Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A useful way of thinking (not only) oceanic waves is to consider them as a superimposition of linear modes: the elevation η of the sea surface is given by:

1: $\eta({\bf x}, t) = \int_{-\infty}^{\infty} \hat{\eta}({\bf k}) e^{i({\bf k \cdot x} - \omega t)} d{\bf k}$

where $\omega$ and ${\bf k}$ are related by a dispersion relation, and $\hat{\eta}({\bf k})$ is a sort of time independent Fourier transform. $\hat{\eta}$ is given by:

2: $\hat{\eta}({\bf k}) = const \int_{-\infty}^{\infty} e^{-i{\bf k \cdot x}}(\eta({\bf x}, 0) + \frac{i}{\omega({\bf k})} \frac{\partial \eta({\bf x}, 0)}{\partial t}) d{\bf x}$

meaning that if we know $\eta({\bf x}, 0)$ and its time derivative we can determine the amplitude of all the normal modes, and neglecting non linear interactions between the modes we know the evolution of the wave. Clear and smart.

What's the simples way to obtain relation 2 from relation 1?

Thanks!

share|improve this question

1 Answer 1

up vote 2 down vote accepted

It's the inverse Fourier transform which is used for all waves, not just ocean waves. The simplest way to obtain (2) from (1) is to guess (2) and prove that it's right.

It's easy to prove (2) given (1). Just calculate $\partial\eta(\vec x, t)/\partial t$. You will get the same formula as (1) except that there will be an extra coefficient of $-i\omega=\omega/i$ inserted in it.

Now you're really ready to prove (2). Substitute (1) instead of $\eta(\vec x,t)$, but use $\vec k'$ for the integration variable in (1), instead of $\vec k$, so that it doesn't get confused with $\vec k$ in (2). The dependence of the integral on $\vec x$ will be $$\exp[-i(\vec k-\vec k')x]$$ and if you integrate this over $d^3 x$, you will get the famous delta-function: $$(2\pi)^3 \delta^{(3)}(\vec k-\vec k')$$ Because we're integrating over $\vec k'$ and the integral contains a delta-function, we may replace $\vec k'$ by $\vec k$ everywhere. So (2) reduces to $$\hat \eta(\vec k) = const \cdot (2\pi)^3\cdot \hat\eta(\vec k)\cdot e^{-i\omega t} (1+1)$$ (in the second term, $\omega/\omega$ and the $i$ factors canceled so we got one) which is indeed correct for $$const =\frac{e^{i\omega t}}{2(2\pi)^3}$$ As you can see, it would be enough to use one of the terms and avoid the time derivative. However, this is just an artifact of your oversimplification of equation (1). The equation (1) as you wrote it only contains terms with "positive values of $\omega$" so $\eta(\vec x)$ clearly can't be real. If you wrote (1) as the real part of what you did, you would get some extra terms with the opposite $\omega$ but the same $k$ which wouldn't cancel, and you would have to get rid of them by hand. Adding the time-derivative with the right coefficient, so that they contribute equally $(1+1)$ to the "right term", is exactly the right way to cancel the wrong term. But I invite you to check it yourself.

share|improve this answer
    
Thanks for the exaustive explaination! –  menta Feb 9 '11 at 22:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.