How does the formula C-6 come about in the following image?

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1. Feb 23, 2016

weezy

2. Feb 23, 2016

Simon Bridge

Via the principle of superposition ... where did it lose you?
C-6 is just summing up all the independent solutions already found...

3. Feb 23, 2016

dextercioby

Do you know anything about Fourier transformation and how you can use it for PDEs?

4. Feb 23, 2016

weezy

I meant why does the superposition look like this? i.e. a Fourier transform?

5. Feb 23, 2016

weezy

I do know about Fourier transforms but am not familiar with their relation to PDEs

6. Feb 24, 2016

Staff: Mentor

A superposition of two plane waves looks like this: $$\psi(\vec r, t) = g_1 e^{i(\vec k_1 \cdot \vec r - \omega_1 t)} + g_2 e^{i(\vec k_2 \cdot \vec r - \omega_2 t)}$$
A superposition of infinitely many (but still "countable" 1,2,3...) plane waves looks like this: $$\psi(\vec r, t) = \sum_{j=0}^\infty g_j e^{i(\vec k_j \cdot \vec r - \omega_j t)}$$
Suppose the $\omega_j$ and the $\vec k_j$ can be related by ("fitted to") a continuous function $\omega(\vec k)$. Likewise for the $g_j$ and the $\vec k_j$, with another function $g(\vec k)$. Then we can rewrite this slightly as $$\psi(\vec r, t) = \sum_{j=0}^\infty g(\vec k_j) e^{i(\vec k_j \cdot \vec r - \omega(\vec k_j) t)}$$
Finally, suppose we have a superposition of an "uncountably" infinite number of waves with a continuous (not discrete) distribution of values of $\vec k$. In this case the sum becomes an integral: $$\psi(\vec r, t) = \int g(\vec k) e^{i(\vec k \cdot \vec r - \omega(\vec k) t)} d^3 k$$
This is a sort of "volume integral" in 3-dimensional $\vec k$-space. The extra constant in front of your equation C-6 is basically a sort of normalization factor.

Last edited: Feb 24, 2016