# Can you see on Miller's planet?

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

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. ($$E=hf$$)

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

There is a very nice explanation by @profrob, and I must ask a follow-up question because there are two point where I am asking for some more details how specifically gravitational time dilation would cause blueshift:

1. Is there an effect of gravitational time dilation that would increase all photon's frequency just because they (and the observer) are in a deep gravitational field? Can someone please explain that effect and how that works? And is it correct to say that this affects all photons, so not just CMB?

2. All along I thought gravitational time dilation was a effect that is only realizable when compared to another frame, far away from the black hole. If you are on Miller's planet, your clock runs normally for you. It is only when you compare it to another clock, you realize that your clock runs much slower on Miller's planet, relative to the other clock far away from the black hole. Isn't this the same way (because of relativity) with photon's energy? Photon's energy in GR is observer dependent. For an observer on Miller's planet, the photon's would be normal (i.e. visible would still be visible), it is just when you would compare it to another photon far away from the black hole that you would see the energy difference?

Now if you go with the calculations to that question, and all photons are blueshifted just because they are in a strong gravitational field (even for a local observer on Miller), then this means not just CMB, but all photons are blueshifted, including visible light photons (which would be blueshifted to non-visible range), but that means it is impossible to see on this planet.

Question:

1. Can you see on Miller's planet?
• @safesphere thank you, can you please help me edit the question, because there is then an effect, which is not real (like the direction dependent one), but then there is another effect (which is not direction dependent), can you please help me how that gravitational time dilation effect works on photons' energy? May 27, 2022 at 1:03
• @safesphere thank you I edited. Can you please see if it is correct? I have a question though. I understand now that it is frame dependent, at the time of emission, in the stronger field I see a photon that is emitted far away to be blueshifted, that is understandable. But how does gravitational time dilation cause this effect? Can this be explained by spacetime curvature? Or does this work like the photon clock example too? May 27, 2022 at 3:24
• @safesphere thank you, this should be the answer. How does time running faster cause the emitting atom to emit a higher frequency photon? Time running faster corresponds to higher frequency? May 27, 2022 at 6:27
• Why the -1 vote? It's a nice question! May 27, 2022 at 8:20
• @safesphere What about photonic energy in inflating, expanding space? May 27, 2022 at 13:13

I think the answer is yes, but the illumination levels would be similar to a heavily overcast day on Earth.

The CMB is a blackbody continuous spectrum, so even though the peak is blueshifted into the ultraviolet, there are still visible photons hitting the planet.

Indeed, a higher temperature blackbody is more intense at all wavelengths than a cooler blackbody of the same emitting area.

I use italics because the blueshifted CMB comes from a tiny hot spot in the sky with a much smaller angular extent than the Sun for instance.

I ran the numbers using this calculator.

According to my answer to the linked question (and recall, these are numbers from a GR ray-tracing simulation published by others), the radiation incident upon Miller's planet is in the form of blackbody radiation at about 700,000 K from an intensely bright spot on the sky, which bathes (one side of) the planet with 400 kW/m$$^2$$ of mainly EUV and UV radiation. If you calculate what fraction of this blackbody flux falls in the visible band between 400nm and 700nm, it is only about 2.2 W/m$$^2$$.

This is several hundred times fainter than direct illumination by direct sunlight. This is similar to the illumination one might experience on a heavily overcast day. Of course the spectrum is very, very different and heavily weighted towards the UV.

• This is a good answer, but not to the question the OP is asking. He is not asking if he can see CMB there. He is asking if he can see light emitted locally or if this light is also blueshifted. May 27, 2022 at 0:14
• Thank you so much! May 27, 2022 at 1:03
• Lets set aside the black-body glow from lava-like temperatures. Fluorescence (even "non-fluorescent materials" can have a little) and other mechanics to convert UV to visible light would add a lot of light. The sky would glow from recombination much like how electric arcs make air glow. The hills would look like a black light party. So most of the visible light would not be "sunlight". Dec 22, 2022 at 1:58

Neglecting any brightness from the low energy part of the blue-shifted CMB (which might be significant, as ProfRob points out), a surface temperature of 890 degrees Celsius corresponds to a red-orange glow which would provide sufficient dim light for vision. Neglecting hot spots and cold spots, it would have a comparable brightness to completely covering the surface of the planet with iPhones with the screens set to a fairly dim setting.

If you were not protected by a reflective suit, you would also spontaneously combust, providing abundant, if short-lived light for anyone in the vicinity.

I'm not sure if approximating the whole planet as an 890 degree blackbody is reasonable - there might be e.g. yellow-hot lava oceans at the equator and polar "ice caps" of black solid rock at the poles.

• Thank you so much! May 27, 2022 at 1:03

You could see perfectly well. If you didn't get fried you could use a lantern to enlight the stuff around you. You could hide in a deep cave to take shelter from the burning CMBR (not sure if that has turned to visible). The visible stars would not be visible anymore. They would have become UV at least. In the cave, just switch on the light, smoke a cigar, and enjoy the daily TV show.

Okay, a little elaboration. Time on the planet runs about 61000 times as slow as the time in the space where the stars shine happily along. This means the radiation they emit (assuming them Sun-like) will have it's black body peak frequency increased by 61000. Look at the Sun's frequency spectrum:

Now the max will shift into the non-visible part. The low frequencies get higher and shift to the visible. But the intensity is much smaller. So the stars will turn very dim.

The light of a flashlight won't be influenced because the light is produced on the planet.

What amazed me was that in the movie there was a fully enlightened Miller. What caused this light? Studio spotlights?

• @safesphere Yeah, these spelling errors drive me crazy. I type on mobol phene... May 27, 2022 at 8:16
• @safesphere Does everything need an explanation? From where did the 5D reversed hypersphere quantum vacuum spacetime connecting two 4D hyperbolic spaces come from? Gods probably! May 27, 2022 at 8:25
• @safesphere The answer is correct, but is based on an opinion with no explanation Would that be the reason for the two downvotes? May 27, 2022 at 14:08
• Thank you so much! This is a nice answer, maybe you could expand some detail on why the non-locally emitted photons (starlight) would not be visible anymore, and why the locally emitted ones (TV) is still visible. May 27, 2022 at 15:31
• What about photonic energy in inflating, expanding space?” - If space expands without acceleration, then the energy of photons is conserved. The cosmological redshift is caused by observing light from a different frame of reference, not by photons losing energy - they don’t. Imagine a mirror a billion light years away that is stationary relative to you (the distance between you and the mirror remains constant). If you emit a photon in an expanding space and this photon reflects back to you from this mirror, then you’d see it returning with the same energy it was emitted - not redshifted. May 28, 2022 at 3:37