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Frank Castle

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Frank Castle

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blue_leaf77

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Yes, you can. In fact, a function of the form $$f(\mathbf{x},t) = Ae^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)}$$ is a fundamental mode (subject to some dispersion relation) of that differential equation.That is, can I assume an ansatz of the form

- #3

Frank Castle

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Yes, you can. In fact, a function of the form

f(x,t)=Aei(k⋅x−ωt)f(\mathbf{x},t) = Ae^{i(\mathbf{k}\cdot \mathbf{x} - \omega t)} is a fundamental mode (subject to some dispersion relation) of that differential equation.

So one can do this kind of "partial" Fourier decomposition in general then? Intuitively, are we making a Fourier decomposition of the function at a particular (fixed) instant in time, and then requiring that the decomposition should retain its form for all ##t## hence arriving at a differential equation (in time) that the Fourier coefficient functions must satisfy in order for this to hold?!

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blue_leaf77

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Yes.are we making a Fourier decomposition of the function at a particular (fixed) instant in time,

So long as the beam propagates in a space free of any obstacles (e.g. interfaces, aperture, etc), the shape of the spatial frequency spectrum should remain constant.then requiring that the decomposition should retain its form for all ttt

- #5

Frank Castle

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So long as the beam propagates in a space free of any obstacles (e.g. interfaces, aperture, etc), the shape of the spatial frequency spectrum should remain constant.

But is one assumes that one can express the solution in terms of a Fourier decomposition $$u(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}\tilde{u}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{x}}$$ then doesn't it follow that (using the wave equation example) $$\nabla^{2}u(\mathbf{x},t)-\frac{1}{v^{2}}\frac{\partial^{2}}{\partial t^{2}}u(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}\left[(-\mathbf{k}^{2})\tilde{u}(\mathbf{k},t)-\frac{1}{v^{2}}\frac{\partial^{2}}{\partial t^{2}}\tilde{u}(\mathbf{k},t)\right]e^{i\mathbf{k}\cdot\mathbf{x}}=0 \\ \Rightarrow \frac{1}{v^{2}}\frac{\partial^{2}}{\partial t^{2}}\tilde{u}(\mathbf{k},t)+\mathbf{k}^{2}\tilde{u}(\mathbf{k},t)=0$$ such that one finds a differential equation that the functions ##\tilde{u}(\mathbf{k},t)## must satisfy in order for this Fourier decomposition to be valid for all times ##t##?!

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blue_leaf77

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I understand your derivation but I cannot get what you want to say with those maths.But is one assumes that one can express the solution in terms of a Fourier decomposition $$u(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}\tilde{u}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{x}}$$ then doesn't it follow that (using the wave equation example) $$\nabla^{2}u(\mathbf{x},t)-\frac{1}{v^{2}}\frac{\partial^{2}}{\partial t^{2}}u(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}\left[(-\mathbf{k}^{2})\tilde{u}(\mathbf{k},t)-\frac{1}{v^{2}}\frac{\partial^{2}}{\partial t^{2}}\tilde{u}(\mathbf{k},t)\right]e^{i\mathbf{k}\cdot\mathbf{x}}=0 \\ \Rightarrow \frac{1}{v^{2}}\frac{\partial^{2}}{\partial t^{2}}\tilde{u}(\mathbf{k},t)+\mathbf{k}^{2}\tilde{u}(\mathbf{k},t)=0$$ such that one finds a differential equation that the functions ##\tilde{u}(\mathbf{k},t)## must satisfy in order for this Fourier decomposition to be valid for all times ##t##?!

- #7

Frank Castle

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I understand your derivation but I cannot get what you want to say with those maths.

My understanding is that this is the differential equation that the Fourier coefficient functions must satisfy in order for the Fourier decomposition to be valid at times ##t## other than the fixed instant in time that we made the Fourier decomposition. That is, we wish to be able to express the solution to $$\nabla^{2}u(\mathbf{x},t)-\frac{1}{v^{2}}\frac{\partial^{2}}{\partial t^{2}}u(\mathbf{x},t)=0$$ as $$u(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}\tilde{u}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{x}}$$ at

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blue_leaf77

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?$$\frac{1}{v^{2}}\frac{\partial^{2}}{\partial t^{2}}\tilde{u}(\mathbf{k},t)+\mathbf{k}^{2}\tilde{u}(\mathbf{k},t)=0$$

The solution takes the form

$$

\tilde{u}(\mathbf{k},t) = \tilde{u}(\mathbf{k},0) e^{-i\omega t}

$$

with ##\omega = kv##.

- #9

Frank Castle

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The solution takes the form

~u(k,t)=~u(k,0)e−iωt\tilde{u}(\mathbf{k},t) = \tilde{u}(\mathbf{k},0) e^{-i\omega t}

with ω=kv\omega = kv.

No, I understand that the solution takes that form. What I'm really trying to check is that I've understood intuitively what's going on, i.e. why we are "allowed" to Fourier decompose in this fashion (with respect to ##\mathbf{x}## only)?!

- #10

blue_leaf77

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The most general form of an electromagnetic radiation can be written to bewhy we are "allowed" to Fourier decompose in this fashion (with respect to xx\mathbf{x} only)?

$$

u(\mathbf{x},t)=\int \int \frac{d^{3}k}{(2\pi)^{3}} d\omega \, \tilde{u}(\mathbf{k},\omega)e^{i(\mathbf{k}\cdot\mathbf{x} - \omega t)}

$$

At this point, I would like to point out that there is certain theorem in calculus which governs whether one can interchange the order of an integration involving more than one variables. Unfortunately I don't remember how this theorem called, but I think for most practical purpose of a localized, square-integrable spectra you can do the integral over ##\omega## first

$$

u(\mathbf{x},t)=\int \frac{d^{3}k}{(2\pi)^{3}} e^{i\mathbf{k}\cdot\mathbf{x}} \left( \int d\omega \, \tilde{u}(\mathbf{k},\omega)e^{- i\omega t} \right)

$$

and you denote the integral inside the bracket as ##\tilde{u}(\mathbf{k},t)##. So, it's allowed.

- #11

Frank Castle

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The most general form of an electromagnetic radiation can be written to be

u(x,t)=∫∫d3k(2π)3dω~u(k,ω)ei(k⋅x−ωt)u(\mathbf{x},t)=\int \int \frac{d^{3}k}{(2\pi)^{3}} d\omega \, \tilde{u}(\mathbf{k},\omega)e^{i(\mathbf{k}\cdot\mathbf{x} - \omega t)}

At this point, I would like to point out that there is certain theorem in calculus which governs whether one can interchange the order of an integration involving more than one variables. Unfortunately I don't remember how this theorem called, but I think for most practical purpose of a localized, square-integrable spectra you can do the integral over ω\omega first

u(x,t)=∫d3k(2π)3eik⋅x(∫dω~u(k,ω)e−iωt)u(\mathbf{x},t)=\int \frac{d^{3}k}{(2\pi)^{3}} e^{i\mathbf{k}\cdot\mathbf{x}} \left( \int d\omega \, \tilde{u}(\mathbf{k},\omega)e^{- i\omega t} \right)

and you denote the integral inside the bracket as ~u(k,t)\tilde{u}(\mathbf{k},t). So, it's allowed.

Ah ok, so if the wave has a

Also, what's the justification for why we don't Fourier transform the temporal part of ##u## also?

- #12

blue_leaf77

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##\omega## is frequency. I don't understand what you are implying. The frequency, which you falsely called phase, do not change for a propagation in an non-absorptive medium.Ah ok, so if the wave has aconstantphase ωω\omega then one can simply write the solution as I put it, however, if the phase changes then we have to use the more general expression that you put. Is this correct?

Didn't I just show you that the temporal part (the integral inside the bracket in the last equation in post#10) can also be integrated, yielding a function of time ##what's the justification for why we don't Fourier transform the temporal part of uuu also?

\tilde{u}(\mathbf{k},t)##?

- #13

Frank Castle

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The frequency, which you falsely called phase

Sorry, I meant frequency.

Didn't I just show you that the temporal part (the integral inside the bracket in the last equation in post#10) can also be integrated, yielding a function of time ~u(k,t) \tilde{u}(\mathbf{k},t)?

Sorry, yes you did. I wasn't being observant enough.

Sorry if this is an obvious question, but I've seen examples in several sets of notes where they just Fourier transform the spatial part, and then use the differential equation to find a differential equation for the Fourier coefficients ##\tilde{u}(\mathbf{k},t)##. Why not just Fourier transform the whole thing (as you did) and assume this as an ansatz and solve the differential equation this way?

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