Solving Maxwell's Equations: Wave Equation in Vacuum

[fluidistic]

If I understood well my professor, he showed that "playing" mathematically with Maxwell's equation $$\frac{\partial \vec E}{\partial t} = c \vec \nabla \times \vec B$$ can lead to the result that $$\frac{\partial \vec E}{\partial t}$$ satisfies the wave equation (only in vacuum).
So what does this mean?! That the derivative of the electric field with respect to time is a wave when we're considering the vacuum?
If so, then I'm not understanding well the meaning of it. I can't imagine really what is the wave...
Any explanation is greatly appreciated.

 

[kloptok]

Well, since the electric field satisfies the wave equation it propagates as a wave. Note that it's not the time derivative satisfying the wave eq, but rather all components of E separately.

Take a look at the wiki article on 'Electromagnetic radiation' for example, I think you'll find lots of good things there.

 

[homology]

... that "playing" mathematically with Maxwell's equation $$\frac{\partial \vec E}{\partial t} = c \vec \nabla \times \vec B$$ can lead to the result ...

Try this for yourself. I'll get you started. If E satisfies a wave equation then we know we need a term like $$\partial^2 E/\partial t^2$$. What do you need to do to obtain such a term. Then look around at Maxwell's other equations and see if there's anything you can use to get the 'rest' of the wave equation.

Let us know if you figure it out :)

 

[jambaugh]

If a function satisfies the wave equation then so does its time derivative. (The wave operator and the partial derivative w.r.t. time commute.)

So saying $$\partial_t F$$ satisfies the wave equation means:

$$[\partial_t^2 - \nabla^2]\partial_t F = 0$$

Using the notation: $$\partial_u \equiv \frac{\partial}{\partial u}$$

Then by the commutativity of these differential operators
$$\partial_t[\partial_t^2 - \nabla^2] F=0$$

In the most general case this means F satisfies an inhomogenous wave equation:
$$[\partial_t^2 - \nabla^2] F=G$$
such that $$\partial_t G=0$$
(G = 0 is a special case!)

But in general if any quantity satisfies the wave equation then so will its time derivative, its derivative in the x-direction or its fifteenth derivative in the z direction. "Being a solution to the wave equation" is not equivalent to "being a (physical) wave".

What I think is the physical significance of the time derivative (and any others) of the E field satisfying the wave equation is that changes in the E field also propagate like waves. Imagine an E field in vacuum and then imagine some perturbation. The perturbation will likewise propagate like a wave.

Ultimately this is all about superposition. Derivatives are limits of difference quotients which is to say limits of linear combinations. The wave function, being linear, admits linear combinations of solutions as solutions and thus derivatives of solutions as solutions.

 

[fluidistic]

Well, since the electric field satisfies the wave equation it propagates as a wave. Note that it's not the time derivative satisfying the wave eq, but rather all components of E separately.

Take a look at the wiki article on 'Electromagnetic radiation' for example, I think you'll find lots of good things there.

Thanks for your comment. I searched and I think I found what you mean. In which case I don't understand what my professor did.

Try this for yourself. I'll get you started. If E satisfies a wave equation then we know we need a term like $$\partial^2 E/\partial t^2$$. What do you need to do to obtain such a term. Then look around at Maxwell's other equations and see if there's anything you can use to get the 'rest' of the wave equation.

Let us know if you figure it out :)

Hmm not sure I get it. You're saying that it's the E field that does satisfy Maxwell's equations? Or the derivative of the E field with respect to time like I thought?
Since $$\frac{\partial \vec E}{\partial t}=c \vec \nabla \times \vec B$$, $$\frac{\partial ^2 \vec E}{\partial t ^2}=c \frac{\partial }{\partial t} (\vec \nabla \times \vec B)$$. Looking at the other evolution Maxwell's equation, I get $$\frac{\partial ^2 \vec E}{\partial t ^2}=\frac{\partial ^2 \vec E}{\partial t ^2}$$ which is of course right but not helpful. Obviouly I misunderstood you.
What my professor did is a bit messy to me but he reached $$\frac{\partial ^2 \vec E}{\partial t^2}=-c^2 \vec \nabla (\vec \nabla \cdot \vec E)+c^2 \triangle \vec E$$.
Then he derives with respect to time and get $$\frac{\partial ^3 \vec E}{\partial t^3}=-c^2 \vec \nabla \left ( \vec \nabla \cdot \frac{\partial \vec E}{\partial t} \right )+c^2 \triangle \frac{\partial \vec E}{\partial t}$$.
Thus he calls $$\vec Y = \frac{\partial \vec E}{\partial t}$$. So that $$\frac{\partial ^2 \vec Y}{\partial t^2}=-c^2 \vec \nabla (\vec \nabla \cdot \vec Y )+c^2 \triangle \vec Y \Rightarrow \frac{\partial ^2 \vec Y}{\partial t^2}=c^2 \triangle \vec Y$$. He justifies why some terms are worth zero and almost each step.
When I look at the last equation I read that $$\vec Y$$ satisfies the wave equation. However $$\vec Y =\frac{\partial \vec E}{\partial t}$$, not $$\vec E$$.
Is there something wrong?Edit: I just see your post jambaugh, very helpful. So can I think the derivative of the E field with respect to time as a perturbation of the E field, which (the derivative of the E field) propagates like a wave?

 

[homology]

differentiate $$\frac{\partial \vec E}{\partial t} = c \vec \nabla \times \vec B$$ with respect to time. The time partial commutes with the curl. So you'll have the curl of $$\partial B/\partial t$$. The change in B with respect to time is caused by a curl of E. Substitute this and simplify the wave equation for E falls right out. Works the same for B. Make sense now? Its really quick.