Complex electric field vectors


by capandbells
Tags: complex, electric, field, vectors
capandbells
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#1
Jul2-11, 04:57 PM
P: 93
1. The problem statement, all variables and given/known data
I'm trying to figure out the heat dissipation in a volume V due to an incident harmonic electric field.
I know
[tex]
Q = \frac{1}{2}\int_V\mathrm{Re}\left(\mathbf{j}^{*} \cdot \mathbf{E}\right) d^3x
[/tex]
[tex]
= \frac{1}{2}\int_V\mathrm{Re}\left(\left(\sigma \mathbf{E}\right)^{*} \cdot \mathbf{E}\right) d^3x
[/tex]

My biggest problem is that I don't know how to evaluate that dot product. If someone could please either explain it or link me to a resource that explains working with complex vectors, I would appreciate it. I haven't taken an EM class yet and I've never had it explained elsewhere how to operate with complex vectors, so I am mostly flailing around with this stuff.
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Redbelly98
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#2
Jul3-11, 02:54 PM
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It depends, is [itex]\sigma[/itex] a scalar or a tensor?

If it's a scalar, this is pretty straightforward. Since [itex]\sigma \mathbf{E}^* = ( \sigma E_x^*, \sigma E_y^*, \sigma E_z^* ) [/itex], try dotting that with E and see what you get.
capandbells
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#3
Jul3-11, 04:10 PM
P: 93
Sigma (the complex conductivity) is a scalar. Anyway, I think this dot product should be
[tex]
\sigma(E_x^2,E_y^2,E_z^2)
[/tex]
but it seems weird to have E^2 as opposed to |E|^2 for complex numbers.

Redbelly98
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#4
Jul3-11, 06:48 PM
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Complex electric field vectors


Quote Quote by capandbells View Post
Sigma (the complex conductivity) is a scalar.
Okay, good.

Anyway, I think this dot product should be
[tex]
\sigma(E_x^2,E_y^2,E_z^2)
[/tex]
That's incorrect. Can you show how you came up with that?
capandbells
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#5
Jul3-11, 07:02 PM
P: 93
[tex]
( \sigma \mathbf{E})^{*} \cdot \mathbf{E} = ( ( \sigma E_x)^{*}, ( \sigma E_y)^{*}, ( \sigma E_x)^{*}) \cdot (E_x, E_y, E_z) = (( \sigma E_x)^{*})^{*}(Ex) + (( \sigma E_y)^{*})^{*}(E_y) +(( \sigma E_y)^{*})^{*}(E_z)
[/tex]
[tex]
= ( \sigma E_x)(E_x) + ( \sigma E_y)(E_y) + ( \sigma E_z)(E_z) = \sigma (E_x^2 + E_y^2 + E_z^2)
[/tex]
Redbelly98
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#6
Jul3-11, 07:12 PM
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Okay.

The way I learned it, a dot product does not involve taking the complex conjugate of the first vector. So actually you would get terms like Ex*Ex, etc., and you'd end up with σ|E|2.

Note that heat dissipation Q should be a real number (right?).
Antiphon
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#7
Jul3-11, 08:48 PM
P: 1,781
The expression for energy invloves a conjugate so that energy is real. The dot product is a vector operation and doesn't care if the components are complex or not.
Reloaded47
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#8
Jul20-11, 11:14 AM
P: 3
lets just go from basics, step by step, assuming everything is complex

(sigma E)* . E
where E(vector) and sigma(scalar) are complex

(sigma)* E* . E = sigma* |E|^2

or

sigma* (Ex*, Ey*, Ez*).(Ex, Ey, Ez) = sigma* [|Ex|^2+|Ey|^2+|Ez|^2] = sigma* |E|^2

must take modulus of the field and its components


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