Wouldn't Miller's planet be fried by blueshifted radiation?

Question

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.

Answer 1

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

Answer 2

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!