Direction of the Force from an EM Wave

[ChrisVer]

Why are you asking what I think?
The definition given by the Wolfram is satisfying ,,, they are vectors:
\vec{E}(t,x,y,z)= (E_{x}(t,x,y,z),E_{y}(t,x,y,z),E_{z}(t,x,y,z))
\vec{B}(t,x,y,z)= (B_{x}(t,x,y,z),B_{y}(t,x,y,z),B_{z}(t,x,y,z))

The propagation of EM waves though need for \vec{E}(t,x,y,z),\vec{B}(t,x,y,z) to be perpendicular to each other... If they have some common component, then this component is not going to contribute to the EM wave (because the poynting vector which I asked for you to check, \vec{S} = \frac{1}{\mu_{0}} \vec{E} \times \vec{B} will be zero).

This in formulas means that if you have:
\vec{E}(t,x,y,z)= (E_{x}(t,x,y,z),0,E_{z}(t,x,y,z))
\vec{B}(t,x,y,z)= (B_{x}(t,x,y,z),0,B_{z}(t,x,y,z))
Then you won't have EM wave...

If though:
\vec{E}(t,x,y,z)= (E_{x}(t,x,y,z),0,E_{z}(t,x,y,z))
\vec{B}(t,x,y,z)= (0,B_{y}(t,x,y,z).0)
Then your EM wave will propagate towards \hat{S}:
\vec{S}=E_{x}(t,x,y,z)B_{y}(t,x,y,z) (\hat{x} \times \hat{y})+ E_{z}(t,x,y,z)B_{y}(t,x,y,z)(\hat{z} \times \hat{y})= E_{x}(t,x,y,z)B_{y}(t,x,y,z) \hat{z} - E_{z}(t,x,y,z)B_{y}(t,x,y,z) \hat{x}

 

[KatamariDamacy]

This is a question that you could only ask if either you still haven't looked at Maxwell's equations, or if you've looked and didn't understand what any of the symbols in them mean.

It's rhetorical. I was pretty clear that I know those E and B fields are three-dimensional vector fields, as usual.

If the latter, speak up and may be able to steer you towards the right introductory texts.

I welcome any texts that can explain those blue and red vectors called "field variations". Variations in what? Position, magnitude, direction, momentum, velocity, force ...?

 

[ChrisVer]

Since they have sketched over them \lambda wavelength, it's clear the axis is position...

 

[KatamariDamacy]

Why are you asking what I think?
The definition given by the Wolfram is satisfying ,,, they are vectors:
\vec{E}(t,x,y,z)= (E_{x}(t,x,y,z),E_{y}(t,x,y,z),E_{z}(t,x,y,z))
\vec{B}(t,x,y,z)= (B_{x}(t,x,y,z),B_{y}(t,x,y,z),B_{z}(t,x,y,z))

That is change in position over time, as I have been suggesting all along. So that little arrow denoting a vector is a reference to their velocity vector. It's not fields that are perpendicular by themselves, it's their velocity vectors that are perpendicular, that's the proper answer.

So when you have linearly polarized EM wave, that is what is "varying" - there is E field oscillating up and down in a vertical plane, and B field oscillating perpendicular to it, left and right in a horizontal plane. Like this:

Isn't that right?

 

[jtbell]

So those blue and red vectors are "field variations". Variations in what? Position, magnitude, direction, momentum, velocity, force ...?

The electric field $\vec E$ and magnetic field $\vec B$ at a specific location are defined in terms of the force that a charge q would experience if it were at that location:

$$\vec F = q(\vec E + \vec v \times \vec B)$$

You can think of the electric field as the "ability to produce a certain amount of force per unit of charge", and the magnetic field as the "ability to produce a certain amount of force per unit of charge and per unit of velocity, modified by the angle between the field and the velocity" (gad, what a mouthful...).

In an electromagnetic wave, the fields vary from one point to another, and from one instant in time to another. That's all Hyperphysics means by "field variations." I would have labeled that graph with simply "electric field" and "magnetic field." The spatial variations are obvious from looking at the graph.

It might be worth noting that this graph only shows the variation of $\vec E$ and $\vec B$ along a single line represented by the axis of the graph. It doesn't show the "volumetric" nature of the wave. I once attempted to describe this (for a plane wave) using a different diagram and a set of instructions:

https://www.physicsforums.com/showthread.php?p=533190#post533190

There's some clarifying discussion about the diagram in this thread:

https://www.physicsforums.com/showthread.php?p=2701711#post2701711

 

[Nugatory]

I welcome any texts that can explain those blue and red vectors called "field variations". Variations in what? Position, magnitude, direction, momentum, velocity, force ...?

The magnitude of both the E and B vectors at the point x and time t is proportional to sin(kx-ct).

That picture is a graph in which the magnitude of the E and B vectors at a fixed time are plotted on two axes as a function of the position on the third axis, the x-axis. There is no spatial distance associated with the difference between the positions of the heads and tails of the blue arrows, those just represent the magnitude of the vector at that point on the x-axis at that time.

Any decent E&M text will see you through this, although I personally like Purcell.

 

[ChrisVer]

That is change in position over time, as I have been suggesting all along. So that little arrow denoting a vector is a reference to their velocity vector. It's not fields that are perpendicular by themselves, it's their velocity vectors that are perpendicular, that's the proper answer.

Where did you see velocity in that quote? Velocity of something, in general, would be a derivative, do you see anything like that? I never mentioned a change of position over time (that would be a \dot{\vec{x}} \equiv \frac{d \vec{x}}{dt})- What I wrote is that the E,B fields are vectors and their components depend on the position and time. As for your scheme, it's the fields at some fixed time and not their velocities.

 

[Dale]

Brightness is a function of photons quantity, that is light intensity. For all we know photons are neutrally charged, we have not measured anything related to their electric or magnetic fields, it's supposed to be zero.

Until you learn classical EM there is no point in discussing quantum mechanical EM. However, briefly, we do know that photons are uncharged and their EM fields are most definitely not zero since they are quantized excitations of the EM fields.

So those blue and red vectors are "field variations". Variations in what? Position, magnitude, direction, momentum, velocity, force ...?

The blue and red arrows represent the direction and magnitude of the magnetic and electric field at the location of the tail the corresponding arrow. The "field variations" are represented by the fact that the corresponding arrows in neighboring locations have different magnitudes and/or directions.

The magnitude and direction of the arrows represents the magnitude and direction of the field. The position of the tail represents the position where the field is being shown. Momentum, velocity, and force are not represented here.

 

[ChrisVer]

Katamari, could you please inform us about your current level of understanding of physics?
It's difficult to try to explain someone something if you don't know their background... and maybe some of our posts would confuse you rather than clearing the topic to you...

 

[Dale]

That is change in position over time, as I have been suggesting all along.

No. It is change in the field over space or over time. If you look at Maxwell's equations you will see terms like $\nabla \cdot E$ and $\nabla \times B$ which represent changes in the E and B fields over space as well as terms like $\partial E/\partial t$ which represent changes in the E and B fields over time.

So that little arrow denoting a vector is a reference to their velocity vector. It's not fields that are perpendicular by themselves, it's their velocity vectors that are perpendicular, that's the proper answer.

No, that is not the proper answer, that is wrong. In a plane wave the E field is perpendicular to the B field. The fields themselves don't even have velocity vectors.

So when you have linearly polarized EM wave, that is what is "varying" - there is E field oscillating up and down in a vertical plane, and B field oscillating perpendicular to it, left and right in a horizontal plane.

Yes, this is correct. In a vertically polarized plane wave the magnitude of the E and B fields oscillate, the E field in a vertical direction and the B field in a horizontal direction.

 

[unter71]

If you look at Maxwell's equations you will see terms like $\nabla \cdot E$ and $\nabla \times B$ which represent changes in the E and B fields over space as well as terms like $\partial E/\partial t$ which represent changes in the E and B fields over time.

http://en.wikipedia.org/wiki/Gauss's_law

Gauss law is essentially equivalent to Coulomb's law, it's closed surface integral of it, and they both basically describe inverse square law. The change or variation they describe is in flux gradient or charge density of the field relative to its source or sink. This magnitude gradient is not time-varying, it is constant and only spatially-varying proportionally to distance squared.

http://en.wikipedia.org/wiki/Ampère's_circuital_law

In the same manner Ampère's law is analogous to loop or line integral of Biot-Savart law. This time the change in flux gradient is not constant but varies with electric current, which means the magnitude of B field is proportional to E field density and velocity. E field magnitude relative to its source or sink remains constant per inverse square law in this equation as well, only its location or velocity changes.

The fields themselves don't even have velocity vectors.

You didn't really embrace the answers given in post #7 and #9. Post #9 points to Poynting vector and why is momentum in the direction of wave propagation, and post #7 explains that more clearly and in more detail. You will not find this explanation in Wikipedia though, it's a bit controversial. In any case, here you can see in even more detail how exactly Poynting vector and momentum is related to Lorentz force and field velocity vectors:

http://www2.ph.ed.ac.uk/~mevans/em/lec14.pdf

As they say, there is a velocity component of E and B field not only in the direction of wave propagation +z, but also +/- x direction for E field, and for B field in +/- y direction. So the answer to your question as to why radiation pressure force is directed in the direction of wave propagation when EM field vectors are perpendicular to that direction is because of Lorentz force. It's because there is a perpendicular velocity component of the fields in the wave equation that is causing Lorentz force perpendicular to it, and so it points back in the direction of wave propagation. But again, there a controversy around it and not everyone agrees. Like these guys here:

http://arxiv.org/pdf/0807.1310v5.pdf

 

[Dale]

here you can see in even more detail how exactly Poynting vector and momentum is related to Lorentz force and field velocity vectors:

http://www2.ph.ed.ac.uk/~mevans/em/lec14.pdf

This reference never speaks of a field velocity, only of a wave velocity, and the other reference only speaks of the velocity of a charge. It is clear that EM waves have velocity, but the fields do not.

You should not confuse a wave velocity with a velocity of the thing that is being waved. For example, in solids you may have a shear wave which propagates quite quickly in, e.g., the z direction while the solid has no motion at all in z and only some small oscillation velocity in the x direction. So wave velocity and material velocity are clearly different things.

Similarly with EM. Although the EM waves have a velocity, the fields themselves do not.

 

[unter71]

I'm not sure if you accept Lorentz force explanation or not. But if you do then you see the velocity has to be perpendicular to the direction of wave propagation in order for Lorentz force to turn that direction back in the direction of propagation.

It seems you don't consider the speed of E and B field "oscillation" to be a velocity, but as far as Lorentz force is concerned it is. There is a spatial rate of change of E field magnitude, whether you call it a velocity or frequency it's a composite vector in the general direction of propagation and the direction perpendicular to it.

 

[Dale]

The only v in the Lorentz force is the v of the charge, not the field.

 

[unter71]

The only v in the Lorentz force is the v of the charge, not the field.

Poynting vector is derived from the Lorentz force law. There is a momentum within EM wave that is already in the direction of propagation before any collision with electrons or protons occur.

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


When the collision occurs the impacted electron will move in +/- x plane up or down depending on whether E field approached it from below or above, and then Lorentz force of the EM wave interacts with Lorentz force of the charge.

 

[Dale]

Sure. None of that implies that the fields have velocity.

Do you understand that there is a distinction between the wave, the charge, and the field?

 

[unter71]

Sure. None of that implies that the fields have velocity.

Do you understand that there is a distinction between the wave, the charge, and the field?

There is no clear distinction. All those equations actually deal with what is called "charge density" and "current density", which is really just a description of a vector field, static and dynamic.

Look at Poynting's theorem:

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

...where ρ is the charge density of the distribution and v its velocity. This charge density oscillates up and down certain number of times per second in EM wave, which can be expressed as velocity perpendicular to the direction of propagation, and produces Lorentz force in the direction of propagation.

Why do you not recognize E field oscillation in a vertical plane to be actual movement of the field going up and down? If this "charge density" of E filed shifts up and down, then isn't that the same thing as to say the whole field moves itself?

 

[Dale]

There is no clear distinction. All those equations actually deal with what is called "charge density" and "current density", which is really just a description of a vector field, static and dynamic.

Frankly, this argument seems to undermine your claim even further. If you use charge and current density then you are even further removed from velocity since there is no unique relationship between velocity and current density.

Do you have a peer reviewed reference that makes this argument that attributes velocity to the field Itself and not just the wave or the charge?

 

[unter71]

Charge density distribution and its velocity are explicit in Poynting's theorem, so if that applies to EM waves and radiation pressure, as they say, then that's already peer reviewed and accepted with the status of being a theorem. Velocity is implicit in "current density" because it's measured in amperes per unit area.

Whether you think of it as movement or oscillation this change in E and B field of EM wave varies spatially in a perpendicular plane along as it moves in the direction of propagation. I think we agree on that. I don't understand why do you hesitate to recognize this perpendicular spatial variation over time can be expressed as having a velocity. It's just a frequency broken down into x, y, z vector components.

 

[rubi]

Poynting vector is derived from the Lorentz force law. There is a momentum within EM wave that is already in the direction of propagation before any collision with electrons or protons occur.

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

You don't need the Lorentz force to come up with the Poynting vector. You can derive Poynting's theorem for free fields and it is just (a part of) the conservation of the stress-energy tensor, which is a consequence of Lorentz-invariance. For a free field, you have $\partial_t u + \nabla \mathrm S = 0$, which is a continuity equation that tells you that the Poynting vector determines the direction of the energy flux (and of course also its magnitude).

 

[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.