# B How do EM waves move?

1. Jun 6, 2016

### Dexter Neutron

An EM wave is nothing but just magnetic and electric fields regenerating each other. Now if a charge oscillates and it produces sinusoidally varying magnetic field which induces an electric field perpendicular to it at the same place.This induced electric field even varies sinusoidally thus again producing a magnetic field still perpendicular to it varying sinusoidally. This process continuous. The question now arises is that all the field are induced perpendicular to each other at the same place so how does the system of electric and magnetic fields move away? i.e. fields inducing at the same place would produce something called stationary EM wave but we actually know that it is not stationary so what causes it to move?

2. Jun 6, 2016

### Delta²

Nope it doesn't happen only in the same place but in all the space (up to infinity) that surrounds that specific region of space. If you are familiar with the integral form of Maxwell-Faraday's Law and Maxwell-Ampere's Law (these two laws are the two of the four Maxwell's equations) you can understand why a time varying field in a specific finite region of space, lets call this region A, generates a time varying field in the whole space that surrounds the region A.

There is also a good diagram/figure that shows the field lines of the B-field and E-Field from a vertical wire that carries time varying current, and how these field lines are generating each other and propagate in space but at the moment I cant find the link to the web page. And I am very bad on drawing figures...

Last edited: Jun 6, 2016
3. Jun 6, 2016

### Simon Bridge

Considering the model you are using ... you are correct: if the electric field varied so as to induce a magnetic field in the same place, then nothing would propagate. Since some EM fields do propagate, the logical conclusion is that the induced fields are not in the same place as the inducing field.

The reason for EM radiation is more simply explained in relativity, with the principle of superposition, rather than in terms of separate electric and magnetic fields generating each other. The electric and magnetic fields are not separate at all, but different ways of looking at the same phenomenon.
http://physics.weber.edu/schroeder/mrr/MRRtalk.html
... theory

http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html
... has an applet where you can see how the moving a charge produces radiation.

4. Jun 6, 2016

### Dexter Neutron

If a region A has a time varying magnetic field along with a conductor(perpendicular to magnetic field) placed in it then an electric field would get induced in the conductor due to change of magnetic flux through the conductor resulting in an emf. In this case the electric field is induced in the same region A and it seems to contradict what you said. What is wrong with this and why it contradicts?

5. Jun 6, 2016

### Delta²

It doesn't contradict actually. If you put a conductor outside the region A you will get an electric field and an emf there too but it will be much smaller.

The electric field that dominates inside region A is the near field, while outside region A the far field dominates. What we call the near field (or the field in the induction zone) is the part of the field that is very strong but falls rapidly as $1/r^2$ as the distance r increases. The far field (or the field in the radiation zone) is the field that is not so strong but falls only as $1/r$

6. Jun 6, 2016

### Simon Bridge

The fact that it is possible to induce an electric field in the same place as the time-varying magnetic field in no way contradicts the argument. There are lots of different kinds of fields, not all of them propagate.

In order to get a self-sustaining propagating field, you need the kind of variation that results in the induced field being slightly in advance of the inducing field... and that the induced field needs to vary in time and space in such a way that it will, in turn, induce the original inducing field only advanced slightly.
warning: this is a very clumsy way to think about it.

7. Jun 6, 2016

### Staff: Mentor

This is not correct. Faraday's law says $\nabla \times E=-\partial B/\partial t$. In English this would be something like "a B field varying in time equals an E field varying in space". The spatial variation is inherent in Maxwell's equations.

8. Jun 6, 2016

### Dexter Neutron

If B is a sinusoidal function then its derivative would also be a sinusoidal function. Consider B to be proportional to $\sin{\omega t}$ where $\omega$ is the angular frequency of the oscillating charge, so the derivative of it with respect to time would be depended on $\cos{\omega t}$. So how could you say that the induced E field would not vary in time?
What does the upside down triangle means?

9. Jun 6, 2016

### Dexter Neutron

That's what I am trying to ask. How to get the variation resulting in induced field being ahead of inducing field because all the variations we know have the induced field at the same place as the inducing field is except the fact that the induced field is perpendicular to inducing field.

10. Jun 6, 2016

### Simon Bridge

You already know how to get the right variation - ie. a sinusoidally varying E field.
There are a bunch of others - like you can send EM pulses through vacuum just by giving a charge a bit of a bump.
You can see this in Maxwell's equations - anything that is a solution to the wave equation will work.

11. Jun 6, 2016

### Simon Bridge

(The upside down triangle is called "nabla", in cartesian coords it is $\partial_x\hat\imath + \partial_y\hat\jmath + \partial_z\hat k$ ... the "times" symbol is a cross product and $\partial_x$ indicates the partial derivative wrt the x coordinate.)

You seem to be forgetting that (a) the electric and magnetic fields are vectors, and that (b) they may vary with space as well as time.

So when you say:
That would be: $\vec B(\vec r, t) = B_0\sin\omega t$ ... is that what you are saying?
That would be a magnetic field whose strength is the same everywhere in space (though it varies in time).
So if you try to solve: $\nabla\times \vec E(\vec r,t) = -\partial_t\vec B(\vec r,t)$ what do you get?

12. Jun 6, 2016

### Dexter Neutron

No that's not what I am talking about. I am just talking about the term that causes it to depend on time( and not taking into context the term that causes it to vary in space) thus causing its derivative to still depend sinusoidally on time(though it may or may not very in space) as to what Dale has said.

13. Jun 6, 2016

### Simon Bridge

OK - so $\vec B(\vec r,t) = B_0f(\vec r)\sin\omega t$(direction?) - leaving the space variation arbitrary.
Now solve the equation: $\nabla\times\vec E = -\partial_t\vec B$

14. Jun 6, 2016

### Delta²

In my opinion the differential form of Maxwell's equations is not the proper if we want to have an intuitive and qualitative understanding of how EM waves are generated and propagate. The integral form is much more suitable for this.

However I 've to admit that the ultimate (if I can call it that way) proof that time-varying E and B fields are waves that travel with the speed of light, comes from starting with Maxwell's equations in differential form and with standard mathematical processing to arrive at the conclusion that the fields satisfy the three dimensional wave equation.

15. Jun 6, 2016

### Staff: Mentor

I didn't say that it does not vary in time. I said that it does vary in space. That is what the upside down triangle means.

Specifically $\nabla \times E$ is a particular type of spatial variation called a "curl". Roughly speaking it describes how the E field twists in space. So Faraday's law says that the amount the E field curls in space is equal to the amount the B field changes in time. This curling in space is how the fields at one point affect the field at the neighboring point.

Last edited: Jun 6, 2016
16. Jun 6, 2016

### Staff: Mentor

This is false, as I already explained. Maxwell's equations explicitly specify the variation in space.

17. Jun 6, 2016

### Staff: Mentor

If you don't take all of Maxwell's equations into context then you are not really talking about an EM wave.