The front of an electromagnetic wave (light propagation)

In summary: Basically, what you're saying is that the magnetic field at a certain point changes depending on where the electric fields are when you take the snapshot.
  • #1
FireBones
103
0
I was looking into propagation of EM waves, and it appears there is an overlooked nuance here. It is often said that EM-waves are self-propagating because a change in the E-field causes a magnetic field nearby, so a constantly changing E-field [i.e. a "vibrating" field] causes a constantly changing M-field.

All well and good, and if you look at the super-standard images showing EM-waves the mathematics works out fine [using the right-hand rule to show at each point the magnetic field you get is in the right direction and with the right size based on the changing E-fields, etc.]

Here is an example of what I mean. Using the right-hand rule for magnetic field generated by a changing electric field gives the correct answer [the field is moving out of the page with a small magnitude]

In the figure below, the gray areas represent regions contributing to a magnetic field at A that comes out of the page. That is, regions to its left where the electric field is decreasing or regions to its right where it is increasing. The other areas are regions that contribute to a magnetic field going into the page.

NormalSelfPropSmall.jpg


But it seems all of that breaks down at the front of the wave. At that point you end up sometimes getting results the opposite of what you would expect (essentially the line integrals used to determine size/direction of EM field are missing part of their loops because the EM perturbation has not reached points farther along.

If you imagine that point A is very near the front of the wave, and thus that there is no electric field disturbance far to the right, then the increasing electric field to its left will cause the magnetic field there to go the wrong way, as it is not balanced by an increasing electric field to its right.

FrontSelfPropSmall.jpg


Normally, the change in an electric field near a point determines the magnetic field due to the attenuation with distance [which is why the gray areas to the left do not overcome the significant white area closer to the point.]

Is the solution here that the front of an EM wave has modulated amplitude? So the change in the white area shown is relatively small compared to the change in the gray area farther to its left? Thus allowing that larger gray area, though it be farther away, to cause a net magnetic field coming out of the page [i.e., the direction it is supposed to be going?]

That is my assumption, but I was hoping for some verification.

Thanks.
 
Physics news on Phys.org
  • #2
Your gray areas don't make any sense to me. If you have a plane wave propagating in the x direction and you take a snapshot of the electric field (in the y direction) over all space at a given instant in time then:
1) you automatically know the magnetic field (in the z direction) over all space at that instant in time.
2) you can use that information about the electric and magnetic fields to determine the fields at any future time.
 
  • #3
DaleSpam said:
Your gray areas don't make any sense to me. If you have a plane wave propagating in the x direction and you take a snapshot of the electric field (in the y direction) over all space at a given instant in time then:
1) you automatically know the magnetic field (in the z direction) over all space at that instant in time.
2) you can use that information about the electric and magnetic fields to determine the fields at any future time.

What about them does not make sense?

If you wish to calculate the magnetic field at a certain point based on the rate of change of electric fields nearby (using Amphere's circuital law), there are certain points where the fluctuation will contribute a positive-z value to the integral and certain regions where it is contribute a negative-z value to the integral.

The gray regions simply show which is which.

For example, if the field is propagating as shown, then points immediately to the left of A have electric fields increasing in the positive-y direction, by amphere's law, that is going to create a contribution in the "into-the-page" direction (right-hand rule). So those regions are white.

My point is that when you do these calculations using the standard pictures given for E-M waves, the calculations give the wrong answer when done at the wave front, as shown in the second image, where it is assumed that the point at A has only recently been reached by the wave.
 
  • #4
FireBones said:
If you wish to calculate the magnetic field at a certain point based on the rate of change of electric fields nearby (using Amphere's circuital law), there are certain points where the fluctuation will contribute a positive-z value to the integral and certain regions where it is contribute a negative-z value to the integral.
OK, this is a little wrong here. If you wish to calculate the magnetic field at a point then you need to use the differential form, not the integral form. See http://en.wikipedia.org/wiki/Maxwell's_equations#Table_of_.27in_vacuum.27_equations for a comparison of both forms. The integral form will, at best, allow you to calculate the integral of the magnetic field along a line. The integral form is useful for finding what happens to a loop of wire or an antenna, not for finding the field at a point.

So, if you look at the differential form of the equations you see that in free space all that is needed to determine the B field at a point are the derivatives (curl and time) of the E field at that point. Distant regions are not relevant.
 
Last edited:
  • #5
DaleSpam said:
OK, this is a little wrong here. If you wish to calculate the magnetic field at a point then you need to use the differential form, not the integral form. See http://en.wikipedia.org/wiki/Maxwell's_equations#Table_of_.27in_vacuum.27_equations for a comparison of both forms. The integral form will, at best, allow you to calculate the integral of the magnetic field along a line. The integral form is useful for finding what happens to a loop of wire or an antenna, not for finding the field at a point.

So, if you look at the differential form you see that in free space all that is needed to determine the B field at a point are the derivatives (curl and time) of the E field at that point. Distant regions are not relevant.

Dale,
Don't the differential forms merely tell you how a magnetic field is changing, not what its actual value is?

Your remark about using the differential forms leads me to ask you a follow-up question. Recall that I'm asking about the very edge of a light wave. Assume basic laws of classical electromagnetics and imagine you are at the point exactly (tc,0,0) from a source of light t seconds after the light was emitted, so you are at the very edge of the wave.

Clearly (classically speaking), E=B=0 here, but we know that E and B are both going to change in the next instant. But how does either change according to the differential forms? All curls and time derivatives are 0 at that point at that time (assuming we want E and B to be differentiable at all). Smoothness kills off all the local first-degree derivatives, but without any of those derivatives around, it is not clear to me how a non-zero magnetic or electric field can ever arise according to the differential forms.

Indeed the need for differentiability at the leading edge suggests there is a modulation of amplitude there because at the leading edge we would all the derivatives of E and B to vanish for the sake of smoothness, but at the same time the leading edge is a node and normally the node of a E-M wave has maximal values for dE/dt and dB/dT.

P.S. I have figured out the solution to my original concern...and I guess I have you to thank because it was only after I started to construct the specific loops i'd use to show the problem that I realized why there is no contradiction here. Still, I would like an answer to the above question of how the E-field and B-field can evolve at a point when all curls and time derivatives of both are 0.
 
  • #6
FireBones said:
Don't the differential forms merely tell you how a magnetic field is changing, not what its actual value is?
Yes, this is correct. In order to determine the actual value you need to also supply the boundary conditions or initial conditions.


FireBones said:
imagine you are at the point exactly (tc,0,0) from a source of light t seconds after the light was emitted, so you are at the very edge of the wave.
This is your boundary condition. At time t there is an E-field which is zero for x>tc and non-zero for x<tc.

FireBones said:
Clearly (classically speaking), E=B=0 here, but we know that E and B are both going to change in the next instant. But how does either change according to the differential forms? All curls and time derivatives are 0 at that point at that time
This is most definitely not the case. Because the field is zero on one side of x=tc and non-zero on the other side the curls are non-zero. Therefore the time derivatives are also non-zero.

FireBones said:
(assuming we want E and B to be differentiable at all). Smoothness kills off all the local first-degree derivatives
There is no requirement that the fields be smooth. But even a smooth field is not a problem. Smoothness does not imply that the higher derivatives are zero, but it does mean that there won't be a clear "leading edge". In that sense your problem statement presupposes a non-smooth field as a boundary condition so you shouldn't be too surprised to run into problems if you try to analyze it as a smooth field.

FireBones said:
Still, I would like an answer to the above question of how the E-field and B-field can evolve at a point when all curls and time derivatives of both are 0.
I hope my answer above resolved your question, but don't hesitate to ask for clarification if it didn't. The key is that the curl is, in fact, non-zero at the leading edge.
 
  • #7
DaleSpam said:
Yes, this is correct. In order to determine the actual value you need to also supply the boundary conditions or initial conditions.

To provide further clarification, we often work with time-harmonic fields, as you have implied in your figures. That is, we assume that the fields and source operate at a discrete frequency and thus have a time dependence of say exp(-i \omega t). Now looking at the differential forms of Maxwell's Equations we can readily see that from the time derivative of the magnetic field that,

[tex]\mathbf{B} = \frac{1}{i\omega} \nabla\times\mathbf{E} [/tex]

The assumption of time-harmonic fields is part of the initial conditions that Dalespam stated were necessary and likewise in the absence of current sources we can find a similar equation that relates the electric field to the curl of the magnetic field. And like Dalespam stated previously, just because a function evaluates to zero does not mean that its derivatives are zero at that point too. Look at the sine function. It is zero at 0, +/- 2\pi, and so on but its derivative, cosine, has a magnitude of one at the same points.
 
  • #8
Now we are getting somewhere! I hope you will indulge just a couple more rounds.

In an earlier post, I claimed the curls at (ct,0,0) were zero, to which you responded:

DaleSpam said:
This is most definitely not the case. Because the field is zero on one side of x=tc and non-zero on the other side the curls are non-zero. Therefore the time derivatives are also non-zero.

I disagree. Keep in mind the original question I had in mind. Some source at the origin initiates an EM wave, and we are looking at the point (tc,0,0). The E-field and B-field are going to be zero for all (x,0,0) where x>tc. Furthermore, the E-field and B-field will be zero at all points (tc,y,0) (for all y) because any such point is at least a distance tc from the source. Indeed, we can go further than that, for any y<>0 there exists a r>0 such that the E-field and B-field are both zero along the path from (tc-r,y,0) to (tc,0,0)
Based on all the above considerations [and the mathematical definition of curl], the curl of E and B must be zero because all first order partial (spatial) derivatives are zero. If the partial of E with respect to x exists at (tc,0,0) it must be zero since (tc + epsilon,0,0) is zero for all epsilon>0 (that is, its right-hand partial is 0). If the curl is defined at all, it must be zero.

DaleSpam said:
There is no requirement that the fields be smooth. But even a smooth field is not a problem. Smoothness does not imply that the higher derivatives are zero, but it does mean that there won't be a clear "leading edge".

By "smooth" here I simply meant the function was infinitely differentiable, in which case the first order derivatives (that is, the derivatives defining the curl) all exist and (by inspection of the stated problem setup) must all be 0.
The key is that the curl is, in fact, non-zero at the leading edge.
I don't see how this is possible. see above argument.
 
Last edited:
  • #9
Born2bwire said:
And like Dalespam stated previously, just because a function evaluates to zero does not mean that its derivatives are zero at that point too. Look at the sine function. It is zero at 0, +/- 2\pi, and so on but its derivative, cosine, has a magnitude of one at the same points.

I never said that the partial derivatives were zero because the function is zero at (ct,0,0).
I said they are zero because the function is also zero on the entire half-plane (ct + epsilon, y, 0), for epsilon>=0.
 
Last edited:
  • #10
FireBones said:
I never said that the partial derivatives were zero because the function is zero at (ct,0,0).
I said they are zero because the function is also zero on the entire half-plane (ct + epsilon, y, 0), for epsilon>=0.

Why would that be true? Why not say that it has to be non-zero because the entire half-plane in the other direction is non-zero and varying? If we approximate the integral by finite differentiation then central differencing would give the first order derivative as being

[tex] \frac{\partial f(x)}{\partial x} \approx \frac{f(x+\delta)-f(x-\delta)}{2\delta} [/tex]

It's a simple exercise to work out that the electric field in a source free region must satisfy

[tex] \nabla\times\nabla\times\mathbf{E} + \frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2} = 0 [/tex]

From here we can readily recognize the wave equation. Since the field is divergence free,

[tex] \nabla^2\mathbf{E} + \frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2} = 0 [/tex]

If we assume that the electric field is polarized, let's say the z direction, then

[tex] \nabla^2E_z + \frac{1}{c^2}\frac{\partial^2 E_z}{\partial t^2} = 0 [/tex]

and so we just arrive at a scalar wave equation that propagates at a phase velocity of c.
 
Last edited:
  • #11
FireBones said:
I disagree. Keep in mind the original question I had in mind. Some source at the origin initiates an EM wave, and we are looking at the point (tc,0,0). The E-field and B-field are going to be zero for all (x,0,0) where x>tc. Furthermore, the E-field and B-field will be zero at all points (tc,y,0) (for all y) because any such point is at least a distance tc from the source. Indeed, we can go further than that, for any y<>0 there exists a r>0 such that the E-field and B-field are both zero along the path from (tc-r,y,0) to (tc,0,0)
Based on all the above considerations [and the mathematical definition of curl], the curl of E and B must be zero because all first order partial (spatial) derivatives are zero. If the partial of E with respect to x exists at (tc,0,0) it must be zero since (tc + epsilon,0,0) is zero for all epsilon>0 (that is, its right-hand partial is 0). If the curl is defined at all, it must be zero.
This is incorrect. The easiest way to disprove it is by counter-example. For simplicity I will use units where c=1 and consider a distance far enough that the wave is approximately a plane wave. If those assumptions bother you then I would recommend working out an example using e.g. a dipole wave, but I will leave that exercise to you.

Suppose we have the wave:
[tex]\mathbf E = (0,H(t-x),0)[/tex]
[tex]\mathbf B = (0,0,H(t-x))[/tex]

Where H is the Heaviside unit step function. Note that for all x>ct E=0, as required in your scenario. We can take the divergences, curls, and time derivatives of E and B as follows:

[tex]\nabla \cdot \mathbf E = 0[/tex]
[tex]\nabla \cdot \mathbf B = 0[/tex]
[tex]\nabla \times \mathbf E = (0,0,-\delta(t-x))[/tex]
[tex]\nabla \times \mathbf B = (0,\delta(t-x),0)[/tex]
[tex]\frac{\partial \mathbf E}{\partial t} = (0,\delta(t-x),0)[/tex]
[tex]\frac{\partial \mathbf B}{\partial t} = (0,0,\delta(t-x))[/tex]

Where [itex]\delta[/itex] is the Dirac delta function. These clearly satisfy Maxwell's equations in vacuum.
FireBones said:
By "smooth" here I simply meant the function was infinitely differentiable, in which case the first order derivatives (that is, the derivatives defining the curl) all exist and (by inspection of the stated problem setup) must all be 0.
I understood your meaning and answered accordingly. Smoothness (infinitely differentiable) is not a requirement of Maxwell's equations as shown above. However, as I mentioned previously it is certainly possible to have smooth functions which also satisfy Maxwell's equations, but then there is not a clear wavefront where E=0 for all x beyond the wavefront. If you want a smooth E field then it must be something like a Gaussian or a pure sin or cos which has infinite extent in space and time.

In your problem definition, since you are considering waves with a defined leading edge and with E=0 beyond that edge, you are necessarily considering only non-smooth (not infinitely differentiable) fields. You cannot logically make conclusions about some premise based on a class of functions which violates the premise.
 
  • #12
Born2bwire said:
Why would that be true? Why not say that it has to be non-zero because the entire half-plane in the other direction is non-zero and varying?

Because there is no necessity for a function varying on an open interval to have a non-zero derivative at a limit point of that open interval.

For example, consider f(x) = x^2 for x<0 and f(x)=0 for x>=0.
That function varies to the left of 0 and is fixed at zero for x=0 or greater, but its derivative is zero at zero.

On the other hand it is mathematically impossible for a differentiable function to be constant on a closed interval without having its derivative be 0 on that closed interval. In this case, take the intervale [(tc,0,0), (tc +1,0,0)]. The y-component of the E-field and the y-component of B-field are both 0 on this interval, so their partial derivative with respect to x must be 0.

Born2bwire said:
If we approximate the integral by finite differentiation then...

The approximation you gave for the derivative is unreliable. For example, consider the function f(x) = Square root of |x|. That function has no derivative at x=0. It's left-hand derivative is unbounded, as is its right-hand derivative, yet the limit you gave would suggest its derivative is approximately 0.

In any event, we don't need to resort to this approximation because the standard definition of derivative suffices to show that a function that is constant on a closed interval must have a derivative of 0 at all points on that interval if it has a derivative at all.
 
  • #13
DaleSpam said:
This is incorrect. The easiest way to disprove it is by counter-example. For simplicity I will use units where c=1 and consider a distance far enough that the wave is approximately a plane wave. If those assumptions bother you then I would recommend working out an example using e.g. a dipole wave, but I will leave that exercise to you.

Suppose we have the wave:
[tex]\mathbf E = (0,H(t-x),0)[/tex]
[tex]\mathbf B = (0,0,H(t-x))[/tex]

Where H is the Heaviside unit step function. Note that for all x>ct E=0, as required in your scenario. We can take the divergences, curls, and time derivatives of E and B as follows:

The heavyside function is not differentiable. I am not dealing with distributions here. Even if we use plane waves (which I prefer not to), it doesn't get around the basic point I have made all along. The question is "How can a point at the edge of a wavefront go from having a zero E-field and a zero B-field to having non-zero fields?" Even with a plane wave, the problem setup necessitates that E(tc,y,0) = 0 = B(tc,y,0) for all y (and hence the partial with respect to y is zero). It also necessitates E(x,0,0)=0 for all x>= tc, so its right-hand derivative is 0. Similarly for B. If E and B are differentiable, their right-hand derivative is their derivative, so the partials with respect to x are 0.


DaleSpam said:
However, as I mentioned previously it is certainly possible to have smooth functions which also satisfy Maxwell's equations.

The heavyside function is not smooth.


DaleSpam said:
In your problem definition, since you are considering waves with a defined leading edge and with E=0 beyond that edge, you are necessarily considering only non-smooth (not infinitely differentiable) fields. You cannot logically make conclusions about some premise based on a class of functions which violates the premise.

Not true at all. A wave can have a leading edge and still be smooth.

Consider f(x,t)=g(x-tc) where g(x) =[tex]\frac{Sin(x^2)}{2x-1}[/tex] for x<0 and g(x)=0 otherwise. That is a smooth wave with a clear leading edge.
 
  • #14
DaleSpam said:
Your gray areas don't make any sense to me. If you have a plane wave propagating in the x direction and you take a snapshot of the electric field (in the y direction) over all space at a given instant in time then:
1) you automatically know the magnetic field (in the z direction) over all space at that instant in time.
2) you can use that information about the electric and magnetic fields to determine the fields at any future time.

By the way, looking back at my original post, I do think the points raised there are still valid.

The gray areas represent regions where the Biot-Savart law (using displacement current related to the change in the electric field) yields a magnetic field at A in the 'out of page' direction. If the front of an expanding light wave looked like what is given in textbooks as the canonical representation of an EM wave, it would break the Biot-Savart law at the ends. In the second image of the OP, the Biot-Savart law would yield a magnetic field going into the page, which is the wrong direction.
 
  • #15
FireBones said:
The heavyside function is not differentiable. I am not dealing with distributions here.
It is not up to you; you cannot simply discard it by fiat. It is a valid solution of Maxwell's equations. Not only that, but it is in principle a physically realizable function because both the energy and the power are finite.

FireBones said:
Not true at all. A wave can have a leading edge and still be smooth.

Consider f(x,t)=g(x-tc) where g(x) =[tex]\frac{Sin(x^2)}{2x-1}[/tex] for x<0 and g(x)=0 otherwise. That is a smooth wave with a clear leading edge.
This is not a smooth function. The second derivative is discontinuous at x=0 and the third derivative involves a Dirac delta function.

In any case, as I said before, smoothness is not a requirement, so let's take your function g as another example. Starting with the wave:
[tex]\mathbf E = (0,g(x-t),0)[/tex]
[tex]\mathbf B = (0,0,g(x-t))[/tex]

We again take the curl, divergence, and time derivatives and find:
[tex]\nabla \cdot \mathbf E = 0[/tex]
[tex]\nabla \cdot \mathbf B = 0[/tex]
[tex]\nabla \times \mathbf E = (0,0,\frac{2 (2 t-2 x+1) (t-x) \cos \left((t-x)^2\right)-2 \sin
\left((t-x)^2\right)}{(2 t-2 x+1)^2})[/tex]
[tex]\nabla \times \mathbf B = (0,-\frac{2 (2 t-2 x+1) (t-x) \cos \left((t-x)^2\right)-2 \sin
\left((t-x)^2\right)}{(2 t-2 x+1)^2},0)[/tex]
[tex]\frac{\partial \mathbf E}{\partial t} = (0,-\frac{2 (2 t-2 x+1) (t-x) \cos \left((t-x)^2\right)-2 \sin
\left((t-x)^2\right)}{(2 t-2 x+1)^2},0)[/tex]
[tex]\frac{\partial \mathbf B}{\partial t} = (0,0,-\frac{2 (2 t-2 x+1) (t-x) \cos \left((t-x)^2\right)-2 \sin
\left((t-x)^2\right)}{(2 t-2 x+1)^2})[/tex]
for x-t<0 and (0,0,0) otherwise.

So the curl of the function is non-zero over the same open region on which the function itself is non-zero despite the fact that it is zero over the region x-t>0. And again, these also clearly satisfy Maxwell's equations in vacuum. In fact, g can be any arbitrary function and it will satisfy Maxwell's equations.
 
  • #16
FireBones said:
By the way, looking back at my original post, I do think the points raised there are still valid.

The gray areas represent regions where the Biot-Savart law (using displacement current related to the change in the electric field) yields a magnetic field at A in the 'out of page' direction. If the front of an expanding light wave looked like what is given in textbooks as the canonical representation of an EM wave, it would break the Biot-Savart law at the ends. In the second image of the OP, the Biot-Savart law would yield a magnetic field going into the page, which is the wrong direction.
First, the Biot-Savart law is an approximation to Maxwell's equations for the magnetostatic case so it doesn't apply for EM waves. Second, the Biot-Savart law uses real currents, not displacement currents (in the magnetostatic case there are no displacement currents).
 

What is an electromagnetic wave?

An electromagnetic wave is a type of energy that is propagated through space via a combination of electric and magnetic fields. It does not require a medium to travel through and can travel at the speed of light.

What is the front of an electromagnetic wave?

The front of an electromagnetic wave refers to the leading edge of the wave as it travels through space. It represents the point of highest energy and is where the electric and magnetic fields are strongest.

How does light propagate?

Light propagates through space as an electromagnetic wave. It is created by oscillations of electric and magnetic fields, with the electric field oscillating perpendicular to the magnetic field.

How does the front of an electromagnetic wave differ from the rest of the wave?

The front of an electromagnetic wave differs from the rest of the wave in terms of its energy and strength of the electric and magnetic fields. It also represents the point where the wave is most concentrated and has the highest intensity.

What factors can affect the front of an electromagnetic wave?

The front of an electromagnetic wave can be affected by factors such as the medium it is traveling through, the wavelength of the wave, and the source of the wave. It can also be affected by any objects or barriers the wave encounters as it propagates through space.

Similar threads

  • Electromagnetism
Replies
4
Views
852
  • Electromagnetism
Replies
2
Views
306
Replies
8
Views
964
  • Electromagnetism
Replies
2
Views
800
  • Electromagnetism
Replies
8
Views
2K
Replies
1
Views
1K
  • Electromagnetism
Replies
8
Views
2K
  • Electromagnetism
Replies
3
Views
926
Replies
61
Views
2K
  • Electromagnetism
Replies
2
Views
800
Back
Top