If I live on a planet that is heavy enough, would the CMB get blue shifted enough in the atmosphere of this heavy planet, due to gravitational blue-shifting, that the CMB would be in the visible spectrum?

How heavy does this planet need to be relative to Earth?

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    $\begingroup$ Did you try to do a back-of-an-envelope-calculation? $\endgroup$
    – Qmechanic
    Mar 26, 2021 at 7:13
  • $\begingroup$ I like the idea of seeing the cosmic radiation with unaided human eye, but the gravitational method is less practical than the Doppler method. Your best bet is to jump aboard a rocket and accelerate up to $\gamma = 1000$. $\endgroup$ Mar 27, 2021 at 12:42

3 Answers 3


The cosmic background radiation has a wavelength about 2000 times longer than visible light. So you would need to be sitting deep in a gravitational well such that local time progresses 2000 times more slowly than distant time. Gravitational time dilation is given by the formula $\sqrt{1-r_\mathrm{S}/r}$ where $r$ is the distance from the center of the (spherical) mass and $r_\mathrm{S} = 2GM/c^2$ is its Schwarzchild radius.

For a time-dilation factor of 1/2000, one would have to be at a radius that exceeds the Schwarzchild radius by one part in 4,000,000. In other words, the object could only be a black hole and you would be barely outside its event horizon. Assuming you are somehow held stationary above the event horizon, you would not survive the gravitational field. The actual mass of the black hole is not specified, it could be large or small.

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    $\begingroup$ The redshift $z$ of the CMB is ~1089 so you "only" need half the time dilation specified above to blueshift the CMB back to the orange glow shown here. Of course, that doesn't make too much difference. According to this answer, a maximally compressed neutron star has $\frac{r_s}{r}\le 0.71$, which equates to $\gamma\approx 1.85$ $\endgroup$
    – PM 2Ring
    Mar 26, 2021 at 7:43
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    $\begingroup$ I thought only tidal forces were survivable or not? Although in this case, you may not survive the acceleration required to stop you dropping into the black hole. $\endgroup$ Mar 26, 2021 at 12:17
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    $\begingroup$ You don't need to be held stationary; "pretend" you're in a stable orbit. Yes, I know you'll have to be a 2-dimensional creature projected onto the sphere of the orbital path to avoid tidal force disaster. $\endgroup$ Mar 26, 2021 at 14:12
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    $\begingroup$ What you're decribing kind of feels like lockdown, to be honest. $\endgroup$
    – Strawberry
    Mar 26, 2021 at 16:10
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    $\begingroup$ You're doing this the hard way. Put the planet in a star system just outside a center-of-the-galaxy black hole. $\endgroup$
    – Joshua
    Mar 26, 2021 at 18:48

The extreme time dilation proposed can only come if you were on a planet that was orbiting a black hole.

The time dilation for a circular orbit around a Schwarzschild black hole is given by $(1 - 3r_s/2)^{1/2}$, where $r_s$ is the Schwarzschild radius.

Thus arbitrarily large time dilations can be achieved as an orbit approaches $1.5 r_s$, and the orbital speed approaches the speed of light.

Such orbits are not stable around a Schwarzschild black hole and in fact there would be no stable circular orbit that could give you the time dilation required. However, stable prograde orbits at $1.5r_s$ could be stable around a spinning black hole and this is the situation envisaged for "Miller's World" in the film "Interstellar". However, the stable orbit needs to be even closer to the event horizon and the black hole needs to be spinning at almost it's maximal value to get to the required time dilation factor.

There is no issue with the "gravity" crushing you, because an orbiting object is in free fall. There could be an issue with tidal forces, but these are not necessarily so extreme if the orbit is around a supermassive black hole exceeding $10^8M_\odot$.

Note that this just considers time dilation and not the Doppler shift. The specific intensity and blueshift of the CMB would be highly beamed in the direction of orbital motion. This additional blueshift means that the orbit need not be quite so extreme to see a spot of visible CMB (though the orbital speed would need to be highly relativistic)


Going by Buchdahl's Theorem.

If the as the radius of a body approaches $R=\frac{9R_s}{8}$ The pressure inside diverges to infinity.

Since the formula for gravitational Time dilation is $\sqrt{1-r_\mathrm{S}/r}$ This gives a maximum of $\sqrt{1-8/9}=1/3$

In comparisoon, the CMB has wavelengths 2000 longer then visible light.


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