Direction of the Force from an EM Wave

[Dale]

Velocity is implicit in "current density" because it's measured in amperes per unit area.

So what is the implicit velocity of a 12 A/m^2 current density? There is no unique velocity. It could be almost anything, depending on the density of charge carriers.

I don't understand why do you hesitate to recognize this perpendicular spatial variation over time can be expressed as having a velocity.

I recognize that. The spatial variation over time is the wave, and that does have a velocity. What I do not recognize and what you have not provided any references to support is your repeated unsubstantiated claim that the field itself has a velocity (not the wave).

Before you reply, you may wish to review the forum rules regarding personal speculation and references from the professional literature.

 

[unter71]

So what is the implicit velocity of a 12 A/m^2 current density? There is no unique velocity. It could be almost anything, depending on the density of charge carriers.

It can not be zero, that's what makes it implicit. It's explicit or uniquely defined when you know charge density. Together, charge density and its velocity define what current density is. This velocity is usually called "drift velocity". Do you know what is charge density in EM wave?

http://en.wikipedia.org/wiki/Current_density

http://en.wikipedia.org/wiki/Drift_velocity

I recognize that. The spatial variation over time is the wave, and that does have a velocity. What I do not recognize and what you have not provided any references to support is your repeated unsubstantiated claim that the field itself has a velocity (not the wave).

http://en.wikipedia.org/wiki/Poynting's_theorem

Poynting's theorem refers to current density or charge density that has specific velocity, where J=pv. When applied to EM waves, if this charge density and its velocity do not correspond to EM fields, then you tell me what is their relation and how do you explain Lorentz force vector points in the direction of propagation if the velocity in Lorentz force equation is not perpendicular to it?


EM wave is a wave exactly because E and B fields oscillate in the direction perpendicular to propagation vector. Their change in position over time is defined like this:

http://en.wikipedia.org/wiki/Poynting_vector#In_plane_waves

Position of E and B fields obviously varies not only along their propagation vector but also in the direction perpendicular to it, hence "wave" and hence "oscillation". So if you don't call this rate of change of position in perpendicular direction a velocity, then how do you call it?

 

[Dale]

It can not be zero, that's what makes it implicit. It's explicit or uniquely defined when you know charge density.

Your idea doesn't hold water. Charge density can be 0 with a non-zero current density. In fact, this is the usual case in wires.

Do you know what is charge density in EM wave?

Yes, 0.

Poynting's theorem refers to current density or charge density that has specific velocity, where J=pv.

That doesn't work in the usual case where ρ=0 and J≠0.

EM wave is a wave exactly because E and B fields oscillate in the direction perpendicular to propagation vector. Their change in position over time is defined like this:

http://en.wikipedia.org/wiki/Poynting_vector#In_plane_waves

Again, this is not controversial. The wave has a velocity, the field does not.

Position of E and B fields obviously varies not only along their propagation vector but also in the direction perpendicular to it, hence "wave" and hence "oscillation". So if you don't call this rate of change of position in perpendicular direction a velocity, then how do you call it?

This is not even true for a plane wave (the fields point in the perpendicular direction, but they only vary in the parallel direction). And you STILL don't have a reference to support assigning a velocity to the field.

 

[ChrisVer]

Yes, the fields do have a velocity (meaning they change with time - \dot{\vec{E}} \ne 0) in the perpendicular direction, however this velocity doesn't account for propagation of energy or whatever...it's not a change of position, it's a change of the field...So I don't understand why are you confusing this...

 

[Dale]

I cannot tell if you agree with me or not. Sometimes the English is confusing so the math is more clear.

For a x-polarized plane wave propagating in the z direction:

$E=(E_0 \cos(\omega t - k z),0,0)$

so $\frac{\partial}{\partial x}E=\frac{\partial}{\partial y} E=0$ and only $\frac{\partial}{\partial z} E \ne 0$

There is also $\frac{\partial}{\partial t}E \ne 0$ but as you correctly mentioned E is not a position so $\frac{\partial}{\partial t}E$ is not a velocity.

 

[ChrisVer]

yes, it's a rate of change of the field (that's why I explained what I meant by velocity) and it's given for a particular POINT in space (it's not a change of position), it's at the point \vec{x}_{0} the Electric field is going to oscillate along the x-axis (in your example), with a rate/"velocity" \partial_{t}E(t,\vec{x}_0) not that it changes anything in position.
Although the pictures show it as a vector in space, it's not such...the lengths of the vectors denote the magnitude of E and not that it is spatially expanded.

 

[rubi]

Introducing charge and current density into this discussion is entirely pointless. Plane waves are solutions to the free Maxwell equations ($\rho = 0$, $\mathrm j = \mathrm 0$).

 

[ChrisVer]

The Lorentz Force gives you the force applied on a charged particle moving at velocity v within a magnetic field B (or and E)... it has nothing to do with the fields themselves.

 

[Dale]

yes, it's a rate of change of the field (that's why I explained what I meant by velocity) and it's given for a particular POINT in space (it's not a change of position), it's at the point \vec{x}_{0} the Electric field is going to oscillate along the x-axis (in your example), with a rate/"velocity" \partial_{t}E(t,\vec{x}_0) not that it changes anything in position.
Although the pictures show it as a vector in space, it's not such...the lengths of the vectors denote the magnitude of E and not that it is spatially expanded.

Yes, I see we agree. Sometimes the math really helps clarify things.