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We sit in a gravitational potential, so there should be a blue shift on the CMB light from the potential of the Milky Way. Is this blue shift dependent on direction? Is it being subtracted from the CMB? Or is it simply to small to be measurable?

John Rennie
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WIMP
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1 Answers1

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The gravitational potential of the Milky Way will cause a blue shift not a red shift. This happens because relative to an observer far from the Milky Way the gravitational potential within it causes a time dilation i.e. here on Earth our clocks run very slightly slower than clocks out in intergalactic space. Since our clocks run slower the frequency of light coming from intergalactic space is slightly increased.

As it happens I have described the time dilation for observers within the Milky Way in my answer to Why isn't the center of the galaxy "younger" than the outer parts? From experimental data we have the following approximate formula for the gravitational potential energy (per unit mass) inside the Milky way:

$$ \Phi = -\frac{GM}{\sqrt{r^2 + (a + \sqrt{b^2 + z^2})^2}} \tag{1} $$

where r is the radial distance, z is the height above the disk, a = 6.5 kpc and b = 0.26 kpc. The time dilation is well described by the weak field equation:

$$ \frac{\Delta t_r}{\Delta t_\infty} = \sqrt{1 - \frac{2\Delta\Phi}{c^2}} $$

According to NASA the Sun lies about 8 kpc from the centre of the Milky Way, so $r = 8$ kpc and according to the Astronomy SE we are about 20 pc away from the plane so $z = 20$ pc. Finally let's guesstimate the mass of the Milky Way as $10^{12}$ solar masses (it's a guesstimate because we don't know how much dark matter the Milky Way contains). Plugging in all these numbers gives us:

$$ \Phi \approx 4 \times 10^{11} \,\text{joules/kg} $$

And plugging this into out equation for the time dilation gives:

$$ \frac{\Delta t_r}{\Delta t_\infty} = 0.999995 $$

Taking the reciprocal of this to estimate the blueshift we find the frequency of the CMB is blue shifted by a factor of $1.000005$.

John Rennie
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