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Plane Waves and the Poynting Vector 
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#1
Jan1511, 03:02 PM

P: 21

Why is it that that poynting vector is independent of distance from the source?
Is it because EM waves are plane waves? Furthermore I do not fully understand why EM waves have to be plane waves. I understand that changing magnetic fields give rise to electric fields and vice versa, but does that constitute a constant E and B field over an infinite plane? My book states that the intensity of the wave decreases as 1/r^2, but I don't see this anywhere in the poynting vector, and doesn't that violate the premise of a plane wave? 


#2
Jan1511, 03:26 PM

Sci Advisor
P: 5,528

EM waves do not *have* to be plane waves; plane waves are a solution to Maxwell's equations, and so arbitrary solutions can be decomposed into a summation of plane waves. The intensity of the electromagnetic field, for a plane wave, does not have 1/r^2 dependence: the intensity of spherical waves do. 


#3
Jan1811, 10:15 AM

P: 21

So if the poynting vector has no information about the source, does that mean the distance from the source doesn't matter when describing Power/Area
When I spoke of plane waves I just wanted to know why is it that EM waves can be described as plane waves. I want to build up some physical inituition of plane waves, but as of now I cannot imagine how EM waves have constant values over their infinite plane. My intuition tells me that it should decrease. Or is this where the "changing E field give rise to B field and visa versa?" 


#4
Jan1811, 10:55 AM

P: 3,898

Plane Waves and the Poynting Vector
I believe when the wave propagate out from a source far away, even though it start out in spherical shape, but when the radius getting very large, the surface approx a straight plane rather than a spherical shape.
From the solution of Maxwell's equation of: [tex]\nabla X \vec E \;=\; \frac{\partial \vec B}{\partial t} \;\hbox { and }\; \nabla X \vec H \;=\; \vec J \;+\; \frac{\partial \vec D}{\partial t} [/tex] E and B are perpendicular to each other and the direction of propagation is perpendicular to both of them. Both E and B are in a plane that propagate in direction of the normal of the plane. Poynting vector: [tex]\vec P = \vec E X \vec H[/tex] at a point. It is calculated from E and B at that point. It say nothing about where the E and B come from. It tell you the EM energy density (W/unit area) and the direction of the energy flow. These are my understanding, please correct me if I am wrong. I am studying Poynting vector also. 


#5
Jan1811, 11:34 AM

Sci Advisor
P: 5,528

This is one way to understand diffraction by edges say an initially flat wavefront is truncated by an aperture or an edge. The wavefront is then given by a summation of multiple plane waves, all traveling is slightly different directions the wavefront spreads out during propagation. 


#6
May1211, 03:49 PM

P: 617

The distance from the source does matter. Any localized source of radiation will create to a good approximation outwardly traveling spherical waves. At larger distances from the source, the same total energy is spread over a larger sphere and must be weaker. That is why we need so many cell phone towers spread all over the map, instead of on huge tower in the ocean, the signal gets too weak if you are too far away. Mathematically this means that the Poynting vector is proportional to 1/r^2 for spherical waves, i.e. radiation caused by small sources.



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