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• Einstein’s Most Famous Equation, E = mc2

Richard Conn Henry 2017 April 7

1

Einstein’s Theory of Relativity is simply a trivial extension, to 4-dimensions, of the famous Theorem of Pythagoras. The Pythagorean theorem itself can easily be seen to be true, just by glancing at these two (equal-area) big squares:

Obviously, the square on the hypotenuse equals the sum of the squares on the other two sides of (any) red triangle. The invention of algebra allowed us to write this famous Pythagorean theorem as an equation: ds2 = dx2 + dy2 Relativity asserts that Pythagoras’s theorem—but in our 4-dimensional universe—is: ds2 = dx2 + dy2 + dz2 − dt2 where ds is the separation in space-and-time of any two events (for example, your snapping your fingers twice, as you wave your arm around)—and asserts that ds is invariant: that is, is the same for any observer of the two events. Note the critical minus sign: this is the only way that time differs from being merely one more space dimension! How do we know that relativity is true? Because—as we will shortly see—it successfully predicts the atomic bomb! Newton taught us that, for a moving mass m, the important quantities are momentum mv and kinetic energy 1/2 mv2 So!—we will be keeping a sharp eye out for Newton’s mv and 1/2 mv2 as we explore the implications of Einstein’s remarkable pseudo-Pythagorean assertions. What Einstein specifically claimed was that ds (the separation of two events) is the same for a stationary observer, and a moving observer. In the following two diagrams, the stationary observer is you, with your x-y-z coordinate system, while the moving observer is your sister (redheart, with her x′-y′-z′ moving coordinate system): 𝑑𝜏 ≡ 𝑑𝑡!

• Einstein’s Most Famous Equation, E = mc2

Richard Conn Henry 2017 April 7

2

We are going to discover that redheart's velocity u (the red arrow) is predicted, from Einstein's Pythagorean claim, to be greater than the velocity v (the black arrow) that you find for her : dx divided by your clock's change, dt. In your stationary coordinate system redheart moves a distance dx, in time dt (by your two synchronized clocks). The two events that we are considering are redheart’s snapping her fingers at one moment in her motion (our first panel), and then snapping her fingers again at some later moment in her motion (our second panel): those two events are what is illustrated in the two panels.

Einstein's claim is that dx2 + dy2 + dz2 − dt2 = ds2 = dx′ 2 + dy′ 2 + dz′ 2 − dt′ 2 i.e, ds is invariant. In our example, with the motion being entirely in the x direction, dy′= dy = 0 and dz′= dz = 0, and so we can write Einstein’s claim as that:

dx2 - (c dt )2 = ds2 = dx′ 2 - (c dt′ )2 The c’s that I inserted (multiplying the dt’s) are to convert the units of time from seconds to meters—required so that the units will be the same for everything in the equation: but the c’s have no physics significance at all. The required value of c was long ago determined from experiment to be about 299,792,458 meters per second—very close to the the speed of light—and this number is now defined to be the exact value (just as there are exactly 3 feet in a yard). Light moves at that speed because light is massless—all massless particles, including for example gluons and gravitons, move at that identical speed, which is set by the remarkable geometry of spacetime. [If we don’t use different units for time (compared with space) then c = 1 and can be omitted from all equations entirely.] Snap test: what is the value of dx′ ? Right! dx′ = zero! For dx′ is the spatial separation of those two finger snaps in redheart’s own coordinate system—which moves right along with her! Since dx′ is zero only in redheart's coordinate system and in no other, redheart's coordinate system is special indeed—so, instead of writing the time separation (in her special coordinate system) as dt′ we choose to write (from now on) instead of dt′, dτ (the Greek letter tau). So, instead of dx2 - c2 dt2 = ds2 = dx′ 2 - c2 dt′ 2 we now have

c2 dτ2 = c2 dt2 - dx2 and this equation will now expose us to the shocking true nature of photons of light. For consider the case that redheart is a photon of light created on the sun, and absorbed (in your retina) on Earth. Distance equals velocity times time: dx = v dt, and so for light, dx = (c dt) and so c2 dτ2 = c2 dt2 - c2 dt2 = 0 and Einstein predicts that dτ which is the time, by the photon's own watch, of the photon's trip—from its creation on the sun to its ceasing to exist in your retina—is: zero! By its own clock, the photon never exists! So what did Albert Einstein think of this? Einstein (1954): "All these fifty years of conscious brooding have brought me no nearer to the answer to the question, 'What are light quanta?' Nowadays every Tom, Dick, and Harry thinks he knows it, but he is mistaken." Ready for one more major consequence of our 'trivial' postulate? Suppose our "photon" from the sun were some kind of "magical" photon that could go at speed C—just a bit faster than 299,792,458 meters per second. Well, look at our latest equation, which, for this case, would state that c2 dτ2 = c2 dt2 - C2 dt2 . On the left hand side we have c2 dτ2 and since any number squared must be positive, the left hand side of our equation is positive, yet the right hand side would be negative for any such magical photons! We not only conclude that there could be no magical photons, we also conclude that no velocities greater than that of light can ever occur: if anyone says that they can, that would be closely analogous to someone—absurdly—announcing that they can go north of the north pole. So! Speed limit is one—that’s why we introduce 299792458: otherwise, your car would go at 0.000000089 (= 60 mph). Now remember that we are seeking the famous equation E = mc2. To get that, we must launch a bunch of algebra. But it is nothing but algebra: no additional physics, of any kind. (But how does Mother Nature know algebra, which we humans (e.g. Omar Khayyam) developed? No one has any idea: but, know it Mother Nature does!) Redheart's rocket’s speed is 𝑣 = 𝑑𝑥/𝑑𝑡 as measured by us on Earth [and 𝑢 = 𝑑𝑥/𝑑𝜏 as measured by redheart herself—the very same distance dx of course, but (as we shall bring out) a quite different time interval dτ]. So, we have:

• Einstein’s Most Famous Equation, E = mc2

Richard Conn Henry 2017 April 7

3

𝑐! !" !"

! = 𝑐! − 𝑣! , so

𝑑𝑡 𝑑𝜏

= 𝑑𝑡 𝑑𝑥

∙ 𝑑𝑥 𝑑𝜏

= 𝑢 𝑣

= 1

1 − 𝑣 !

𝑐!

≡ 𝛾 = 1 − 𝑣!

𝑐!

!!! = 1 +

1 2 𝑣!

𝑐! + 3 8 𝑣!

𝑐! +⋯

That last step is a binomial expansion (invented by Isaac Newton). Since frequently the velocity v is very much smaller than c, that last term (plus all the higher-order terms: +⋯) can often be ignored. (Be sure to memorize the above definition of γ! Also, note that when v = 0, γ is one—but that as v approaches c, γ approaches infinity!) Now, return to our equation 𝑚!𝑐!𝑑𝜏! = 𝑚!𝑐!𝑑𝑡! −𝑚!𝑑𝑥! where I have now multiplied both sides of the equation by m2 (we seek E = m c2, so m, the mass of redheart (or of redheart plus her rocket) must be introduced somehow!) From that equation,

𝑚!𝑐! = 𝑚!𝑐! !" !"

! −𝑚! !"

!"

! or 𝑚!𝑐! = 𝑚𝑐𝛾 ! −𝑚!𝑢!

Notice that u = γ v [and also that dt = γ dτ] — if you insert u = γ v into the last equation, it dissolves. Try it!

𝑚!𝑐! = 𝑚𝑐 1 + 1 2 𝑣!

𝑐! + 3 8 𝑣!

𝑐! +⋯

!

−𝑚!𝑢!

𝑚!𝑐! = ! ! 𝑚𝑐! + 𝟏

𝟐 𝒎𝒗𝟐 + !

! 𝑚 !

!

!! +⋯

! − 𝛾 𝒎𝒗 ! or, if we now abandon Newton and, with Albert Einstein,

we redefine momentum to be: p = mu = γmv and total energy to be: 𝑬 = 𝑚𝑐! + ! ! 𝑚𝑣! + !

! 𝑚 !

!

!! +⋯ ≡ 𝑚𝑐! + 𝐾

(with K being the kinetic energy) then

𝐸! = 𝑐𝑝 ! + 𝑚𝑐! ! or 𝐸! = 𝑝! +𝑚! You verified all of that? OK! If redheart is not moving at all, her momentum p is zero, and we have, at long last:

𝐸 = 𝑚𝑐! ( or, E = m ) Simply our deriving E = mc2—the most famous formula in all of physics—from a tiny tweak of the Pythagorean theorem—is not enough to have us also understand how the atomic bomb results. So let’s go through that historic application of this most famous equation, so as to fully understand what E = mc2 really means in practice.

If a neutron hits a nucleus of 235U (mass mU) that nucleus decays into a nucleus of 92Kr , plus a nucleus of 141Ba , plus 3 neutrons—and those 3 neutrons can then go on to hit 3 more 235U’s, producing a chain reaction—and thus a bomb. The Uranium, if stationary, had total energy E = mU c2. The resulting Kr and Ba nuclei are observed to each have, in addition to their own internal rest energies mKr c2 and mBa c2 huge kinetic energies K. The total energy after the decay, ignoring the low-mass neutrons, is, according to our above-expounded special relativity,

EKr + EBa = (mKr c2 + KKr) + ( mBa c2 +