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Q & A: E=mc^2

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Most recent answer: 10/22/2007
Q:
Dear Sir: I am an armchair physicist with a Masterís degree in Metallurgical Engineering. My question pertains to Einsteinís famous E=mc2 equation: Why doesnít E = ONE-HALF mc2, which is the sum of the momenta of all of the particles moving at the speed of light? Any elementary physics student knows that kinetic energy (which, when 100% kinetic, there is zero potential energy) = the sum of the momenta of the masses moving at velocity v. In other words, the integral of mv of the masses is 1/2 mv2. Wouldnít it follow that when all of a given massís energy is completely converted to energy, that it is ALL kinetic energy -- whose sum of all of the elementary constituent particles moving at the velocity of light would equal 1/2 mc2. In light of string theory, I would think this would be an obvious result. Please tell me why I am wrong -- why am I off by a factor of 2? I have asked physics professors at a number of universities, who give me typical BS answers as "well, itís just a constant of the universe" (it isnít), or "itís been proven experimentally" (it hasnít -- even man-made nuclear explosions have resulted in only a few percentage points of the entire mass being converted to energy, and how do we absolutely know that high energy matter-antimatter annihilations have resulted in mc2 rather than 1/2 mc2?). If you can explain the answer, I am intelligent enough that you can explain it in technical terms. If you cannot explain the answer, it this a bit of original thinking that has not been previously addressed? Sincerely, Joseph
- Joseph Felice (age 51)
Engineering Consultant , Chicago, Illinois, USA
A:
It's true that in elementary physics courses one is taught that E=pv/2, where E is kinetic energy, v is speed, and p is momentum. The problem is that this equation is an approximation, only valid for speeds low compared to c, the speed of light.

Why do we believe instead the special relativistic expression E=mc^2 (where E= total energy and m is the mass which appears in the relation p=mv)? There are many reasons, but perhaps the most compelling are the direct experimental tests. I'm not sure why you don't believe the enormous number of precise matter-antimatter annihilation results. In addition there are many, many collision results which fit the relativistic mass-energy-momentum relations but not the Newtonian ones. Even such mundane devices as high-power electron microscopes cannot be properly designed without making relativistic calculations.

As for the distinction between kinetic energy and rest energy, it does tend to blur away when you consider the different levels of constituents of some material, just as you suggest. Regardless of such verbal labels, however, the real connections between the quantities follow relativistic rules and not the Newtonian approximations.

Mike W.

p.s. I f you have some specific reason to doubt all the results of experiments involving high-energy particles, we'd be happy to continue this discussion, since we want to be brief but not blow anybody off with mere B.S.

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Here are some enticing graphics of high-energy data to whet your appetite, illustrating the idea of mass and energy interchangeability in high-energy physics experiments.

The first is a picture of the rate at which collisions between electrons and positrons will happen as a function of how much total energy there is in the collision. These experiments have nearly all been performed in a laboratory system in which the e+ and e- have equal and opposite momenta, but the reactions have to happen according to any observer, moving no matter how fast relative to the laboratories. The fact that there is structure in this plot indicates that interesting things are going on. The peaks in the rate correspond to tuning the kinetic energies of the beams so that the total energy of the collision matches mc^2 of some exciting particle, like the Z, or one of the Upsilon system, or a J/psi or other psi particles, or the rho or the omega (you don't have to know much about any of these things, other than that they have masses and you can produce them if you had an e+e- collision with enough energy). These data come from which compiles the results of many different experiments.



The flip-side of the coin is that the total energy-momentum four-vectors of the decay products of a particle can be added up to calculate the mass of the parent particle. Below is a histogram of masses calculated from combinations of momenta of particles which are produced in very high-energy proton-antiproton collisions in the CDF experiment at Fermilab. A requirement is made that these particles had to be produced at some distance away from the main interaction point, or there would be too much random stuff obscuring the results. (unlike e+e- collisions, p-pbar collisions do not involve using all the energy of the incoming proton/antiproton pair. Only one quark collides at a time, or a gluon, and the rest of the proton/antiproton just goes down the beampipe). These are peaks corresponding to the lowest-mass charm mesons, and the mass of a particle does not depend on how fast it is going (this is a more modern convention -- we speak of a particle's mass like many people speak of its "rest mass". If we want to talk about something which varies with momentum, we'll call it "energy" -- if we insist that E=mc^2 all the time, one of our words, be it "mass" or "energy" becomes redundant. Instead, particle physicists treat m as a constant, and the real formula is E^2 = (mc^2)^2 + (pc)^2, where p is the momentum.)


These masses are calculated with the above relationship E^2 = (mc^2)^2 + (pc)^2, taking advantage of the fact that we know the rest masses of the decay products and their momenta. The fact that the peaks are found in these plots indicates that we get the same answer all the time for the invariant mass of the charm meson (or, in the e+e- collisions above, the resonances). The mesons in the second case are produced with all different speeds and directions, and it is the success of the relationship between energy and mass which allows us to say something about the mass of the charm meson regardless of how fast it's going. We know about the masses of the decay products by measuring the relationship between the momentum and kinetic energies in a mass-spectrometer-like apparatus. We can give a particle a well-defined kinetic energy by allowing it to travel a fixed distance in an electromagnetic field, and we can measure the momentum of the particle by measuring the radius of curvature of its path in a known magnetic field. All of the calculations of energy, mass, and momentum match the observed data only if the special relativistic formulas are used.

Tom

(published on 10/22/2007)

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