Fact!

Relativity is violated through and through when temperature approaches
absolute zero.

Not only that, but values of X and T are different. Goodbye Lorentz transforms.

Good bye E=m.c², welcome to E(T) = m(T).c²

Bodkin, would you like to elaborate?

On Thursday, August 26, 2021 at 1:25:40 PM UTC-7, Richard Hertz produced a fresh brain fart:
> snip imbecilities<

Radiant heat is a consequence of electromagnetic radiation: Fact! (Maxwell)..

EM radiation is related to temperature: Fact! (Planck).

u = (8πhʋ^3)/c^3 . 1/(e^(hʋ/kT)-1) (BBC radiation, Planck, Dec. 1900)

u is then the energy distribution law is (in units Joule.m^-3.Hz^-1).

Monochromatic Irradiance or Spectral Flux Density at the aperture in the cavity: Radiance of a surface per unit
frequency or wavelength per unit solid angle. A directional quantity and differs from Planck's formula by (c/4𝜋).
This is sometimes confusingly called "spectral intensity".

Lʋ(ʋ,T) = (c/4𝜋).Wu(ʋ,T) = 2hʋ^3.c^-2/[e^(hʋ/kT)-1] [units: Watt.m-2.Hz-1.sr-1]

E(T) = m(T).c² has to be adjusted to these laws, isn't it?

Try to use it at 2 Kelvin degrees.

On Thursday, August 26, 2021 at 1:38:50 PM UTC-7, crank Richard Hertz strained and emitted yet another brain fart:

> E(T) = m(T).c² has to be adjusted to these laws, isn't it?

Nope, cretinoid, there is no "adjustment", contrary to your raving insanity..

Total energy for closed systems is conserved. You are grossly ignorant of the basic laws of nature.

Richard Hertz <hertz778@gmail.com> wrote:
> Fact!
>
> Relativity is violated through and through when temperature approaches
> absolute zero.
>
> Not only that, but values of X and T are different. Goodbye Lorentz transforms.
>
> Good bye E=m.c², welcome to E(T) = m(T).c²
>
> Bodkin, would you like to elaborate?
>
>

Nothing to elaborate on. Did you want to elaborate?

--
Odd Bodkin -- maker of fine toys, tools, tables

Richard Hertz wrote:

> Fact!
>
> Relativity is violated through and through when temperature approaches
> absolute zero.

relativity is the least of your worries when the temp goes to Kelvin zero.

u(ʋ,T) = (8πhʋ^3)/c^3 . 1/(e^(hʋ/kT)-1) (BBC radiation, Planck, Dec. 1900)

u(ʋ,T) is the energy distribution law is (in units Joule.m^-3.Hz^-1).

Being Mo the mass of a black body cavity with a volume V, both values declared at 300°K , then

U(T) = V . ∫ (8πhʋ^3)/c^3 . 1/(e^(hʋ/kT)-1) dʋ is the energy that exist within a BBC of volume V (ʋ between 0 and ∞)

U(300) = V . ∫ (8πhʋ^3)/c^3 . 1/(e^(hʋ/300.k)-1) dʋ

Mo(300) = M´o + U(300)/c² , where M´o is temperature invariant.

U(2) = V . ∫ (8πhʋ^3)/c^3 . 1/(e^(hʋ/2.k)-1) dʋ

Mo(2) = M´o + U(2)/c²

As Mo(300) - Mo(2) = U(300)/c² - U(2)/c² > 0, then an amount of mass has been lost:

dMo = U(300)/c² - U(2)/c²

So, it's true: Mass decreases if temperature decreases.

On Thursday, August 26, 2021 at 3:22:21 PM UTC-7, Richard Hertz persevered in his imbecility:
>
> So, it's true: Mass decreases if temperature decreases.

So, it is true, you are a crank. But se knew that.

Now, this is an exercise for beginners. Let:

1. The thermal energy taken out of the system to be W
2. The momentum of the particles making up the system be p before W is extracted and p' after W is extracted.
3. m the invariant mass

Then, conservation of energy shows that:

\sqrt{(pc)^2+(mc^2)^2}=\sqrt{(p'c)^2+(mc^2)^2}+W

Therefore, the effect of cooling off the system is that p'<p. Richard, you are an ignorant crank and there is no way around this FACT!

Exercise for the crank Richard Hertz: at what value for W, the motion of the system particles ceases?

On Thursday, August 26, 2021 at 4:07:22 PM UTC-7, Dono. wrote:
> On Thursday, August 26, 2021 at 3:22:21 PM UTC-7, Richard Hertz persevered in his imbecility:
> >
> > So, it's true: Mass decreases if temperature decreases.
> So, it is true, you are a crank. But se knew that.
>
> Now, this is an exercise for beginners. Let:
>
> 1. The thermal energy taken out of the system to be W
> 2. The momentum of the particles making up the system be p before W is extracted and p' after W is extracted.
> 3. m the invariant mass
>
> Then, conservation of energy shows that:
>
> \sqrt{(pc)^2+(mc^2)^2}=\sqrt{(p'c)^2+(mc^2)^2}+W
>
> Therefore, the effect of cooling off the system is that p'<p. Richard, you are an ignorant crank and there is no way around this FACT!
>
>
>
> Exercise for the crank Richard Hertz: at what value for W, the motion of the system particles ceases?

Temp is the above band of EM heat energy. Below light band is not a temp...
Your light bulb doesn't push air like heat pushes a rocket...

Mitchell Raemsch

On Thursday, August 26, 2021 at 8:07:22 PM UTC-3, Dono. wrote:

<snip>

> Now, this is an exercise for beginners. Let:
>
> 1. The thermal energy taken out of the system to be W
> 2. The momentum of the particles making up the system be p before W is extracted and p' after W is extracted.
> 3. m the invariant mass
>
> Then, conservation of energy shows that:
>
> \sqrt{(pc)^2+(mc^2)^2}=\sqrt{(p'c)^2+(mc^2)^2}+W
>
> Therefore, the effect of cooling off the system is that p'<p. Richard, you are an ignorant crank and there is no way around this FACT!

Fucking retarded!. The lame equation that you wrote (learn how to write it correctly) is related to the cooling effect of lasers on gasses,
which was proven with great success in 1985 by Steven Chu, while working at Bell Labs. He, along with Cohen-Tannoudji and Phillips,
who were working on the same field in the ´80s, won the 1997 Nobel Prize in Physics for their work.

It's about removing of part of the avg. KE = 1/2.m.v² of molecules of gas by hitting as many of them as they could by with using photons to make v' < v, decreasing the avg. momentum of molecules going toward the laser beam due to collisions between photons and incoming
molecules (they lose energy by absorption of photon's momentum and further fluorescence). Chu, using sodium atoms, achieved a record temperature of 43 microkelvins. But this was on gasses only.

The real deal is to cool of solid matter with much more advanced techniques.. Read here the timeline of developments as of today, idiot.

https://en.wikipedia.org/wiki/Laser_cooling

Read about cooling off matter as it's happening in these days:
https://ece.engin.umich.edu/stories/cooling-off-with-lasers

What I wrote is not about cooling off matter in gaseous state, but solid matter which doesn't move. And it's not related to any violation
of the principle of conservation of energy, which is conserved (the BBC isn't floating by itself in vacuum, retarded).

> Exercise for the crank Richard Hertz: at what value for W, the motion of the system particles ceases?

I didn't have to do any calculation, cretin. It's all out there, if you know how to search about it.

http://hyperphysics.phy-astr.gsu.edu/hbase/optmod/lascool.html

Excerpt:
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

If we take a sodium atom as an example, and assume that a number of sodium atoms are freely moving in a vacuum
chamber at 300K, the rms velocity of a sodium atom from the Maxwell speed distribution would be about 570 m/s.

Then if a laser is tuned just below one of the sodium d-lines (589.0 and 589.6 nm, about 2.1 eV), a sodium atom
traveling toward the laser and absorbing a laser photon would have its momentum reduced by the amount of the
momentum of the photon. It would take a large number of such absorptions to cool the sodium atoms to near 0K
since one absorption would slow a sodium atom by only about 3 cm/s out of a speed of 570 m/s.

A straight projection requires almost 20,000 photons to reduce the sodium atom momentum to zero.

The change in speed from the absorption of one photon can be calculated from

Δp/p = p(photon)/mv = Δv/v
Δv = p(photon)/m

That seems like a lot of photons, but according to Chu, a laser can induce on the order of 10^7 absorptions per second
so that an atom could be stopped in a matter of milliseconds.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

So, answering your stupid question, in the case of Chu's experiment (1985) a total energy W = 42,000 eV (6.73.10^-15 J)
were required to reach 43 µK° in a cavity trap with sodium gas that had atoms moving at about 570 m/s at 300K°, in msec.

Each photon absorbed causes a decrease in particle's momentum Δp = m Δv = p(photon).

If you want to know the final value of v' ≈ 0, use Maxwell-Boltzmann distribution. Here is a guide,

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.01%3A_Gas_Phase_Kinetics/3.1.02%3A_Maxwell-Boltzmann_Distributions

But remember: an amount of mass dMo = U(300)/c² - U(2)/c² has been lost in a BBC cooling off by itself.

So, it's true: Mass decreases if temperature decreases, and all of this in a stand-alone basis, without external help.

Now, suffer and retort, cretin.

On Thursday, August 26, 2021 at 8:07:22 PM UTC-3, Dono. wrote:

<snip>

> Now, this is an exercise for beginners. Let:
>
> 1. The thermal energy taken out of the system to be W
> 2. The momentum of the particles making up the system be p before W is extracted and p' after W is extracted.
> 3. m the invariant mass
>
> Then, conservation of energy shows that:
>
> \sqrt{(pc)^2+(mc^2)^2}=\sqrt{(p'c)^2+(mc^2)^2}+W
>
> Therefore, the effect of cooling off the system is that p'<p. Richard, you are an ignorant crank and there is no way around this FACT!

Fucking retarded!. The lame equation that you wrote (learn how to write it correctly) is related to the cooling effect of lasers on gasses,
which was proven with great success in 1985 by Steven Chu, while working at Bell Labs. He, along with Cohen-Tannoudji and Phillips,
who were working on the same field in the ´80s, won the 1997 Nobel Prize in Physics for their work.

It's about removing of part of the avg. KE = 1/2.m.v² of molecules of gas by hitting as many of them as they could by with using photons to make v' < v, decreasing the avg. momentum of molecules going toward the laser beam due to collisions between photons and incoming
molecules (they lose energy by absorption of photon's momentum and further fluorescence). Chu, using sodium atoms, achieved a record temperature of 43 microkelvins. But this was on gasses only.

The real deal is to cool of solid matter with much more advanced techniques.. Read here the timeline of developments as of today, idiot.

https://en.wikipedia.org/wiki/Laser_cooling

Read about cooling off matter as it's happening in these days:
https://ece.engin.umich.edu/stories/cooling-off-with-lasers

What I wrote is not about cooling off matter in gaseous state, but solid matter which doesn't move. And it's not related to any violation
of the principle of conservation of energy, which is conserved (the BBC isn't floating by itself in vacuum, retarded).

> Exercise for the crank Richard Hertz: at what value for W, the motion of the system particles ceases?

I didn't have to do any calculation, cretin. It's all out there, if you know how to search about it.

http://hyperphysics.phy-astr.gsu.edu/hbase/optmod/lascool.html

Excerpt:
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

If we take a sodium atom as an example, and assume that a number of sodium atoms are freely moving in a vacuum
chamber at 300K, the rms velocity of a sodium atom from the Maxwell speed distribution would be about 570 m/s.

Then if a laser is tuned just below one of the sodium d-lines (589.0 and 589.6 nm, about 2.1 eV), a sodium atom
traveling toward the laser and absorbing a laser photon would have its momentum reduced by the amount of the
momentum of the photon. It would take a large number of such absorptions to cool the sodium atoms to near 0K
since one absorption would slow a sodium atom by only about 3 cm/s out of a speed of 570 m/s.

A straight projection requires almost 20,000 photons to reduce the sodium atom momentum to zero.

The change in speed from the absorption of one photon can be calculated from

Δp/p = p(photon)/mv = Δv/v
Δv = p(photon)/m

That seems like a lot of photons, but according to Chu, a laser can induce on the order of 10^7 absorptions per second
so that an atom could be stopped in a matter of milliseconds.

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

So, answering your stupid question, in the case of Chu's experiment (1985) a total energy W = 42,000 eV (6.73.10^-15 J)
were required to reach 43 µK° in a cavity trap with sodium gas that had atoms moving at about 570 m/s at 300K°, in msec.

Each photon absorbed causes a decrease in particle's momentum Δp = m Δv = p(photon).

If you want to know the final value of v' ≈ 0, use Maxwell-Boltzmann distribution. Here is a guide,

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.01%3A_Gas_Phase_Kinetics/3.1.02%3A_Maxwell-Boltzmann_Distributions

But remember: an amount of mass dMo = U(300)/c² - U(2)/c² has been lost in a BBC cooling off by itself.

So, it's true: Mass decreases if temperature decreases, and all of this in a stand-alone basis, without external help.

Now, suffer and retort, cretin.

On Thursday, August 26, 2021 at 9:30:04 PM UTC-7, Richard Hertz wrote:
> On Thursday, August 26, 2021 at 8:07:22 PM UTC-3, Dono. wrote:
>
> <snip>
> > Now, this is an exercise for beginners. Let:
> >
> > 1. The thermal energy taken out of the system to be W
> > 2. The momentum of the particles making up the system be p before W is extracted and p' after W is extracted.
> > 3. m the invariant mass
> >
> > Then, conservation of energy shows that:
> >
> > \sqrt{(pc)^2+(mc^2)^2}=\sqrt{(p'c)^2+(mc^2)^2}+W
> >
> > Therefore, the effect of cooling off the system is that p'<p. Richard, you are an ignorant crank and there is no way around this FACT!
> The lame equation that you wrote (learn how to write it correctly) is related to the cooling effect of lasers on gasses,

The equation is general, crank. Wipe the spittle from your chin.

<rest of imbecillic ranting by the resident crank Richard Hertz mercifully snipped>

On Thursday, August 26, 2021 at 1:25:40 PM UTC-7, Richard Hertz wrote:
> Fact!
Fact, you are a cretin
> Relativity is violated through and through when temperature approaches
> absolute zero.

Mass is invariant, imbecile

On Friday, August 27, 2021 at 2:23:14 AM UTC-3, Dono. wrote:

<snip>

> > > Then, conservation of energy shows that:
> > >
> > > \sqrt{(pc)^2+(mc^2)^2}=\sqrt{(p'c)^2+(mc^2)^2}+W
> > >
> > > Therefore, the effect of cooling off the system is that p'<p. Richard, you are an ignorant crank and there is no way around this FACT!

> > The lame equation that you wrote (learn how to write it correctly) is related to the cooling effect of lasers on gasses,
> The equation is general, crank. Wipe the spittle from your chin.
>
> <rest of imbecillic ranting by the resident crank Richard Hertz mercifully snipped>

Cretin, I answered your question with data on gasses that made three persons to won a Nobel Prize!

Your stupid formulation involves relativistic effects for particles in inertial motion, so it's NOT general!

I used it correctly and supply a perfect answer, by quoting work done by others, and i DIDN'T PRETEND to
post any of it as derivations of my own, utter cretin!!

Your formula IS WRONG, and can't be applied to matter at rest, IDIOT. Read the content of my links before answering.

On Friday, August 27, 2021 at 2:29:02 AM UTC-3, Dono. wrote:

<snip>

> > Relativity is violated through and through when temperature approaches absolute zero.
> Mass is invariant, imbecile

Then you have to shove E=mc² up into your ass, cretin.

Planck's formula proves that Eo = Mo.c² = Mo(T).c² if you include the mass of radiation within a BBC!

So, dMo = U(300)/c² - U(2)/c² in the example I chose, for changes between 300 K and 2 K, is LOST at the BBC.

Where did it go?. This is a different question.

Richard Hertz <hertz778@gmail.com> wrote:
> On Friday, August 27, 2021 at 2:29:02 AM UTC-3, Dono. wrote:
>
> <snip>
>
>
>>> Relativity is violated through and through when temperature approaches absolute zero.
>> Mass is invariant, imbecile
>
> Then you have to shove E=mc² up into your ass, cretin.

The E there is rest energy.
If you want the full energy, the correct expression is E =
sqrt[(mc^2)^2+(pc)^2]. I’m sure you knew that.

>
> Planck's formula proves that Eo = Mo.c² = Mo(T).c² if you include the
> mass of radiation within a BBC!
>
> So, dMo = U(300)/c² - U(2)/c² in the example I chose, for changes between
> 300 K and 2 K, is LOST at the BBC.
>
> Where did it go?. This is a different question.
>
>
>

--
Odd Bodkin -- maker of fine toys, tools, tables

On Thursday, August 26, 2021 at 10:32:02 PM UTC-7, Richard Hertz wrote:
>
> Your formula IS WRONG, and can't be applied to matter at rest,

Keep digging yourself deeper , crank. Into shit.

On Thursday, August 26, 2021 at 10:58:46 PM UTC-7, Richard Hertz wrote:
> On Friday, August 27, 2021 at 2:29:02 AM UTC-3, Dono. wrote:
>
> <snip>
> > > Relativity is violated through and through when temperature approaches absolute zero.
> > Mass is invariant, imbecile

> Planck's formula proves that Eo = Mo.c² = Mo(T).c² if you include the mass of radiation within a BBC!

Crank

The complete formula is sqrt{(pc)^2+(mc^2)^2} . Your crass ignorance is hilarious.

Richard Hertz wrote:

> Fucking retarded!. The lame equation that you wrote (learn how to write
> it correctly) is related to the cooling effect of lasers on gasses,
> which was proven with great success in 1985 by Steven Chu, while working
> at Bell Labs. He, along with Cohen-Tannoudji and Phillips,
> who were working on the same field in the ´80s, won the 1997 Nobel Prize
> in Physics for their work.

nonsense, you can't cut a nobel prize into pieces. One man one prize.

On Friday, August 27, 2021 at 11:02:13 AM UTC-3, Beek Sneijder wrote:

<snip>

> Richard Hertz wrote:
>
> > Fucking retarded!. The lame equation that you wrote (learn how to write
> > it correctly) is related to the cooling effect of lasers on gasses,
> > which was proven with great success in 1985 by Steven Chu, while working
> > at Bell Labs. He, along with Cohen-Tannoudji and Phillips,
> > who were working on the same field in the ´80s, won the 1997 Nobel Prize in Physics for their work.

> nonsense, you can't cut a nobel prize into pieces. One man one prize.

The Nobel Prize money is split between those who we awarded the same year (1, 2 or 3).

The 2nd. Nobel Prize in Physics (1902) was awarded to Lorentz and Zeeman.

https://www.nobelprize.org/prizes/lists/all-nobel-prizes-in-physics/

Quote:

"The Nobel Prize in Physics has been awarded 114 times to 216 Nobel Laureates between 1901 and 2020."
"John Bardeen is the only Nobel Laureate who has been awarded the Nobel Prize in Physics twice, in 1956
and 1972. This means that a total of 215 individuals have received the Nobel Prize in Physics."

On Friday, August 27, 2021 at 7:51:33 AM UTC-7, Richard Hertz wrote:
> On Friday, August 27, 2021 at 11:02:13 AM UTC-3, Beek Sneijder wrote:

> "The Nobel Prize in Physics has been awarded 114 times to 216 Nobel Laureates between 1901 and 2020."
> "John Bardeen is the only Nobel Laureate who has been awarded the Nobel Prize in Physics twice, in 1956
> and 1972. This means that a total of 215 individuals have received the Nobel Prize in Physics."

There is no risk that the cretinisms you post will get you any prize, Richard. But they will get the scorn that you well deserve as a top crank.

Richard Hertz wrote:

>> > Fucking retarded!. The lame equation that you wrote (learn how to
>> > write it correctly) is related to the cooling effect of lasers on
>> > gasses,
>> > which was proven with great success in 1985 by Steven Chu, while
>> > working at Bell Labs. He, along with Cohen-Tannoudji and Phillips,
>> > who were working on the same field in the ´80s, won the 1997 Nobel
>> > Prize in Physics for their work.
>
>> nonsense, you can't cut a nobel prize into pieces. One man one prize.
>
> The Nobel Prize money is split between those who we awarded the same
> year (1, 2 or 3).

Not split, but re-doubled.

On Thursday, August 26, 2021 at 1:25:40 PM UTC-7, Richard Hertz wrote:
> Fact!
>
> Relativity is violated through and through when temperature approaches
> absolute zero.
>
> Not only that, but values of X and T are different. Goodbye Lorentz transforms.
>
> Good bye E=m.c², welcome to E(T) = m(T).c²
>
> Bodkin, would you like to elaborate?

If temperature is EM and EM has no mass
how could mass change?

Mitchell Raemsch

On Friday, August 27, 2021 at 2:52:35 PM UTC-3, mitchr...@gmail.com wrote:

<snip>

> If temperature is EM and EM has no mass
> how could mass change?
>
> Mitchell Raemsch

Temperature is not EM. There is a relationship between EM radiation within a Black Body Cavity,
derived by Planck in 1900, which is widely used today and remain undisputed by science:

u(ʋ,T) = (8πhʋ^3)/c^3 . 1/(e^(hʋ/kT)-1) (BBC radiation, Planck, Dec. 1900)

u(ʋ,T) is the energy distribution law is (in units Joule.m^-3.Hz^-1) within the BBC, which is at equilibrium
(T has been set to a constant value for long time)

Being Mo the mass of a black body cavity with a volume V, both values measured at 300°K , then

U(300) = V . ∫ (8πhʋ^3)/c^3 . 1/(e^(hʋ/300.T)-1) dʋ is the energy that exist within a BBC of volume V (ʋ between 0 and ∞)

The amount of mass for the radiation contained within the BBC, using Einstein's dM = E/c² is U(300)/c²

Then, assuming that M'o is the mass of the BBC without the energy within, is temperature invariant (mass of the black body), then

Mo(300) = M´o + U(300)/c²

Mo(300) is the sum of the temperature invariant M´o (black body object) PLUS the mass of the energy contained within the BBC.

So, it's not that EM has mass but that EM radiation enclosed in a BBC at thermal equilibrium has, due to the 1905 formula dM = E/c².

And, generalizing,

Mo(T) = M´o + U(T)/c² , being T in Kelvin degrees.

On Friday, August 27, 2021 at 11:52:49 AM UTC-7, Richard Hertz wrote:
> On Friday, August 27, 2021 at 2:52:35 PM UTC-3, mitchr...@gmail.com wrote:
>
> <snip>
> > If temperature is EM and EM has no mass
> > how could mass change?
> >
> > Mitchell Raemsch
> Temperature is not EM. There is a relationship between EM radiation

What about microwave heat that has a temperature?
No you moron. No... you don't have anything at all there.
Heat is still EM energy. But if light has no mass
how can heat's temperature be any change in mass?

Mitchell Raemsch