- #1
ian_dsouza
- 44
- 3
- Homework Statement
- This is a problem in cosmology but I'll boil it down to the math. $n(\vec{x})$ is the (spatial) number density of particles. By the Fourier transform, $$n(\vec{x}) = \int\mathrm{d}^3k\ n_\vec{k}e^{i\vec{k}\cdot\vec{x}}$$. Also, $a(\vec{x})$ is the field representing the particle (a concept stemming from quantum field theory). By the Fourier transform, $$a(\vec{x}) = \int\mathrm{d}^3k\ a(\vec{k}) e^{i\vec{k}\cdot\vec{x}}$$.
It is known that $$n_\vec{k} = \omega a^2(\vec{k})$$, where $$\omega = \frac{k}{R}$$ and $R$ is a function of time only. I want to find an expression for $n(\vec{x})$ in terms of $a(\vec{x})$ and I was wondering if someone could help me out.
- Relevant Equations
- $$n(\vec{x}) = \int\mathrm{d}^3k\ n_\vec{k}e^{i\vec{k}\cdot\vec{x}}$$
$$a(\vec{x}) = \int\mathrm{d}^3k\ a(\vec{k}) e^{i\vec{k}\cdot\vec{x}}$$
$$n_\vec{k} = \omega a^2(\vec{k})$$
$$\omega = \frac{k}{R}$$
$$n_\vec{k} = \omega a^2(\vec{k})\tag{1}$$
One way is to write the inverse Fourier transforms of the terms above. So, eqn (1) becomes
$$\int\mathrm{d}^3x\ n(\vec{x})e^{-i\vec{k}\cdot\vec{x}} = \omega \int\mathrm{d}^3x^\prime\ a(\vec{x^\prime})e^{-i\vec{k}\cdot\vec{x^\prime}} \int\mathrm{d}^3x^{\prime\prime}\ a(\vec{x^{\prime\prime}})e^{-i\vec{k}\cdot\vec{x^{\prime\prime}}}$$
But I don't know how to proceed from here.
On the other hand, I could multiply both sides of eqn (1) by $$e^{i\vec{k}\cdot\vec{x}}e^{i\vec{k^\prime}\cdot\vec{x}}$$ and first writing $$a^2(\vec{k}) = a(\vec{k})a(\vec{k^\prime})$$ (don't know if this is right though).
$$n_\vec{k}e^{i\vec{k}\cdot\vec{x}}e^{i\vec{k^\prime}\cdot\vec{x}} = \omega a(\vec{k})a(\vec{k^\prime})e^{i\vec{k}\cdot\vec{x}}e^{i\vec{k^\prime}\cdot\vec{x}}$$
Integrating with respect to $\vec{k}$ and $\vec{k^\prime}$
$$\int\mathrm{d}^3k\ n_\vec{k}e^{i\vec{k}\cdot\vec{x}}\int\mathrm{d}^3k^\prime\ e^{i\vec{k^\prime}\cdot\vec{x}} = \omega \int\mathrm{d}^3k\ a(\vec{k})e^{i\vec{k}\cdot\vec{x}}\int\mathrm{d}^3k^\prime\ a(\vec{k^\prime})e^{i\vec{k^\prime}\cdot\vec{x}}$$
$$n(\vec{x})\delta^3({\vec{x}}) = \omega a^2(\vec{x})$$
But this isn't the answer I am looking for either and I suspect I am doing something wrong.
One way is to write the inverse Fourier transforms of the terms above. So, eqn (1) becomes
$$\int\mathrm{d}^3x\ n(\vec{x})e^{-i\vec{k}\cdot\vec{x}} = \omega \int\mathrm{d}^3x^\prime\ a(\vec{x^\prime})e^{-i\vec{k}\cdot\vec{x^\prime}} \int\mathrm{d}^3x^{\prime\prime}\ a(\vec{x^{\prime\prime}})e^{-i\vec{k}\cdot\vec{x^{\prime\prime}}}$$
But I don't know how to proceed from here.
On the other hand, I could multiply both sides of eqn (1) by $$e^{i\vec{k}\cdot\vec{x}}e^{i\vec{k^\prime}\cdot\vec{x}}$$ and first writing $$a^2(\vec{k}) = a(\vec{k})a(\vec{k^\prime})$$ (don't know if this is right though).
$$n_\vec{k}e^{i\vec{k}\cdot\vec{x}}e^{i\vec{k^\prime}\cdot\vec{x}} = \omega a(\vec{k})a(\vec{k^\prime})e^{i\vec{k}\cdot\vec{x}}e^{i\vec{k^\prime}\cdot\vec{x}}$$
Integrating with respect to $\vec{k}$ and $\vec{k^\prime}$
$$\int\mathrm{d}^3k\ n_\vec{k}e^{i\vec{k}\cdot\vec{x}}\int\mathrm{d}^3k^\prime\ e^{i\vec{k^\prime}\cdot\vec{x}} = \omega \int\mathrm{d}^3k\ a(\vec{k})e^{i\vec{k}\cdot\vec{x}}\int\mathrm{d}^3k^\prime\ a(\vec{k^\prime})e^{i\vec{k^\prime}\cdot\vec{x}}$$
$$n(\vec{x})\delta^3({\vec{x}}) = \omega a^2(\vec{x})$$
But this isn't the answer I am looking for either and I suspect I am doing something wrong.