# E=mc2 passes tough test

Status
Not open for further replies.
D

#### drwayne

##### Guest
"In a fitting cap to the World Year of Physics 2005, MIT physicists and colleagues from the National Institute for Standards and Technology (NIST) report the most precise direct test yet of Einstein's most famous equation, E=mc2. <br />And, yes, Einstein still rules." <br /><br />Rest of the story:<br /><br />http://www.physorg.com/news9248.html <div class="Discussion_UserSignature"> <p>"1) Give no quarter; 2) Take no prisoners; 3) Sink everything."  Admiral Jackie Fisher</p> </div>

J

#### jatslo

##### Guest
We are still talking of a medium that is not empty; I would like to see these experiments conducted with inflationary vacuums. Even at cryogenic temperatures; there is resistance on non-zero masses. Still a localized event. <img src="/images/icons/smile.gif" /> I would suggest projecting an inflationary vacuum in front of the non-zero mass, that is if you want cause and effect to equal zero, or instantaneous. I can execute the procedure for you, if you would like to fly me to a lab. <img src="/images/icons/wink.gif" />

V

#### vidar

##### Guest
Einstein used the classic formula E=(mc^2)/2 , for v=c.<br />Somehow he concluded that the total energy is twice the kinetic energy for any mass.<br />In brief, how does the experiment prove that?<br />

S

#### Saiph

##### Guest
The full equation is E^2=m^2c^4 + p^2c^2.<br /><br />P is momentum in this case, m is rest mass, and E is the total energy of the particle.<br /><br />For a photon, despite having m=0, the fact that it has energy shows that it also has momentum (that term doesn't disappear).<br /><br />When V=0 for a particle of matter, P=0,a nd the equation is thus E^2=m^2*c^4, or simply E=mc^2.<br /><br />I don't know where you get the 1/2 term. <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

V

#### vidar

##### Guest
Ek = (mv^2)/2<br /><br />That is the classic equation for kinetic energy.<br />http://en.wikipedia.org/wiki/Energy#Kinetic_energy<br /><br />Einstein used v=c and somehow figured that the total energy is double the kinetic energy.<br />I have not seen any reson and proof of that.<br />(Neither did the Nobel Prize committee.)<br />

R

#### raghara2

##### Guest
You accelreate point up, you accelerate point down. Energy is 2*(1/2)*m*c^2. The funny thing is I could describe a theory that would alow mass to have more energy than that, and still be in correlation with experimental results.<br /><br />

S

#### Saiph

##### Guest
Thats not at all how E=mc^2 is derived.<br /><br />In that v does not equal C for a particle with mass.<br /><br />The kinetic energy term is included in the p^2*c^2 term in the full equation I provided above.<br /><br />The full equation shows that there are two components to the total energy in the system, that of motion (the momentum 'p' term) and that of the matter itself, the 'm' term.<br /><br />This also explains how high energy particle interactions do NOT conserve mass. Some is converted to energy (or vice versa) during particle collisions.<br /><br />Only by using the full equation for E, can you track it all down and have the calculations give the correct results. <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

V

#### vidar

##### Guest
That’s right; the mathematical maze might be more advanced.<br />The formula might be derived more complex. However, the result is the same.<br />Einstein simply doubled Newton’s classic formula, and got most people to believe he got a Nobel Prize for that.<br />

S

#### Saiph

##### Guest
?!<br /><br />It's not just a doubling of any newtonian equation. Heck, all of relativity, when used to calculate low speed situations, reduces to newtonian mechanics. So it's an extension of newton, but it isn't laughable.<br /><br />Second, Einstein did <i>not</i> recieve the Nobel Prize for <i>any</i> work on relativity. Even when he passed away, relativity was (and still sorta is) controversial, at that time not as rigorously tested as people would like (but passing every test they <i>could</i> do at the time).<br /><br />He won the nobel prize for the photo-electric effect experiment he did the same year he published Special Relativity. <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>

V

#### vidar

##### Guest
Saiph wrote:<br />It's not just a doubling of any newtonian equation. Heck, all of relativity, when used to calculate low speed situations, reduces to newtonian mechanics. So it's an extension of newton, but it isn't laughable.<br />--------------------------<br /><br />Einsteinâ€™s E=mc^2 is simply doubling Newtonâ€™s E=1/2 mc^2<br />Itâ€™s so easy to see when keeping the focus outside all the mathematical mazes. <br /><br />Furthermore, Einsteinâ€™s equations of relativity are simply pasting the Lorentz transformation to classic factors of time, length and mass.<br />http://www.physics.northwestern.edu/Phyx103/web/extra/einstein-eqns.pdf<br />http://www.vidargander.com/public/eformula.pdf<br /><br />I think its time to call off that bluff in order to get science back on track.<br />

A

#### alokmohan

##### Guest
Newtons equation are suffficent upto solar system.But when dealing with large bodies You require Einstein.Einstein didnot get nbel prize for his gtwo relativty theory.Saiph is accurate.

H

#### heyscottie

##### Guest
Where are you going with this, vidar? Are you trying to say that relativity is so simple even a moron could have discovered it, or that it is patently false?<br /><br />Because each of these ideas are incorrect, of course...

V

#### vidar

##### Guest
I think the a formula for total-energy must be more complex than E=mc^2<br />It hasnâ€™t ever been proven to be more true than E= 4/3 mc^2 or E= 1/2 mc^2<br /><br />I think the use of Lorentz transformation (Y) isnâ€™t more enlightening than confusing.<br />I donâ€™t think its any legitimate reason to deny masses the speed of light.<br />

M

#### MeteorWayne

##### Guest
Actually, it's been proven much closer than what you suggest, from what I know.<br /><br />E=MC^2 works to explain particle accelrator results, and certainly certain big booms we've set up based on the equation. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

V

#### vidar

##### Guest
There is no evidence close to determine if there is a factor 0.5 1.0 or 1.3 to the E= x mc^2 equation.<br />My guess that its really E= pi mc^2 is good as any.<br />

M

#### MeteorWayne

##### Guest
Well, I think there is.<br />When I have time I will research the subject.<br /><br />Your "My guess that its really E= pi mc^2 is good as any" does not agree with the facts as I understand them. <br /><br />You can't give anything that shows any violation from the basic E=MC^2, can you? <br /><br />W <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

V

#### vidar

##### Guest
E= mc^2 is a violation of the classic and well proven E=1/2 mc^2

M

#### MeteorWayne

##### Guest
<font color="yellow">"E= mc^2 is a violation of the classic and well proven E=1/2 mc^2 "<br /></font><br /><br /><br />Thanks, appreciate the softball pitch.<br /><br />Now that you've made that statement, please provide one iota of evidence that supports that assertion. <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

O

#### origin

##### Guest
<font color="yellow">E= mc^2 is a violation of the classic and well proven E=1/2 mc^2</font><br /><br />I guess it was not so obvious if it is a violation of the newtonian law.<br /><br /><font color="yellow">There is no evidence close to determine if there is a factor 0.5 1.0 or 1.3 to the E= x mc^2 equation. <br />My guess that its really E= pi mc^2 is good as any.</font><br /><br />From the article, which you apparently did not read - <br />"In the test, we at MIT measured m, or rather the change in m associated with the energy released by a nucleus when it captures a neutron," <br /><br />In other words the mass lost produced enrgy that was equivelant to E=mc^2. <br /><br />That certainly sounds like evidence that the equation is correct!<br /> <div class="Discussion_UserSignature"> </div>

O

#### origin

##### Guest
<font color="yellow"> vidar said:<br />Einstein used the classic formula E=(mc^2)/2 , for v=c.<br /><br />and<br /><br />E= mc^2 is a violation of the classic and well proven E=1/2 mc^2</font><br /><br />Both of these statements are incorrect.<br /><br />Einstein did not use the kinetic energy formula as the starting point.<br />Special relativity is centered on the premise that the speed of light is constant. Experiments had been run (such as the Michelson-Morley experiment) that showed the speed of light was constant and independant of the observers velocity, Einstein simply accepted the fact of these observations.<br />The mass and the associated relativistic momentum is where the derivation of the famous E=mc^2 comes from. The classical formula for F=MA does come into play because the force of the relativistic momentum is integrated to give the Energy of the system. I beleive most 3rd semester physics books will cover this. The calculus is easy the algebra gets pretty messy.<br /><br />As for the statement that Special relativity violates classical mechanics:<br />The results from the integration for the velocities at c give E=mc^2.<br />The results ALSO give the results for v<<c, and those results are E= (mv^2)/2.<br /><br />The results fo Special Relativity don't violate classical mechanics they complement them. At low velocities like 10 million miles/hr classical mechanics are a very good approximation it is only at high velocities that relativistic mechanics must be used.<br /><br /> <div class="Discussion_UserSignature"> </div>

V

#### vidar

##### Guest
I very much regret that I fell for answering in the same way the question:<br />â€œYou can't give anything that shows any violation from the basic E=MC^2, can you?â€<br /><br />Still, the facts are that the factors suggested to the E=mc^2 equation are Newton (0.5), Einstein (1.0) and Max Abraham (3/4) and Friedrich HasenÃ¶hrl (3/8). Science isnâ€™t even close to determine which one is closest to reality, because science is much too concerned about the much more important factor; c^2, - and they arenâ€™t even close.<br /><br />Several stories have appeared to support the 1.0 factor. There was the atom-clocks in a plane half a century ago. That was impressive at the time, but now that most kids wear atom clocks on their wrists, and do fly occasionally, the â€˜proofâ€™ is almost forgotten. There was some story that the 1.0 factor to explain an irregular orbit of Mercury, until someone mentioned that the Sun probably was a cause. Nowadays, signal phase irregularities for satellites to mobile phones are somehow proof for the 1.0 factor, while other satellitesâ€™ irregularities arenâ€™t mention in that way. <br /><br />Sure it sounds convincing that a handful scientists locked up in an underground facility has managed to measure release some energy from atoms and neutrons. However, enormous quantities of energy are added to the elements in order to make them interact. There really is no wonder at all, that additional energy is released from those elements. <br /><br />The E=x mc^2 equation is about universal energy. How can anyone claim that some measures on single neutron represent and prove something huge as that? It seems like we just have another â€˜atom-clockâ€™ story going on.<br />

C

#### CalliArcale

##### Guest
* Most kids do not wear atom clocks on their wrists. Atomic clocks are not really suitable for wristwatch applications. You are probably thinking of the things you can buy in the store which call themselves "atomic clocks". They're not really; they're just picking up the radio signals that the US Navy sends out from *their* atomic clock and periodically resetting themselves to match. So you can't use those kinds of "atomic clocks" to try to duplicate the experiments done in the 1950s. You need an *actual* atomic clock, not a regular quartz clock that just has a computer that periodically checks for the US Naval Observatory's radio signal and resets itself to match.<br /><br />* The irregular orbit of Mercury is properly predicted if you use Einstein's theory of relativity. The Sun is indeed the cause. This does not contradict Einstein; the Sun's mass is what Einstein said was causing the effect.<br /><br />* Relativistic corrections are required for handheld GPS units to correctly make use of the signal from the GPS satellites. All satellites' signals are affected by relativity, but for most spacecraft, this doesn't affect their functioning. It's an issue for any spacecraft used for navigation or really accurate timekeeping, but otherwise can be safely ignored. That's not to say it doesn't happen with other spacecraft.<br /><br />* It is indeed amazing that a bunch of scientists in an underground facility can measure the release of energy from atoms and neutrons. Yes, energy goes into it, but the key is that they are getting more energy back than they should, and there is less mass afterwards. Can you explain that, or how Einstein's equations correctly predict the amounts?<br /><br />I think you are dismissing some very serious pieces of evidence without bothering to really understand them. But if you seriously think that children are walking around with atomic clocks on their wrists, then perhaps that shouldn't be surprising. <div class="Discussion_UserSignature"> <p> </p><p><font color="#666699"><em>"People assume that time is a strict progression of cause to effect, but actually from a non-linear, non-subjective viewpoint it's more like a big ball of wibbly wobbly . . . timey wimey . . . stuff."</em>  -- The Tenth Doctor, "Blink"</font></p> </div>

M

#### MeteorWayne

##### Guest
kids don't wear atom clocaks on their arms. They wear watches that calibrate themselves to the WWVB transmissions fro Boulder, CO.<br />Those track "atomic" time. Straw Man.<br /><br />The fact that certain satellites need einsteinian correction and others don't has to do with what the satellites do.<br /><br />If Einstein was ignored our entire GPS system would not work. They deal with time.<br /><br />What other satellites irregularitues are you referring to, or are you just making that up to strengthen your "case"?<br /><br />It is not just that energy is added in collider experiments, it's that using e=mc^2 lets you calculate the in and out amounts of mass and energy, and the fact is, it works.<br /><br />Please provide some data that shows otherwise.<br /><br /><br /> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

3

#### 3488

##### Guest
Hi MeteorWayne,<br /><br />If I am not mistaken, the energy used to accellerate the particles in the collider, has to come<br />from somewhere else?<br /><br />If I am correct, energy is neither created or destroyed, just transferred from one state to another??<br /><br />Ie the collider, the energy used to accellerate the particles could come from Fossil Fuels,<br />solar energy, nuclear, etc. <br /><br />Or am I missing something here?<br /><br />I must admit, I really, truly am ignorant on this subject.<br /><br />Andrew Brown. <div class="Discussion_UserSignature"> <p><font color="#000080">"I suddenly noticed an anomaly to the left of Io, just off the rim of that world. It was extremely large with respect to the overall size of Io and crescent shaped. It seemed unbelievable that something that big had not been visible before".</font> <em><strong><font color="#000000">Linda Morabito </font></strong><font color="#800000">on discovering that the Jupiter moon Io was volcanically active. Friday 9th March 1979.</font></em></p><p><font size="1" color="#000080">http://www.launchphotography.com/</font><br /><br /><font size="1" color="#000080">http://anthmartian.googlepages.com/thisislandearth</font></p><p><font size="1" color="#000080">http://web.me.com/meridianijournal</font></p> </div>

O

#### origin

##### Guest
<font color="yellow">Still, the facts are that the factors suggested to the E=mc^2 equation are Newton (0.5), Einstein (1.0) and Max Abraham (3/4) and Friedrich HasenÃ¶hrl (3/8). Science isnâ€™t even close to determine which one is closest to reality, because science is much too concerned about the much more important factor; c^2, - and they arenâ€™t even close. </font><br />Good grief! I don't mean to be rude but you don't know what you are talking about. As I pointed out in my earlier post, Einstein and Newton agree on the formula for KE, it is only at relativistic energies that Einstein's formula is necessary.<br /><br /><font color="yellow">Sure it sounds convincing that a handful scientists locked up in an underground facility has managed to measure release some energy from atoms and neutrons. However, enormous quantities of energy are added to the elements in order to make them interact. There really is no wonder at all, that additional energy is released from those elements.</font><br />Do you really think that the scientist forgot about the kenetic energy of the constituents? Come on.<br />Did you know that the mass of a U235 atom weighs more than the fission products that are produced from fission. Did you know that if you add together the energy of the gamma rays and the kenetic energy of the fission products that it equals the missing mass by the relationship E=mc^2?<br /><br /><font color="yellow">Nowadays, signal phase irregularities for satellites to mobile phones are somehow proof for the 1.0 factor, while other satellitesâ€™ irregularities arenâ€™t mention in that way. </font><br />I assume you are talking about the GPS system and the satelites, which has nothing to do with E=mc^2 or even special relativity. The problem is addressed in general relativity. Time goes slower in higher gravity, and since the gravity is stronger at the surface of the earth than at the satelites this time dialation must be taken into account <div class="Discussion_UserSignature"> </div>

Status
Not open for further replies.