What would result from compressing free protons?

degeneracy

Postpocelipse said:

What would result from compressing free protons?

By what means?

// By what means?

sounds like expectation value

dv said:

Postpocelipse said:

What would result from compressing free protons?

By what means?

Hypothetically by thought experiment. Wondering what the feedback of exaggerating the repulsive element would result in. How is the spacetime managed between two protons that are forced into confinement.

SCIENCE said:

degeneracy

??

Postpocelipse said:

What would result from compressing free protons?

Hypothetically by thought experiment.

That’s not going to fly. Even for a thought experiment, we’ll need a mechanism if we are give meaningful answers.

dv said:

Postpocelipse said:

What would result from compressing free protons?

Hypothetically by thought experiment.

That’s not going to fly. Even for a thought experiment, we’ll need a mechanism if we are give meaningful answers.

So start with +charge. What is in action when two free protons act on each other to put distance between them?

It might only confuse the thread but at this point the simplest means I have of breaking it down is to observe that bosons seemingly provide a spatial definition to physics while fermions provide a time definition. As approaching the question from both directions is overly complex I’m just starting with the characteristics of protons in search of a mechanism.

Electromagnetism

dv said:

Electromagnetism

Can you have electromagnetism without fermionic presence?

You mean in the universe?

dv said:

You mean in the universe?

More so just between the subject free protons.

The only result I can think of in a hypothetical scenario of this sort is that the protons charge increases with greater forced proximity and no means of refracting the action through a fermionic outlet. Even if that has any accuracy it doesn’t tell me much as yet.

Postpocelipse said:

The only result I can think of in a hypothetical scenario of this sort is that the protons charge increases with greater forced proximity and no means of diffracting the action through a fermionic outlet. Even if that has any accuracy it doesn’t tell me much as yet.

fixed

> What would result from compressing free protons?

That can be done, for example in particle colliders. The net result is that the energy put into the compression manifests as new particles such as electron-positron pairs as the protons fly apart again.

SCIENCE said:

// By what means?

sounds like expectation value

{golf clap}

MartinB said:

SCIENCE said:

// By what means?

sounds like expectation value

{golf clap}

I looked up “expectation value”, but I’m none the wiser:

“In quantum mechanics, the expectation value is the probabilistic expected value of the result (measurement) of an experiment. It is not the most probable value of a measurement; indeed the expectation value may have zero probability of occurring. It is a fundamental concept in all areas of quantum physics.”

Why would anyone expect a result that had zero probability?

Ensemble average.

Imagine a system that for example can only be up or down but are so with equal probability.

The expected value is zero, even though that is not a possible result of an individual measurement.

MartinB said:

Ensemble average.

Imagine a system that for example can only be up or down but are so with equal probability.

The expected value is zero, even though that is not a possible result of an individual measurement.

Thanks MB – that makes sense now (although it still seems like a pretty misleading term)

I’m just glad Bren isn’t here to have a Bayes at me about saying ‘ensemble’.

mollwollfumble said:

> What would result from compressing free protons?

That can be done, for example in particle colliders. The net result is that the energy put into the compression manifests as new particles such as electron-positron pairs as the protons fly apart again.

It is not clear to be that In proton-proton collisions, protons are compressed.

Unclear, much about this thread is.

Does ‘compression’ mean “reducing proton-proton separation” or does it mean “reducing the volume of the proton itself”?

MartinB said:

Unclear, much about this thread is.

Does ‘compression’ mean “reducing proton-proton separation” or does it mean “reducing the volume of the proton itself”?

I’d take it to mean reducing separation, with the follow on question of what effect this has on the proton itself.

The Rev Dodgson said:

MartinB said:

Unclear, much about this thread is.

Does ‘compression’ mean “reducing proton-proton separation” or does it mean “reducing the volume of the proton itself”?

I’d take it to mean reducing separation, with the follow on question of what effect this has on the proton itself.

Yes thanks Rev. That is pretty well what I am looking for. I guess I’d have to assume that heat would be a product but in a purely bosonic field I don’t know how that develops.

The expected value or expectation value is standard terminology in probability theory. Everyone eventually gets over the fact that the expected value might actually be impossible. I’ve never seen anyone get into trouble because of this terminology, unlike the word “ensemble”, which implies false things.

Brendon said:

The expected value or expectation value is standard terminology in probability theory. Everyone eventually gets over the fact that the expected value might actually be impossible. I’ve never seen anyone get into trouble because of this terminology, unlike the word “ensemble”, which implies false things.

waves

cheers Bro..

Well, if it makes you feel better as soon as I said it I wish I hadn’t :-)

But really, Postpoc, can you explain what you meant by compressing the protons?

Do you envision putting a proton in a little vise?

dv said:

But really, Postpoc, can you explain what you meant by compressing the protons?

Do you envision putting a proton in a little vise?

More of a force crush idea. Degeneracy involves collision and if there is a limit asserted by planck quanta that predicts that beyond a certain proximity forcing protons into proximity is analogous to kinetic collision then degeneracy would be the result for the protons, which, with a confined quantity of free protons would result in quark plasma. That could be as far as the question goes but I’ll consider it further.

Brendon said:

The expected value or expectation value is standard terminology in probability theory. Everyone eventually gets over the fact that the expected value might actually be impossible. I’ve never seen anyone get into trouble because of this terminology, unlike the word “ensemble”, which implies false things.

Hi, Brendon! Thanks for dropping by. I hope we see more of you around here.

…

Here’s a classic example of an impossible expected value. If you toss a standard 6 sided die each of the numbers 1 to 6 have equal probability of occurring, so the expected value is simply the mean of {1, 2, 3, 4, 5, 6} = 21/6 = 3.5. Of course, it’s not possible to get 3.5 when you roll a die, but with enough rolls the mean of the values rolled should approach 3.5.

Postpocelipse said:

What would result from compressing free protons?

as a thought experiment?

the proton for whatever must occupy a volume that for whatever MUST occupy that space

if you could crush the proton then something would happen

I’d say that looking at a neutron star would provide answers

if you could crush the proton then it might change state to some other particle/s that occupies a smaller volume???

wookiemeister said:

Postpocelipse said:

What would result from compressing free protons?

as a thought experiment?

the proton for whatever must occupy a volume that for whatever MUST occupy that space

if you could crush the proton then something would happen

I’d say that looking at a neutron star would provide answers

if you could crush the proton then it might change state to some other particle/s that occupies a smaller volume???

Have to wonder if free protons could be cooled to a h2 superfluid type state also………

Postpocelipse said:

The only result I can think of in a hypothetical scenario of this sort is that the protons charge increases with greater forced proximity and no means of refracting the action through a fermionic outlet. Even if that has any accuracy it doesn’t tell me much as yet.

No, the charge of a proton is constant. But as you push them closer together the electromagnetic repulsive force between them grows according to the inverse square law. If protons were ideal classical particles, and there were no such thing as the nuclear forces, they’d obey the inverse square law exactly (by definition), but observation shows that that doesn’t actually happen. When the protons are really close together the nuclear forces start to become significant (relative to the electromagnetic force); also, quantum effects restrict the degrees of freedom of the system of interacting protons.

I don’t know what you mean by “a fermionic outlet”. Protons are fermions, although they are composite particles, not fundamental fermions.

According to the Standard Model of particle physics, the fundamental particles of matter (quarks and leptons) are all fermions, and the fundamental forces between them are mediated by certain fundamental bosons collectively known as gauge bosons.

Quantum Field Theory says that these fundamental fermions and bosons are actually quantized excitations of underlying fields that pervade all of spacetime, eg an electron is a quantized “blip” in the electron field, and a photon is a quantized blip in the electromagnetic field.

As I said earlier, the proton is not a fundamental particle – it’s essentially a composite of 3 quarks, but those quarks interact with each other in a rather complex fashion via the electromagnetic force and the strong and weak nuclear forces.

When you try to compress a bunch of protons, things get rather complicated. Given enough energy, the separation between a pair of protons can be made small enough that the strong force attraction between them temporarily overcomes their electromagnetic repulsion and they fuse together to form a helium-2 nucleus. But this combination is highly unstable and it generally falls apart almost immediately. But very rarely a weak force reaction happens before the nucleus manages to fall apart: one of the up quarks in one of the protons in the nucleus changes into a down quark, turning the proton into a neutron, which means the helium-2 has turned into deuterium, which is stable. An electron and an antineutrino are also released by this weak force event.

For diagrams and further details see Proton–proton chain reaction

So it’d be somewhat easier if we modify your thought experiment to be about compressing a bunch of electrons, which are fundamental fermions. But even so, electrons feel the electromagnetic and weak nuclear forces, so you can never have a real electron in total isolation: it will always have a bunch of “virtual” photons and weak force bosons associated with it.

When you try to bring two electrons close together they will resist you (mostly) due to the electromagnetic repulsion between them. That electromagnetic force is communicated between the two electrons by the electromagnetic field. That field is quantized, so effectively the two electrons are communicating their charge to each other via a photon-based interaction.

PM 2Ring said:

Postpocelipse said:

I don’t know what you mean by “a fermionic outlet”. Protons are fermions, although they are composite particles, not fundamental fermions.

oops…… got them round the wrong way but thanks for the explanation. Still going through it.

thanks again PM. That was the information I was after…..

PM 2Ring said:

When you try to compress a bunch of protons, things get rather complicated. Given enough energy, the separation between a pair of protons can be made small enough that the strong force attraction between them temporarily overcomes their electromagnetic repulsion and they fuse together to form a helium-2 nucleus. But this combination is highly unstable and it generally falls apart almost immediately. But very rarely a weak force reaction happens before the nucleus manages to fall apart: one of the up quarks in one of the protons in the nucleus changes into a down quark, turning the proton into a neutron, which means the helium-2 has turned into deuterium, which is stable. An electron and an antineutrino are also released by this weak force event.

For diagrams and further details see Proton–proton chain reaction

Mirror-image dyslexia strikes again. :(

When a proton is transformed into a neutron a positron and a neutrino are released.

Sorry about that.

SCIENCE said:

degeneracy

Eventually. But first you’d get fusion and protons being converted into neutrons.

Of course, it’s pretty hard just to contain a large quantity of free protons in a small volume, and if you try to compress them the electrostatic repulsion soon becomes enormous. It’s a lot easier to compress a bunch of protons if they’re mixed with electrons, like they are in a star core.

But we have have studied proton compression a bit – particle colliders like the LHC routinely fire focused beams of high energy protons at each other. As Molly said, when you throw protons at each other with that much energy you tend to get all sorts of particles being produced, essentially converting some of the kinetic energy of the protons into matter (and antimatter).

Fusion researchers have also studied proton collision, but they tend to focus on more practical reactions like deuterium-deuterium and deuterium-tritium collisions.

Postpocelipse said:

Have to wonder if free protons could be cooled to a h2 superfluid type state also………

I assume you meant to say He II superfluid . Superfluidity is a boson thing, so it’s not easy for protons to do that: they’d need to pair up somehow, which isn’t so easy for free protons in a vacuum.

The core difference between fermions and bosons is their intrinsic angular momentum, aka quantum spin. Bosons have integer spin, IOW, their spin is an integer multiple of ħ (h-bar, the reduced Planck’s constant = h/2π), whereas fermions have half-integer spin, i.e., their spin is ħ multiplied by a whole number plus a half, eg, ħ/2, 3ħ/2, etc.

This difference in spin might not seem important, but it reflects a significant difference in the symmetry of the wave function of these particles: bosons have symmetrical wave functions, fermions have anti-symmetrical wave functions. And that difference has a profound effect on how they behave. Wikipedia has a nice diagram illustrating simple symmetrical and anti-symmetrical wave functions here .

The quantum spin of a composite particle (or quasiparticle) is simply the sum of the spins of its constituents. So if a composite particle consists of an even number of fermions it has an overall integer quantum spin and thus has bosonic characteristics.

So it’s easy for ⁴He which consists of 6 fermions (2 protons + 2 neutrons + 2 electrons) to form a superfluid because it behaves like a boson, but ³He only contains 5 fermions, an odd number, so it behaves like a fermion. However, if you cool it down even lower than the ⁴He superfluid point then the ³He can go into a state where pairs of the atoms have correlated momentum. Those pairs aren’t really bound together into a “normal” particle, but they can be considered to be a quasiparticle (called a Cooper pair ), and of course, this quasiparticle has bosonic properties, as it consists of a pair of fermions, so the ³He turns into a superfluid.

But to get back to superfluid protons. As I said earlier, we can’t do it with protons, but we could do it with hydrogen atoms, which are bosons. Of course, hydrogen atoms tend to combine into hydrogen molecules, which are also bosons. But hydrogen is a moderately reactive element, which means its electrons don’t stay put, and as a molecule loses or gains an electron it swaps between being bosonic & fermionic. However, by encouraging the hydrogen molecules to link up into pairs we can get a stable bosonic form, which will go superfluid at the right temperature, as mentioned here .

http://fivethirtyeight.com/datalab/the-stanley-cup-center-of-gravity-is-somewhere-in-lake-huron/

P.S. Ever hear the one about the statistician who drowned crossing a river that was 3 feet deep, on average?

bump for Postpocelipse

oh! cheers PM…