4. Maxwell's equations can be used to describe...

4. Maxwell's equations can be used to describe electromagnetic fields. (a) From Maxwell's equations, show that the electromagnetic wave obeys a wave equation. Use the vector identity ( abla imes abla mathbf{A}= abla( abla cdot mathbf{A})- abla^{2} mathbf{A} ). (15 marks) (b) Show that the solution above is a wave and thus, that the wave velocity of light is a constant. (15 marks)

Transcript

00:01 In this question we're using maxwell's equations to derive that there are vacuum waves of electromagnetic radiation in maxwell's theory.

00:12 So the equations are that the divergence of e is equal to rho over epsilon nought, the divergence of b is equal to zero, the curl of e is equal to minus the time derivative of b, and that the curl of h...

00:43 Actually what we should say is that this is the divergence of d, and the curl of h is equal to j plus the time derivative of d.

01:00 These are maxwell's equations.

01:02 Now d is equal to epsilon nought e, and b is equal to mu nought h.

01:14 So we can rewrite these in terms of e and b.

01:20 But first of all, in the vacuum, in a vacuum we have that the density, the charge density, is equal to zero, and the current density is equal to zero.

01:32 Now we can rewrite these.

01:33 Now if the divergence of d is zero, the divergence of e is zero over epsilon nought, which is zero.

01:44 The divergence of b is always zero, because there are no magnetic monopoles.

01:50 The curl of e is minus the time derivative of b.

02:02 And then the last one, the curl of h, well h is b over mu nought.

02:08 So we can multiply by mu nought to...

02:16 Well we can divide both sides by mu nought, and we can get...

02:23 Well multiplying both sides by mu nought, we get the equation in terms of b.

02:31 So we get...

02:32 But then d is epsilon nought e.

02:34 So we need to divide by...

02:39 We need to multiply by epsilon nought as well when we put in an e.

02:42 So we get a mu nought epsilon nought e d, but it's the time derivative of e.

02:55 Right.

02:56 So these are our four equations, one, two, three, and four, which apply in a vacuum.

03:06 Now what i'm going to do is i'm going to take a...

03:12 Now we have this vector identity, which is not written out fully, but it's that the curl of the curl of vector a is equal to the gradient of the divergence of a minus the laplacian of a.

03:30 So this is just an identity.

03:36 So what we're going to do is we're going to take the curl of equation two.

03:40 So let's take the curl of equation two, and that gives us the gradient of the divergence of e minus the laplacian of e is equal to minus d by dt of the curl of b, where we can interchange the time and the space derivatives...

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