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Gauge transformation

  1. Aug 4, 2013 #1
    I understand ##\vec A\rightarrow\vec A+\nabla \psi\;## as ##\;\nabla \times \nabla \psi=0##[tex]\Rightarrow\;\nabla\times(\vec A+\nabla \psi)=\nabla\times\vec A[/tex]

    But what is the reason for
    [tex]V\;\rightarrow\;V+\frac{\partial \psi}{\partial t}[/tex]

    What is the condition of ##\psi## so

    [tex]\nabla \left(V+\frac{\partial \psi}{\partial t}\right)=\nabla V[/tex]

    Thanks
     
  2. jcsd
  3. Aug 4, 2013 #2

    king vitamin

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    Gold Member

    Recall Faraday's law:

    $$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} = - \frac{\partial}{\partial t}\left( \nabla \times \vec{A} \right)$$

    So it's tempting to make the definition

    $$ \vec{E} = - \frac{\partial \vec{A}}{\partial t} $$

    but now we make your transformation

    $$ \vec A\rightarrow\vec A+\nabla \psi\ $$
    $$ \vec{E} \rightarrow \vec{E} - \frac{\partial \nabla \psi}{\partial t} $$

    so the electric field now fails to be gauge invariant. So we introduce another gauge-dependent parameter to "fix" this:

    $$ V\;\rightarrow\;V-\frac{\partial \psi}{\partial t} $$

    and change our definition of E

    $$ \vec{E} = - \frac{\partial \vec{A}}{\partial t} - \nabla V $$

    Now the electric and magnetic fields are gauge invariant, as required.
     
  4. Aug 4, 2013 #3
    From
    $$ \vec{E} \rightarrow \vec{E} - \frac{\partial \nabla \psi}{\partial t} $$
    $$ V\;\rightarrow\;V-\frac{\partial \psi}{\partial t} $$
    How do you get to
    $$ \vec{E} = - \frac{\partial \vec{A}}{\partial t} - \nabla V $$
    Thanks
     
  5. Aug 4, 2013 #4

    WannabeNewton

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    Given a vector potential ##A##, we can write Faraday's law as ##\nabla \times (E + \partial_{t}A) = 0##. Now it is a fundamental result of differential topology that a curl free vector field in a 'nice enough region' (I'll spare you the details of what 'nice enough region' means) can be written as the gradient of some smooth scalar field ##V## i.e. ##E + \partial_{t}A = -\nabla V##; ##V## is of course the scalar potential. Hence the electric field can be written as ##E = -\nabla V - \partial_{t}A##.

    Now there is a gauge freedom in the potential fields. Notice that if we define a new vector and scalar potential by ##A' = A + \alpha, V' = V + \beta##, where ##\alpha## is a vector field and ##\beta## a scalar field, such that the new pair gives us the same physical electromagnetic field then we must have that ##\nabla \times \alpha = 0## and ##\nabla \beta + \partial_{t}\alpha = 0##. The first condition implies that ##\alpha = \nabla \lambda## for some smooth scalar field ##\lambda##. The second condition consequently says that ##\nabla(\beta + \partial_{t}\lambda) = 0## meaning that the expression in the parenthesis is only a function of time ##\beta + \partial_{t}\lambda = \gamma(t)## i.e. ##\beta =- \partial_{t}\lambda ## where we have absorbed ##\gamma(t)## into ##\lambda##.

    In other words, we can make a gauge transformation of the form ##A \rightarrow A + \nabla \lambda , V\rightarrow V - \partial_{t}\lambda## and still get the same physical electromagnetic field.
     
    Last edited: Aug 4, 2013
  6. Aug 4, 2013 #5
    Thanks

    What is the meaning of Gauge? What is Gauge freedom, Gauge invariant?
     
  7. Aug 4, 2013 #6

    WannabeNewton

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    Hi. In the above context, a gauge is just a particular choice of ##A,V##. The potential fields have a gauge freedom in the sense that any two gauges ##A,V## and ##A',V'## such that ##A' = A + \nabla \lambda## and ##V' = V - \partial_{t}\lambda##, will result in the same physical EM field. A gauge invariant quantity is one that does not depend on the choice of gauge.
     
  8. Aug 4, 2013 #7
    Do you mean ##\beta## is a function of time ONLY where we let ##\beta=-\frac{\partial \lambda}{\partial t}##. Since ##\beta## is a time function ONLY, ##\nabla \beta=0##?

    Am I understanding this correctly?

    So for ##\nabla \lambda\;\neq\;0\;and\;\beta=\frac{\partial\lambda}{\partial t}##, then ##\lambda=f(x,y,z)+g(t)## so ##\nabla \lambda=\nabla (f(x,y,z)),\; \beta=\frac{d g(t)}{d t}##

    Thanks
     
    Last edited: Aug 4, 2013
  9. Aug 4, 2013 #8

    WannabeNewton

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    Nono I'm saying that since ##\nabla (\beta + \partial_{t} \lambda) = 0##, all of ##\beta + \partial_{t} \lambda## is a function of time only i.e. ##\beta + \partial_t \lambda = \gamma(t)## hence ##\beta = \partial_{t}\lambda - \gamma(t)##. Then I just absorbed ##\gamma(t)## into ##\lambda## i.e. I defined a new ##\lambda'## by ##\lambda' = \lambda + \int _{0}^{t}\gamma(t')dt'## then ##-\partial_{t}\lambda = -\partial_{t}\lambda'-\gamma(t)\Rightarrow \beta = -\partial_{t}\lambda'##.
     
  10. Aug 4, 2013 #9
    I got it, I mistakenly thinking ##\nabla \times \vec \alpha=0\;\Rightarrow\; \vec \alpha## is position independent. But actually it only means ##\vec \alpha## is irrotational, but it can still be a function of 3 space coordinates.

    Thanks
     
  11. Aug 5, 2013 #10

    vanhees71

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    2016 Award

    The electromagnetic field in terms of the four-potential reads (using Heaviside-Lorentz units with [itex]c=1[/itex])
    [tex]\vec{E}=-\vec{\nabla} \Phi-\partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.[/tex]
    From the second equation we see that the magnetic components don't change, if I use
    [tex]\vec{A}'=\vec{A}-\vec{\nabla} \chi[/tex]
    with an arbitrary scalar field [itex]\chi[/itex] instead of [itex]\vec{A}[/itex] as vector potential. Now we look for a scalar potential [itex]\Phi'[/itex] such that
    [tex]\vec{E}=-\vec{\nabla} \Phi'-\partial_t \vec{A}'=-\vec{\nabla} (\Phi' -\partial_t \chi)-\partial_t \vec{A}.[/tex]
    The expression in the parentheses must be [itex]\Phi[/itex], and thus
    [tex]\Phi'=\Phi+\partial_t \chi.[/tex]
    Thus the gauge transformation of the potentials must be
    [tex]\Phi'=\Phi+\partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi.[/tex]
     
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