KE=m*(c^2)/2 ?

[Capiert]

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

 

Wed 2018 03 28 02:48 PS Wi 6.9 C

 

https://en.m.wikipedia.org/wiki/Annus_Mirabilis_papers

 

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

 of the Michelson–Morley experiment

 

 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 The fact that the energy withdrawn from the body becomes (light's_speed's) energy of radiation evidently makes no difference, so that we are led to the more general conclusion that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. ?

 

 

 

 

[Bender]

1 hour ago, Capiert said:

Maybe somebody can help point me in the right direction,

 because I can't confirm him. ?

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.  

[Sensei]

2 hours ago, Capiert said:

Maybe somebody can help point me in the right direction,

 because I can't confirm him. ?

Suppose so you have Uranium-235 atom.

It has mass m_u

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

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.

 

m_uc^2 = m_thc^2 + m_hec^2 + 4.68 MeV

Reverse equation:

m_uc^2 - ( m_thc^2 + m_hec^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

[math]KE = m_0c^2\gamma - m_0c^2[/math]

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 March 31, 2018 by Sensei

[Strange]

1 hour ago, Capiert said:

Maybe somebody can help point me in the right direction,

 because I can't confirm him. ?

Learn some basic physics. Learn to write properly.

[Capiert]

4 hours ago, 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.  

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.

 

 

[Strange]

Just now, Capiert said:

Strange,

 KE is about as basic to physics,

 as it gets.

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!

[Capiert]

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?

[Sensei]

45 minutes ago, 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.?

 

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

and Special Relativity [math]KE = m_0c^2\gamma - m_0c^2[/math]

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 March 31, 2018 by Sensei

[Capiert]

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.

[Strange]

29 minutes ago, Capiert said:

What should I be asking you about KE

 to get you to recognize my perspective?

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.

10 minutes ago, Capiert said:

It really looks like this guy Einstein

 mixed things up!

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

[Sensei]

31 minutes ago, 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.

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

 

Edited March 31, 2018 by Sensei

[Capiert]

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 March 31, 2018 by Capiert

[Strange]

11 minutes ago, Capiert said:

because nu is a photo film distance;

 NOT the (real) inverse wavelength

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?

[Bender]

41 minutes ago, Capiert said:

the photon energy formula also looks wrong.

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]

[Sensei]

1 hour ago, Capiert said:

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

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

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

But it used in many places in quantum physics.

Starting from photoelectric effect:

[math]h f - W = \frac{1}{2} m_e v^2[/math]

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.

[math]v = \sqrt{ \frac{ 2 ( h f - W) }{ m_e} }[/math]

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 March 31, 2018 by Sensei

[Capiert]

1 hour ago, Strange said:

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

nu (ν) is frequency.

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).

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?

[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).

 

https://en.m.wikipedia.org/wiki/Wavenumber

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

 

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:

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 March 31, 2018 by Capiert

[Strange]

36 minutes ago, Capiert said:

Nu is a wave number,

 NOT frequency.

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):

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

37 minutes ago, Capiert said:

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

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

 

[Capiert]

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.

[Strange]

Reported

for

trolling

 

 

[hypervalent_iodine]

!

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