E = ?

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posted on Nov, 12 2002 @ 10:35 AM
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on a previous thread "speed of light broken with a basic lab kit", William said:

"Let's see it... start a new thread."

so i am ....

what he refers to is proof of my statement that:

E /= mc^2

contrary to the assertion by FreeMason.

so, i dutifully wandered into the library, grabbed a book on nuclear physics and hey-presto, there it is.

E^2 = p^2 c^2 + m^2 c^4

however, it is better explained in:

B. R. Martin & G. Shaw, "Particle Physics", 2nd Ed., John Wiley & Sons, 1997. (part of the blue Wiley series of physics texts, which i highly recommend as baseline textbooks, BTW)

so, what i present here is paraphrased therefrom and is essentially nothing new in the world of science. i post it here merely to settle (or perhaps start) an arguement. without further ado, page 3:

we start from the assumption that a particle with momentum p in free space is described by a de Broglie wavefunction with frequency

v = E/h

and wavelength

L = h / p

(apologies for poor use of variable names here, but its just alphanumeric i'm afraid...)

the corresponding wave equation depends on the assumed relation between energy E and momentum p. non-relativistically,

E = p^2 / (2 m)

and the wavefunction obeys the non-relativistic Schrodinger equation. relativistically, however,

E^2 = p^2 c^2 + m^2 c^4

where m is the rest mass ....

etc. etc.

here end any quoting from the text. now, its obvious that the relativistic and non-relativistic equations are close but distinctly different. the non-relativistic E = mc^2 is often used, but as shown is clearly false.

to be explicit in pursueing my original statement, i said that - contrary to FreeMason's statement - E /= mc^2, and the above arguement clearly demonstrates. the non-relativistic equation is an approximation that is often acceptable for classical mechanics, but not for quantum mechanics nor anything requiring relativity. ergo, FreeMason's statement was incorrect.

my apologies if this little discussion has left any ATS readers a trifle bored ....


- qo.




posted on Nov, 12 2002 @ 11:04 AM
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Allright.... very nice... thank you.

Lemme get some coffee and consider this. I've seen these before and understand your direction.... but somehow, it seems that some evil person has seen fit to give me decafinated coffee this morning and the typical fog has not yet lifted.



posted on Nov, 12 2002 @ 11:14 AM
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lmfao ... my working day is coming to a close, and the fog still hasn't lifted. i can't stand the taste of tea or coffee, so no caffeine for me.


the above is standard physics. as i've said, its ripped straight from a textbook.

- qo.



posted on Nov, 12 2002 @ 11:23 AM
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yes but.....don't you think its spooky.....you know....that his names "freemason" hmmmm? like, maybe all this faster than light stuff is all an iluminate plot to like, ~control~ light..................................



posted on Nov, 12 2002 @ 11:32 AM
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lmfao ..... *waves fingers @ lupe in a spooky way* who knows .......

- qo.



posted on Nov, 12 2002 @ 11:54 AM
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users.aber.ac.uk...

The above link is a very nice summary of one physicist's point of view on issue with currently accepted physics. A much better resource on the rethink some people are doing on specific aspects of relativity and quantum physics.

Interesting read from the Skeptic Overlord's bookmarks.... highly recommended




posted on Nov, 12 2002 @ 06:56 PM
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Looks interesting, I'll have to go back sometime when I can concentrate better (caffeine is the tool of satan).

I just hope my primitive understanding of science is enough to grasp at least some of the concepts. Can you recommend any other science related sites?



posted on Nov, 12 2002 @ 07:43 PM
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Ok for starters I think you missed some important points....thought maybe I'm just too tired and don't care...however...

point number one: Your E/h=v is wrong it's (Change in E)/h=v there is a difference because the actuall frequency is v=c/L(lambda) lambda being of course wavelength. Therefore we find that Change in E is equal to hc/L.

Now as I said...I'm not really going to debate this so you correct any points I make and explain why what you've stated is correct.

I'd like to point out that you ARE solving for E^2 when in fact we are calculating E. I don't feel like sitting and doing the math, but let's work the E^2 equation and see what it boils down to when you are solving for just E.

Oiy I'd be glad to solve E for mc^2 if you just give me the damn numbers....you have p's and mv's where they seemingly should not be.....come back, post a list of what your letters all mean...and I'll be glad to solve E=mc^2.

What is p? Is it simply mv?

And make sure the E^2 is not CHANGE IN ENERGY, because that has NOTHING to do with E=mc^2

Actually let's see what this does:

E^2 = ((E/c^2)v)^2c^2+m^2c^4
E=((1/c^2)v)^2c^2+m^2c^4
E=(v^2/c^2)+m^2c^4
E=(v^2+m^2c^4/c^2)
E=(v^2+m^2c^2)
E-v^2=m^2c^2

Oiy I'm just looking this up at an outside source, remember, IT IS!!! Einstein who figured this out...


Here's a source, and it says as I said earlier, it is a proportion....not an exact one...but still E=mc^2
.... www.phys.lsu.edu...

Actually I do want to state...I don't believe man that E^2=p^2c^2+m^2c^4 has any relation to what E=mc^2 is talking about. But I am just uncertain of what p is...you said p has something to do with frequency and wavelength...but in what way? P=v/L?
AH HA! I think I see something...

E^2=p^2c^2+m^2c^4
p=h/L
E^2=(h/L)^2c^2+m^2c^4
E^2=(hc/L)^2+m^2c^4
E=(hc/L)+mc^2
hc/L = change in E
So what this equation you gave is saying is that E=(change in energy)+mc^2 which must have something to do with nuclear reactions because....look at your source.
I do believe I am not wrong in my equation there but do point out mistakes...I suck at algebra


Sincerely,
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[Edited on 13-11-2002 by FreeMason]

[Edited on 13-11-2002 by FreeMason]



posted on Nov, 13 2002 @ 04:38 AM
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well i'm gald that you're at least willing to think about this.

first off, as you requested, here is a list of variable definitions:

E = energy
- you debate whether this should be change in energy, but we'll leave it as is.

h = plank's constant

v = f = frequency
- i introduce f here since v can be confused with velocity

c = speed of light in vacuum

L = wavelength

m = mass / rest mass
- this distinction is a matter of relativity, and changes depending on the regime to which the equation applies

p = momentum

now, lets start from the top:

"point number one: Your E/h=v is wrong it's (Change in E)/h=v there is a difference because the actuall frequency is v=c/L(lambda) lambda being of course wavelength. Therefore we find that Change in E is equal to hc/L."

well, i disagree that E should be change in energy. as a laser physcisit i often use E = hf, where E is the change in energy. however, the change in energy relates to the change in energy of the states being considered - in this case the states between which a transition is being made and from which a photon is being emitted. so, in general, E = hf, where E is the energy of the particle. the rest of your statement is correct:

E/h = f
E = hf
f = c/L
E = hc/L

Now:

"E^2 = ((E/c^2)v)^2c^2+m^2c^4
E=((1/c^2)v)^2c^2+m^2c^4"

incorrect. starting from:

E^2 = p^2 c^2 + m^2 c^4

i can then only assume you have assumed:

p = mv (where v here is velocity)
E = mc^2
m = E/c^2
p = (E/c^2)v

giving:

E^2 = ((E/c^2)v)^2 c^2 + m^2 c^4

for a start i am not sure that p = mv is applicable to quantum particles. i'd have to go away and serious check that. in any case, the calculation you make after should be:

E^2 = ((E/c^2)v)^2 c^2 + m^2 c^4
E^2 = ((E^2 v^2)/c^2) + m^2 c^4

not E=((1/c^2)v)^2c^2+m^2c^4 as you gave.

finally, i would like to point out a quote from your own source: www.phys.lsu.edu...

"In order to understand the law of the equivalence of mass and energy, we must go back to two conservation or "balance" principles which, independent of each other, held a high place in pre-relativity physics."

the key being 'pre-relativitiy'. on a philisophical (as well as scientific note) this arguement actually boils down to a question of semantics and pendantry. your statement, E = mc^2 is of course true, PROVIDED you stipulate the approximations, assumptions and limitations of the statement, which you did not. hence, having stated an absolute, it was easy for me to contradict it. the above proof that E^2 = p^2 c^2 + m^2 c^4 is besides the point, as is used merely as an example of how your statement can be false.

- qo.



posted on Nov, 13 2002 @ 07:59 PM
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Yes it seems it CAN be false BUT!!! Surely you will agree that E=mc^2 DOES show the relation between mass and energy


And in any case I AM not sure if momentum actually matters in something as we are discussing either.

Actually, I find it interesting about our derivations of E.

Oh wait, hmm I said exactly what you said? Hmm oh ok, the E=hc/L is my claim that the Es in question are in fact two different Es.

I believe that E=mc^2 or E^2=blah blah are referring to the initial energies? (maybe I'm wrong?)

Where as the E=hc/L is refering (I think
) to the final energy, yeah I think maybe that is right? Hrrmmm...dealing with electrons emitted by a wavelength of light hitting a metal plate?

Oh well...that's off the top of my head...I could look it up, but I'm just looking for something to get our minds thinking, this post has been quite fun
What do you think about the Es in question??? Any further thoughts on the differences I claim?

Sincerely,
no signature

Hmm this should be an interesting thread for anyone still in high school
I know my high school chem sucked



posted on Nov, 14 2002 @ 06:02 AM
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Originally posted by FreeMason:

"Yes it seems it CAN be false BUT!!! Surely you will agree that E=mc^2 DOES show the relation between mass and energy
"

well, yes it does, but only under strict approximations - in particular non-relativism. the point i was trying to get across originally is that a lot of what einstein said was wrong. he corrected a lot of it himself, but we're still trying to correct his theory of gravity, for example.

"Where as the E=hc/L is refering (I think
) to the final energy, yeah I think maybe that is right? Hrrmmm...dealing with electrons emitted by a wavelength of light hitting a metal plate?"

hmmm ... firing electrons at a metal plate, which releases photons? sounds a little like thermionic emmission, but i studied that many years ago. in any case, if a photon is emitted at the end of the process - the ~final~ energy as you put it - then this is the energy of the photon. E. in that sense it is a "change in energy" because there will be some energy loss from the incoming electron due to interaction with the metal, etc, or some such. however, the statement holds that the energy is the energy of the photon. the fact that it can be considered as a "change in energy" is due to the application and the situation being considered.

"Hmm this should be an interesting thread for anyone still in high school
I know my high school chem sucked
"

chemistry!?!?! how dare you man! this is physics, not chemistry! *fume fume fume*


- qo.



posted on Nov, 14 2002 @ 09:57 PM
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Ok guys strait from the horses mouth.

Einstein Speaking on E=mc^2



posted on Nov, 15 2002 @ 12:08 AM
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LoL Chem and Physics blur for a while in this area till you progress in either


However no not emissions of photons, but of electrons. Hitting the metal plate with a certain wavelength of light will "knock" loose an electron. Yeah...I'm not sure if there is another use for it, but you'd use the change in energy to find the energies of an electron in it's quantum level
(initial) and in where ever it ends up (final) Yes?

Oh whatever....point is it is the emission of an electron, but I think you think I said photon
.


Originally posted by quiet one
chemistry!?!?! how dare you man! this is physics, not chemistry! *fume fume fume*

Hmm more the reason for me to get 3 majors


Sincerely,
no signature



posted on Nov, 15 2002 @ 02:10 AM
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Now using basic physics formula we can prove an intersting theory.

Now E=Energy
E=E
E=mc^2
E=0.5mv^2 So

mc^2 = 0.5 mv^2 The ms cancel out (Basic algebra)
c^2=0.5v^2
v^2 / c^2 = 2
v/c=2^0.5
c/v=1/2^0.5
Now if you know your exact values you will know that a/2^0.5 is the same as Sin45

So

c/v=Sin45
v=c/Sin45
v=424264068.7metres per second

Intersting huh

Feel free check my working it all works out

So clearly physics and math do no mix

By the way i have another proof

mass=(voltage*Amperage*Displacement*Velocity^2)/(Gravity * height*(intial velocity+acceleration*time))






posted on Nov, 15 2002 @ 02:13 AM
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Sorry there is a problem in my working where it says a/2^0.5 it should say 1/2^0.5 which is = Sin 45

other than that it is all correct



posted on Nov, 15 2002 @ 06:09 AM
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Originally posted by Toltec:

"Ok guys strait from the horses mouth.
Einstein Speaking on E=mc^2"

two problems with this:

1) i don't have a soundcard in this machine (i'm at work) so could only read the text.

2) this whole arguement started from my statement that einstein was wrong 9 times out of 10.

still, interesting little site. i might just rip the sound files and take them home ...

- qo.



posted on Nov, 15 2002 @ 06:11 AM
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Originally posted by FreeMason:

"LoL Chem and Physics blur for a while in this area till you progress in either
"

well i already have my masters and i'm working on my PhD, so i think i can recognise physics why i see it ....


"However no not emissions of photons, but of electrons. Hitting the metal plate with a certain wavelength of light will "knock" loose an electron."

my apologies, that is indeed what you said and that is indeed thermionic emission as i thought.

- qo.



posted on Nov, 15 2002 @ 06:16 AM
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dellarb, nice try, and the answer isn't far off what it should be (~ 3x10^8 m/s) however:

E = 0.5 m v^2 only relates to kinetic energy

E = m c^2 refers to the total energy of some mass (a particle, say, for simplicity) which includes potential energy and rest mass.

as such i do not believe your basic equation:

0.5 m v^2 = m c^2 to be valid.

besides which, as already shown on this thread, if v is in the region your calculations indicate then we're in a relativistic regime, and hence:

E /= m c^2

voiding your calculation.

- qo.


TN1

posted on Dec, 19 2002 @ 12:49 PM
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Quit one you it seems that you know much more and you Do!!I don't understend how the others can have doubts and cause a debate if they don't know the subject.Well for the others .....You can estimate the momentum and therefore the energy of a non-relativistic particle by using Heisenberg's uncertainty principle
pDx=h/2 and you will find that the energy of an orbiting electron around the nucleus is 1/137 of the speed of light for some elements not always ...Which means that the most of the physics laws are correct ,probably all of them but there is space for regeneration and improvement !!Did you see or experienced greater than light speeds???I don't think so and you because the speed of light is decreasing slightly form time to time ,if you want to find the results you have to wait another 10 billion years to observe a speed much less than this one...This does not imply that there are no particles travelling with more speed!!!



posted on Dec, 20 2002 @ 06:03 AM
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Well, the "h" in h/2 that's the hamiltonian opperator right? I need to look such things up. But the Energy of an electron in an Orbital that would be based on what quantum level it is at?

so surely it can not be just one figure for any orbital. If you can call them orbitals.

I'm beginning to think, that since we can not determine where an electron is precisely anyways...that it does not orbit the atom even in the way now thought of, but rather it fills the entire "orbital" it exists in all positions. I thought of this when looking at a "Plasma globe" so its probably half-cocked but if you figure that the probability of an electron being in the orbital, as being its strength at that point of the orbital, then I figure covalent bonds would look much like the plasma globe


Where the most probability is just where the two electrons are the strongest. And in effect there really is no motion...just we can use motion to figure stuff out. Eh...I'm just rambling and throwing out new theories, to see what dog chews them up


Sincerely,
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