Undergrad Invert a 3D Fourier transform when dealing with 4-vectors

[AndrewGRQTF]

I am having trouble following a step in a book. So we are given that $$\varphi (x) = \int \frac {d^3k}{(2\pi)^3 2\omega} [a(\textbf{k})e^{ikx} + a^*(\textbf{k})e^{-ikx}] $$
where the k in the measure is the spatial (vector) part of the four-momentum k=($\omega$,$\textbf{k}$) and the k in the exponentials is the four-momentum. Note that $x = (t,\textbf{x})$ and that $kx$ = $k^\mu x_\mu = -\omega t + \textbf{k}\cdot \textbf{x}$.

After applying $\int d^3x e^{-ikx}$ to both sides, my book says that we get
$$\int d^3x e^{-ikx} \varphi (x) = \frac{1}{2\omega} a(\textbf{k}) + \frac {1}{2\omega} e^{2i\omega t}a^*(-\textbf{k})$$
$$\int d^3x e^{-ikx} \frac {\partial}{\partial t}\varphi (x) = \frac{-i}{2} a(\textbf{k}) + \frac {i}{2} e^{2i\omega t}a^*(-\textbf{k})$$

My question is: where does the $e^{2i\omega t}$ on the right side of the last two equations come from? The result I get has no $e^{2i\omega t}$.

Also, can someone tell me how the positive/negative signs work in Fourier transforms? Like, if I instead applied $\int d^3x e^{ikx}$ to both sides I would get rid of the integration on the right, so isn't that also inversion of the transform?

 

[phyzguy]

Well, I think I can answer the first question. If I expand: [a(\vec{k})e^{ikx}+a^∗(\vec{k})e^{−ikx}], I get: [a(\vec{k})e^{i\vec{k}\cdot \vec{x}}e^{-i \omega t}+a^∗(\vec{k})e^{-i\vec{k}\cdot \vec{x}}e^{i \omega t}]. When I multiply this by e^{i \omega t}, the first term has no t dependence, and the second term has a factor of e^{2 i \omega t}.

 

[AndrewGRQTF]

Well, I think I can answer the first question. If I expand: [a(\vec{k})e^{ikx}+a^∗(\vec{k})e^{−ikx}], I get: [a(\vec{k})e^{i\vec{k}\cdot \vec{x}}e^{-i \omega t}+a^∗(\vec{k})e^{-i\vec{k}\cdot \vec{x}}e^{i \omega t}]. When I multiply this by e^{i \omega t}, the first term has no t dependence, and the second term has a factor of e^{2 i \omega t}.

So when changing the dummy integration variable in the first equation, we change the $\textbf{k}$ to another variable but keep $\omega$ the same because it is not an integration variable?

 

[AndrewGRQTF]

Well, I think I can answer the first question. If I expand: [a(\vec{k})e^{ikx}+a^∗(\vec{k})e^{−ikx}], I get: [a(\vec{k})e^{i\vec{k}\cdot \vec{x}}e^{-i \omega t}+a^∗(\vec{k})e^{-i\vec{k}\cdot \vec{x}}e^{i \omega t}]. When I multiply this by e^{i \omega t}, the first term has no t dependence, and the second term has a factor of e^{2 i \omega t}.

I now understand my mistake. I was using the same letter for two things, and that was the source of my miscalculation.

Thank you.