# Gauge transformation

1. Aug 4, 2013

### yungman

I understand $\vec A\rightarrow\vec A+\nabla \psi\;$ as $\;\nabla \times \nabla \psi=0$$$\Rightarrow\;\nabla\times(\vec A+\nabla \psi)=\nabla\times\vec A$$

But what is the reason for
$$V\;\rightarrow\;V+\frac{\partial \psi}{\partial t}$$

What is the condition of $\psi$ so

$$\nabla \left(V+\frac{\partial \psi}{\partial t}\right)=\nabla V$$

Thanks

2. Aug 4, 2013

### king vitamin

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

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.

3. Aug 4, 2013

### yungman

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

4. Aug 4, 2013

### WannabeNewton

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
5. Aug 4, 2013

### yungman

Thanks

What is the meaning of Gauge? What is Gauge freedom, Gauge invariant?

6. Aug 4, 2013

### WannabeNewton

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.

7. Aug 4, 2013

### yungman

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
8. Aug 4, 2013

### WannabeNewton

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 $-\partial_{t}\lambda = -\partial_{t}\lambda'-\gamma(t)\Rightarrow \beta = -\partial_{t}\lambda'$.

9. Aug 4, 2013

### yungman

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

10. Aug 5, 2013

### vanhees71

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