# Proof for E=mc^2...

by Cheman
Tags: emc2, proof
 P: 236 This is an apparent proof for e=mc^2 which i copied of the internet, but I have some problems with it: "An object that emits a ray of light will recoil at a velocity v, given by: v = EM ....................[1] where E is the energy of the emission and M is the mass of the object emitting the light. Also, the recoil distance x, of the object, will be given by: x = vt ....................[2] where v is the velocity of the object and t is time. Assuming that the beam travels a distance L, and because the light beam moves at the speed of light, the time t, taken for the beam to travel across L, is given by: t = L/c ...................[3] where c is the speed of light. Substituting [1] and [3] into [2] gives: x = (E/M).(L/c) ...........[4] Now because the center of mass doesn't move, we can say: Mx = mL ...................[5] (but I'm not sure why this is, although I know that m represents the apparent mass of the light). Now by combining [4] and [5] you should be able to get E = Mc^2, but I just can't seem to get it!" Firstly, why is v= EM? Secondly, surely the distance travelled by the light is NOT equal to L, since the object is recoiling in the other direction so this distance will be shorter? And thirdly, why is Mx= mL? Any clarification on this proof would be appreciaited. Thanks.
 P: 2 i don't understand how u get [1]. can u explain
 P: 2 i don't understand how u get [1]. can u explain
 Emeritus Sci Advisor PF Gold P: 10,429 Proof for E=mc^2... This is complete garbage. The first equation, v = Em, is not even dimensionally correct. No offense, but whoever wrote this doesn't have a clue what he/she is talking about. If you want a derivation for E=mc2, consult just about any book on special relativity. If you would like me to post one, let me know. - Warren
 P: 236 I've found this website; it has a slight alteration on some of the mathematics in the other "proof". (which I personally thought was a bit iffy when i saw it) This site even has diagrams! Lol. http://www.drphysics.com/syllabus/energy/energy.html However, there are still bits of this I don't understand: 1) Why does the momentum of light = E/c 2) Surely the distance travelled by the light is NOT equal to L, since the object is recoiling in the other direction so this distance will be shorter? 3) And thirdly, why is Mx= mL? Thanks. PS - Chroot, that would be very nice if you would please.
 P: 2,954 Cheman - Please show us this web site that you got this from. Regarding p = E/c for light. This can be deduced in a few ways. One of which is from E2 - (pc)2 = m02c4. Use this for a single photon and the fact that the proper mass, m0, is zero. That yields p = E/c. This relationship can be derived from the principles of electromagnetism by requiring both energy and momentum be conserved. I think I did this out on my website but geocities doesn't seem to be working now. Pete
 P: 236 The site link to a similar proof is there. Could somebody please help with my ques?
P: 2,954
 Quote by Cheman 2) Surely the distance travelled by the light is NOT equal to L, since the object is recoiling in the other direction so this distance will be shorter?
The distance is so small that its ignored.
 3) And thirdly, why is Mx= mL?
It represents the conservation of the center of mass of the system which must remain fixed. In fact Einstein used E = mc2 to show that the center of mass remains fixed in his paper The Principle of the Conservation of Motion of the Center of Gravity and the Inertia of Energy, A. Einstein, Annalen der Physik 20 (1906): 627-233.

Note: Einstein mentions the work of Poincare in that paper. Poincare spoke of the mass of light as does the link you provided. Einstein writes
 Although the simple formal considerations that have to be carried out to prove this statement are in the main already contained in a work by H. Poincare, for the sake of clarity I shall not base myself upon that work.

Pete
 P: 236 ok - but what does the Mx= mL exactly mean? Whats it got to do with conservation of centre of mass?
 Sci Advisor P: 905 Centre of mass location of 2 masses m_1 and m_2 is (m_1x_1+m_2x_2)/(m_1+m_2). For this not to change requires$$m_1\Delta x_1=-m_2\Delta x_2$$. Get it now? BTW, this is not a proof, but only demonstrates consistency. BTW, it has a glaring hole, as it relies on an infinitely rigid cylinder. IOW, when the light is emitted, the whole cylinder cannot instantaneously begin to move, in fact the other end only "feels" the movement after a delay equal to the time of transit of a sound wave through the cylinder material; this is very much longer time than that needed for the light pulse to traverse it. Improved versions of this gedanken experiment are available in better text books.
P: 2,954
 Quote by krab Improved versions of this gedanken experiment are available in better text books.
There is an article in the American Journal of Physics which surveys most, if not all, of these gedanken experiments. That article is

Inertia of energy and the linerated photon, Adel F. Antippa, Am. J. Phys., Vol. 44, No. 9, September 1976.

Pete
 P: 3 L could be the distance traveled in recoil, theoretically, IF the energy conveyed that caused the recoil were purely directed along the reverse vector to the direction of primary projection and IF no energy were wasted or transferred along any other axis - and of course assuming away things like friction, resistance, and shifts in gravity as the energy travels. I'm beginning to wonder about c (light speed). The E-Mc^2 is predicated on light speed being constant. NO longer sure abou that. From our perspective, sitting in our gravity well (Earth), sure, looks constant, even with the tools we have that measure to the nth decimal point. But what if you measured light speed from the surface of Mercury (if you weren't melted by being that close to the sun and stuff)? Far closer to a star and on a smaller planetary mass, would light seem to be traveling at the exact same speed. Hey NASA!
P: 3,967
 Quote by djm229 I'm beginning to wonder about c (light speed). The E-Mc^2 is predicated on light speed being constant. NO longer sure abou that. From our perspective, sitting in our gravity well (Earth), sure, looks constant, even with the tools we have that measure to the nth decimal point. But what if you measured light speed from the surface of Mercury (if you weren't melted by being that close to the sun and stuff)? Far closer to a star and on a smaller planetary mass, would light seem to be traveling at the exact same speed. Hey NASA!

Longer answer: Sitting on the surface on a neutron star, free falling towards a black hole, floating about in space, inside an accelerating rocket, inside a high speed centrifuge, the speed of light in a vacuum is always c if measured locally in a small enough region.
 P: 3 So, that being said, doesn't that mean that E=Mc^2 can't be right, because it's only RELATIVELY correct? If our tech is moved away from our location relative to the Earth and Sun etc., shouldn't it be made to deal with the shift in gravatic consequence? I just mean: are we building spacecraft that go to Jupiter and beyond with an inherant flaw due to simply trusting E=Mc^2?
PF Gold
P: 4,087
 I just mean: are we building spacecraft that go to Jupiter and beyond with an inherant flaw due to simply trusting E=Mc^2?
Why not ? Several thousand nuclear warheads have been built on the same assumption.
 P: 3 Yes, but they'll nuke us all relatively close to Earth.
P: 2
 Quote by chroot This is complete garbage. The first equation, v = Em, is not even dimensionally correct. No offense, but whoever wrote this doesn't have a clue what he/she is talking about. If you want a derivation for E=mc2, consult just about any book on special relativity. If you would like me to post one, let me know. - Warren
You are right v=Em is wrong and it cant be derived the true derevation is that v=E/Mc.

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