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Local Gauge Symmetry ?

  1. Apr 10, 2008 #1
    Local Gauge Symmetry ??

    Trying to understand local gauge symmetry

    I have an undergraduate degree in physics, so I know basic quantum mechanics, but that's all.

    Still, I'm trying to understannd the concept of local gauge symmetry.

    I would appreciate if someone could tell me what's wrong with my point of view.

    So here is my understanding of it:
    As an example, consider the classical Newtonian equation for the gravitational force between masses m1 and m2 at position x1 and x2 on the x-axis.
    It is F = G m1 m2 / (x2 - x1)^2 .

    Now if we replace x1 and x2 with x1+k and x2+k, we get the same equation and the same force F.
    So here we have a global gauge symmetry.
    What are the physically real value of x1 and x2 ?
    They don't have physically real values, cause they are not measurable.

    So let's go one step further and replace x1 and x2 with x1+k1 and x2+k2.
    We now have a local gauge symmetry.
    That's fine except now we don't get the same equation, nor the same force.
    So let's add a new field or force to compensate for that.
    But that new field is arbitrary, since it is function of the arbitrary values for k1 and k2.
    And that new field has no physical reality, it is not measurable (how could it be, since it is arbitrary?)

    So how can we switch to local gauge symmetry if it changes measurable values?
    How does postulating a new compensating field help anything?
    I don't understand what is gained by doing this.
    All this doesn't make sense to me.
  2. jcsd
  3. Apr 10, 2008 #2


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    Your example does not quite work because x1 and x2 are measurable quantities. They are the position of the two objects. And you also haven't shown exactly how you would introduce a gauge field to make the symmetry local.

    The whole approach is more transparent in a Lagrangian approach. But it's also possible to do part of it in the equations of motion.

    It would be much better if you would start with ordinary quantum mechanics and consider the invariance of the theory under the change of overall phase of wavefunctions

    [tex] \psi \rightarrow e^{i \alpha} \psi [/tex]

    If you make this space dependent you must introduce a gauge field.

    To answer your question, when we introduce a gauge field, we also introduce interaction terms for the gauge field itself so the gauge field becomes a dynamical entity. Then the end result is that the gauge field produces a force between the particles of your theory. In fact, making the phase invariance described above local leads to electromagnetism, which is quite remarkable.
  4. Apr 10, 2008 #3
    When I say x1 is not measurable I mean it does not have a definite value. What is x for the Sun? It depends on the origin of our coordinate system.
    I know you are right about that, but I don't see how this works. Is the force produced by the gauge field a real force, or just a pseudo-force to compensate for our arbitrary phase change (pseudo-force not needed if we don't change the phase?)
  5. Apr 10, 2008 #4


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    Local gauge symmetry applies to quantum theory.
    Trying to use it in Newtonian gravity leads to your confusion.
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