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Symmetries and Conversation Laws

  1. Mar 3, 2009 #1
    According to Noether Theorem, Every symmetry corresponds to a conversation Law in the nature.
    For example: If rotational symmetry exits, Angular momentum is conserved.

    How can I be sure that which symmetry corresponds to which conversation Law?
    Can you tell me the conversation Laws under these symmetries:
    Charge symmetry,
    Parity symmetry,
    Time symmetry,
    Flavor symmetry,
    Helical symmetry,
    Scale symmetry.
  2. jcsd
  3. Mar 3, 2009 #2


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    Time symmetry is conservation of Energy
  4. Mar 3, 2009 #3

    Vanadium 50

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    That's not what it says. It makes statements about specific kinds of symmetries in the Lagrangian.

    Before going too far down this path, do you understand Noether's Theorem? Could you prove it if you had to? If the answer is no, probably the best we can do is to say some symmetries have this property, and to leave "some" vague.
  5. Mar 3, 2009 #4


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    I believe Noether's theorem applies to continuous symmetries.
    Most of the symmetries you list are discrete, internal symmetries.
  6. Mar 3, 2009 #5


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    so as the others have said, you should definitely go through and be sure you understand Noether's theorem. You also seem to apply the word "symmetry" and "conservation law" interchangeably in your question, thus jumping the gun! For example, I don't really know what "charge symmetry" is when read literally, but "charge" is CONSERVED due to the gauge symmetry of E&M!

    So let me answer your question by brute force:

    Charge conservation is a result of the gauge symmetry (to be technically correct, charge conservation actually follows from the GLOBAL part of this symmetry, for those who are in the know).

    Parity is a discrete spacetime symmetries and Noether's theorem does not apply in the naive way, and mentioned above.

    Flavor symmetry gives conservation of flavor quantum number. For example, things like "baryon number" and "lepton number".

    Helical symmetry is angular momentum, and is a subset of the rotational symmetry already mentioned.

    Scale symmetry is not actually a symmetry in most cases! But when it is a symmetry, it generates conserved charges called "Virasoro charges", but this is very advanced stuff and is not relevant for most phenomenology. The keyword to understand these charges is "Conformal Field Theory".

    OK, there is a brute-force answer. The reason I know which charge goes with which symmetry is because I use Noether's theorem to CALCULATE it! That answers the other part of you question.

    Hope that helps!
  7. Mar 3, 2009 #6
    @malawi_glenn, @Vanadium 50, @clem

    I should have said that Noether Theorem concerns continuous transformations on the scalar fields.
    But doesn't it mean that for the discrete symmetries Noether theorem leaves us alone??

    After infinitesimal translations to Lagrangian or generalized coordinates, i can show that Noether current or momentum conserved.
    Is there any other proof of the theorem i should know?? If there is, it maybe helpful for me.


    Thank you for detailed answer.
    I have to say that interchangeability between "symmetry" and "conservation law" confuses me a lot.
    Am i right if i would say that Gauge transformations are invariant because of charge symmetry ??
    Charge is conserved under Gauge symmetry ??
    Don't you think there's a conceptual confussion??

    Scale symmetry was the most important part of my question. Because i wonder why it is not included in the elementary field theory since it is an appearant symmetry in the universe.

    Can we derivate other fields (including supersymmetric and unknown fields too) after some kind of scale transformations??
    We know the supersymmetry transformation between boson and fermion fields. Can it be defined as a scale transformation??
    Is it what they called supersymmetric conformal field theory??
  8. Mar 3, 2009 #7
    When proving Noether's theorem, the way you calculate the conserved quantity (constant of the motion) that corresponds to the continuous symmetry is by differentiating the action wrt a variable that parametrizes the symmetry. With discrete symmetries, there's no variable, Noether's theorem can't be applied directly.

    Scale symmetry is only a symmetry when your fields are massless. If your fields are massive, you can have a symmetry that preserves the lagrangian by rescaling spacetime, all particle masses, and all dimensional coupling constants.
  9. Mar 3, 2009 #8
    The way I remember it, scale symmetry is badly broken in the real world. Beginning with chiral symmetry breaking breaking in QCD and the special scale at which the coupling constant grows big (even with massless quarks), but with QED as well, chemical bonds set spatial scales, and a simply scaled structured, like say a chair, would collapse under its own weight, if for instance you would scale the planet and and the chair a factor (say) [itex]10^{6}[/itex]
  10. Mar 4, 2009 #9


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    No, the Lagrangian is invariant under a Local U(1)-gauge transformation of the fields (your generalized coordinates), and what is conserved, the Noether charge, is the Electric Charge.
  11. Mar 4, 2009 #10

    So I must say;

    "Lagrangian is invariant under a Local U(1)-gauge transformations because of charge symmetry"

    Do you mean that there's no concept called gauge symmetry?? or they mean another thing when they are talking about gauge symmetry??

    Is it wrong to say "Charge is conserved under Gauge symmetry" ??
    Last edited: Mar 4, 2009
  12. Mar 4, 2009 #11


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    NO NO NO

    You have to say:

    The Lagrangian is invariant under a Local U(1)-Gauge transformation, the Noethers Theorem then gives you that Electric Charge is conserved.

    Why is it so hard to read what we write?? "charge symmetry" is a sloppy usage of words.

    What is symmetric is the Lagrangian under a Local U(1)-gauge transformation. This symmetry will give you charge conservation. Not vice versa or something in between.
  13. Mar 4, 2009 #12
    I don't really want to confuse the issues even more, but can we please be accurate and say that charge conservation is due to the global U(1) symmetry? Apart from the global subgroup, gauge symmetries are not real symmetries; gauge degrees of freedom do not relate different physical states.
  14. Mar 4, 2009 #13


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    yes, of course, we can even be that accurate :-)
  15. Mar 4, 2009 #14


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    malawi_glenn and genneth did a good job explaining the gauge symmetry part. That's a subtle issue about the global vs gauge symmetry - it has lead me to make more than one mistake here, on a QFT exam solution set (yikes!), and worst of all, in my own research! So don't feel too bad if you don't understand it. :wink:

    Moving on to "the most important part of [ophase's] question":

    WARNING: this might get complicated. If it confuses you, I'm sorry. Don't worry too much about it.

    Scale symmetry [itex](x\rightarrow\lambda x)[/itex] is a very tricky subject. CLASSICALLY, this is a symmetry of any theory of massless fields, but the problem is that QUANTUM MECHANICALLY, this symmetry is broken! These kinds of symmetries are called "anomalous symmetries".

    The way this materializes in the real world is that if you start with a "scale-invariant" theory that has no masses in it, the problem is that the couplings themselves depend on the energy! Example: if you measure the electromagnetic charge of the electron (e) at rest, you get:

    [tex]\alpha=\frac{e^2}{4\pi\epsilon_0\hbar c}=\frac{1}{137}[/tex]

    However, if you measure the electromagnetic charge of the electron moving with an energy of 100 GeV, you get:

    [tex]\alpha=\frac{e^2}{4\pi\epsilon_0\hbar c}=\frac{1}{128}[/tex]

    So we find that the coupling gets LARGER as the energy increases! This breaks the scale invariance and is due to this "quantum anomaly" I mentioned.

    So even massless theories typically do not have scale invariance, except for some VERY special examples, but they are rare.

    I do not understand your question about SUSY. Supersymmetry is another spacetime symmetry, different from rotations, translations and scale transformations. Maybe you can clarify?

    Anyway, I hope this helps and doesn't confuse. But this is why scale symmetry is treated a little differently than ordinary translations/rotations.

    Note Added: Hey, while proofreading this post, I got a sudden rush of deja vu! ophase: did we have this conversation before?! :confused:
  16. Mar 4, 2009 #15


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    I just wanted to point out that what you say is quite right. However, in all of these cases, you are talking about MASSIVE field theories (except for your coupling growing big example). In MASSLESS QED, for example, there are no bound states. You can see this by two arguments:

    1. Brute force: after a hard-work calculation, you find the bond lengths are inversely proportional to the mass of the electron, so as that mass goes to zero, the bond lengths go to infinity and there are no bound states.

    2. By dimensional analysis: a massless theory has no scale, and therefore you cannot write down anything with dimensions of length. Therefore there cannot be a bound state.

    So your chemistry argument fails in the case of massless electrons for example (not a "real world" scenario, of course, but I'm a theorist and that's never stopped me before! :wink:). However, even massless QED has the running coupling, as I said above.

    Anyway, just to clarify our positions.
  17. Mar 4, 2009 #16
    Indeed we agree, and as a matter of fact, I appreciate the clarification.
  18. Mar 4, 2009 #17

    If one have a discrete symmetry and Noether theorem can't be applicable how can one find out the conserved quantity?


    In the classification of symmetries, I never heard of a branch called real symmetries. Can you give us a more detailed definition of Real Symmetries?


    That's great to find some scale symmetric relations in QED. But I don't understand how did you calculate the constant while electron moving at 100 GeV.

    My question about SUSY: Why should it be a spacetime symmetry?? (I know the additional fields like W,D,F to satisfy the algebra and it disturbs me a lot)
    Isn't it possible to construct (let's say) a scale symmetry between the fields??

    Do you mean i'm the only one here asking silly questions after our past thread and you'r saying like "i recognize you wherever you go regardless of time" :)

    https://www.physicsforums.com/showthread.php?t=205735 ??

    I'll have more questions soon
    Last edited: Mar 4, 2009
  19. Mar 4, 2009 #18

    Vanadium 50

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    He just told you - it doesn't apply in this case.
  20. Mar 4, 2009 #19
    Ok, i edited it.. I think it's more clear now... Since english is not my native language, these happens. Sorry
  21. Mar 4, 2009 #20


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    Calculate the scattering cross section of electrons fusing to make a Z-boson, which then decays to muons (for example), when the CoM energy of the electrons is at the Z mass (91 GeV, which is close enough to 100). This answer is proportional to the fine structure constant (squared, actually).

    Then measure said cross section at LEP. This gives you your value for the coupling.

    I'm oversimplifying, but that's the idea.

    SUSY is a spacetime symmetry because it has nontrivial relations with the other spacetime symmetry generators such as

    [tex][Q,\bar{Q}]_+\sim P[/tex]

    So it is a spacetime symmetry.

    I'm not understanding what's disturbing you. What's "W"? Presumably, "D,F" are the auxiliary fields you introduce in SUSY - they're only introduced to make the MATH easier, you don't really need to introduce them, they're not new "physical" fields.

    The scale symmetry is also a spacetime symmetry, but as I said in my last post, it is violated in all but the rarest of quantum field theories. If you try to write a "scale-invariant" theory down, you will find that quantum corrections break the scale symmetry. There's nothing to be done with that! SUSY does not have this problem. Nor does the usual Poincare invariance (translations/rotations).

    I seriously thought I wrote these words down before, perhaps with some other post that wasn't you. Maybe I'm just on drugs...

    Anyway, ask away!
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