Hi there. I have a couple of questions regarding the derivation of the Boltzmann equation in Dodelson for photons when given scalar overdensity perturbations to the FRW metric.
To start with, let ##\Theta(\vec{x})## denote the temperature perturbations to the Bose-Einstein distribution of the CMB gas and let ##\Theta(\vec{x}) = \int \frac{d^3 k}{(2\pi)^3}e^{i\vec{k}\cdot \vec{x}}\Theta(\vec{k})## denote its Fourier decomposition. Dodelson states, in the second paragraph of p.101, that "The wavevector ##\vec{k}## is pointing in the direction in which the temperature is changing, so it is perpendicular to the gradient..." but I'm not completely sure on what he means by this.
Is he simply referring to each ##\vec{k}## being perpendicular to the gradient of the associated Fourier modes given by ##\Theta_{\vec{k}}(\vec{x}) = \Theta(\vec{k})e^{i\vec{k}\cdot \vec{x}}##? But then ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x}) = i\vec{k}\Theta_{\vec{k}}(\vec{x})## so each Fourier mode is such that ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})## is parallel to ##\vec{k}##, not perpendicular to it. In fact for each mode ##\vec{k}##, ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})## itself points in the direction in which the temperature changes so, by Dodelson's own remark "The wavevector ##\vec{k}## is pointing in the direction in which the temperature is changing", it must be that ##\vec{k}## is parallel to ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})##; his second remark "...so it is perpendicular to the gradient..." makes no sense in light of that and in fact it would be the level curves of ##\Theta_{\vec{k}}(\vec{x})## that ##\vec{k}## is perpendicular to.
My second question is with regards to his statement atop p.99 that "In the absence of a bulk velocity for the electrons (##v_b = 0##), the [Compton] collision terms serve to drive ##\Theta## to ##\Theta_0##". Here ##\Theta_0## is the monopole moment of ##\Theta##. But if we already have ##v_b = 0## for the average velocity of the electrons then why would there be e.g. a dipole moment ##\Theta_1## of ##\Theta## to be driven to ##\Theta_0## under efficient Compton scattering? If ##v_b = 0## then wouldn't we already have only a monopole distribution for the temperature perturbations?
The phrase "...the [Compton] collision terms serve to drive ##\Theta## to ##\Theta_0##" makes it seem like there is, amongst higher moments, an existing dipole moment even when ##v_b = 0## and that eventually it gets suppressed by the monopole moment after many Compton scatterings but if ##v_b = 0## then how can there be a preferred direction for there to even be a dipole moment to be suppressed to start with? Wouldn't it make more sense to say that ##v_b \neq 0## and ##\Theta_1 \neq 0## at first but that efficient Compton scatterings serve to drive both ##\Theta \rightarrow \Theta_0## and ##v_b \rightarrow 0##?
To start with, let ##\Theta(\vec{x})## denote the temperature perturbations to the Bose-Einstein distribution of the CMB gas and let ##\Theta(\vec{x}) = \int \frac{d^3 k}{(2\pi)^3}e^{i\vec{k}\cdot \vec{x}}\Theta(\vec{k})## denote its Fourier decomposition. Dodelson states, in the second paragraph of p.101, that "The wavevector ##\vec{k}## is pointing in the direction in which the temperature is changing, so it is perpendicular to the gradient..." but I'm not completely sure on what he means by this.
Is he simply referring to each ##\vec{k}## being perpendicular to the gradient of the associated Fourier modes given by ##\Theta_{\vec{k}}(\vec{x}) = \Theta(\vec{k})e^{i\vec{k}\cdot \vec{x}}##? But then ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x}) = i\vec{k}\Theta_{\vec{k}}(\vec{x})## so each Fourier mode is such that ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})## is parallel to ##\vec{k}##, not perpendicular to it. In fact for each mode ##\vec{k}##, ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})## itself points in the direction in which the temperature changes so, by Dodelson's own remark "The wavevector ##\vec{k}## is pointing in the direction in which the temperature is changing", it must be that ##\vec{k}## is parallel to ##\vec{\nabla}\Theta_{\vec{k}}(\vec{x})##; his second remark "...so it is perpendicular to the gradient..." makes no sense in light of that and in fact it would be the level curves of ##\Theta_{\vec{k}}(\vec{x})## that ##\vec{k}## is perpendicular to.
My second question is with regards to his statement atop p.99 that "In the absence of a bulk velocity for the electrons (##v_b = 0##), the [Compton] collision terms serve to drive ##\Theta## to ##\Theta_0##". Here ##\Theta_0## is the monopole moment of ##\Theta##. But if we already have ##v_b = 0## for the average velocity of the electrons then why would there be e.g. a dipole moment ##\Theta_1## of ##\Theta## to be driven to ##\Theta_0## under efficient Compton scattering? If ##v_b = 0## then wouldn't we already have only a monopole distribution for the temperature perturbations?
The phrase "...the [Compton] collision terms serve to drive ##\Theta## to ##\Theta_0##" makes it seem like there is, amongst higher moments, an existing dipole moment even when ##v_b = 0## and that eventually it gets suppressed by the monopole moment after many Compton scatterings but if ##v_b = 0## then how can there be a preferred direction for there to even be a dipole moment to be suppressed to start with? Wouldn't it make more sense to say that ##v_b \neq 0## and ##\Theta_1 \neq 0## at first but that efficient Compton scatterings serve to drive both ##\Theta \rightarrow \Theta_0## and ##v_b \rightarrow 0##?
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