Maxwell’s Equations Wave Solutions

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In summary, in the conversation, the speakers discuss the solutions to Maxwell's wave equations and the relationship between the electric and magnetic fields. They question whether the fields must be in phase and perpendicular to each other, and if this applies to all solutions or just plane waves. They also discuss the linear property of superpositions and its effect on the fields. Ultimately, it is concluded that for an individual plane wave, the fields must be perpendicular, but for a superposition of waves, the fields may not necessarily follow this property.
  • #1
Isaac0427
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Hi,

I have two questions about the solutions to Maxwell’s wave equations:

1. I always hear that E and B must be in phase. Why is this, mathematically? And wouldn’t this also be a solution:
By=B0sin(kz)cos(wt)
Ex=E0cos(kz)sin(wt)
In which case E and B are out of phase.

2. In a vacuum with no charges or currents, would all solutions of Maxwell’s equations (including super positions of plane waves) involve E and B being perpendicular to each other? Also, on a related note, in the case of no charges or currents, do all solutions have to be superpositions of plane waves?

Thanks in advance!
 
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  • #2
Isaac0427 said:
And wouldn’t this also be a solution:
By=B0sin(kx)cos(wt)
Ex=E0cos(kx)sin(wt)
What do you get when you plug it back into Maxwell’s equations?
 
  • #3
Dale said:
What do you get when you plug it back into Maxwell’s equations?
Ah, I see where I went wrong. I used E=cB which is not valid for my “solution” that I know know is not a solution. So they must be in phase. Is there a mathematical proof for this? I haven’t really gotten a solid proof for that through the derivations that I have seen.

EDIT: I saw my flaw in the second part of the post that I have just deleted.

As to number two, do the E and B fields have to be perpendicular under the conditions I described?
 
  • #4
Isaac0427 said:
As to number two, do the E and B fields have to be perpendicular under the conditions I described?

https://simple.wikipedia.org/wiki/Maxwell's_equations Look at the Maxwell equations. In case of no currents, the partial time derivatives of both B and D are parallel to the curl of E and H, respectively. Differentiation with respect to time do not change the direction of the vector quantity, and D is parallel to and B is parallel to H in isotropic media, so B is perpendicular to E.
 
  • #5
I think only in the far field region E and B are perpendicular to each other and perpendicular to the wave propagation vector.
In the near field region there can be longitudinal components.
 
  • #6
ehild said:
https://simple.wikipedia.org/wiki/Maxwell's_equations Look at the Maxwell equations. In case of no currents, the partial time derivatives of both B and D are parallel to the curl of E and H, respectively. Differentiation with respect to time do not change the direction of the vector quantity, and D is parallel to and B is parallel to H in isotropic media, so B is perpendicular to E.
Can we guarantee for a vector X that ##(\nabla\times\vec{X})\cdot \vec{X}=0##?
 
  • #7
Delta² said:
Can we guarantee for a vector X that ##(\nabla\times\vec{X})\cdot \vec{X}=0##?
It is true for the plane electromagnetic waves propagating in free space and represented with vectors of form ##\vec A = \vec A_0 \exp (i(ωt-\vec k \cdot \vec r))##
 
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  • #8
ehild said:
It is true for the plane electromagnetic waves propagating in free space and represented with vectors of form ##\vec A = \vec A_0 \exp (i(ωt-\vec k \cdot \vec r))##

Ok fine for that but the problem is that it seems it doesn't pass through the linear property, that is if vectors X and Y satisfy it then for the vector X+Y we get:
##(\nabla\times(\vec{X}+\vec{Y}))\cdot(\vec{X}+\vec{Y})=(\nabla\times\vec{X})\cdot\vec{X}+(\nabla\times\vec{X})\cdot\vec{Y}+(\nabla\times\vec{Y})\cdot\vec{X}+(\nabla\times\vec{Y})\cdot\vec{Y}=(\nabla\times\vec{X})\cdot\vec{Y}+(\nabla\times\vec{Y})\cdot\vec{X}## which I can't find a reason why it should be always zero.
 
  • #9
Delta² said:
I can't find a reason why it should be always zero.
It isn't. The curl of an arbitrary three-dimensional vector field at a point is not necessarily orthogonal to the vector field at that point; ##\vec{F}=y\hat{x}-x\hat{y}+\hat{k}## is an example.

However, here we aren't asking about an arbitrary vector field; we're asking only about only vector fields that can be written as a superposition of plane-wave solutions of Maxwell's wave equation. As @ehild shows, the E and B fields must be perpendicular for an individual plane wave, but @Isaac0427 was asking if this is also true for superpositions including those for which the wave vectors ##\vec{k}## are not parallel?
 
  • #10
Nugatory said:
As @ehild shows, the E and B fields must be perpendicular for an individual plane wave, but @Isaac0427 was asking if this is also true for superpositions including those for which the wave vectors ##\vec{k}## are not parallel?
The superposition of two waves with non-parallel wave vectors is not a single wave. It is two waves.
 
  • #11
I am just saying that since the linear property doesn't seem to hold (at least not always), the superposition (which in simple words is addition) of plane waves will not necessarily yield a vector field whose curl is perpendicular to the vector field itself.

One case though I can think that the superposition of plane waves retains the property is when they are of the form

##\vec{A_n}=\lambda_n\vec{A_0}e^{i(\vec{k_n}\cdot\vec{r}-\omega_n t)}## that is all ##\vec{A_n}## are collinear to ##\vec{A_0}##
 
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  • #12
ehild said:
The superposition of two waves with non-parallel wave vectors is not a single wave. It is two waves.
But, a superposition of two waves does create a net wave effect on the electromagnetic field, right? And, in that net effect, without charges or currents, will the electric and magnetic fields be perpendicular?
 
  • #13
To clarify my confusion, Wikipedia says at the bottom of this section:
In addition, E and B are perpendicular to each other and to the direction of wave propagation, and are in phase with each other.
My question is if this applies to all solutions to the vacuum equations (which are necessarily superpositions of plane waves, I believe), or just for plane wave solutions.
 
  • #14
Isaac0427 said:
To clarify my confusion, Wikipedia says at the bottom of this section:

My question is if this applies to all solutions to the vacuum equations (which are necessarily superpositions of plane waves, I believe), or just for plane wave solutions.
E and B are perpendicular to each other and to the wave vector as well in case of plane waves. Linearly polarized waves with the same frequency, same wave vector and direction of polarization, but different amplitudes and phase constants make also a linearly polarized wave, with E and B perpendicular to each other. In general, the electric and magnetic fields are not necessarily perpendicular.
 
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  • #15
ehild said:
E and B are perpendicular to each other and to the wave vector as well in case of plane waves. Linearly polarized waves with the same frequency, same wave vector and direction of polarization, but different amplitudes and phase constants make also a linearly polarized wave, with E and H perpendicular to each other. In general, the electric and magnetic fields are not necessarily perpendicular.
But my question is specifically about vacuum solutions, i.e. superpositions of plane waves. I do understand that a general E and B field are not perpendicular.
 
  • #16
Isaac0427 said:
But my question is specifically about vacuum solutions, i.e. superpositions of plane waves. I do understand that a general E and B field are not perpendicular.
Superposition of plane waves of non-parallel wave vectors is not a plane wave. Also, their B and E vectors are not perpendicular.
 
  • #17
ehild said:
Also, their B and E vectors are not perpendicular.
Can you explain to me why this is. It just seems like it should be in my head but I am having trouble expressing my thoughts in math.
 
  • #18
I found out a derivation in my textbook (it is written in greek) that proves that superposition of plane waves (of parallel wave vectors) that satisfy Maxwell's Equations in free space results in E and B waves that remain perpendicular to each other and perpendicular to the wave vector. The plane waves need not have the same frequency or the same polarization vector. I can post it here if you interested, it uses that if ##\vec E_n=\vec E_{(0,n)}e^{i(\vec k_n \cdot \vec r-\omega_n t)}## then it follows (by plugging it in Maxwell's Faraday equation) that it will be for the corresponding ##B_n##, ##\vec B_n=\frac{1}{\omega_n}(\vec k_n \times \vec E_{(0,n)}e^{i(\vec k_n \cdot \vec r-\omega_n t)}## so that the ratio ##\frac{|\vec E_n|}{|\vec B_n|}=\frac{\omega_n}{|\vec k_n|}=c## remains constant for all n.
 
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  • #19
Yes, it is true if the wave vectors are parallel, the field vectors are parallel with the same plane, normal to the wave vector. The curl of a vector-vector function of form ##\vec A = \vec {A_0}e^{j(\vec k \cdot \vec r - \omega t)} ## with ##\vec A ## perpendicular to ##\vec k##, is in Cartesian coordinates ##∇ \times\vec A = (∂_y A_z- ∂_z A_y)\vec i +(∂_z A_x- ∂_x A_z)\vec j +(∂_x A_y- ∂_y A_x)\vec k = j \left(\vec k \times \vec A\right)##, so it is perpendicular to the wave vector. In case of a single plane wave, curl of E is parallel to B, and the curl of B is parallel to E, so both field vectors are all in the plane normal to the wave vector. If the electric /magnetic field is the sum of plane waves with parallel wave vectors, the resultant electric field is also in that plane normal to k, and they are normal to each other.
 
  • #20
Delta² said:
I found out a derivation in my textbook (it is written in greek) that proves that superposition of plane waves (of parallel wave vectors) that satisfy Maxwell's Equations in free space results in E and B waves that remain perpendicular to each other and perpendicular to the wave vector. The plane waves need not have the same frequency or the same polarization vector. I can post it here if you interested, it uses that if ##\vec E_n=\vec E_{(0,n)}e^{i(\vec k_n \cdot \vec r-\omega_n t)}## then it follows (by plugging it in Maxwell's Faraday equation) that it will be for the corresponding ##B_n##, ##\vec B_n=\frac{1}{\omega_n}(\vec k_n \times \vec E_{(0,n)}e^{i(\vec k_n \cdot \vec r-\omega_n t)}## so that the ratio ##\frac{|\vec E_n|}{|\vec B_n|}=\frac{\omega_n}{|\vec k_n|}=c## remains constant for all n.
What about the general case of the wavevectors not necessarily being parallel. Is there an explanation for that?
 
  • #21
Isaac0427 said:
What about the general case of the wavevectors not necessarily being parallel. Is there an explanation for that?
Nope. It just says that in the presence of a medium other than vacuum the ratio ##\frac{\omega_n}{|\vec k_n|}## might not be constant (that is the speed of the wave can depend on the frequency of the wave) so E and B might not be perpendicular in this case.
 
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  • #22
My question is, in the general solution to Maxwell’s wave equation in a vacuum, which is a superposition of plane waves, are E and B perpendicular to each other? And why?
 
  • #23
Isaac0427 said:
My question is, in the general solution to Maxwell’s wave equation in a vacuum, which is a superposition of plane waves, are E and B perpendicular to each other? And why?
I think we answered your question here (forget about my post regarding the linear property not retained, it is retained afterall) the only case we didn't cover is that where the wave vectors of the plane waves are not parallel. Probably it holds in that case too but can't find it in my textbook, neither I can think myself something to make the math work in this case.
 
  • #24
Delta² said:
the only case we didn't cover is that where the wave vectors of the plane waves are not parallel.
Right, I completely understood the case for parallel wavevectors without this thread. The case of not parallel wavevectors is really what I am interested, precisely because I cannot find the answer anywhere.

Thank you very much for your help though.
 
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  • #25
Isaac0427 said:
Right, I completely understood the case for parallel wavevectors without this thread. The case of not parallel wavevectors is really what I am interested, precisely because I cannot find the answer anywhere.

Thank you very much for your help though.
If the polarization direction is the same (parallel) for all the plane waves (that is the plane waves have the form I give at post #11) then the math work, even if the wave vectors ##\vec k_n## are not parallel to each other.
But in the most general case with different (not parrallel ) polarization vectors and not parallel wave vectors, I just can't see how the math work sorry. But seems that it holds here too because I know spherical waves satisfy the wave equation in vacuum and spherical waves can be expressed as superposition of plane waves with non parallel wave vectors and non parallel poralization vectors.
 
  • #26
Isaac0427 said:
What about the general case of the wavevectors not necessarily being parallel. Is there an explanation for that?
Explanation for what? You have Maxwell equations and the wave vector is arbitrary in free space. They are linear equations, so the general solution can be any linear combination of the individual plane waves. For a certain wave vector k1 E1 and B1 perpendicular , for an other one, k2 , you have E2 and B2 E2 , perpendicular to each other. If k1 and k2 are not parallel E1+E2 is not perpendicular to B1+B2 in general.
 
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  • #27
ehild said:
Explanation for what? You have Maxwell equations and the wave vector is arbitrary in free space. They are linear equations, so the general solution can be any linear combination of the individual plane waves. For a certain wave vector k1 E1 and B1 perpendicular , for an other one, k2 , you have E2 and B2 E2 , perpendicular to each other. If k1 and k2 are not parallel E1+E2 is not perpendicular to B1+B2 in general.
Is there a mathematical explanation for this, though?

Here’s my thinking:
Let’s use natural units here, so ##\left| \pmb{E}_n \right|=\left| \pmb{B}_n \right|## (this is not necessary but it makes it easier to think about). Say that ##\pmb{E}_1## makes an angle θ with ##\pmb{E}_2##, the same angle ##\pmb{B}_1## makes with ##\pmb{B}_2##. If this were the case, it would seem as though the angle between ##\pmb{E}_1## and ##\pmb{E}_1+\pmb{E}_2## would be the same angle as the angle between ##\pmb{B}_1## and ##\pmb{B}_1+\pmb{B}_2##. Since ##\pmb{E}_1## is perpendicular to ##\pmb{B}_1##, ##\pmb{E}_1+\pmb{E}_2## must be perpendicular to ##\pmb{B}_1+\pmb{B}_2##, right? Where did I go wrong?
 
  • #28
Isaac0427 said:
Is there a mathematical explanation for this, though?

Here’s my thinking:
Let’s use natural units here, so ##\left| \pmb{E}_n \right|=\left| \pmb{B}_n \right|## (this is not necessary but it makes it easier to think about). Say that ##\pmb{E}_1## makes an angle θ with ##\pmb{E}_2##, the same angle ##\pmb{B}_1## makes with ##\pmb{B}_2##.
Do not forget that E1 and E2 are in planes perpendicular to their wave vectors, and so are the B vectors. If the wave vectors are not parallel, you are in 3 dimension. The The field vectors can be in different planes. Consider a simple example:
The wave vector of one plane wave is K1=(0;0;1). E1=(1;0;0) B1= K1xE1=(0;1;0)
The wave vector of an other wave is K2=(0;1;0). E2=(0;0;1), B2= K2xE2=(1;0;0)
E1+E2=(1;0;1), B1+B2=(1;1;0), Are they perpendicular?
 
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  • #29
ehild said:
Do not forget that E1 and E2 are in planes perpendicular to their wave vectors, and so are the B vectors. If the wave vectors are not parallel, you are in 3 dimension. The The field vectors can be in different planes. Consider a simple example:
The wave vector of one plane wave is K1=(0;0;1). E1=(1;0;0) B1= K1xE1=(0;1;0)
The wave vector of an other wave is K2=(0;1;0). E2=(0;0;1), B2= K2xE2=(1;0;0)
E1+E2=(1;0;1), B1+B2=(1;1;0), Are they perpendicular?
Ah, ok. Thank you very much.
 

Related to Maxwell’s Equations Wave Solutions

1. What are Maxwell’s Equations Wave Solutions?

Maxwell’s Equations Wave Solutions are a set of four mathematical equations that describe the behavior of electric and magnetic fields in space. They were developed by James Clerk Maxwell in the 19th century and are considered one of the cornerstones of classical electromagnetism.

2. What is the significance of Maxwell’s Equations Wave Solutions?

Maxwell’s Equations Wave Solutions have had a profound impact on our understanding of electricity, magnetism, and light. They have been used to make predictions about the behavior of electromagnetic waves, leading to advancements in technology such as radio, television, and wireless communication.

3. How do Maxwell’s Equations Wave Solutions relate to light?

Maxwell’s Equations Wave Solutions describe the fundamental connection between electricity and magnetism, and how they give rise to electromagnetic waves. Light is an example of an electromagnetic wave, and Maxwell’s Equations help us understand its behavior and properties.

4. Can Maxwell’s Equations Wave Solutions be solved analytically?

Yes, Maxwell’s Equations Wave Solutions can be solved analytically, meaning that they can be solved using mathematical equations. However, in some cases, the equations may be too complex to solve analytically and require numerical methods.

5. How are Maxwell’s Equations Wave Solutions used in practical applications?

Maxwell’s Equations Wave Solutions have been used in various practical applications, such as designing antennas, developing telecommunications technology, and studying the properties of materials. They are also fundamental in understanding and developing technologies related to optics, such as lasers and fiber optics.

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