# Sinusoidal Electric Field

1. Sep 30, 2008

### Apteronotus

In a paper I am reading it states that
since the electric potential (and field) have sinusoidal time dependence, then
$$\Phi(\textbf{x},t)=\Phi(\textbf{x})e^{i\omega t}$$​

Why would this equation be true?

$$\Phi(\textbf{x},t)=Im(\Phi(\textbf{x})e^{i\omega t})$$​

2. Sep 30, 2008

### Staff: Mentor

Using imaginary exponentials to describe oscillatory or wave motion of real quantities is an extremely common practice, because it simplifies many mathematical manipulations.

It's so common that most authors (except in introductory textbooks) assume that the reader knows that you have to take either the real part or the imaginary part at the end of a calculation, to get the actual physically observable behavior.

3. Oct 1, 2008

### Apteronotus

I'm not entirely sure that your statement is true in this particular case. Since upon taking the derivative of the potential with respect to $$t$$ the author arrives at:

$$\frac{\partial\Phi(x,t)}{\partial t}(x,t)=i\omega\Phi(x)e^{i\omega t}$$

$$\Phi(x,t)=Im(\Phi(x)e^{i\omega t})$$

or

$$\Phi(x,t)=Re(\Phi(x)e^{i\omega t})$$

the $$i$$ term would not be appearing in the answer to the partial derivative.

I am wondering if there is any physics that states that ...
a sinusoidal time dependent field can be represented by a complex function.

4. Oct 1, 2008

### Staff: Mentor

Electric potential is a real quantity, so it must actually be either a (real) sine or cosine. This method generalizes it to a complex exponential to simplify some of the math. At the end you "un-generalize" it by taking the real or imaginary part depending on whether you started out with a cosine or a sine. This procedure is so common that few authors spell out all the steps explicitly.

Starting with a real function in the form of a sine, for example, and making all the steps explicit:

$$\Phi = A \sin (\omega t)$$

$$\Phi = Im (A e^{i \omega t})$$

$$\frac{d \Phi}{dt} = Im (i \omega A e^{i \omega t})$$

$$\frac{d \Phi}{dt} = Im (i \omega A (\cos (\omega t) + i \sin (\omega t)))$$

$$\frac{d \Phi}{dt} = Im (i \omega A \cos (\omega t) - \omega A \sin (\omega t))$$

$$\frac{d \Phi}{dt} = \omega A \cos (\omega t)$$

In this case people often say simply, "let $\Phi = A e^{i \omega t}$". This is sloppy and incorrect, strictly speaking, but people do it anyway.

5. Oct 1, 2008

### marcusl

Note also that, by convention, physics and electrical engineering texts usually use Real parts rather than Imaginary.

6. Oct 1, 2008

### Staff: Mentor

Right, I usually start with cosines so I use the real part. I did this example with the imaginary part because the first post used it.

7. Oct 1, 2008

### Apteronotus

Yes jtbell,
I see now what you alluded to earlier. I took quite a stumble on this. Thank you very very much.