In Interstellar, wouldn't Miller's planet be fried by blueshifted radiation? The 61,000x time dilation multiplier would make even cosmic background radiation photons into extreme UV photons.

I was told that this is off-topic for Science Fiction & Fantasy, so I'm posting it here.

For those not in the know - Miller's world orbits an almost maximally spinning massive black hole at just barely above its Schwarzschild radius. This results in extreme time dilation - 1 hour on Miller's world passes in about 7 years according to a distant observer.

  • $\begingroup$ @J.G. Miller's planet is orbiting a supermassive BH which has very high spin, so the innermost stable prograde orbit is close to the event horizon, much closer than the ISCO of a Schwarzschild BH of the same mass. $\endgroup$
    – PM 2Ring
    May 23, 2022 at 8:26
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    $\begingroup$ This would work better as a Physics.SE question if you described the scenario in some detail instead of assuming any reader is familiar with the source media. $\endgroup$ May 23, 2022 at 20:06
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    $\begingroup$ Take a step back and explain the point of the original question… I personally read this as an SF&F Question, but who am I? I suggest the point is not whether in reality, a Miller Planet would be fried by blueshifted radiation but whether in fiction, any reader could either accept or refute that idea. I suggest almost no readers of fiction - and not many more calling themselves physicists - would recognise a Miller Planet if one smacked into Earth tomorrow. Write what you can, paying less attention to detractors than you've so far been inclined! $\endgroup$ May 24, 2022 at 20:24
  • $\begingroup$ The planet would probably also be torn apart by tidal forces and fried by the radiation from the accretion disk. But why do we know that the planet was very close to the horizon? $\endgroup$
    – RC_23
    Jul 1, 2023 at 15:27
  • $\begingroup$ Because otherwise it would have been impossible to achieve the degree of time dilation that MIllers planet had $\endgroup$ Jul 1, 2023 at 15:29

2 Answers 2


Miller's world would be fried by a strong flux of extreme ultraviolet (EUV) radiation. The cosmic microwave background (CMB) would be blueshifted by gravitational time dilation and then would be very strongly blueshifted and beamed coming from the direction of orbital motion. The overall effect would be a very strong dipolar distribution of temperature that is then distorted by the curved ray paths close to the black hole, whose shadow would fill nearly half the sky.

However, the size of the ultra-blueshifted spot is correspondingly very small. A detailed numerical calculation$^\dagger$ comes up with an equilibrium temperature for Miller's world of 890 $^{\circ}$C (Opatrny et al. 2016), with a flux of about 400 kW/m$^2$ from an EUV blackbody(!) arriving from the CMB "hotspot". I guess you would classify this as "fried"$^{\dagger\dagger}$. It is hotter than Mercury anyway.

$\dagger$ According to Opatrny et al. the peak blueshift in the direction of orbit is $275000$ - i.e. wavelengths are shortened by a factor of $275000+1$. Since temperature goes as redshift, then a tiny spot on the sky is an intensely bright (brightness goes as $T^4$) blackbody source of soft X-rays and EUV radiation. The source size is of order angular radius $1/275000$ radians due to Doppler beaming. Back of the envelope - the source is 130 times hotter than the Sun but covers a $(1200)^2$ times smaller solid angle in the sky. Thus the power per unit area received should be $130^4/1200^2 = 200$ times greater than from the Sun. This is in pretty good agreement with Opatrny et al.'s calculation that also claims to take into account the lensing effects.

$\dagger\dagger$ Apparently, the typical temperature in a frying pan is 150-200$^{\circ}$C

  • $\begingroup$ Something confusing in your calculation. Solar irradiation on Earth is roughly 1 kW/m^2 . Miller's Planet getting 400X that sure sounds like instant death for any organic animal. I suppose 1200 K surface temperature would be instant death too. $\endgroup$ May 23, 2022 at 14:17
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    $\begingroup$ @CarlWitthoft it isn't my calculation, but I am confused about what you are confused about. 400 kW/m$^2$ would indeed be classified as "fried" in my book. Which is what I have said. Is it that you think the equilibrium T isn't high enough? Recall that flux $\propto T^4$. $\endgroup$
    – ProfRob
    May 23, 2022 at 15:03
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    $\begingroup$ How does this work? The microwave background is an extremely low power density. Even x61,000, how does it get enough total Watts to fry the planet? Does it have to do with the total surface over which the beam comes from? $\endgroup$ May 24, 2022 at 4:19
  • $\begingroup$ @The_Sympathizer the peak blueshift is 275000. Temperature increases with blueshift. Brightness goes as $T^4$. I have added a back of the envelope calculation. $\endgroup$
    – ProfRob
    May 24, 2022 at 8:23
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    $\begingroup$ @ProfRob Sorry - I just felt you understated the degree [pun intended] of frying that would happen $\endgroup$ May 24, 2022 at 10:33

For a 2.7 K blackbody (like the CMB) the calculator at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radfrac.html gives me 3 $\mu$W/m$^2$. This is how much Earth is heated by the CMB — not a lot!

There are two effects acting on a body in a deep gravity well that increase this radiation:

  • Blueshift: Time going 61,000$\times$ slower would increase the observed frequency of the photons by the same amount. The energy of photons is proportionally increased. ($E=hf$)
  • Time dilation: We see 61,000$\times$ more photons arrive per second because our seconds are 61,000$\times$ longer.

Together that's a multiplier of $61000\times\ 61000 \approx 4\times 10^9$ giving us 12 kW/m$^2$. That's almost 10 times the radiation Earth gets from the Sun (1.4 kW/m$^2$). That's like being 3 times closer to the Sun — closer than Mercury.

But Miller's planet is not just standing still in a deep gravity well. It's moving unbelievably fast too. According to How fast is Miller's planet orbiting Gargantua in the movie Interstellar? it's moving at 55% the speed of light.

For simplicity we can imagine the planet just moving in a straight line. Then we have two effects from this high speed:

  • Photons on the "front" side will again be blueshifted. According to https://www.atnf.csiro.au/people/Tobias.Westmeier/tools_redshift.php: $$ 1+z_{pec}=\sqrt{\frac{1+\beta}{1-\beta}} $$ where $\beta=0.55$. That gives each photon 86% more energy.
  • A moving planet hits more particles (photons) per second than a planet that stands still. (It's a bit like standing vs running in rain.) If we only consider light from the front and ignore relativity, the planet is clearing a volume of $A \times c \times 1.55$ compared to $A \times c$ for a stationary planet. So the front side hits 55% more photons.

These two effects together increase the radiation to $12 \times 1.86 \times 1.55 = 35$ kW/m$^2$ on the front side, which is like being 5 times closer to the Sun than Earth. Very hot!

  • $\begingroup$ I know this is not in perfect agreement with arxiv.org/abs/1601.02897, so I'm wrong somewhere. But it's much simpler and still gets the planet fried. :) $\endgroup$ May 24, 2022 at 8:49
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    $\begingroup$ It's because the relevant blueshift factor isn't 61,000. It's 4.5 times bigger due to the orbital motion. $\endgroup$
    – ProfRob
    May 24, 2022 at 9:06
  • $\begingroup$ Oh, thanks! And it would be 4.5 less on the other side, right? But even if I just multiply by 4.5, that still only gets me to 54 kW/m^2. $\endgroup$ May 24, 2022 at 9:35
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    $\begingroup$ It would be 4.5 squared. But you probably can't do the calculation like this because the radiation is no longer isotropic. $\endgroup$
    – ProfRob
    May 24, 2022 at 10:42
  • $\begingroup$ Sorry, I know you must be right, but it beats me how it gets squared. Now that I looked a bit more it also beats me how it gets to 4.5. I only get 1.86 for blueshift at 0.55 c. But I can't picture additional time dilation from this movement. We know the time dilation is 61,000. The movement causes more photons to be hit on the front side, but only an extra 55%? (If I'm getting this totally wrong, a downvote is fine too, I just wanted to try to work it out.) $\endgroup$ May 24, 2022 at 13:36

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