From: Chris (Avatar) 12/09/00 13:24:45 Subject: Deriving Einstein's mass-energy relation post id: 133203 This posting is in response to the current threads about Einstein's mass-energy relation and its place in SRT. This is intended merely to be informative, not argumentative, so please only post questions about the derivation here. Other argument postings should be addressed to the particular thread to which they refer (for clarity). Deriving Einstein's mass-energy relation is no easy exercise. If you start with just mass you're going to flounder. I think the point here is that the mass-energy relationship doesn't necessarily follow straight from the Lorentz transform, but it is within the predictive ability of SRT. There are a few ways of deriving E=mc2, some much more elegant than the one below. I present this one because it is the most "nuts and bolts" and requires the least fancy math (it does require some understanding of SRT and also some basic integral calculus). For the working, you can skim it by just following the yellow thread. Additional working is provided in red, and comments are in white (note I'll use the symbol åa®b f(x) for the integral of f(x) from a to b - I don't have an integral symbol): I start with a relation for kinetic energy (KE) as being produced by the action of a force (F) over a distance (s): KE = åo®s F.ds The force F may be defined in terms of the impulse it imparts (since we're concerned with Kinetic energy here): F = d(m.v)/dt (Note that this is just the relativistic form of Newton's second law F=ma). KE = åo®s d(mv)/dt . ds KE = åo®mv v.d(mv) Substitute the relativistic expression for m in terms of mo: KE = åo®v v.d(mo.v / Ö [1 - v2/c2]) Now we perform the integral (those who want to work through the integral continue through the red section, others may skip to the next yellow bit): Integrate by parts: åx.dy = xy - åy.dx Now KE = mo.v / Ö [1 - v2/c2] * v - åo®v (mo.v . dv / Ö [1 - v2/c2]) KE = mo.v2 / Ö [1 - v2/c2] - mo.åo®v (v.dv / Ö [1 - v2/c2]) åo®v (v.dv / Ö [1 - v2/c2]) = - (c2 - v2) / Ö [1 - v2/c2] |o®v = - c2 (Ö [1 - v2/c2] ) |o®v = - c2 Ö [1 - v2/c2] + c2 Substitute into KE expression to obtain: KE = mo.v2 / Ö [1 - v2/c2] + mo ( c2 Ö [1 - v2/c2] - c2 ) From: Chris (Avatar) 12/09/00 13:26:25 Subject: re: Deriving Einstein's mass-energy relation post id: 133206 Integrated expression for KE: KE = mo.v2 / Ö [1 - v2/c2] + mo.c2 Ö [1 - v2/c2] - mo.c2 KE = {mo.v2 / Ö [1 - v2/c2] + mo.c2 Ö [1 - v2/c2] } - mo.c2 The part in these brackets { } reduces to mc2. To work through the reduction, read the next red bit - or skip through to yellow: { } = mo.v2 / Ö [1 - v2/c2] + mo.c2 Ö [1 - v2/c2] = mo.v2 + mo.c2 [1 - v2/c2] / Ö [1 - v2/c2] = mo ( v2 + c2 [1 - v2/c2] ) / Ö [1 - v2/c2] = mo ( v2 + [c2 - v2] ) / Ö [1 - v2/c2] = mo ( c2 ) / Ö [1 - v2/c2] = mc2 …… for m = mo / Ö [1 - v2/c2] (relativistic mass expression) Now substitute back into KE expression: KE = { mc2 } - moc2 mc2 = moc2 + KE Which is equivalent to the energy expression: E = Eo + KE which tells us that an object's energy is its initial or rest energy plus energy of motion (KE). Equating, we see that: E = mc2 and Eo = moc2 Which gives us the total energy in terms of relativistic mass, and then the rest energy in terms of rest mass. For the full mass-energy relation, start with the Lorentz transform on a mass m: m = mo / Ö [1 - v2/c2] Square both sides: m2 = mo2 / [1 - v2/c2] Multiply RHS by c2/ c2: m2 = mo2c2 / [c2 - v2] Rework: m2 [c2 - v2] = mo2c2 m2c2 - m2v2 = mo2c2 Substitute relativistic momentum p = mv: m2c2 - p2 = mo2c2 m2c2 = mo2c2 + p2 Multiply through by c2: m2c4 = mo2c4 + p2c2 Substitute in E=mc2 (worked above): E2 = mo2c4 + p2c2 Done. Note that in deriving this equation from the Lorentz transform on mass, we needed to already know that E=mc2 in order to (a) multiply through by c2 and (b) substitute in E at the end. We relied on the first derivation to furnish the full equation of the second derivation. This shows that the mass-energy relation is not a necessary or simple result of a Lorentz transform (or a mass in motion). Hope this helps! Chris From: Robert ® 12/09/00 13:33:00 Subject: re: Deriving Einstein's mass-energy relation post id: 133215 Note that in deriving this equation from the Lorentz transform on mass, we needed to already know that E=mc2 I thought so - no wonder it was so hard - I could get (mc2)2 on the LHS but I needed the E=mc2 relation to get the E in. Thanks for that, anyhow. So the statement: KE = indefintegral(p . dt) is false? From: Chris (Avatar) 12/09/00 13:45:24 Subject: re: Deriving Einstein's mass-energy relation post id: 133230 Yes. :o) From: Chris (Avatar) 12/09/00 14:41:13 Subject: re: Deriving Einstein's mass-energy relation post id: 133308 Hi Nat This post is all about Einstein's famous equation E=mc2. This equation says that there is some equivalence between mass and energy. What I've done in this post is derive the equation so that interested people with enough math background can see where it comes from. Unfortunately that means that some people won't get much benefit from it because they don't have the math. I guess the best you can get from this at the moment is that the equation has real meaning, and takes a good understanding of math and relativity to derive. Also (if you wanted to) one day you may be able to work through this for yourself. Cheers Chris This forum is un-moderated. The views and opinions expressed are those of the individual poster and not the ABC. The ABC reserves the right to remove offensive or inappropriate messages.