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When a continuous global symmetry is broken, we lose one degree of freedom.
We also gain a massless spin 0 boson, called a Goldstone boson.
We then say the number of degrees of freedom is the same.

Here is where I have a problem. Doesn't a spin 0 massless particle moving in d+1 dimensions have d degrees of freedom? (I'm fine with an answer in 3+1 dimensions if it makes things simpler, as that's mainly the case I deal with.)

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    $\begingroup$ Why do you say that breaking a continuous global symmetry results in the loss of one degree of freedom? What is your definition of degree of freedom? $\endgroup$
    – user2309840
    Commented Jul 19, 2017 at 17:25
  • $\begingroup$ Haven't thought of that. I suppose if you were to think of the symmetry group as being dictated by N parameters, then losing a degree of freedom would be setting 1 parameter and not changing it. $\endgroup$
    – Omry
    Commented Jul 19, 2017 at 17:48
  • $\begingroup$ Your first and second statements (sentences) are flat wrong. Symmetry breaking involves loss of symmetry relations among degrees of freedom, or a changeover of the types of relationships (for the spontaneous breaking case.) The Goldstone modes are evolutes of extant degrees of freedom upon flipping from a linear to a nonlinear symmetry realization mode. Where did your unsound statements come from? Please cite source. $\endgroup$
    – Cosmas Zachos
    Commented Jul 19, 2017 at 18:47
  • $\begingroup$ @CosmasZachos Don't really have a source, as that was my understanding from a course on particle physics. What do you mean when you say "a changeover of the types of relationships [Among degrees of freedom]"? Could you link me a source on the subject? $\endgroup$
    – Omry
    Commented Jul 19, 2017 at 19:48
  • $\begingroup$ Well, the chapter on SSB and the Goldstone theorem of your text should cover it... Try WP. As you tweak parameters of a theory, the symmetry transformation laws may change from linear to nonlinear (in leading order in the fields). $\endgroup$
    – Cosmas Zachos
    Commented Jul 19, 2017 at 20:00

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You have to be careful about what sort of degrees of freedom you are counting.

When you expect that a field in $d$ spatial dimensions has $d$ degrees of freedom you are trying to associate dynamics along the independent directions of space. Of course these represent degrees of freedom but then the number would be infinity. Strictly speaking any field has infinite degrees of freedom, even if it is restricted to the line. A continuous function lives in an infinite dimensional space.

When people count degrees of freedom in the context of Goldstone theorem they refer to the internal degrees of freedom, such as electric charge, color, spin, etc.

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    $\begingroup$ I see the source of my confusion. I was thinking particles, not fields. I am still not sure why it counts as one degree of freedom though. $\endgroup$
    – Omry
    Commented Jul 19, 2017 at 17:49

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