Welcome back to this lecture about the Lawrence Cage, so in the previous lecture, we have transformed
the Maxwell equations in their most general case to four new equations.
So these are based on the vector potential and on the electrostatic potential, which is a scalar function.
Now, you can see two of these equations have the same mathematical shape we have to square operate
to acting on the potential than we have here, some derivative here, time derivative here.
It's the gradient, basically the space derivative.
And then this bracket expression here in both cases and on the right hand side, we have to charge density
and here we have the current density.
So these equations rely not on the electric field, but and not on the magnetic field, but they rely
on the vector potential and on the electrostatic potential and we have already already discussed this
in electrostatic and in magnetic statics, that these two potentials are not the observables.
So this means we can use certain gauges and we can change these potentials basically in a certain manner
without affecting the magnetic and the electric field.
And maybe you remember this, for example, we have this expression here, this Maxwell equation, where
the divergence of the magnetic field is zero.
So this means we have the possibility to express the magnetic field in terms of the rotation of a.
So this is like an indirect definition of A, we have B is equal to rotation of A, we do not have the
expression A is equal to something.
So this means this A is not uniquely defined.
We can basically add any function to a whose rotation is zero because this leaves B unchanged.
So this means he is not unique and it can be gauged and the same applies to fire.
So it would be really good if we could use a gauge where this expression in the brackets would become
zero, because this would make our equations and our lives much easier.
And it turns out we can do this.
There is a certain gauge and it's very famous.
It's called the Lawrence Gage, which is exactly that this bracket expression becomes zero.
But of course, we need to verify that this is a valid gauge.
We cannot do any transformation to a because there are some transformations that change the observables,
for example, B, in this case, for example, if we would add some function to A whose rotation gives
some value, then of course this value is added up to B and then B is changed.
So this would not be a valid gauge.
So we need to really confirm that as Lawrence is allowed and it's valid that it leaves B and PHI invariant.
And first of all, I want to tell you how you can gauge these two potentials A and find the answer is
you can gauge or you can change A in this way that you add a function that is a gradient of a scalar
so you can take any scalar function.
It does not matter, can be any scalar dependent on the position R and to time T and then you calculate
the gradient of this function.
And in any case you can add it to a this is of course because the rotation of a gradient of some function
is always zero.
And then for five you can change your fire to a Newfie by subtracting.
Or you could also write adding the time derivative of a different function.
And if you want to use the same function in both cases, then you have to write it down like this.
And this can easily be verified.
You can check this for yourself.
So we must make sure that the electric field and the magnetic fields stay unchanged and that these transformations
so we can write down that our electric field is this expression from the from the previous lecture.
So this was the Maxwell equation that we had transferred.
So we have the time and time derivative of our new vector potential in the gradient of our new scalar
potential.
And then we use these two expressions here and set them in.
So we get here an additional term, which is a time derivative of this gradient.
And here we get the same thing.
We get a gradient of the time derivative.
And since these two derivatives are exchangeable because the position and the time are independent variables,
these this term here and this term, they compensate.
So we are left with this term in this term, which is E, so the new E is equal to the old E and then
for a magnetic field.
I already mentioned this briefly.
This is very, very easy.
You have the new magnetic field is the rotation of the new vector potential, which is of course the
rotation of the old vector potential, plus the rotation of the verdie and gradient of ky. and a great
and the rotation of a gradient of any function is always zero.
So this means it's NewBay is the rotation of the old EI.
So it's the old B.
OK, so now we know how we can transform the vector potential and the electrostatic potential.
And so this means when we take this term here and the brackets and apply this transformation, we get.
Yeah, this expression.
But just with this new electrostatic potential and this new vector potential.
So now let's use these two expressions here for the transformed potentials and put them in and see what
it looks like.
So, of course, from this term and from this term, we get the old expression.
So that's exactly the same as this one, but with the old vector potential and the old scalar potential.
And we get to new terms from this one and that one here.
So we get here at a time derivative of the gradient of chi and here we get the rotation of the rotation
of the gradient of chi.
So what this means is we need to choose Chi in a certain way so that this whole Brexiteer is equal to
this Brackett's.
And we do not really have to say what Chi looks like, we just have to realize that this is, of course,
possible.
So we have a certain solution to our problem, a certain calculated fire and a certain calculated A..
And now we have to choose chi accordingly so that this bracket is the same magnitude, but the opposite
side to this bracket here.
And it's possible we will do this.
And this means this whole expression becomes zero if Chi is chosen accordingly.
So that means it is indeed possible to gauge this whole term away.
And so our equations become really simple.
It's just a scratch.
Operator Phi is equal to one of absolute zero and the great operator acting on A is equal to zero times
current density.
So what we have done in this lecture is we have started from the Maxwell's equation, we have transformed
them and then we have applied this Lawrence and this led to this to these new Maxwell equations.
And now you can really see what is the advantage.
We have now two equations, this first and this fourth one here that are not coupled anymore.
So you just have to hear the charge density.
And here you have the I'm sorry, here you have to charge density and here you have to current density.
And from them you can calculate the electrostatic potential and the vector potential.
And when you have these, you can just take them and calculate the magnetic field and the electric field.
And here on the very left, it was much more difficult because here you had a couple of equations,
for example, here, when you wanted to calculate the magnetic field, you also had to know the electric
fields.
And here it's viceversa and you wanted to calculate the electric field.
You had to know the magnetic fields.
So this could be an advantage.
And we will discuss one example in the following.
So in the next lecture, we will discuss -- potentials, which basically take our knowledge from
electrostatic and magnetic statics and apply them to time dependent problems based on the solution we
have just established.