# Maxwell Eqns - Prob converting from fundamental to time-harmonic form

1. Nov 16, 2008

### JamesGoh

Hi all

Im trying to derive the time-harmonic form of the Maxwell equations from the original, time-dependent form, however Im not sure if my working and logic is correct

e.g. for

$$\nabla\times E = -dB/dt$$

I want to be able to get

$$\nabla\times \vec{E}$$ = $$-j\omega\vec{B}$$

where

$$E$$ = time-varying form of E-field
$$B$$ = time-varying form of applied magnetic field (is this the correct assumption ?)
$$\vec{E}$$ = time-harmonic form of E field
$$\vec{B}$$ = time-harmonic form of applied magnetic field

The textbook has given the relationship between the time-varying and time-harmonic forms of the field as follows

$$E =$$ Re($$\vec{E}$$ej*$$\omega$$*t) (1.1)
$$B =$$ Re($$\vec{B}$$ej*$$\omega$$*t) (1.2)

where Re() = real part of complex number

However, if this is the case, $$E = \vec{E}cos(\omega*t)$$ will be true and given that there is a time-derivative to the RHS of the time-varying equation (which is going to produce $$dB/dt = -\omega*\vec{B}sin(\omega*t)$$), it would be impossible to get the time-harmonic form, as you have cosine on one side and sine() on the other.

The textbook simply substitutes the time-harmonic version over its time-varying version and differentiates the

$$B =$$ $$\vec{B}$$ej*$$\omega$$*t

term and

eliminates the common e(j*$$\omega$$*t) on both sides, however I cannot seem to see the link between this and the equations linking the time-varying and time-harmonic forms, What could I possibly be doing wrong or misunderstand ?

Last edited: Nov 16, 2008
2. Nov 17, 2008

### JANm

Shouln't your omega be a vector?

3. Nov 17, 2008

### JamesGoh

Whats your reasoning behind why omega should be a vector ?

4. Nov 18, 2008

### Andy Resnick

I wonder if the textbook is causing confusion by not explaining how an arbitrary time-varyng function can (usually) be represented as a sum of time-harmonic functions. OTOH, the relationships you gave for time-varying vs. time-harmonic are very simple, so perhaps not.

In any case, the time-harmonic form is usually preferred becasue it's a more general solution ot the wave equation.

5. Nov 18, 2008

### JANm

E is a scalar from equation 1.1; because it is the real part of something... But it should be a vector if a crossproduct can be made. That is my confusion 1

Could it be that j is not the current density vector, but just the other way of defining imaginaries (used by physici) so the direction i, at right angle with the real axis?

Then omega indeed no vector yet it has the possibility to vary, and in the first equation I see a total derivative of the magnetic field... It should be a partial derivative to time...

6. Nov 18, 2008

### JamesGoh

yeah j = sqrt(-1). Im more used to j then i since I come from an Elec Eng background

7. Nov 18, 2008

### JamesGoh

Ok I realised just then from re-reading the book that I made a slip-up in the definitions

$$\vec{E}$$ = E-field in terms of spatial coordinates (e.g. cartesian system) only
$$\vec{B}$$ = Applied magnetic field in terms of spatial coordinates only

To avoid more potential confusion * = scalar multiplication

Also, the textbook does not show how arbitrary time-varyng function can be represented as a sum of time-harmonic functions.

8. Nov 19, 2008

### gabbagabbahey

I think better notation for the time-harmonic forms would be:

$$\vec{E}(\vec{r},t)=\Re[ \widetilde{E}(\vec{r},t)]=\Re[\vec{E}_0(\vec{r}) e^{j \omega t}]$$

$$\vec{B}(\vec{r},t)=\Re[ \widetilde{B}(\vec{r},t)]=\Re[\vec{B}_0(\vec{r}) e^{j \omega t}]$$

Where $\vec{E}_0(\vec{r})$ and $\vec{B}_0(\vec{r})$ are time-independent E- and B-fields AND $$\widetilde{E}(\vec{r},t)=\vec{E}_0(\vec{r}) e^{j \omega t}$$ and $$\widetilde{B}(\vec{r},t)=\vec{B}_0(\vec{r}) e^{j \omega t}$$ are the complex time-harmonic E- and B-fields

The reason that you can represent any time dependent electric and magnetic field by an infinite sum these time-harmonic fields is that the set of these time-harmonic fields over all possible frequencies $\omega$ is a complete set of orthogonal functions , and in general you can represent any continuous, differentiable function by a linear combination of orthogonal functions like this (similar to a Fourier series)....This is why studying the time-harmonic fields is a good indicator of the behavior of any general time-dependent field.

....Now, when you apply maxwell's law's to the time-harmonic fields, you actually want to deal with the complex time-harmonic fields $$\widetilde{E}(\vec{r},t)$$ and $$\widetilde{B}(\vec{r},t)$$....that should give you your desired forms of the maxwell equations.

9. Nov 19, 2008

### cmos

My apologies, but I'm having trouble seeing exactly what your question is; maybe you can clarify specifically what you are uncomfortable with. Regardless, let me see if I can help. You seem to have problem with the RHS of the Faraday law. In the general form, you (correctly) found the RHS to give you
$$\frac{dB}{dt}=-\omega B _0 sin(\omega t)$$
Now consider the time-harmonic expression for B
$$B=Re( B _0 e^{i \omega t} )$$
Then the time-derivative is
$$\frac{dB}{dt}=Re( i \omega B _0 e^{i \omega t} )$$
Isn't this the same as the first equation?

Is that what you were asking?

10. Nov 20, 2008

### JamesGoh

Um basically what I want to do is derive the time-harmonic form of maxwell's equations from the fundamental form strictly using (1.1) and (1.2).

My problem came from the fact that to get the fundamental form of the field, you have to take the real part of the time-independent part when multiplied with the phasor.

11. Nov 20, 2008

### cmos

When you say "time-harmonic" the implicit assumption is that the (magnetic) field can be written in the following form:
(1) $$B=B_0 cos \omega$$
where B0 is a function of position only. The same idea follows for the electric field.

Concentrating on the magnetic field, the foregoing equation can be written as
(2) $$B=Re(B_0 e^{i \omega t})$$
For convenience, we often write simply:
(3) $$B=B_0 e^{i \omega t}$$
where it is understood that we will take the real part at the end.

So, for example, Faraday's law:
(4) $$\nabla\times E=-\frac{dB}{dt}$$
We may plug in the time-harmonic form of the fields on both sides and it is understood that at the very end we will take the real part of the entire equation.

Also, substitute (3) into the RHS of (4). This will give you the time-harmonic form of Faraday's law. Is this what you were asking?

12. Nov 21, 2008

### JamesGoh

Yep that makes sense. thanks heaps everyone :)

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