Question: If there was a star in the universe that was so immensely hot that its wavelength was that of the Planck distance. (Obviously this is purely theoretical) say we managed to add more energy to make it hotter, what would happen to the wavelength, it can't get smaller, but it has to? If you can answer this, I will vote for you, get thinking scientists!

Jack Miller answered on 16 Jun 2013:

Hi Emperor9271,

Wow, what a good question! First off, a disclaimer: I’m not really an astrophysicist, but I’m sure the others will quickly correct me if I say anything they disagree with!

The short answer is that it wouldn’t be a star anymore — nor, in fact, would it be to start with. Stars are defined primarily by a few properties — such as mass, radius, temperature and luminosity (brightness). Stars are very good “black bodies”, which means that their colour is determined primarily by their temperature alone — there’s a distribution in the wavelength of light they emit, with a large peak that decreases (i.e. gets bluer) with temperature. The temperature of the star is, itself, linked to other things — most notably mass and what elements are being used to fuel the star by the process of nuclear fusion in its core. It’s actually quite hard to calculate exactly this relationship, and there are various approximate forms that give reasonably accurate (for astrophysics) answers. Likewise, there’s a relationship between stellar radius and mass, and all of these things together give a mathematical description of a star.

If you take the classical understanding of how stars work, and turn “the dial up to 11”, i.e. work out the temperature that your ‘star’ would have to be in order to emit light with a wavelength on the Planck scale, you’ll find probably that a) your ordinary calculator can’t hack it, and b) you’ll get a truly enormous number. This would correspond to a star with an insanely high mass. If you then work out the pressure that all this mass would cause in the star’s core, you’d quickly work out that it was a very large number indeed — far larger than the limits for ordinary “main sequence” stars, or indeed neutron stars (which are even denser than stars made out of ordinary matter) — and, instead of having a hypothetical star, you should really be thinking about a hypothetical black hole. At this point, your black hole obviously doesn’t radiate much — hence the reason they’re black, although there’s a complicated process called Hawking radiation by which black holes gradually “evaporate” and lose mass by quantum effects — but we can still ask your question of what would happen as you add mass to it. Boring answer: you get a bigger black hole. Add more mass to that? Bigger black hole.

In fact, there exist very, very large black holes at the centre of galaxies (including ours!) called supermassive black holes, with masses on the order of 20 billion times that of the sun (which itself has a mass of around 10^30 kg). Your hypothetical star-that-became-a-black-hole-as-soon-as-you-constructed-it would have a mass that, I think, if my back-of-the-envelope calculations are roughly correct, would probably not be more than about ten orders of magnitude greater than one of these supermassive black holes, though I’m still fairly convinced it could never form in nature. Why? As black holes get larger, their mass and their radius increases. As their radius increases (roughly proportionally with their mass), the more Hawking radiation you get, and when talking about very large structures it’s been theorised that this outward facing radiation becomes significant — disturbing the clouds of gas that (at the centre of a galaxy) “feed” this otherwise evaporating black hole.

Lastly, I’d better say what Hawking radiation actually is, as it’s rather cool: the vacuum of space (or anywhere else, for that matter) isn’t quite as static and boring as you might think it is. It’s entirely possible for it to fluctuate, creating particle-antiparticle pairs that briefly form and then quickly annihilate each other. If one of these particle-antiparticle pairs appears close to the event horizon of a black hole — the point beyond which nothing, not even light, can escape — one of the pair might fall into the black hole whilst the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole). By this process, the black hole loses mass, and, to an outside observer, it would appear that the black hole has just emitted a particle. Pretty neat, huh?

Sorry for an absolutely enormous wall of text — I hope it makes sense!

Thanks for an excellent question,

— Jack