- #1

ChrisVer

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In first order, this is given by [itex] \beta \approx 1.2 \times 10^{-3}[/itex] or [itex]u_{CMB} \approx 360 km/s [/itex].

I have one question here:

Why is the velocity given as such?

In first approximation we have [itex]T' (\theta)_{observed} \approx T_{cmb} (1 + \beta \cos \theta)[/itex]

and this depends on [itex]\theta[/itex], the direction in which we are looking at CMB. So I'd say that the velocity of the earth wrt to the CMB rest frame, should also depend on [itex]\theta[/itex]:

[itex]1.2 \times 10^{-3} = |\frac{\Delta T}{T}| = |\frac{T - T'}{T}| = |1- T'/T| = |\beta \cos \theta|[/itex]

Also, I'm not sure how can this affect the measurements of distances with the Hubble's law:

[itex] u = H_0 D[/itex]

Is it because the [itex]u[/itex] will have to get the contribution from [itex]u_{CMB}[/itex]? but this is crazy, because the far-away galaxy we are observing, will also have to move with some velocity wrt CMB... How is Hubble's law then affected?