How exactly do fundamental particles decay before the heat death of the universe?
How exactly do fundamental particles decay before the heat death of the universe?
Hi Everyone
A while ago, I heard on a BBC documentary (I can't remember which one) that by the time the heat death of the universe happens, all fundamental particles will have decayed, leaving behind only photons. Is this correct?
As I understand it, under the circumstances that we are all familiar with, fundamental particles such as electrons and quarks are supposed to be near-perfectly stable. So if they do have finite lives, they must be extremely long.
If this is the case then can anyone describe to me the exact processes by which electrons and quarks will eventually decay into photons before the heat death of the universe?
Thank you very much.
Kind regards
Tim
A while ago, I heard on a BBC documentary (I can't remember which one) that by the time the heat death of the universe happens, all fundamental particles will have decayed, leaving behind only photons. Is this correct?
As I understand it, under the circumstances that we are all familiar with, fundamental particles such as electrons and quarks are supposed to be near-perfectly stable. So if they do have finite lives, they must be extremely long.
If this is the case then can anyone describe to me the exact processes by which electrons and quarks will eventually decay into photons before the heat death of the universe?
Thank you very much.
Kind regards
Tim
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Re: How exactly do fundamental particles decay before the heat death of the universe?
The BBC might know more than me, but I have given this some thought and now believe in a very different Universe.
In my Universe ground potential is falling which in turn makes it appear as if the Universe is expanding, my reasoning is that potential is E/q or energy per charge. The Sun and the Earth to a lesser extent are hot bodies suspended in cold space, energy is leaking away into space but charge is presumably preserved, therefore E/q is falling. This gives the impression of expanding space.
So what does falling ground potential mean?
It is a little complicated, but essentially the relativistic mass of protons at ground potential are falling whilst the relativistic mass of electrons are rising, this is fortunately happening very slowly, but given enough time, all electrons and the protons in your world will reach the same mass at which point your entire universe will annihilate in a flash.
Contrary to the standard model theory, I believe the Universe goes in a cycle where the annihilation of one proton and electron becomes a new free proton that starts a new Universe with the gentle fusion of two protons.
Here is a brief story
https://www.gammaspectacular.com/steven ... _clock.php
and if you want nitty gritty details my papers are here.
https://www.researchgate.net/publicatio ... _potential
At least I am not asking anyone to believe in a miracle.
In my Universe ground potential is falling which in turn makes it appear as if the Universe is expanding, my reasoning is that potential is E/q or energy per charge. The Sun and the Earth to a lesser extent are hot bodies suspended in cold space, energy is leaking away into space but charge is presumably preserved, therefore E/q is falling. This gives the impression of expanding space.
So what does falling ground potential mean?
It is a little complicated, but essentially the relativistic mass of protons at ground potential are falling whilst the relativistic mass of electrons are rising, this is fortunately happening very slowly, but given enough time, all electrons and the protons in your world will reach the same mass at which point your entire universe will annihilate in a flash.
Contrary to the standard model theory, I believe the Universe goes in a cycle where the annihilation of one proton and electron becomes a new free proton that starts a new Universe with the gentle fusion of two protons.
Here is a brief story
https://www.gammaspectacular.com/steven ... _clock.php
and if you want nitty gritty details my papers are here.
https://www.researchgate.net/publicatio ... _potential
At least I am not asking anyone to believe in a miracle.
Steven Sesselmann | Sydney | Australia | https://gammaspectacular.com | https://beejewel.com.au | https://www.researchgate.net/profile/Steven-Sesselmann
Re: How exactly do fundamental particles decay before the heat death of the universe?
Hi StevenSesselmann wrote: ↑24 Sep 2024, 21:35The BBC might know more than me, but I have given this some thought and now believe in a very different Universe.
In my Universe ground potential is falling which in turn makes it appear as if the Universe is expanding, my reasoning is that potential is E/q or energy per charge. The Sun and the Earth to a lesser extent are hot bodies suspended in cold space, energy is leaking away into space but charge is presumably preserved, therefore E/q is falling. This gives the impression of expanding space.
So what does falling ground potential mean?
It is a little complicated, but essentially the relativistic mass of protons at ground potential are falling whilst the relativistic mass of electrons are rising, this is fortunately happening very slowly, but given enough time, all electrons and the protons in your world will reach the same mass at which point your entire universe will annihilate in a flash.
Contrary to the standard model theory, I believe the Universe goes in a cycle where the annihilation of one proton and electron becomes a new free proton that starts a new Universe with the gentle fusion of two protons.
Here is a brief story
https://www.gammaspectacular.com/steven ... _clock.php
and if you want nitty gritty details my papers are here.
https://www.researchgate.net/publicatio ... _potential
At least I am not asking anyone to believe in a miracle.
Thank you very much indeed for your fascinating reply.
I will need to do a lot of thinking in order to attempt to fully understand what you have said.
Does this mean that you don't believe in the Big Bang?
What is your explanation for the red shift of distant galaxies? Do you believe in what some people refer to as "tired light"?
Thank you very much.
Kind regards
Tim
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Re: How exactly do fundamental particles decay before the heat death of the universe?
Not exactly, my theory has a long bang, in fact for someone somewhere it is banging right now !!Does this mean that you don't believe in the Big Bang?
Let me explain in more detail, in my theory the passage of time == observers change in potential I am using the term observer loosely here as I consider any particle or body as it's own observer. This will become clearer as you understand how potential curves with respect to the observer, essentially my definition of observer is the point on the potential where the tangent is horizontal ie. derivative = zero .
In the case of us humans on Earth this tangent sits at 930 million volts it is
\[ \frac{mass \ per \ nucleon \ of \ Ni62}{charge} = \frac{930 MeV}{e} \]
But our universe is around 13-14 Billion years old, and if our potential has been falling we might ask the question, where did it all start ?
Another way to ask the same question is how high can potential go?
Well, let's do a gedanken experiment... Take two electrodes and transfer some electrons from the anode to the cathode, this will create a potential difference, now take it to the extreme and transfer all electrons to the cathode and all protons to the anode 😱.
The difficulty in achieving this experiment does not matter, what matters is that potential has reached a theoretical limit.
How much is this limit ?
Since the important factor is the ratio of electrons to protons and not the absolute number, we can simplify our two electrodes down to a single atom of hydrogen and we do know it's energy with great precision 938.272... MeV/c2
It is therefore reasonable to assume that 938.272 million volts is the absolute limit for a particle at rest, which also happens to agree woth our observations, no one has seen a fermion with higher mass than a proton and when we try smashing stuff together at higher energies we just get more protons.
I like to define the beginning of my universe as a lonely proton with potential 938 MV with respect to its massless anti particle the electron, it's potential is unchanging and for all intents and purposes it does (can) not experience time nor does it have space.
These lonely timeless and spaceless universes are created in extremely large numbers and occasionally two such universes combine, and when they do it results in a huge potential drop \( p + p > deuterium + energy\)
This rapid drop in potential represents a huge leap in time and with it the new world (deuterium atom) has also gained some space.
You might see where this is going, next the deuterium atom meets a lone proton and becomes Tritium which again releases energy and creates more space. This fusion process continues for around 13.6 billion years, stars and planets form as time goes by, until one day a planet forms of atoms that are essentially iron, this planet has a surface potential of 930 million volts.
As you can see we are still in the early days, the age of our Universe in terms of potential is \( \frac{8MV}{938MV} = 0.8.5% \)
Projecting the future...
As time goes on, stars burn fuel and nuclei continue fusing and falling to lower and lower potential and as they do their electron counter part slowly become heavier, until one day when local potential reaches \( \frac{938MV}{2} = 468 MV \) and all hell breaks loose literally.
We know what happens when a particle and an antiparticle of the same mass come in contact with each other, they annihilate violently and emit dimetric gamma rays.
Now the casual reader may visualise an atom somewhere in the universe annihilating and thinking, so what?
No, this is the wrong view, because specific to that observers universe every proton and every electron annihilates simultaneously, talk about going out with a bang !
By no coincidence this happens exactly on the Schwartzchild radius, it is the exact point where potential is 468 MV.
Important to understand that this absolute annihilation of all matter only happens in the world of the observer crossing the SR radius, no such event is experienced by observers at higher potential, we can only observe a region of strong gravitational attraction, but one day we will of course be sucked in to experience the fireworks.
These black holes reside in the centre of galaxies and I presume they all have huge galactic jets like quasars, where the energy of the inn falling matter is ejected, however because these jets project along our time axis (past <-> future) we can not observe the jets from our own galaxy. However due to the relativistic curvature of our universe we can observe quasars if they are extremely distant.
These jets of energy shooting out from the center of our galaxy (into the past and the future) have roughly enough energy to make as far as the edge of our galaxy, leading me to think that these jets are the source of all protons.
This in turn leads me to think that galaxies recycle all their on material indefinitely and that time begins with the fusion of two protons and ends with a big bang and then starts all over again.
Steven Sesselmann | Sydney | Australia | https://gammaspectacular.com | https://beejewel.com.au | https://www.researchgate.net/profile/Steven-Sesselmann
Re: How exactly do fundamental particles decay before the heat death of the universe?
Hi Steven
Thank you very much indeed fro your very informative reply.
To be honest, it actually goes somewhat over my head because I am actually somewhat of a newbie to all of this. I would need to study it for a long time to get my hear around it.
I would be very grateful if you could explain to me in a beginner-friendly way what ground potential actually is.
Thank you very much.
Kind regards
Tim
Thank you very much indeed fro your very informative reply.
To be honest, it actually goes somewhat over my head because I am actually somewhat of a newbie to all of this. I would need to study it for a long time to get my hear around it.
I would be very grateful if you could explain to me in a beginner-friendly way what ground potential actually is.
Thank you very much.
Kind regards
Tim
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Re: How exactly do fundamental particles decay before the heat death of the universe?
It's whatever your potential is.I would be very grateful if you could explain to me in a beginner-friendly way what ground potential actually is.
If you were floating somewhere out in space it would be your mass energy divided the number of charges, however you and I (I assume) are on Earth in an accelerating field so it will be the entire mass energy of Earth divided by all of it's charges, I counted them all and it works out to 930 million volts 🤣
Steven
Steven Sesselmann | Sydney | Australia | https://gammaspectacular.com | https://beejewel.com.au | https://www.researchgate.net/profile/Steven-Sesselmann
Re: How exactly do fundamental particles decay before the heat death of the universe?
Hi Steven
Thank you very much for your answer.
I would be very grateful if you could provide me with the equation that connects the mass of an electron to the ground potential.
Thank you very much.
Kind regards
Tim
Thank you very much for your answer.
I would be very grateful if you could provide me with the equation that connects the mass of an electron to the ground potential.
Thank you very much.
Kind regards
Tim
- Sesselmann
- Posts: 1264
- Joined: 27 Apr 2015, 11:40
- Location: Sydney
- Contact:
Re: How exactly do fundamental particles decay before the heat death of the universe?
Sure...
\[ \gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-\frac{V_g^2}{\Phi_0^2}}} \]
First we have this expression where \( V_g \) is the observer (ground) potential and \( \Phi \) is absolute potential (potential of a proton)
Now I claim it is possible to relate the mass of an electron \(V_e\) electron potential to \(V_g\) ground potential as follows,
\[ V_e\ \gamma = \frac{\Phi_0 - V_g}{2} \]
What this equation says is that the electron mass increases with the passage of time, i.e. the electron is lighter than a proton because it exists in a deep potential energy well with respect to you the observer (old school expression would be mass defect).
Let us proceed and solve this, but first for simplicity we shall define the electron potential and ground potential in terms of a ratio of absolute potential of matter as alpha and beta.
\[ \frac{V_e}{\Phi _0} = \alpha\]
\[ \frac{V_g}{\Phi_0} = \beta\]
\[ replace \ gamma \]
\[ V_e \frac{1}{\sqrt{1-\beta^2}} = \frac{\Phi_0 - Vg}{2}\]
\[ switch\ sides\]
\[ V_e = \frac{\Phi_0 - V_g}{2} \sqrt{1-\beta^2} \]
\[ simplify\]
\[ V_e = \frac{\Phi_0}{2}(1-\beta)\sqrt{1-\beta^2} \]
\[rearrange\]
\[ 2(\frac{V_e}{\Phi_0}) = (1-\beta)\sqrt{1-\beta^2} \]
\[ replace\ ratio\ (V_e/\Phi_0)\ with\ \beta \]
\[ 2\alpha = (1-\beta)\sqrt{1-\beta^2} \]
\[factorise\]
\[ 4 \alpha^2 = (1-\beta)^2(1-\beta^2)\]
\[ conclusion\]
\[ (1-\beta)^2(1-\beta^2)-4\alpha^2 = 0 \]
Which gives four solutions,
\(V_g = 930.377 \ MV\) << This is the only sensible solution
\(V_g = -938.272 \ MV\)
\(V_g = 942.219+6.81798i \)
\(V_g = 942.219-6.81798i \)
So we have derived the potential of Ni62 (930.417 MV) to an accuracy of 0.999957 using nothing but the mass of an electron a proton and my formula, how can anyone ignore this?
It is by no coincidence that Ni62 is the isotope with the highest binding energy and lowest mass per nucleon, because ground potential is by definition the lowest point on the potential curve and as we understand, the Earth's interior is for the most part Iron.
I rest my case...
\[ \gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-\frac{V_g^2}{\Phi_0^2}}} \]
First we have this expression where \( V_g \) is the observer (ground) potential and \( \Phi \) is absolute potential (potential of a proton)
Now I claim it is possible to relate the mass of an electron \(V_e\) electron potential to \(V_g\) ground potential as follows,
\[ V_e\ \gamma = \frac{\Phi_0 - V_g}{2} \]
What this equation says is that the electron mass increases with the passage of time, i.e. the electron is lighter than a proton because it exists in a deep potential energy well with respect to you the observer (old school expression would be mass defect).
Let us proceed and solve this, but first for simplicity we shall define the electron potential and ground potential in terms of a ratio of absolute potential of matter as alpha and beta.
\[ \frac{V_e}{\Phi _0} = \alpha\]
\[ \frac{V_g}{\Phi_0} = \beta\]
\[ replace \ gamma \]
\[ V_e \frac{1}{\sqrt{1-\beta^2}} = \frac{\Phi_0 - Vg}{2}\]
\[ switch\ sides\]
\[ V_e = \frac{\Phi_0 - V_g}{2} \sqrt{1-\beta^2} \]
\[ simplify\]
\[ V_e = \frac{\Phi_0}{2}(1-\beta)\sqrt{1-\beta^2} \]
\[rearrange\]
\[ 2(\frac{V_e}{\Phi_0}) = (1-\beta)\sqrt{1-\beta^2} \]
\[ replace\ ratio\ (V_e/\Phi_0)\ with\ \beta \]
\[ 2\alpha = (1-\beta)\sqrt{1-\beta^2} \]
\[factorise\]
\[ 4 \alpha^2 = (1-\beta)^2(1-\beta^2)\]
\[ conclusion\]
\[ (1-\beta)^2(1-\beta^2)-4\alpha^2 = 0 \]
Which gives four solutions,
\(V_g = 930.377 \ MV\) << This is the only sensible solution
\(V_g = -938.272 \ MV\)
\(V_g = 942.219+6.81798i \)
\(V_g = 942.219-6.81798i \)
So we have derived the potential of Ni62 (930.417 MV) to an accuracy of 0.999957 using nothing but the mass of an electron a proton and my formula, how can anyone ignore this?
It is by no coincidence that Ni62 is the isotope with the highest binding energy and lowest mass per nucleon, because ground potential is by definition the lowest point on the potential curve and as we understand, the Earth's interior is for the most part Iron.
I rest my case...
Steven Sesselmann | Sydney | Australia | https://gammaspectacular.com | https://beejewel.com.au | https://www.researchgate.net/profile/Steven-Sesselmann
Re: How exactly do fundamental particles decay before the heat death of the universe?
Hi Steven
Thank you very much indeed for this.
I will need to try and get my head around it.
Kind regards
Tim
Thank you very much indeed for this.
I will need to try and get my head around it.
Kind regards
Tim
- Sesselmann
- Posts: 1264
- Joined: 27 Apr 2015, 11:40
- Location: Sydney
- Contact:
Re: How exactly do fundamental particles decay before the heat death of the universe?
Tim,
There are some important steps you need to complete in order to understand the equation.
1) Why should rest potential have a limit?
Answer: Because potential is the ratio of electrons to protons and it can not be more than one (100% protons)
2) Why should \( (v/c) \) be the same as \( (V/\Phi) \)
Answer: Because both are dimensionless ratios of one
3) Why should factoring the electron potential by gamma equal half the difference between the proton potential and ground potential ?
Answer: Because the assumption is that electrons and protons are a particle pairs, where one of the particles has an apparent mass loss due to it's relative position with respect to the observer.
If you can rationalise these three points you will understand my equation and to solve it I suggest using Wolfram Alpha as it involves rather many steps.
Steven
There are some important steps you need to complete in order to understand the equation.
1) Why should rest potential have a limit?
Answer: Because potential is the ratio of electrons to protons and it can not be more than one (100% protons)
2) Why should \( (v/c) \) be the same as \( (V/\Phi) \)
Answer: Because both are dimensionless ratios of one
3) Why should factoring the electron potential by gamma equal half the difference between the proton potential and ground potential ?
Answer: Because the assumption is that electrons and protons are a particle pairs, where one of the particles has an apparent mass loss due to it's relative position with respect to the observer.
If you can rationalise these three points you will understand my equation and to solve it I suggest using Wolfram Alpha as it involves rather many steps.
Steven
Steven Sesselmann | Sydney | Australia | https://gammaspectacular.com | https://beejewel.com.au | https://www.researchgate.net/profile/Steven-Sesselmann
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