If I traveled fast enough, my current understanding is that visible light would be blueshifted to the blue/UV range, but also that microwaves and longer wavelength waves would be blueshifted into the visible range. Is this true? If I were to go fast enough, would I be able to see the CMB? In reality, doesn't the Earth hurtle around the Sun quite quickly? So either this effect doesn't exist, or you'd need to go many orders of magnitude faster (an estimation perhaps?)

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    $\begingroup$ You can see the CMB now, we have pretty good pictures of it. If you were to travel really fast in one direction, you could see a tiny speck of the CMB blue shifted into the visible but since the CMB has a temperature that is about 1000 times lower than that of visible light, you would have to go very close to the speed of light. $\endgroup$
    – CuriousOne
    Commented Jan 3, 2015 at 7:01
  • $\begingroup$ @CuriousOne you should make this an answer! :D $\endgroup$ Commented Jan 3, 2015 at 7:28
  • $\begingroup$ Nah, there are much more detailed real answers, already. Thanks for suggesting, though. $\endgroup$
    – CuriousOne
    Commented Jan 3, 2015 at 7:40
  • $\begingroup$ @CuriousOne The "speck" actually turns out to be 0.2 degrees across. Surprising. $\endgroup$
    – ProfRob
    Commented Jan 3, 2015 at 11:20
  • $\begingroup$ @RobJeffries: I agree. To me that's surprisingly large at that boost, but then, I never made an attempt to calculate it. On the other hand, being less than half of the moon... which to most will probably be surprisingly small? Great answers, by the way! Thanks to you and John Rennie! $\endgroup$
    – CuriousOne
    Commented Jan 3, 2015 at 14:52

2 Answers 2


You do see the Doppler shift of the Earth's motion with respect to the CMB. It imprints a dipole component on the CMB temperature map.

The motion of the Earth around the Sun is a modest 30 km/s, but even this doppler shifts inferred temperatures by a factor $\simeq v/c$ and needs to be taken out of the CMB analysis. A larger effect is the motion of the Sun with respect to the Galaxy/local group and with respect to the Hubble flow around us, which is 10 times bigger. This makes one hemisphere of the CMB about 1 part in a thousand hotter than the hemisphere the Earth is receding from.

To make the CMB visible we need the blackbody radiation temperature to be $>3000\ K$, requiring a blueshift of a factor 1100 from the current $2.7\ K$. This puts the peak of the Planck function at a wavelength of 966nm, outside the visible range, but the Wien tail of the distribution should be comfortably visible.

Using the relativistic redshift formula: $$ \frac{c + v}{c-v} = 1100^2$$ gives a required velocity of $0.99999835c$.

For this redshift, the specific spectral intensity of the radiation (in the direction of motion, expressed per unit frequency) is increased by a factor of $1100^3$. Thus the specific spectral intensity becomes $3.7\times10^{-18} \times 1100^3 = 4.9\times10^{-9}\ Wm^{-2} Sr^{-1} Hz^{-1}$ at 966 nm and about half this at 550nm where the eye's response peaks. For comparison, the specific intensity of the solar photosphere is $2.6\times 10^{-8}\ Wm^{-2} Sr^{-1} Hz^{-1}$ at 550 nm i.e. only $\sim 10$ times brighter. So, even accounting for the fact that the radiation from the Sun is better matched to the eye's response, this still means that the CMB would be extremely bright in the forward travelling direction. In fact you could play with the numbers and probably find you could get away with a smaller doppler boost and still be able to see the Wien tail of the CMB, but I'll leave it to someone with more knowledge of eye physiology and sensitivity to red wavelengths to work out the details; it would involve the convolution of two sharply changing functions of wavelength (the Wien tail and the eye response).

EDIT: Note to take into account Curious One's excellent comment. The doppler boosted CMB will be seen as an intense spot on the sky. The boosted radiation occupies an opening angle $\sim 2/\gamma$ radians, where the $\gamma = (1-v^2/c^2)^{-1/2} = 550$. i.e. a spot of width 0.2 degrees (about half the full moon) on the sky.

  • $\begingroup$ I have a follow up question. (Assuming there was nothing BUT the CMB radiation), even if you were travelling at the speed of light, I assume you could never “reach” the CMB, due to the theory that objects are receding at a rate much faster than the speed of light. If that’s the case, wouldn’t the CMB still be highly redshifted even if you were travelling at nearly the speed of light? $\endgroup$ Commented Jul 18, 2020 at 18:48
  • $\begingroup$ @keithknauber it is redshifted in the cosmological co-moving frame. But you would be moving relativistically with respect to that. $\endgroup$
    – ProfRob
    Commented Jul 18, 2020 at 20:40
  • $\begingroup$ Yes, my question is, how does relativity apply here? When I think of this question in terms of length contraction, it is also a conundrum. The “distance” to the CMB can decrease, but it seems it could only decrease a certain amount given that the space from whence CMB photons are arriving has expanded at faster than the speed of light. $\endgroup$ Commented Jul 18, 2020 at 20:48

The effect you describe exists and is called the dipole anisotropy. A review of its measurement is given in the paper Planck 2013 results. XXVII. Doppler boosting of the CMB: Eppur si muove. The dipole anisotropy allows us to calculate the Earth's peculiar velocity, that is the velocity relative to the comoving frame, and it turns out we're scooting through space at about 370 km/sec. For comparison the Earth's orbital velocity around the Sun is about 30 km/sec.

To blue shift the CMB into the visible range will require relativitic velocities, and therefore you need to use the relativistic doppler equation:

$$ \frac{f_s}{f_0} = \sqrt{\frac{1 + \beta}{1 - \beta}} \tag{1} $$

where $\beta = v/c$. The frequency peak of the CMB is around $160$ GHz and green light has a frequency of about $5 \times 10^{14}$ Hz so that makes the ratio $f_s/f_0 \approx 3400$. Substituting this value in equation (1) and solving for $\beta$ I get (it's 7 a.m. here so you might want to independantly verify the sums :-)

$$ v = 0.9999998 c $$


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