Help with evaluating this Fourier transform

In summary: It's not so much you're applying the properties separately, but you're applying them consecutively. So, for example, say you have the pair ##f## and ##F##, then ##g(t) = f''(t) \leftrightarrow G(\omega) = \omega^2 F(\omega)##. Then you can apply the next property to the pair ##g## and... get lost in all the integrals.
  • #1
user1139
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Homework Statement
Help with evaluating Fourier transform
Relevant Equations
The definition of Fourier transform (F.T.) that I am using is given as:
$$f(\vec{x},t)=\frac{1}{\sqrt{2\pi}}\int e^{-i\omega t}\tilde{f}(\vec{x},\omega)\,\mathrm{d}\omega$$
The definition of Fourier transform (F.T.) that I am using is given as:
$$f(\vec{x},t)=\frac{1}{\sqrt{2\pi}}\int e^{-i\omega t}\tilde{f}(\vec{x},\omega)\,\mathrm{d}\omega$$

I want to show that:
$$\frac{1}{c\sqrt{2\pi}}\int e^{-i\omega t}\omega^2 e^{ikx}(\vec{x}\times\vec{p}_{\omega})\,\mathrm{d}\omega=-\frac{\vec{x}}{c}\times\frac{d^2}{dt^2}\,\vec{p}(t-x/c)$$

To show the above, two F.T. must occur to get the second derivative w.r.t. time and the ##(t-x/c)## term. However, I am not sure how to proceed since obtaining either one will induce another F.T. into the second equation. How should I continue?
 
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  • #2
Thomas1 said:
To show the above, two F.T. must occur to get the second derivative w.r.t. time and the $(t-x/c)$ term. However, I am not sure how to proceed since obtaining either one will induce another F.T. into the second equation. How should I continue?
I'm not sure what you mean here. And what is ##\vec{p}_\omega##?
 
  • #3
##\vec{p}_{\omega}## is some function of ##\omega## which was not specified in the question. From my understanding, if I want to find let's say the derivative portion, I need to F.T. or inverse F.T.. Since there is currently only one F.T. I will need to manually insert in another F.T.. As such, there will now be two F.T..
 
  • #4
Thomas1 said:
##\vec{p}_{\omega}## is some function of ##\omega## which was not specified in the question.
So is ##\vec p## in the result supposed to be the Fourier transform of ##\vec p_\omega##?

From my understanding, if I want to find let's say the derivative portion, I need to F.T. or inverse F.T..
Can you write out what you mean using math? It's not clear what you mean by "find the derivative portion."
 
  • #5
vela said:
So is ##\vec p## in the result supposed to be the Fourier transform of ##\vec p_\omega##?

No, it's just a function of ##\omega##.

Can you write out what you mean using math? It's not clear what you mean by "find the derivative portion."
What I meant was
$$\int e^{-i\omega t}\frac{d^2}{dt^2}(\vec{x}\times\vec{p}_{\omega})\,\mathrm{d}\omega=-\omega^2(\vec{x}\times\vec{p}_{\omega})$$
 
  • #6
Thomas1 said:
No, it's just a function of ##\omega##.
That can't be right. You're integrating over ##\omega##, so the righthand side can't depend on ##\omega##.

What I meant was
$$\int e^{-i\omega t}\frac{d^2}{dt^2}(\vec{x}\times\vec{p}_{\omega})\,\mathrm{d}\omega=-\omega^2(\vec{x}\times\vec{p}_{\omega})$$
Same problem with that expression. On the lefthand side, you're integrating with respect to ##\omega##, which is a dummy variable, so it can't appear in the result on the righthand side.

But I think I finally understand what your question is. Say you have some function ##f(t)## and its Fourier transform ##F(\omega)##. Then you have
$$f'(t) = \frac{d}{dt} f(t) = \frac{d}{dt} \int F(\omega)e^{-i\omega t}\,d\omega = \int F(\omega)\frac{d}{dt}e^{-i\omega t}\,d\omega = \int [-i\omega\, F(\omega)] e^{-i\omega t}\,d\omega.$$ The last integral can be thought of as the Fourier transform of the function ##G(\omega) = -i\omega F(\omega)##, and it pairs with ##g(t) = f'(t)##. The integral doesn't go away when you translate ##d/dt## in the time domain into the factor ##-i\omega## in the frequency domain.
 
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  • #7
vela said:
$$f'(t) = \frac{d}{dt} f(t) = \frac{d}{dt} \int F(\omega)e^{-i\omega t}\,d\omega = \int F(\omega)\frac{d}{dt}e^{-i\omega t}\,d\omega = \int [-i\omega\, F(\omega)] e^{-i\omega t}\,d\omega.$$ The last integral can be thought of as the Fourier transform of the function ##G(\omega) = -i\omega F(\omega)##, and it pairs with ##g(t) = f'(t)##. The integral doesn't go away when you translate ##d/dt## in the time domain into the factor ##-i\omega## in the frequency domain.

Indeed, that was what I tried to reason out. I still can't see a way to motivate ##e^{ikx}## into the F.T. and thereafter, assuming we managed to do so, I suppose we can do a combined F.T. into what was wanted?
 
  • #8
##k## and ##\omega## are related, right?
 
  • #9
Yes, they are. They are related by ##\omega=kc##.

Actually, is it correct to do the F.T. separately i.e. one for the derivative and the other for ##e^{ikx}## then use both results to do the combined F.T.?
 
  • #10
It's not so much you're applying the properties separately, but you're applying them consecutively. So, for example, say you have the pair ##f## and ##F##, then ##g(t) = f''(t) \leftrightarrow G(\omega) = \omega^2 F(\omega)##. Then you can apply the next property to the pair ##g## and ##G##.
 
  • #11
I always thought that to perform a F.T. you need to have the integral. Since there is only one integral, why is it that we can think of the F.T. to be done as consecutively rather than as combined?
 
  • #12
On the right side of the expression in the OP , express ## p(t-x/c) ## as a F.T. integral in ## \tilde{p}(\omega) ##. You can then differentiate w.r.t. time (twice) and bring it inside the integral.
 
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Related to Help with evaluating this Fourier transform

1. What is a Fourier transform?

A Fourier transform is a mathematical operation that decomposes a function or signal into its constituent frequencies. It is commonly used in signal processing, image processing, and other fields of science and engineering.

2. Why is a Fourier transform useful?

A Fourier transform allows us to analyze a signal in the frequency domain, which can provide valuable insights and information that may not be apparent in the time domain. It is also a powerful tool for filtering and extracting specific frequencies from a signal.

3. How is a Fourier transform calculated?

A Fourier transform is calculated by taking the integral of a function over all frequencies. This can be done analytically or numerically using various algorithms and methods.

4. What types of signals can be transformed using a Fourier transform?

A Fourier transform can be applied to any signal that is finite and has a well-defined frequency spectrum. This includes continuous signals, discrete signals, and even non-periodic signals.

5. Can a Fourier transform be reversed?

Yes, a Fourier transform can be reversed using an inverse Fourier transform. This allows us to reconstruct the original signal from its frequency components, making it a powerful tool for signal processing and analysis.

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