EM wave orthoganality

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So im trying to work through the proof why why the direction of proporgation, the E field and B field are all orthogonal to one another.

What i have is...

[tex]E=E_{0}e^{i(k\ \bullet \ r-\omega t)}[/tex]
[tex]B=B_{0}e^{i(k\ \bullet \ r-\omega t)}[/tex]

[tex]\nabla \times E= -\frac{dB}{dt} \Rightarrow k \times E_{0}= \omega B_{0}[/tex]

[tex]\nabla \times B= \mu_{0}\epsilon_{0}\frac{dE}{dt} \Rightarrow k \times B_{0}= \mu_{0}\epsilon_{0}\omega E_{0}[/tex]

and i can see from this why k, B and E must be orthogonal. What im having difficulty with is how to get from the left to the right...

Any ideas?
 
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Answers and Replies

  • #2
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So im trying to work through the proof why why the direction of proporgation, the E field and B field are all orthogonal to one another.

and i can see from this why k, B and E must be orthogonal. What im having difficulty with is how to get from the left to the right...

Any ideas?
It is interesting and very useful to think about two signals passing one another in a coax cable (e.g., RG-8) without interference. I show cartesian coordinates below, but in a coax, E is radial, and H is azimuthal. Also, for RG-8, E/H = 50 ohms

Ex = E0 exp[j(k z - w t)]
Hy = H0 exp[j(k z - w t)]

and
Ex = E0 exp[j(-k z - w t)]
Hy = -H0 exp[j(-k z - w t)]

So one pulse goes from left to right, and the other from right to left, without a "collision". There are instruments (directional couplers) that can measure the amplitude AND direction of each pulse simultaneously.

Edit: changed k x to k z four places
 
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  • #3
jtbell
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What im having difficulty with is how to get from the left to the right...
You mean how to work out the derivatives?
 
  • #4
Andy Resnick
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So im trying to work through the proof why why the direction of proporgation, the E field and B field are all orthogonal to one another.

What i have is...
<snip>

and i can see from this why k, B and E must be orthogonal. What im having difficulty with is how to get from the left to the right...

Any ideas?
Have you tried taking a dot product?

Edit- I just noticed the form you wrote for E and B- spherical wavefronts. That can complicate the maths, so it's easier to use plane waves.
 
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  • #5
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For a plane wave, we can use the vector identity (ignoring any currents J)
div(E x H) = H curl E - E curl H
= -E dD/dt - H dB/dt = -1/2 (d/dt)(e0E2 + u0 H2). (= energy loss per meter). So
S= E x H is a vector (called the Poynting vector), whose direction is the direction, and magnitude is the average power flow, in watts per square meter. Both E and H are orthogonal to direction of power flow.
 
  • #6
gabbagabbahey
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Edit- I just noticed the form you wrote for E and B- spherical wavefronts. That can complicate the maths, so it's easier to use plane waves.
Actually, the OP is using plane waves (and just not specifying the direction of propagation in Cartesian coords, explicitly). Spherical waves have a much more complicated form.
 
  • #7
gabbagabbahey
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So im trying to work through the proof why why the direction of proporgation, the E field and B field are all orthogonal to one another.

What i have is...

[tex]E=E_{0}e^{i(k\ \bullet \ r-\omega t)}[/tex]
[tex]B=B_{0}e^{i(k\ \bullet \ r-\omega t)}[/tex]

[tex]\nabla \times E= -\frac{dB}{dt} \Rightarrow k \times E_{0}= \omega B_{0}[/tex]

[tex]\nabla \times B= \mu_{0}\epsilon_{0}\frac{dE}{dt} \Rightarrow k \times B_{0}= \mu_{0}\epsilon_{0}\omega E_{0}[/tex]

and i can see from this why k, B and E must be orthogonal. What im having difficulty with is how to get from the left to the right...

Any ideas?
If you are having trouble working out the derivatives, it will help you to look at my responses to question 2. in this thread since you are essentially both trying to work out the same curls....Are you also having trouble with the time derivatives?
 
  • #8
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That thread has proved helpful - i have cracked the derivatives. Thanks.

Another question though, k is the wavenumber right? Is the wavenumber equivalent to the direction of proporgation?

I mean, from the solution k, E and B must be orthogonal, but how does this relate to the direction of proporgtaion?
 
  • #9
gabbagabbahey
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That thread has proved helpful - i have cracked the derivatives. Thanks.

Another question though, k is the wavenumber right? Is the wavenumber equivalent to the direction of proporgation?

I mean, from the solution k, E and B must be orthogonal, but how does this relate to the direction of proporgtaion?
You're welcome!:smile:

The scalar [itex]k[/itex] is the wavenumber, the vector [itex]\vec{k}[/itex] which appears in your plane waves (it's a good idea to get into the habit of explicitly writing vectors with some sort of marking to distinguish them from scalars!) , is defined as the product of the wavenumber with the unit vector in the direction of propagation. In other words, [itex]\vec{k}[/itex] points in the direction of propagation and its magnitude is the wavenumber.
 

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