# On Electromagnetic Waves

1. Dec 12, 2008

### _F_

Hi all,

Can someone please elaborate on some of the following ideas so that I can get these down correctly?

1. is an electromagnetic wave a form of a self-perpetual motion machine?

2. since electric and magnetic fields contain energy, is this the hope that people (Tesla!) have for wireless energy transmission?

Thank you.

2. Dec 12, 2008

### olgranpappy

no.

i dont know.

3. Dec 12, 2008

### _F_

why not?

okay.

Thank you.

4. Dec 12, 2008

### Staff: Mentor

Obviously because a perpetual motion machine violates the conservation of energy.

Don't follow this line of conversation any more, discussion of such crackpot theories is useless and is prohibited by the https://www.physicsforums.com/showthread.php?t=5374".

Last edited by a moderator: Apr 24, 2017 at 9:09 AM
5. Dec 12, 2008

### Staff: Mentor

A "perpetual motion machine" usually means a device that can supply energy indefinitely without an external energy source. You can certainly extract energy from an electromagnetic wave, but you have to have an energy source in order to produce the wave in the first place. And when you extract energy from the wave, you absorb the wave, so if you want the process to continue, you have to generate new waves.

6. Dec 13, 2008

### _F_

I'm not trying to argue that you can use electromagnetic waves to harness any type of perpetual motion. I was just asking if an elecromagnetic wave is a type (maybe "type" is the wrong word here?) of perpetual motion?

Or maybe I'm mistaken. Do electromagnetic waves die out over some distance?

In space there is no friction, so anything with any initial velocity would go on forever and thats not really the type of perpetual motion I'm talking about. But in air or some other medium where there is friction,... perhaps electromagnetic waves die out? if not, why not?

7. Dec 13, 2008

### Hootenanny

Staff Emeritus
Indeed, in dielectric media electromagnetic waves are attenuated and so do "die out".

8. Dec 13, 2008

### olgranpappy

No, it is not. Maybe you are thinking that because the electric field induces the magnetic field that this back and forth motion is "perpetual". However, any undamped (and thus highly idealized) classical motion appears "perpetual" in this sense. Consider, for example, a steel marble rolling back and forth in the bottom of a glass bowl. Without damping there is no reason why the marble should ever stop rolling back and forth. But so what? This marble and bowl system is no more a perpetual motion machine than your wave of light.

9. Dec 13, 2008

### Mårten

Really interesting. We were just discussing this here, when looking at Maxwell 3 and Maxwell 4, and how the fields are inducing each other back and forth, in a sort of (but maybe not really) perpetual way. Still, I can't see that "_F_" had his question answered, "Do electromagnetic waves die out over some distance". Think of this:

1) EM waves are emitted from a source, travelling through vacuum. Do they ever die out? Do they faint with distance? If so, what damps the wave? Where in Maxwell's equations can I see this damping effect?

2) Concentrating on the electric field here. It will according to Coulomb's law decrease with distance squared. But is the inducing back-and-forth-mechanism counteracting this decreasing?

Like _F_ I'm also eager to know!

10. Dec 13, 2008

### Staff: Mentor

When you get far enough away from the source, the amplitude of the wave decreases as 1/r, if there is nothing to absorb it. The irradiance of the wave (power per unit area) is proportional to the square of the amplitude, so it decreases as $1/r^2$. This is what one expects from energy conservation: the area of a sphere centered on the source increases as $r^2$, so the total energy that passes through successive concentric spheres remains constant.

You don't see this directly in Maxwell's equations, but rather in their solution for e.g. the [oscillating] Hertzian dipole. Equations 1090 and 1091 at the link show the E and B fields in the "far field" of such a dipole.

You could say that, yes. The average E for the oscillating dipole decreases as 1/r instead of $1/r^2$, that is, less rapidly.

11. Dec 15, 2008

### Redbelly98

Staff Emeritus