# Direction of the Force from an EM Wave

1. Aug 8, 2014

### Staff: Mentor

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?

2. Aug 8, 2014

3. Aug 9, 2014

### KatamariDamacy

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?

4. Aug 9, 2014

### 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
$$\vec{\Pi}=\frac{\vec{E} \times \vec{B}}{c}.$$
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 $\vec{k}$, you can show that the total momentum of the wave packet is mostly into the direction of $\vec{k}$.

5. Aug 9, 2014

### KatamariDamacy

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?

6. Aug 9, 2014

7. Aug 9, 2014

### Staff: Mentor

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.

Last edited: Aug 9, 2014
8. Aug 9, 2014

### Staff: Mentor

Thanks, Jt. That makes sense.

9. Aug 9, 2014

### vanhees71

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

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

$$\vec{P}=\frac{1}{c} \vec{E} \times \vec{B}=\frac{1}{c} \vec{E}_0 \times \vec{B}_0 \cos^2(\omega t-\vec{k} \cdot \vec{x}).$$
Now from the above calculations we have
$$\vec{E}_0 \times \vec{B}_0=\vec{E}_0 \times (\hat{k} \times \vec{E}_0)=\hat{k} \vec{E}_0^2 - \vec{E}_0 (\hat{k} \cdot \vec{E}_0) = \hat{k} \vec{E}_0^2.$$
As you see, the momentum of the electromagnetic field is directed along $\hat{k}$, 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 $\vec{k}_0$ you get something nearly as a plane wave, and then the main momentum is again nearly in the direction of $\vec{k}_0$.

10. Aug 9, 2014

### nsaspook

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

Last edited: Aug 9, 2014
11. Aug 9, 2014

### Staff: Mentor

What do you mean by "volumetric" here?

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

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).

Last edited: Aug 9, 2014
12. Aug 10, 2014

### KatamariDamacy

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

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.

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?

13. Aug 10, 2014

### Orodruin

Staff Emeritus
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.

14. Aug 10, 2014

### vanhees71

As I've shown in my posting explicitly, the three vectors $\vec{k}$, $\vec{E}_0$ , and $\vec{B}_0$ 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:

$$\vec{B}_0=\hat{k} \times \vec{E}_0, \quad \hat{k} \times \vec{B}_0=-\vec{E}_0.$$

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

15. Aug 10, 2014

### KatamariDamacy

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?

16. Aug 10, 2014

### ChrisVer

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

17. Aug 10, 2014

### Staff: Mentor

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.

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

18. Aug 10, 2014

### KatamariDamacy

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?

19. Aug 10, 2014

### Staff: Mentor

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.

20. Aug 10, 2014

### KatamariDamacy

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

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?

Last edited: Aug 10, 2014