds.
  4. In the end, you have to do the same thing for QFT, and that might be technically and conceptually more challenging. You might e.g. prefer dimensional to cutoff regularisation because Lorentz invariance is manifest, but in the end you have to carefully check that your procedure doesn't introduce some strange behaviour. Sometimes that's easy, sometimes it's hard -- an obvious case would be anomalies in a quantised theory, which took some time to understand. One complication is that an unrenormalised theory in itself is ill-defined in that e.g. amplitudes are given by divergent integrals. In that sense, the renormalised theory is the real one, while the starting point is more of a heuristic motivation. On the other hand, field content and symmetries (IF your renormalisation procedure can maintain them) are carried over.
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ can't you realize that if already in classical mechanics, taking a limit process does not preserve physics claims, then you have no reason to go in the other direction either? As in, if a physics claim is true in a limit case/regularized case, there is no reason to believe it holds true in the physical case. Just think of "this object is not accelerated" as a physics claim, obviously true for an infinite mass, never true for a finite mass if there is some interaction term $\endgroup$ – sure Jun 5 at 9:41
  • $\begingroup$ @sure I don't unserstand your comment. (a) In general, you cannot just take over some statement from the "limit system" to the original one. That's pretty obvious, and I don't see where I would claim something like that. (b) My point rather is: You need to understand the system you're lookign at and the "limit" you want to take, and then you can (in principle) determine what's conserved, what's not, whether the limit is a good approximatio to some actual situation etc. $\endgroup$ – Toffomat Jun 5 at 9:51
  • $\begingroup$ (a) this is exactly what you do in QFT, (b) tell me what physics claims are preserved from a non-renormalized QFT model to a renormalized one, or the other way around. Which physics claims that hold true in a renormalized QFT does not hold true in the non-renormalized one? $\endgroup$ – sure Jun 5 at 9:53
  • $\begingroup$ (a) I don't think you really work in the unrenomalised theory. (b) I have amended the answer. I should point out that if you really care about rneomalisation in QFT, the long part in your question about classical mechanics is somewhat of a distraction. $\endgroup$ – Toffomat Jun 5 at 10:14
  • $\begingroup$ (a) that's what you do, the feynman rules and so on are given in this case. (b) that's completely empty, you still do not know what is preserved or not, especially since a lot of symmetries are "hidden", and that's only for symmetries in this case, not even speaking of the physics claims themselves $\endgroup$ – sure Jun 5 at 10:18
-1
$\begingroup$

I don't feel your question is formed in the right way. I think all the physicists know something is wrong with our current status of artificial law in physics. It doesn't work universally. All the laws explain only the part of the reality of nature. The physics law made by a human is lack of something. And we don't know what it is so they do the hard work to improve.

But the surprise is it could explain many things including the conservation of quantities. Even though the law is not perfect, it has some usage so we need to do more study. By doing the study, we may find out the deeper truth someday.

Your question, why the law doesn't know what symmetry is conserved prior, isn't the right way to look at it. We have this very poor, limited theory but it can explain the conservation of quantities by doing some operations looking artificial. That's very interesting, meaningful, and surprise. If we look at it hard enough, we may find out the breakthrough. And that's the point of physics education.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.