Sensei

I am just interpreting what you wrote in image, and showing you that there is basic unit mismatch.

M has unit kg

P has unit N/m^2

 

kg/(N/m^2) doesn't match N..

I see no evidence of the constructuion method any more than I do in my quoted examples of later constructions.

 

Imagine such device:

 

Each block of rock used to build has average 2.5 tons.

We can lift it up on little "boat", place rock on it, surround it by wood walls, fill container with water, and block is 1 level up. It's moved to it's place. Water is released. Boat goes down. New block of rock is placed on boat, and it's filled by water again, and cycle is repeated.

We can imagine row of such water-lifters, around whole pyramid.

Obviously when one level is finished, new lifters have to be build on above level.

 

We would have to pump water to the top most lifter. Egyptians were using pumps since ever to irrigate grain fields, so technology is present.

Water released from lifter in above level, can flow to below lifter, so it's reused, and less water have to be pumped.

Please click the button marked QUOTE. You might be amazed at what happens.

---- I don’t have any. Where on earth is?

 

I am telling this to you for months...

 

Here:

 

 

Basically you select some text, and press that button, and you have quoted text...

Everybody on this forum is using it hundred times per day.. Except you.

 

There's a big difference between reading the heiroglyphs and knowing what the symbols mean.

 

Have you bothered reading Rosetta Stone article, I provided?

There are 3 languages used on it, and more or less, the same content.

So far in 150 years Egyptology has failed to identify any of the symbols of the ancient Egyptians from the simplest to the most complex. Almost no writing of any sort exists and they don't understand it.

 

What a nonsense you are writing..

 

How we would know names of kings and pharaohs without being able to identify symbols.. ?

 

Have you heard about Rosetta stone?

http://en.wikipedia.org/wiki/Rosetta_Stone

You could alternatively take broken/not used old electronics, like video player, amplifier etc. and find needed parts in it.

From going to electronic shop for breadboard and electronic elements?

Not me, Hawking radiation means: gather lots of baryons into a black hole, wait until it evaporates (massless Hawking radiation) - and there is this number of baryons less in the universe.

Hawking radiation is hypothetical.

 

There is considered hypothetical decay of proton - usually into positron and neutral pion, which quickly decays into two photons.

 

Do you realize it would have to violate either Baryon number conservation and Lepton number conservation?

Neutrinos evacuate a significant portion of the fusion energy?

Neutrinos are taken into account. See below "Sum of energies". I am adding just once 0.42 MeV, assuming 50% of energy is taken by neutrino.

 

73.46% Hydrogen and 24.85% Helium from photosphere are pretty accurate measurements (similar content has Jupiter), as they're obtained from spectral lines of light from the Sun.

What is inside is prediction from models, and will remain this way. After all we won't send there any device to check it experimentally.

Two He-3 in the core are very unlikely to collide, which has to happen for them to fuse. The reaction rate is concentration-dependent. It's probably small enough it can be ignored, but I don't know for sure.

That's true it's unlikely to collide.

 

But as I showed at the beginning of post, it must happen [latex]9.17*10^{37}[/latex] times per second anyway. Otherwise Earth would be able to get 1367 J of energy per meter^2 per second.

 

Helium-4 fusion is even more unlikely to happen, because it requires 3 atoms at the same time (Be-8 has more mass-energy than 2 He-4 alone, and additionally decays to 2 He-4 + 92 keV, very rapidly):

 

[latex]_2^4He + _2^4He + _2^4He \rightarrow _6^{12}C + \gamma + 7.275 MeV[/latex]

 

More likely is fusion between He-3 and He-4 (if your suggestion about larger amount of He-3 in core is true):

 

[latex]_2^3He + _2^4He \rightarrow _4^{7}Be + \gamma + 1.5861 MeV[/latex]

 

Beryllium-7 is unstable and decays to Lithium-7 via electron capture:

 

[latex]_4^7Be + e^- \rightarrow _3^7Li + V_e + 0.861893 MeV[/latex]

 

This is exactly what is triggering Chlorine based neutrino detector (neutrinos from proton-proton have up to 0.42 MeV too low to trigger this kind of detector).

 

The abundance you cite is for earth, where He-4 is (and has been) in constant production via alpha decay.

True.

 

There's no reason to assume that the sun (or any star) has a similar composition.

We're ashes remaining from older generation of stars. He-3 can probably only two (or three) ways to be created: by decay from Tritium, or by fusion between proton and Deuterium.

The wikipedia page says that 60% of the innermost region is He, not of the whole sun.

I didn't calculate from whole Sun. But multiplied by 0.34 (34%) first (calculated 60% from just core).

0.34 * 0.6 = 0.204

 

I think your outer regions calculation is moot. The core isn't convective, so the He produced there shouldn't get to the outer regions.

Yes, I am aware of no convection (according to theory/model).

But I don't care about convection in this equation at all..

 

If you start reaction with x particles (no matter if it's star or chemical reaction at home/lab), and then it's growing to x+y after t seconds. x remain constant. y/t is rate of production.

x is initial amount of Helium

x+y is current amount of Helium

 

Meaningless amount of He-4 fused further (according to wiki CNO cycle is 0.8% of energy source at the moment).

 

How much of the current mass in the core is He-3? The wikipedia article doesn't say the 60% is all He-4.

I have no idea.

But 2 He-3 would fuse quickly to He-4 and two protons at the beginning of life of star, so it might be meaningless..

That's why it's so rare isotope (1.34(3)×10^−6)

Hello!

 

Each [latex]m^2[/latex] of surface of Earth is receiving 1367 W (ignoring atmosphere influence).

 

We know inverse-square law:

 

[latex]P = \frac{P_0}{4*\pi*r^2}[/latex]

 

After reversing it, we're receiving:

 

[latex]P_0 = P*4*\pi*r^2[/latex]

 

Distance to the Sun is approximately [latex]r = 150*10^9[/latex] meters.

 

After plugging numbers we get:

 

[latex]P_0 = 1367*4*\pi*(150*10^9)^2=3.8651*10^{26} W[/latex]

 

As you can see this pretty much agree with wikipedia Sun page

http://en.wikipedia.org/wiki/Sun

(Luminosity on the right table)

 

It's energy Sun is emitting every single second.

 

One of possible fusion path is:

 

Fusion of 4 protons to 2 Deuterium:

[latex]p^+ + p^+ \rightarrow D^+ + e^+ + V_e + 0.42 MeV[/latex] (neutrino takes part of energy)

[latex]p^+ + p^+ \rightarrow D^+ + e^+ + V_e + 0.42 MeV[/latex]

[latex]e^+ + e^- \rightarrow \gamma + \gamma + 1.022 MeV[/latex]

[latex]e^+ + e^- \rightarrow \gamma + \gamma + 1.022 MeV[/latex]

 

Fusion of 2 Deuterium to 2 Helium-3:

[latex]p^+ + D^+ \rightarrow _2^3He + \gamma + 5.49 MeV[/latex]

[latex]p^+ + D^+ \rightarrow _2^3He + \gamma + 5.49 MeV[/latex]

 

Fusion of 2 Helium-3 to 1 Helium-4:

[latex]_2^3He + _2^3He \rightarrow _2^4He + p^+ + p^+ + 12.86 MeV[/latex]

 

Sum of energies:

E = 0.42 MeV + 1.022 MeV + 1.022 MeV + 5.49 MeV + 5.49 MeV + 12.86 MeV = 26.304 MeV

 

Conversion to Joules:

[latex]26.304 MeV * 10^6 * 1.602*10^{-19} = 4.2139008*10^{-12} J[/latex]

 

Divide energy Sun has to emit by single fusion cycle to get quantity of reactions per second:

[latex]\frac{3.8651*10^{26} W}{4.2139008*10^{-12} J} = 9.17*10^{37}s^{-1}[/latex]

 

Additionally we can calculate quantity of neutrinos:

[latex]9.17*10^{37}*2 = 1.8344*10^{38}[/latex]

[latex]\frac{1.8344*10^{38}}{4*\pi*(150*10^9)^2}=6.488*10^{14} m^{-2}=64.88 \frac{bln}{cm^2}[/latex]

Which also nicely fits with wiki page neutrino flux 65 bln.

http://en.wikipedia.org/wiki/Neutrino

 

Mass of the Sun is [latex]1.98855*10^{30} kg[/latex]

 

Single atom of Hydrogen has mass = [latex]\frac{\frac{1.007825}{1000}}{6.022141*10^{23}}=1.6735*10^{-27} kg[/latex]

Single atom of Helium has mass = [latex]\frac{\frac{4.0026}{1000}}{6.022141*10^{23}}=6.6465*10^{-27} kg[/latex]

 

If we know mass of object, and what it's made of, we can calculate quantity of atoms.

 

According to this website:

http://en.wikipedia.org/wiki/Solar_core

Sun's core has 34% of mass of the Sun. And it's made of Helium in ~60%, and ~40% Hydrogen.

Phrase "so the innermost portion of the Sun is now roughly 60% helium" can be found on net and Sun wiki page

http://en.wikipedia.org/wiki/Sun

 

[latex]1.98855*10^{30} kg * 0.34 * 0.4 = 2.7*10^{29} kg / mass_H = 1.616*10^{56}[/latex] Hydrogen atoms

[latex]1.98855*10^{30} kg * 0.34 * 0.6 = 4.0566*10^{29} kg / mass_{He} = 6.1*10^{55}[/latex] Helium atoms

 

If core has 34% of mass then in outer regions of the Sun there will be:

[latex]1.98855*10^{30} kg * 0.66 * 0.7346 = 9.6412*10^{29} kg / mass_H = 5.761*10^{56}[/latex] Hydrogen atoms

[latex]1.98855*10^{30} kg * 0.66 * 0.2485 = 3.2614*10^{29} kg / mass_{He} = 4.907*10^{55}[/latex] Helium atoms

 

In total we can find now

[latex]6.1*10^{55}+4.907*10^{55}=1.101*10^{56}[/latex] Helium atoms

 

Initial quantity of Helium atoms according to theories was 27.4%, so:

[latex]1.98855*10^{30} kg * 0.274 = 8.1978*10^{55}[/latex] Helium atoms

 

Time needed to fuse can be calculated by f.e.:

[latex] t = \frac{Q_{current}-Q_{init}}{rate}[/latex]

(Q - quantity)

 

[latex] t = \frac{1.101*10^{56}-8.1978*10^{55}}{9.17*10^{37}}=3.07*10^{17} seconds = 9.7[/latex] bln years

 

Why so large discrepancy from 4.56 bln years?

Where is error?

Abundance of Helium-4 in the core is lower than 60% as all materials are suggesting?

 

And this graph would make Sun even more older than that:

The smaller luminosity (power), the smaller amount of Helium-4 produced per second, and star must be even older..

 

Faster rate of production = more energy emitted to cosmos = vaporization of planets..

 

ps. In older version of calc, I was taking into account also lost of mass due to flares etc. but don't want to be boring you too much..

----I am afraid that you misunderstood my (+mx, mx). They are particles of mater and antimatter.

In your example : Ey = (m1 m2) * C^2 , equation may reduced in Ey = m3*C^2, which is nothing else but Einstein equation. In this case m3, that is difference of masses (m1-m2), must be a mass created by equal fifty-fifty from particles of mater and anti mater. Only they are copable to create photons. Other ways you cant accelerate mass m3

with C velocity.

You misunderstood my objections.

 

Photon, or kinetic energy of particles is causing gravitation also.

Gravitation doesn't disappear, just because unstable isotope decayed, as long as energy is still there.

Gamma photon will be absorbed by other particles near it.

my question is how Jesus Could be the son of David and he is not the son of Joseph. Jesus was born from the Holy Spirit , he is not the son of Joseph the husband of Mary, even if Joseph is the son of David, what is the connection between Joseph (a son of David) and Jesus.

You're treating word "son" quite literally.

 

These statements are perfectly fine (according to science):

 

"We're sons and daughters of the Sun"

 

"We're children of the Milky Way galaxy"

 

Our bodies are ashes of dead supernova star.

 

"We're children of the God" (if we treat God = Universe = Energy of the all particles)

---- All experiments that have, in ultimate result, only photons, pair (-e ,+e ) , (v, +~v) ---and whatever (+mx, -mx) may be considered as pairs structured by : ((-e/-g) + (+e/+g)) or ((-e / +g) + (+e /-g)). For outside, they do not display electric charge, but only electromagnetic variable field (wave), results of bi-charges in rotational movements.They do not display gravity.

 

We have isomer isotope with mass m₁. Then it's decaying by isomeric transition to "plain" isotope with mass m₂.

Energy of gamma photon is [latex]E_y = (m_1-m_2 ) * c^2[/latex]

Before transition/decay higher mass of isomer was influencing gravity by small factor.

After decay total energy is still the same.

Before decay we had little gravity influence, after we don't have just because photon was emitted?

Makes no sense.

 

If gamma photon will be absorbed by some near material (and don't let escape Earth), and heat it, mass-energy of the system will remain constant. And we will have still the same gravitation as before.

No one can claim electrons are smaller than protons.

Of course anybody can claim that. But have to prove it in experiment.. And it's different story.

 

You should have open mind for any new experimental evidence that might come up in the future.

So basically what you are saying is when a down quark decays, it produces an up quark, an electron, and an anti-neutrino.

Correct.

 

And when an up quark decays, it produces a down quark, and another electron and antineutrino.

No.

It produces down-quark, positron and neutrino.

Positron is anti-particle of electron (opposite positive charge).

Neutrino is anti-particle of anti-neutrino (or vice versa).

 

This happens *exclusively* in proton-rich nucleus.

 

If particle and anti-particle meet together, they annihilate. Which for positron and electron means that 2 or more gamma photons will be created in their place.

Like Feynman lines, showing how udd is changing to udu + e- + V

----- I dont know how happens this miracle that d gave birth (u + e- + V)

There is reverse of it in proton-rich nucleus. See my explanation here

http://www.scienceforums.net/topic/85575-electrons-smaller-than-protons-but-have-equivalent-charge/?p=826742

 

That's quark-model (or any other sub-particles model) must agree with experimental data (decay of neutron, decay of proton-rich nucleus, decay of neutron-rich nucleus etc), than Universe must agree with theory.

If there is disagreement, it's always fault of theory (for well established experimental data).

 

You know that Pion+ can decay to Pion0 and positron and neutrino, right?

----- I have not crystallized in myself about structure of neutrino and Antineutrino

In this case neutrino v = (-e / +g) + (+e /+g) and antineutrino ~v = (-e /-g) + (+e / -g)

If this rule go in contradiction with other cases, this means --- something is in core wrong in my hypothesis.

In decays of more complex mesons and baryons you will probably find more examples of disagreements.

 

Annihilation producing 3+ photons:

[latex]e^+ + e^- \rightarrow \gamma + \gamma + \gamma + ...[/latex]

(more photons)

 

How do you solve it?

 

Photon with energy [latex]E_1[/latex] is absorbed by material, and two new photons with energies [latex]E_2 + E_3 = E_1[/latex] are created.

 

How do you solve it?

 

What is difference between neutrino with max. 18.6 keV (from Tritium decay f.e.) from neutrino with 0.862 MeV (from Beryllium-7 decay) ?

 

What does mean +g and -g ? If +g-g = 0, their "charges" cancel together? And there is no gravitation from such particle?

For outside particles -- right. For inside particles ---- no .

So, in your model gravity is not caused by any particle with G=0 "outside".

Which means photon,

pion+,

pion0

Although they have rest-mass..

 

And positron possessing G=+2, will cancel electron G=-2, gravitation ?

 

Sorry, but experimental evidences disagree with your theory..

Quarks give protons their charge, but what give electrons their charge? That is more of what I want to know, now that I know that electrons are not made of the same thing as protons. Also, something has to be in the electrons, they can't be just particles built out of nothing. Maybe the building blocks for down quarks?

 

Down quark is decaying to up quark and electron and antineutrino:

 

[latex]d \rightarrow u + e^- + \bar V_e[/latex]

 

It happens in neutron-rich nucleus, or bare neutron.

[latex]n^0 \rightarrow p^+ + e^- + \bar V_e + 0.782 MeV[/latex]

(neutron has half-life ~10 minutes, and mean-life ~15 minutes)

 

The first isotope that's decaying this mode is Tritium (3rd isotope of Hydrogen, and the first unstable atom):

[latex]T^+ \rightarrow _2^3He + e^- + \bar V_e + 18.6 keV[/latex]

(Tritium has half-life 12.32 years)

 

As you can easily see on the right side there is energy that varies depending on which isotope decayed. It's not constant. It's not even constant for electron and antineutrino. Once electron is accelerated more, other time reverse, and neutrino is taking more energy.

 

Don't be "fooled" that down quark is "made of" electron and up quark.

In proton-rich nucleus exactly reverse reaction happens:

 

[latex]u \rightarrow d + e^+ + V_e[/latex]

 

Up quark is decaying to down quark, positron and neutrino.

 

f.e. Carbon-11 will decay to Boron-11, positron and neutrino:

[latex]_6^{11}C \rightarrow _5^{11}B + e^+ + V_e + 0.960408 MeV[/latex]

Thanks everyone! i will buy a used one

I don't recommend you buying used one.

 

Personally have dual-setup Samsung SyncMaster SA300 S24B150BL 24" LED.

It's 16:9, Full HD. With good quality.

 

On 1st I see stock market on-line charts, on 2nd I work.

 

It looks like this:

Sensei

What is in that case structure of mesons:

neutral pion ---- = 2* γ = 2* (( - e / - g) + ( + e / + g)) = 2 * (Mun1+ Mun2)

---- = (- me + (+ me)) = (2*(-e/-g)) + ( 2* (+e/+g)) = 2 * Mun1 + 2 * Mun2

Summa active inside of (e) = 4 Summa active for outside = 0

Summa active inside of (g) = 4 Summa active for outside = 0

 

charged pion+ ----- = μ+ + mνμ~ = (me+ + mνe + mνμ~) + mνμ =

= 2*(M2~) + (M1 +M3) + (M2~ +M4~) + (M1 +M3) =

= (2*( +e/+g ) + ((-e/-g) +(+e/-g)) +((+e/+g) + (-e/+g))) + ((-e/+g)... +(+e/+g))

Summa active inside of (e) = 8 Suma active for outside in static status =2*(+e)

Suma active inside of (g) = 8 summa active for outside in static status = 4*g

 

 

= ( (2*Mun1) + (Mun2 +Mun4) + (Mun1 +Mun3) ) + (Mun1+Mun3)

( - e + ~νe + νμ ) + ( νμ )

= 4*Mun1 + 1*Mun2 + 2*Mun3 + Mun4 =

= 4* ( -e/-g) + 1*( +e/+g) + 2+ (+e/+g) + (-e/+g)

Sorry, but WTF does it mean?

 

I have bloody no idea.. (and you probably as well)

 

Maybe image would be more helpful in visualizing it.. Like Feynman lines, showing how udd is changing to udu + e- + Ve..

 

Pion+ is made of 16 sub-particles?

And pion0 is made of 8 sub-particles?

 

You know that Pion+ can decay to Pion0 and positron and neutrino, right?

 

- substructure of photon = (-e / - g) + (+e / + g) => Σ(e) = 0, Σ(g) = 0

- substructure of neutrinos, = ( - e / - g) + ( + e / -g ) => Σ(-e) =0. Σ(-g) =2

- substructure of electron/positron,= ((- e /- g) + (- e / - g)) / ((+e /+g) +(+e /+g))

If pion+ will decay to pion0, and then pion0 further to 2 gammas in your model I see total g = 0

[latex]\pi^+ \rightarrow \gamma + \gamma + e^+ + Ve[/latex]

 

Meantime if it will decay through the most common decay path:

[latex]\pi^+ \rightarrow \mu^+ + V_m[/latex]

[latex]\mu^+ \rightarrow e^+ + V_e + \bar V_m[/latex]

with overall formula:

[latex]\pi^+ \rightarrow e^+ + V_e + \bar V_m + V_m[/latex]

If I calc right, it's also g=0,

but above you wrote [latex]\pi^+[/latex] has g=4 " summa active for outside in static status = 4*g"

What I am missing?

 

If Pion0 will decay to y + e+ + e-, I see that pion 0 has 8 your sub-particles, photon has 4, electron has 4, and neutrino has 4. 8 mismatch 12.. Right?

 

What does mean +g and -g ? If +g-g = 0, their "charges" cancel together? And there is no gravitation from such particle?

 

ps. Why are you wasting time coloring these quotes, instead of putting them in quotes?

Simply write [ quote ] ..... and end it with [ /quote ]

can be done with keyboard, no need to use mouse..

Plug energy monitor (it's quite cheap, something like $20)

Like this

15 ukp cost

https://energenie4u.co.uk/index.php/catalogue/product/ENER007

 

And you will know exactly how much device in your house is taking energy.

 

If you will record 120 W (as Studiot worst case) and drop to 20 W, it means 100 J is saved per second.

100 J/s * 3600 s * 4 (h) * 30 (days) = 43.2 MJ

1 kWh = 3.6 MJ, so

43.2 MJ / 3.6 MJ = 12 kWh saved per month. 144 kWh per year.

 

I am using 330 kWh per month. So 12 kWh from 330 kWh is just 3.7% less... That would be 2.5$ less on bill..

 

Never mind power, health is more important.. CRT is accelerating electrons and hitting them on screen, creating x-rays. That's why it has so bold lead glass screen for protection. But it can't absorb everything with 100% accuracy.

Spin is always derived from the tensor rank of the field, as ajb says.

 

You're kinda turning everything upside down.

 

Spins are measured by device, for particles we are able to.

http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment

or Zeeman effect

http://en.wikipedia.org/wiki/Zeeman_effect

 

From this experimental knowledge there has been created theory and conservation rules.

 

Basing on conservation/formulas we can predict what spin will have to have particle that we have not detected yet. I don't have to search database of isotopes to check particular one if it has even quantity of protons and even quantity of neutrons - it will have nuclear spin 0 for sure.