Direction of the Force from an EM Wave

[Drakkith]

Quick question. If the EM field vectors are perpendicular to the direction that an EM wave is traveling, how can light push objects away from the light source, such as in the use of a solar sail?

 

[nsaspook]

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

 

[KatamariDamacy]

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

It says: ...E and B are always perpendicular to each other and the direction of propagation... and their time and position dependences are:

Both E and B fields are three-dimensional or volumetric, uniformly so. How then they can be "perpendicular" to anything, especially E field with its spherical symmetry? What are those unique directions or vectors of E and B fields that are perpendicular, what defines them?

Those two equations seem to suggest E and B fields of EM wave do not originate at the same point, but are actually separate entities each with its own origin. Is this recognized by any other theory or expressed through any other equation?

 

[vanhees71]

Of course, you cannot use plane waves to formally answer your question. Plane waves do not exist in nature but only wave packets/trains with finite total energy and momentum. The total momentum is given by the Poynting vector. In Heaviside-Lorentz units the momentum density of such a field is
[tex]\vec{\Pi}=\frac{\vec{E} \times \vec{B}}{c}.[/tex]
If the wave packet travels in good approximation in a single direction (i.e., if the spatial Fourier spectrum is sharply peaked around a wave vector [itex]\vec{k}[/itex], you can show that the total momentum of the wave packet is mostly into the direction of [itex]\vec{k}[/itex].

 

[KatamariDamacy]

If the wave packet travels in good approximation in a single direction (i.e., if the spatial Fourier spectrum is sharply peaked around a wave vector [itex]\vec{k}[/itex], you can show that the total momentum of the wave packet is mostly into the direction of [itex]\vec{k}[/itex].

So the momentum varies with frequency. Only, the higher the frequency the less momentum vector points in the direction of propagation and more it points perpendicular to it, and yet we measure otherwise. Don't we?

 

[Drakkith]

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

Thanks, but I'm afraid I don't quite understand. Can you elaborate a bit on how this answers my question?

 

[jtbell]

Quick question. If the EM field vectors are perpendicular to the direction that an EM wave is traveling, how can light push objects away from the light source, such as in the use of a solar sail?

This is a kind of hand-wavy argument that I think gives the correct result in the end:

Suppose ##\vec E## is in the +x direction and ##\vec B## is in the +y direction. Let a positive charge be initially at rest.

##\vec E## exerts a force ##\vec F_E = q \vec E## which is in the +x direction, and the charge starts moving in the +x direction with an initial velocity ##\vec v##.

Now it starts to feel the effect of the magnetic field which exerts another force ##\vec F_B = q \vec v \times \vec B## which is perpendicular to both ##\vec v## and ##\vec B##. By the right-hand rule, this force is in the +z direction.

After a little while, ##\vec E## switches direction (-x) and so does the component of velocity along the x-axis.^(note) But ##\vec B## also switches direction (-y) so the right-hand rule still gives a magnetic force in the +z direction.

For a negative charge, the electric force and the perpendicular velocity are in the opposite direction (-x), but the minus sign on q makes the magnetic force still be in the +z direction.

So the "thrust force" is basically the magnetic force on the particle.

----

^(note)This is a bit hand-wavy because the perpendicular velocity doesn't switch direction at the same time that ##\vec E## and ##\vec B## do. (Think of driving a mass hanging on a spring with an oscillating force. Sometimes you're pushing "with" the motion and sometimes "against" it.) So it seems that sometimes the magnetic force should be in the -z direction.

Also, once the particle starts moving in the +z direction (as well as in the +x direction) the direction of the velocity starts to vary in a complicated way.

Nevertheless, I expect that the net effect of the magnetic force over a long time should turn out to be in the +z direction.

 

[Drakkith]

Thanks, Jt. That makes sense.

 

[vanhees71]

Let's also argue with the Poynting vector or the momentum-density of the electromagnetic field. It's given by
[tex]\vec{\Pi}=\frac{\vec{E} \times \vec{B}}{c}.[/tex]
Let's assume a plane wave first. We make the ansatz
[tex]\vec{E}=\vec{E}_0 \cos(\omega t - \vec{k} \cdot \vec{x}),
\vec{B}=\vec{E}_0 \cos(\omega t - \vec{k} \cdot \vec{x}).[/tex]
Each field component obeys the wave equation, as follows from the free Maxwell equations by some simple algebra:
[tex]\frac{1}{c^2} \partial_t^2 \vec{E}-\Delta \vec{E}=0,[/tex]
and this gives
[tex]\omega^2=c^2 \vec{k}^2,[/tex]
i.e., the dispersion relation for em. waves in a vacuum.

Now this doesn't solve the Maxwell equations completely, i.e., the vectors [itex]\vec{E}_0[/itex] and [itex]\vec{B}_0[/itex] are not independent of each other. Since both fields are source free, first of all we have
[tex]\vec{\nabla} \cdot \vec{E}=0 \; \Rightarrow \; \vec{k} \cdot \vec{E}_0=0, \quad \vec{\nabla} \cdot \vec{B}=0 \; \Rightarrow \; \vec{k} \cdot \vec{B}_0=0.[/tex]
Both, the electric an magnetic field components are perpendicular to the wave vector [itex]\vec{k}[/itex].
further we have Faraday's Law
[tex]\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B},[/tex]
from which you get
[tex]\vec{k} \times \vec{E}_0=-\frac{\omega}{c} \vec{B}_0.[/tex]
From this we get
[tex]\vec{B}_0=-\hat{k} \times \vec{E}, \quad \text{where} \quad \hat{k}=\frac{\vec{k}}{|\\vec{k}|}.[/tex]
Then there's the Ampere-Maxwell Law,
[tex]\vec{\nabla} \times \vec{B}=\frac{1}{c} \partial_t \vec{E} \; \Rightarrow \; \vec{k} \times \vec{B}_0=\frac{\omega}{c} \vec{E}_0[/tex]
or
[tex]\vec{E}_0=\hat{k} \times \vec{B}_0.[/tex]
To see that this is compatible with the previous equation, we calculate
[tex]\hat{k} \times \vec{B}_0=-\hat{k} \times (\hat{k} \times \vec{E}_0) = -\hat{k} (\hat{k} \cdot \vec{E}_0) + \vec{E}_0 \hat{k}^2=\vec{E}_0,[/tex]
i.e., we have a valid solution of the Maxwell equation in terms of plain waves.

[tex]\vec{P}=\frac{1}{c} \vec{E} \times \vec{B}=\frac{1}{c} \vec{E}_0 \times \vec{B}_0 \cos^2(\omega t-\vec{k} \cdot \vec{x}).[/tex]
Now from the above calculations we have
[tex]\vec{E}_0 \times \vec{B}_0=\vec{E}_0 \times (\hat{k} \times \vec{E}_0)=\hat{k} \vec{E}_0^2 - \vec{E}_0 (\hat{k} \cdot \vec{E}_0) = \hat{k} \vec{E}_0^2.[/tex]
As you see, the momentum of the electromagnetic field is directed along [itex]\hat{k}[/itex], and that momentum (per unit time and unit area) is transferred to an object that absorbs this radiation completely (if it's completely reflected you get twice as much according to energy-momentum conservation).

As I stressed in my previous mail: Such a plane wave doesn't exist in nature, but you can use the above calculation to make wave packets of finite total energy and momentum content and work with them in terms of a Fourier transformation. If the Fourier coefficients are sharply peaked around a certain value [itex]\vec{k}_0[/itex] you get something nearly as a plane wave, and then the main momentum is again nearly in the direction of [itex]\vec{k}_0[/itex].

 

[nsaspook]

Thanks, but I'm afraid I don't quite understand. Can you elaborate a bit on how this answers my question?

It seems that's been done by others but here's a visual (Projection of Circular Motion) and the original paper that can elaborate better than I.

http://en.wikisource.org/wiki/On_the_Transfer_of_Energy_in_the_Electromagnetic_Field

Starting with Maxwell's theory, we are naturally led to consider the problem: How does the energy about an electric current pass from point to point — that is, by what paths and according to what law does it travel from the part of the circuit where it is first recognisable as electric and magnetic to the parts where it is changed into heat or other forms?

The aim of this paper is to prove that there is a general law for the transfer of energy, according to which it moves at any point perpendicularly to the plane containing the lines of electric force and magnetic force, and that the amount crossing unit of area per second of this plane is equal to the product of the intensities of the two forces, multiplied by the sine of the angle between them, divided by 4\pi, while the direction of flow of energy is that in which a right-handed screw would move if turned round from the positive direction of the electromotive to the positive direction of the magnetic intensity. After the investigation of the general law several applications will be given to show how the energy moves in the neighbourhood of various current-bearing circuits.
...
I have therefore given several cases in considerable detail of the application of the mode of transfer of energy in current-bearing circuits according to the law given above, as I think it is necessary that we should realize thoroughly that if we accept Maxwell's theory of energy residing in the medium, we must no longer consider a current as something conveying energy along the conductor. A current in a conductor is rather to be regarded as consisting essentially of a convergence of electric and magnetic energy from the medium upon the conductor and its transformation there into other forms. The current through a seat of so-called electromotive force consists essentially of a divergence of energy from the conductor into the medium. The magnetic lines of force are related to the circuit in the same way throughout, while the lines of electric force are in opposite directions in the two parts of the circuit — with the so-called current in the conductor, against it in the seat of electromotive force. It follows that the total E.M.I. round the circuit with a steady current is zero, or the work done in carrying a unit of positive electricity round the circuit with the current is zero. For work is required to move it against the E.M.I. in the seat of energy, this work sending energy out into the medium, while an equal amount of energy comes in in the rest of the circuit where it is moving with the E.M.I. This mode of regarding the relations of the various parts of the circuit is, I am aware, very different from that usually given, but it seems to me to give us a better account of the known facts.

 

[jtbell]

It says: ...E and B are always perpendicular to each other and the direction of propagation... and their time and position dependences are:

Both E and B fields are three-dimensional or volumetric, uniformly so.

What do you mean by "volumetric" here?

How then they can be "perpendicular" to anything, especially E field with its spherical symmetry?

There is no spherical symmetry here. These are standard equations for a plane wave.

What are those unique directions or vectors of E and B fields that are perpendicular, what defines them?

The vector ##\vec k## points in the direction of propagation. ##\vec E## and ##\vec B## are each perpendicular to ##\vec k##. ##\vec E## and ##\vec B## are also perpendicular to each other. These perpendicularity conditions are not contained in the two equations above, but are specified separately. They follow from the requirement that ##\vec E## and ##\vec B## must satisfy Maxwell's equations. See for example this page:

http://farside.ph.utexas.edu/teaching/em/lectures/node48.html

specifically equations (445), (446) and (448).

 

[KatamariDamacy]

What do you mean by "volumetric" here?

Three-dimensional, they occupy a volume, rather than a point, a line or a plane.

There is no spherical symmetry here. These are standard equations for a plane wave.

If there was no symmetry there would not be any attraction or repulsion, Coulomb's and Lorentz force equations your referred to would not work. And if the symmetry was not spherical (toroidal for magnetic field) there wouldn't be "4Pi * r^2" in those equations, but there is.

What symmetry or geometry do you think those E and B fields have? They sure must have some geometry otherwise they couldn't possibly ever be "perpendicular" to anything.

The vector ##\vec k## points in the direction of propagation. ##\vec E## and ##\vec B## are each perpendicular to ##\vec k##. ##\vec E## and ##\vec B## are also perpendicular to each other. These perpendicularity conditions are not contained in the two equations above, but are specified separately. They follow from the requirement that ##\vec E## and ##\vec B## must satisfy Maxwell's equations. See for example this page:

http://farside.ph.utexas.edu/teaching/em/lectures/node48.html

specifically equations (445), (446) and (448).

Only lines and planes can be perpendicular to other lines or planes. I see those equations, but I don't see what lines/vectors or planes of E and B field they are referring to when they say they are perpendicular. Maybe their velocity or momentum vectors are perpendicular?

 

[Orodruin]

You are confusing having a spatial symmetry with E and B fields at a given point in space. At each point in space, the E field has a direction and a magnitude and so does the B field. In a free EM wave, the E and B fields will be perpendicular at each point in space.

That only lines or planes can be perpendicular is just wrong. Two vectors are perpendicular if their inner product is zero. In fact, perpendicularity of lines and planes may be defined using their tangent/normal vectors.

 

[vanhees71]

As I've shown in my posting explicitly, the three vectors [itex]\vec{k}[/itex], [itex]\vec{E}_0[/itex] , and [itex]\vec{B}_0[/itex] build a right-handed set of orthogonal coordinates.

BTW in this posting I made some sign mistakes in typing my calculations into the forum. Unfortunately, I cannot edit this posting anymore. So here are the correct equations:

[tex]\vec{B}_0=\hat{k} \times \vec{E}_0, \quad \hat{k} \times \vec{B}_0=-\vec{E}_0.[/tex]

In the evaluation of the Poynting vector the signs are correct.

 

[KatamariDamacy]

At each point in space, the E field has a direction and a magnitude and so does the B field.

What's the magnitude of E field in EM wave of color red, for example? Same as in blue EM wave? I know about velocity, acceleration, force or momentum direction, what direction of E and B fields are you talking about, can you name it? What is it, what defines it?

 

[ChrisVer]

E,B fields are vectors . As vectors they have direction.
If for example your wave propagates along the z-axis (that is [itex]\hat{k}=\hat{z}[/itex]) then the E,B fields will be on the xy plane (E might have x and B might have y...)

 

[Dale]

What's the magnitude of E field in EM wave of color red, for example? Same as in blue EM wave?

The difference between a red and a blue EM wave is the frequency, not the direction. A blue EM wave has a high frequency and a red EM wave has a low frequency. If you trace a ray from the blue/red object to the eye then the E and B fields of that wave will both be perpendicular to that ray and also perpendicular to each other at every point along the path. The eye does not distinguish polarization, but if the wave is vertically polarized then the E field will be vertical and the B field horizontal, and vice versa if the wave is horizontally polarized.

I know about velocity, acceleration, force or momentum direction, what direction of E and B fields are you talking about, can you name it? What is it, what defines it?

At each point the E field has a magnitude and a direction. At each point the B field has a magnitude and a direction. They are vector fields. That is what it means for a field to be a vector field. There is no special name given to the direction of the fields, but they have one at each point nonetheless.

I recommend starting here:
http://hyperphysics.phy-astr.gsu.edu/hbase/emcon.html#emcon

 

[KatamariDamacy]

If for example your wave propagates along the z-axis (that is [itex]\hat{k}=\hat{z}[/itex]) then the E,B fields will be on the xy plane (E might have x and B might have y...)

Define "will be". Maybe E and B fields will exist only in their two-dimensional plane? Maybe E and B fields will move in their plane and so their velocity vectors will be perpendicular? Or what?

 

[Dale]

KatamariDamacy, I will again remind you to be inquisitive rather than argumentative. If you wish to learn you are welcomed, but there is no point in attempting to argue about such basic things. Please study the material provided and ask questions about parts of the material that you do not understand.

 

[KatamariDamacy]

The difference between a red and a blue EM wave is the frequency, not the direction. A blue EM wave has a high frequency and a red EM wave has a low frequency. If you trace a ray from the blue/red object to the eye then the E and B fields of that wave will both be perpendicular to that ray and also perpendicular to each other at every point along the path. The eye does not distinguish polarization, but if the wave is vertically polarized then the E field will be vertical and the B field horizontal, and vice versa if the wave is horizontally polarized.

Do you confirm then the magnitude of E and B fields is the same for every EM wave?

At each point the E field has a magnitude and a direction. At each point the B field has a magnitude and a direction. They are vector fields. That is what it means for a field to be a vector field. There is no special name given to the direction of the fields, but they have one at each point nonetheless.

At each point from where? Is magnitude relative to distance, squared? -- Yes, vector fields are three-dimensional and have infinitely many directions, that is vectors, one at each any arbitrary point in space. So, if at each point the E field has a magnitude and a direction, then which one of all those directions do you call "the direction", and what's unique about it?

 

[Orodruin]

At each particular point, the E field points in the direction it is pointing. You seem to want to define a direction of the entire EM field as a whole taken over the entire space, this is physically not very relevant. The acceleration of a charge will depend on the field strength at the point in space where it is located.

 

[KatamariDamacy]

At each particular point, the E field points in the direction it is pointing.

At each point in space vectors in E vector field point to the center of the field, or to directly opposite direction in case of positive polarity, that's why we say electric fields have spherical symmetry. Obviously then vector fields have nothing to do with E and B fields being "perpendicular" in EM wave.

 

[Orodruin]

It has everything to do. It follows from Maxwell's equations. EM waves are not generated by stationary charges and you seem to be basing all of your intuition about E and B fields on your experience from those.

Just as a central field is a possible solution to Maxwell's equations, corresponding to a point charge, EM waves are also a solution, corresponding to an EM wave in free space with the E and B field perpendicular to the direction of propagation. This is really not a matter of discussion but a well established fact. If there are specific points in this that you are having trouble understanding, feel free to ask about them, but it really is not something you can change by arguing.

 

[Dale]

Do you confirm then the magnitude of E and B fields is the same for every EM wave?

No. For visible light the magnitude is related to the brightness. The brighter the light the greater the magnitude.

At each point from where, is magnitude relative to distance, squared? Yes, vector fields are three-dimensional and have infinitely many directions, that is vectors, one at each any arbitrary point in space. So, if at each point the E field has a magnitude and a direction, then which one of all those directions do you call "the direction", and what's unique about it?

Please read the material provided and ask questions about the material.

 

[ChrisVer]

how can I define a verb? Will be means their vectors will lie on the xy-plane and they'll be perpendicular (at least the parts of them which cause the energy propagation- lead to EM wave propagation =read about Poynting Vector). That means exactly what I wrote, if E is on x, then B will be on y axis... you can roll them around (eg circular polarization)...
E,B velocity vectors? What is that?
E,B are vectors themselves... a vector has a magnitude and a direction. E,B will be perpendicular.

 

[KatamariDamacy]

Just as a central field is a possible solution to Maxwell's equations, corresponding to a point charge, EM waves are also a solution, corresponding to an EM wave in free space with the E and B field perpendicular to the direction of propagation.

What do you think is their geometry? Do you think E and B are three-dimensional fields, or some kind of one-dimensional vectors, or what?

 

[ChrisVer]

1 dimensional vector= number...
A vector field:
http://mathworld.wolfram.com/VectorField.html
(fast definition)

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

 

[KatamariDamacy]

No. For visible light the magnitude is related to the brightness. The brighter the light the greater the magnitude.

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.

Please read the material provided and ask questions about the material.

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwavecon.html#c1

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

 

[KatamariDamacy]

1 dimensional vector= number...
A vector field:
http://mathworld.wolfram.com/VectorField.html
(fast definition)

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

Do you think E and B are one-dimensional vectors, two-dimensional vector fields, or three-dimensional vector fields?

 

[Nugatory]

What do you think is their geometry? Do you think E and B are three-dimensional fields, or some kind of one-dimensional vectors, or what?

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.

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

 

[ChrisVer]

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

The propagation of EM waves though need for [itex] \vec{E}(t,x,y,z),\vec{B}(t,x,y,z)[/itex] 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, [itex]\vec{S} = \frac{1}{\mu_{0}} \vec{E} \times \vec{B}[/itex] will be zero).

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

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

 

[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 [itex]\lambda[/itex] 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:
[itex] \vec{E}(t,x,y,z)= (E_{x}(t,x,y,z),E_{y}(t,x,y,z),E_{z}(t,x,y,z)) [/itex]
[itex] \vec{B}(t,x,y,z)= (B_{x}(t,x,y,z),B_{y}(t,x,y,z),B_{z}(t,x,y,z)) [/itex]

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