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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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  • Yes, sorry, I meant blue-shift. But the frequency-shift is due to the integral over the curve on which the photons traveled. And the potential is time-dependent, see Sunyaev-zel'dovich. So it isn't clear to me why there's no assymmetry? – WIMP Oct 25 '18 at 07:28
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    @WIMP The Sunyaev–Zel'dovich effect describes the Compton scattering of CMB photons in galaxy clusters. It's hard to see how it's applicable to the effect of the Milky Way's gravity on the CMB. The gravitational potential is a scalar. To get from intergalactic space to the solar system incurs the same change of PE regardless of the direction. There may well be other effects related to the amount of matter through which the CMB has passed, but then we avoid the galactic plane when making CMB measurements for precisely this reason. – John Rennie Oct 25 '18 at 07:34
  • I'm an idiot, sorry, I meant Sachs-Wolfe. Forget about the name for a moment, I mean you calculate the blue-shift by integrating over the geodesic equation in a potential that is neither spherically symmetric nor time-independent. How do you know the result is isotropic (as seen from Earth). You just seem to assume this, but it's not clear to me. – WIMP Oct 25 '18 at 14:46
  • @WIMP I assume you mean the late time integrated SW effect. This happens when a gravitational well changes while the CMB photons are passing through it, specifically the well becomes shallower due to accelerated expansion of the universe. This doesn't happen in the Milky Way because the Milky Way is not undergoing accelerated expansion. Indeed, the time scale needed for light to cross the Milky Way is only $10^5$ years, and on this timescale the spacetime geometry around the Milky Way is effectively is constant. – John Rennie Oct 25 '18 at 15:16
  • More on the Sachs-Wolfe effect here. – John Rennie Oct 25 '18 at 15:17
  • I understand you say the time-dependence is negligible. Leaves the question how do I see that there's no angular dependence in the red/blueshift given that the potential depends on the angle? – WIMP Oct 25 '18 at 17:05
  • The potential is a scalar. It's defined at a point - at the position of the Earth for the purposes of this discussion. It is not dependent on angle. – John Rennie Oct 25 '18 at 17:17
  • You miss the point: Blue-shift is calculated from integrating the equation for the photons from infinity to the point of measurement. You are using an equation I know is correct for a spherically symmetric case. How do you know it has no angular dependence? – WIMP Oct 26 '18 at 04:53
  • @John Rennie Hi John, I suggest you to change units of $\Phi$. It is not an energy, but a speed squared. – Elio Fabri Oct 26 '18 at 06:33
  • @ElioFabri It's an energy per unit mass isn't it? I guess I should make that clearer. – John Rennie Oct 26 '18 at 06:43
  • @John Rennie Yes, so the right unit is J/kg or $\rm m^2!/s^2$. After all, in your formula for blueshift, dividing $\Phi$ by $c^2$ you get a pure number. – Elio Fabri Oct 26 '18 at 13:33