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Posted

Kelvin 1900 vs M&M & (K)E is m*(c^2)/2

 
Wed 2018 03 28 02:48 PS Wi 6.9 C
 
 
In 1900, Lord Kelvin, in a lecture titled
 "Nineteenth-Century Clouds
 over the Dynamical Theory
 of Heat
 and Light",[5]
 
 suggested
 that physics had
 
 no satisfactory explanations
 for the results
 
 and for black body radiation. 
 
As introduced, special relativity provided an account for the results of the Michelson–Morley experiments. Einstein's theories for the photoelectric effect extended the quantum theory which Max Planck had developed in his successful explanation of black body radiation.
 
---
"It's about half!":
 
(K)E=m*(c^2)(/2)
 
PS: How did Einstein know
 the energy was m*(c^2);
 & NOT KE=m*(c^2)(/2).
I say he (=Einstein) did NOT know(!):
 he only guessed
 or approximated!
 
(But he=Einstein
 is very decided
 in using "2" "halves" of (K)E [#L]
 in the early part of the paper;
 & pushing the Lorentz contraction thru
 to make (up) the
 (non_sense; non_followable, quantum_jump)
 so_called following "conclusion?"
 (i.e. without "K"E).
 

If a body gives off the (kinetic) energy (K)E=m*(c^2)/2 [#L] in the form of radiation, its mass should diminish by (K)E*(2)/c2. The fact that the energy withdrawn from the body becomes (light's_speed's) energy of radiation evidently makes no difference, so that we are led to the more general conclusion that

The mass of a body is a measure of its (kinetic) energy_content; if the energy changes by (K)E, (then) the mass changes in the same sense by (K)E*(2)/(9 × 1020), the energy being measured in ergs, and the mass in grammes.

(PS: Also typical confusion,

 preferring Einstein's original units;

 NOT (promoting) standard (modern) SI units,

 such as Joules & Kg.)

 

See

http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

---

Sat 2018 03 31 04:28.. PS Wi 6.2 C

I repeat,
 I doubt Einstein's calculation(s)
 because he gave the clue
 to their basis
 being
 the KE's mass,
 but he pushed (=promoted)
 his Lorentz contraction 1st
 with=as priority
 before mentioning the KE relevance.
Instead I would prefer
 to start with the KE relevance
 to mass
 m=KE/(v*va)
 
 & work backwards
 (wrt Einstein),
 to show
 he's wrong (when)
 starting without the half
 in the 1st place,
 as his fatal error.
 
 
Einstein's syntax is also a bit reversed
 (to my preference
 for larger postscript having more;
 but seems (a typical) standard).
E.g.
Initial_energies=postscripts 0
final_energies=postscript 1
Resting coordinate energy E
Moving coordinate energy H.
[At 1st I (wrongly) thought (his energy) L
 could have represented wavelength
 (or a wave number)
 but it's NOT true.]
 
Somewhere I've done
 similar KE comparisons
 to show KE is (already) relativistic,
 (e.g. (extrap(olate)able) using simple fractions of c=1,
 or physicists beta=v/c
 e.g. let v=0, 1/2, 1/4 (of c).
The (non_linear) relativism
 becomes obvious
 with the 1/4 vs 3/4
 ratios
 for initial_speed vi
 & final_speed vf,
 when
 v=vf-vi
 in the form
 KE=m*((vf^2)-(vi^2))/2.
 
Let vi=0, & vf=1/2 (of c)
KE=m*(((1/2)^2)-(0^2))/2
KE=m*((1/4)-(0^2))/2
KE=m/8.
Let vi=1/2, & vf=1 (of c)
KE=m*((1^2)-((1/2)^2))/2
KE=m*(1-(1/4))/2
KE=m*(3/4))/2
KE=m*3/8.
Please notice,
 both together
 add to m*4/8
 =m/2
 i.e. "half"
 which is the maximum
 for those cases.
That was only a split
 into 2
 producing
 1/8th & 3/8th
 of m.
If we split into 4,
 vf=1/4, 2/4, 3/4, 4/4,
 then the finals are
 1/32,
 4/32
 9/32
16/32
 where (the KE's are (related so:))
 m*1/32 (for) vi=0, vf=1/4
 m*3/32 vi=1/4, vf=2/4
 m*5/32 vi=2/4, vf=3/4
 m*7/32 vi=3/4, vf=4/4.
Please notice
 those differ by
 1 (wrt 0), 2, 2, 2, 1
 of 32 (=(c^2)/2).
Important is
 their sum is m*16/32
 =m/2
 = half of m, again.
 
(E.g. KE=m*(c^2)/2.)
 
So that is how
 CO(k)E (conservation of (kinetic) energy)
 is (non_linearly) guaranteed.
 
I would then have to question
 what Einstein is doing
 with [his] coe,
 because
 he seems to be in error
 (when missing the half).
 
I mean, that
 is how (kinetic) energy is calibrated
 i.e. related.
 
Getting more than half (m)
 is abracadabra,
 i.e. energy creation.
 
 =Wishful thinking.
 
Maybe somebody can help point me in the right direction,
 because I can't confirm him. ?
 
 

 

 

Posted
  On 3/31/2018 at 4:13 AM, Capiert said:
Maybe somebody can help point me in the right direction,
 because I can't confirm him. ?
Expand  

Go to the library and pick up a physics text book. There is no half in the equation because when you derive it properly, it is not there. The E in the equation also does not represent kinetic energy.

Your calculations are nothing but numerology and thus meaningless and irrelevant to this topic.  

Posted (edited)
  On 3/31/2018 at 4:13 AM, Capiert said:
Maybe somebody can help point me in the right direction,
 because I can't confirm him. ?
Expand  

Suppose so you have Uranium-235 atom.

It has mass mu

It's decaying to Thorium-231 and Helium-4, with rest-masses mth and mhe.

U-235 -> Th-231 + He-4 + 4.68 MeV

And it releases approximately 4.68 MeV energy (it's called decay energy). In this case (alpha decay) it (decay energy) is in kinetic energy of newly created particles.

(small) part of that energy is taken by Thorium-231, and (larger) part of that energy is taken by highly accelerated Helium-4 nucleus.

 

muc^2 = mthc^2 + mhec^2 + 4.68 MeV

Reverse equation:

muc^2 - ( mthc^2 + mhec^2 ) = 4.68 MeV

 

Rest-mass of Uranium has been converted to kinetic energies of newly created particles.

 

Rest-mass of Uranium, rest-mass of Thorium, and rest-mass of Helium can be measured independently many different ways.

Decay energy can be calculated using above equation.

Kinetic energy of newly created particles can be measured (they're (traces) visible by naked eye in e.g. Cloud Chamber).

 

Kinetic energy in Special Relativity is

KE = m_0c^2\gamma - m_0c^2

not what you wrote in title!

 

ps. Multiplication by c^2 just coverts units from unit used to express mass to unit used to express energy.

ps2. Why am I giving over and over again the same example of radioactive decay process? Because it's easy to see how rest-mass is converted to energy, without bothering much about original Einstein's work.

 

Edited by Sensei
Posted
  On 3/31/2018 at 5:17 AM, Bender said:

Go to the library and pick up a physics text book. There is no half in the equation because when you derive it properly, it is not there. The E in the equation also does not represent kinetic energy.

Your calculations are nothing but numerology and thus meaningless and irrelevant to this topic.  

Expand  

I'm sorry Bender,

 but I don't think you got my point.

(The rest mass energy is in no way correctly calibrated

 to KE,

 thus can NOT be energy.

Einstein was discussing vis Viva=2*KE;

 NOT KE!).

I know there is no half in Einstein's equation,

 everybody knows that.

But I doubr Einstein

 got it right

 because he broke continuity.

He broke the rules.

(What is for you properly?

Starting with an assumption?)

 

Sensei, KE can still calculate the mass,

 but it's a different value

 than Einstein's;

 & perhaps has the guarantee that it's correct

 because it does NOT break the rules.?

 

Strange,

 KE is about as basic to physics,

 as it gets.

You guys try to escape (=avoid) that,

 (basic physics).

That's what bluffs me.

You prefer 1 guys assumptions,

 inspite of the hints

 against it.

 

 

Posted
  On 3/31/2018 at 10:07 AM, Capiert said:

Strange,

 KE is about as basic to physics,

 as it gets.

Expand  

All the more reason you should learn about it instead of posting ignorant crap.

And, if I were a moderator, I would ban you for failing to write like an adult!

Posted

Strange, maybe I haven't found the right question to make it click for you?

What should I be asking you about KE

 to get you to recognize my perspective?

Posted (edited)
  On 3/31/2018 at 10:07 AM, Capiert said:

Sensei, KE can still calculate the mass,

 but it's a different value

 than Einstein's;

 & perhaps has the guarantee that it's correct

 because it does NOT break the rules.?

Expand  

 

Difference between Classical Physics kinetic energy KE=\frac{1}{2} * m * v^2

and Special Relativity KE = m_0c^2\gamma - m_0c^2

is only visible at significant velocities (at the moment unavailable for humans, except ions in particle accelerators), while approaching to speed of light in vacuum.

 

Einstein with his 1st wife Mileva, work on Photoelectric effect, they used classical physics kinetic energy equation, not special relativity kinetic energy equation.

 

ps. Probably nobody reading your posts, have any idea, what are you talking about generally..

ps2. You should never replace v by c in classical physics kinetic energy equation!

 

Edited by Sensei
Posted

Sensei,

 the photon energy formula also looks wrong.

It's missing the "half"!

I.e. The same failure,

 but in reverse!

It really looks like this guy Einstein

 mixed things up!

He makes 1 severe error,

 & then compesates for it elsewhere.

 

The momentum is

 mom=h*nu (wavenumber nu; NOT frequency f!)

 for (Einstein's) Planck's constant h.

Energy=mom*va.

E=h*f*c/2,

 where the average_speed va=(vi+vf)/2

 is between initial_speed vi=0 m/s

 & the final_speed vf=c, light's_speed.

c=f*lambda

 for wavelength lambda.

Posted
  On 3/31/2018 at 10:18 AM, Capiert said:

What should I be asking you about KE

 to get you to recognize my perspective?

Expand  

You should explain what it is you don't understand. You should write your questions in proper English, like an adult, not like a pretentious teenager.

  On 3/31/2018 at 10:39 AM, Capiert said:

It really looks like this guy Einstein

 mixed things up!

Expand  

The problem is in your failure to understand. Not in well-tested physics.

Posted (edited)
  On 3/31/2018 at 10:39 AM, Capiert said:

 

The momentum is

 mom=h*nu (wavenumber nu; NOT frequency f!)

 for (Einstein's) Planck's constant h.

Energy=mom*va.

E=h*f*c/2,

 where the average_speed va=(vi+vf)/2

 is between initial_speed vi=0 m/s

 & the final_speed vf=c, light's_speed.

c=f*lambda

 for wavelength lambda.

Expand  

You permanently ignore conversion of units.. and that's just for a start...

 

Edited by Sensei
Posted (edited)

Yes Sensei, I wanted to say that formula is still wrong.

E=h*nu*c/2

& still won't give you the frequency f,

 because nu is a photo film distance;

 NOT the (real) inverse wavelength.

lambda=((2*d*nu)^0.5)/m

 wrt Wiki's diffraction grating syntax

 slit to slit width d,

 order m, e.g. 1.

Edited by Capiert
Posted
  On 3/31/2018 at 11:00 AM, Capiert said:

because nu is a photo film distance;

 NOT the (real) inverse wavelength

Expand  

Well, if you use the wrong number in the equation then you will get the wrong result.

nu (ν) is frequency. lambda (λ) is wavelength.

None of these equations use "photo film distance" (whatever that means). Why would the position of a photographic film change the energy of a photon?

Posted
  On 3/31/2018 at 10:39 AM, Capiert said:

the photon energy formula also looks wrong.

Expand  

I think this is the essence of the problem.

Whether something "looks" wrong is irrelevant. What matters is whether it can be used to make predictions which are then verified experimentally. Einstein's theories excell at that. If you make any predictions with your versions, your results will be too small with a factor of about 2.

What also matters is communication, such as not

Starting (a paragraph)

In the middle of a sentence [collection of words]

Posted (edited)
  On 3/31/2018 at 11:00 AM, Capiert said:

Yes Sensei, I wanted to say that formula is still wrong.

Expand  

You're "victim of not proper education"..

You claim that photon energy formula y E=h v=\omega \hbar is incorrect..

But it used in many places in quantum physics.

Starting from photoelectric effect:

h f - W = \frac{1}{2} m_e v^2

Photon is disappearing from system, instead free electron is ejected from metal. And its kinetic energy (can be calculated from velocity) is reminder of energy needed to eject it.

v = \sqrt{ \frac{ 2 ( h f - W) }{ m_e} }

C'mon! This can be proven simply using white light source, prism, and metal such as Cesium, Potassium, Sodium, Lithium..

Then annihilation, then pair production, etc. etc.

 

 

Edited by Sensei
Posted (edited)
  On 3/31/2018 at 11:15 AM, Strange said:

Well, if you use the wrong number in the equation then you will get the wrong result.

nu (ν) is frequency.

Expand  

As you can see above,

 I think Sensei (also) complained enough

 about those units

 NOT being correct.

Nu is a wave number,

 NOT frequency.

 

  Quote

lambda (λ) is wavelength.

None of these equations use "photo film distance" (whatever that means).

Expand  

The (diffracted) distance where the spectral lines are found on film,

 versus (=wrt) straight thru without diffraction (grating, or prism).

  Quote

Why would the position of a photographic film change the energy of a photon?

Expand  

[Reflected photons can loose some of their energy

 as a bounce,

 but your question is wrongly put

 for what you want to know.]

The film does NOT change the photon energy per se,

 (in what we're interested here).

The amount, that the light is bent or deflected (i.e. diffracted)

 indicates the wavelength.

We know c=f*L(ambda);

 & the energy is proportional to the frequency.

Assuming c is constant.

So f=c/Lambda;

 energy is inversely proportional to the deflected distance on the photo film.

But it is NOT correct to assume

 that (either inverse or not inverse) distance nu (which I simply call x (=deflected distance)

 wrt y_distance (diffraction_grating to film, perpendicular_distance))

 is the wavelength.

Nu is NOT the wavelength;

 & it (=nu) is NOT the inverse_wavelength either

 (which physicists call wavenumber).

 

In spectroscopy, "wavenumber" {\tilde {\nu }} (=nu_bar) often refers to a frequency which has been divided by the speed of light in vacuum:

 
{\displaystyle {\tilde {\nu }}={\frac {\nu }{c}}={\frac {\omega }{2\pi c}}.}

The historical reason for using this spectroscopic wavenumber rather than frequency is that it proved to be convenient in the measurement of atomic spectra: the spectroscopic wavenumber is the reciprocal of the wavelength of light in vacuum:

\lambda _{{{\rm {vac}}}}={\frac  {1}{{\tilde  \nu }}},

which remains essentially the same in air, and so the spectroscopic wavenumber is directly related to the angles of light scattered from diffraction gratings and the distance between fringes in interferometers, when those instruments are operated in air or vacuum. Such wavenumbers were first used in the calculations of Johannes Rydberg in the 1880s. The Rydberg–Ritz combination principle of 1908 was also formulated in terms of wavenumbers. A few years later spectral lines could be understood in quantum theory as differences between energy levels, energy being proportional to wavenumber, or frequency. However, spectroscopic data kept being tabulated in terms of spectroscopic wavenumber rather than frequency or energy.

Edited by Capiert
Posted
  On 3/31/2018 at 12:23 PM, Capiert said:

Nu is a wave number,

 NOT frequency.

Expand  

Why not (a) learn to write properly and stop behaving like a spoiled teenager and (b) LEARN some basic physics:

  Quote

The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ):

Expand  

https://en.wikipedia.org/wiki/Photon#Physical_properties

  On 3/31/2018 at 12:23 PM, Capiert said:

In spectroscopy, "wavenumber" (=nu_bar) often refers to a frequency which has been divided by the speed of light in vacuum

Expand  

1) We are not talking about spectroscopy

2) That is nu-bar, not nu

3) Learn some basic physics

4) Stop writing like a child

 

Posted

The frequency

 f=c/lambda

 (should be)

 f=c*m/((2*d*x)^0.5).

As you can see

 that's NOT f=c/nu

 if nu is x (=photo film displacement).

You might not be discussing spectroscopy

 but I am;

 & a (search for a) consistent (kinetic) energy;

 NOT a mixture of energy & vis viva.

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