count_of_monte_carlo

joined 2 years ago
MODERATOR OF
[–] count_of_monte_carlo@lemmy.world 2 points 4 weeks ago (1 children)

A black hole is believed to contain a singularity with all of the mass as a single point. So this is well past the point of baryonic matter and in a region where our physics models break down.

If you just take the total mass of a black hole and divide it by the volume of the Schwarzschild radius (aka event horizon) you get a density MUCH greater than a neutron star. This isn’t a useful measure of the black hole density though, since all of the mass is at a single point of presumably infinite density.

[–] count_of_monte_carlo@lemmy.world 30 points 4 weeks ago (5 children)

Neutron stars are the most compact form of matter that we know about; they’re even denser than the nucleus of an atom.

Neutrons, protons and electrons are fermions, meaning they must obey the Pauli exclusion principle. No two neutrons (or protons or electrons) can be in the same quantum mechanical state. If you take ordinary star matter (plasma made of dissociated protons and electrons) and squeeze it, eventually the electrons will nearly overlap in their states. You’d have two electrons with nearly the same energy, spin and location. They cannot overlap though, so this creates a repulsive force that prevents the matter from further compression; this is called the electron degeneracy pressure.

If the compressive force overcomes this pressure, then the electrons can capture on the protons to form neutrons. Neutrons and protons also have degeneracy pressures, but they can be packed much more densely than electrons. This is because their wavelength is shorter. The wavelength of a massive particle is inversely proportional to its mass, and protons and electrons are about 2000 times the mass of electrons. So compressed ordinary matter will inevitably become pure neutrons, simply because this is the most compact form.

A pure electron or pure proton star wouldn’t be as compact because both are charged particles so there would be Coulomb repulsion (this isn’t an issue in ordinary matter since the number of electrons and protons is roughly equal). You’d also need to somehow separate the electrons from the protons, and this isn’t a process that would naturally occur in a collapsing star.

They would die immediately.

Breaking atomic and molecular bonds is what ionizing radiation such as gamma rays or neutrons does to a human body, so this scenario would be sort of like an extreme, unrealistically high dose of radiation.

Immediately after all bonds are broken, atoms would react and start to form new bonds. In some cases these bonds would reform correctly- for example, water being remade as H2O. In many cases alternative bonds would form which could make substances incompatible with life- say H2O2, aka hydrogen peroxide. Basic cellular and nerve function wouldn’t work, hence the instant death.

Lepton number is an observationally conserved quantity. As far as I know there’s no fundamental reason for it to be conserved (and indeed there are searches for physics beyond the standard model that would violate it) but it’s been found to generally be conserved in reactions so far. Lepton particles have a lepton number of +1, lepton antiparticles have -1.

There’s a similar conserved quantity known as the baryon number, with a similar definition. Protons and neutrons (baryons) have values of +1, anti-protons and anti-neutrons are -1.

An example: consider the beta- decay of a neutron, baryon number +1 and lepton number 0. It emits a proton (baryon number +1), an electron (lepton +1), and an electron anti-neutrino (lepton -1). Total lepton number of the decay products is 1-1=0, so the value is conserved.

Locking because the discussion isn’t consistent with community rules.

While /c/askscience doesn’t have a rule against soliciting medical advice, I’d strongly recommend against trusting the comments from random internet strangers. Especially those without authoritative sources to back them up.

Unfortunately the only way to get a reliable answer to your question is to speak to a medical professional.

[–] count_of_monte_carlo@lemmy.world 11 points 6 months ago (1 children)

I’ll echo the other replies that the gravitational waves from black hole mergers have been detected by LIGO. In fact, the 2017 Nobel Prize in physics was awarded to members of this collaboration specifically for this feat.

We haven’t (yet) seen a pair of black holes collide using light directly, but the gravitational waves have been perfectly consistent with general relativity calculations. Here’s a video from LIGO that shows what one of these simulations looks like, for a simulation that reproduces a detected gravitational wave.

As an aside, right around the time the LIGO team was awarded the Nobel prize, they detected the collision of a pair of neutron stars. They alerted the astronomy community to the direction they saw the signal from, and within a day there were telescope observations of light from the kilonova that resulted from the collision. Ultimately various sensors recorded optical light, infrared, ultraviolet, gamma rays, and radio waves being emitted from the explosion. The hope is that someday we’ll get lucky enough to see similar photon signatures from a black hole merger!

For physics specifically, a bachelor’s degree probably won’t be enough to get a job in physics.

You might be able to get a job as a technician in a lab, but they typically will look for people with a master’s degree for those roles. With just a bachelor’s , you’d need to get your foot in the door by already having some relevant experience, which is a possibility if you get some research experience in college and pivot that into an internship or something. But it would definitely require effort and luck.

I’m not sure that’s a good comparison. The kill mechanism from a neutron bomb is the deposition of ionizing radiation in the body, but the microwave radiation is non-ionizing.

You’ve gotten some good answers explaining that heat changes the density, and therefore the index of refraction of air.

Fun fact: Schlieren Imaging allows one to photograph shockwaves by relying on the same effect. As a shockwave travels through air, it creates a region of high density, which can be imaged with this technique.

[–] count_of_monte_carlo@lemmy.world 14 points 10 months ago (1 children)

in the photon's frame of reference

There are no valid inertial frames for an object moving at the speed of light. The idea that “a photon doesn’t experience time” is a common, but misleadingly incorrect statement, since we can’t define a reference frame for it. Sometimes this misconception can be useful for conveying some qualitative ideas (photons don’t decay), but often it leads to contradictions like your question about Hawking Radiation for black holes.

[–] count_of_monte_carlo@lemmy.world 10 points 10 months ago

Yes, the wavelength of photons will be preserved if they travel through non-expanding space. If the photon is emitted by a source that’s in motion with respect to a detector, there could still be redshift or blueshift from the relativistic Doppler effect. This would only depend on the relative velocity between the emitter and observer, and not on the distance the photon traveled between them.

view more: next ›