Einstein's Derivation of E=Mc^2
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- Опубликовано: 21 июл 2023
- Einstein's original thought experiment and derivation that a photon has (inertial) mass equal to its energy (E=Mc^2). Taken from his original 1905 paper: "Does the Inertia of a Body Depend on its Energy-Content?"
Original paper (translated to English from German): www.fourmilab.ch/etexts/einst...
Just as a historical note, Thompson, Lorentz, Poincaré, et al. had a bunch of theories about "electromagnetic mass", "transverse mass", "longitudinal mass", etc. which were a complicated mess and... wrong, so they don't get mentioned very often in undergrad physics classes.
A funny thing about E=mc² is that Thompson almost derived it from modeling an electic charge moving through a dialectric medium. And Poincaré derived E=mc² just from the recoil of light bouncing around inside a box of perfect mirrors. It's basically this same setup as Einstein used in this video. Everyone forgets about Poincaré. (Actually, a *lot* of stuff in Special Relativity is the same as it is in the Lorentz-Poincaré flavor of aether, but Einstein could do it without requiring an aether. Length contraction was Fitzgerald's idea, time dilation is Poincaré's, "Relativistic mass" had been experimentally measured with cathode rays by Kaufman in 1904, etc. etc. Nobody working in the field was surprised by Einstein's results.)
Thanks for the historical info. As usual, and in accordance with Stigler's Law of Eponymy, only the person who finds the last piece of the puzzle is remembered by history...
super
Duper!
I think it's interesting that the momentum goes to zero. If the momentum goes to zero, doesn't that mean the photon goes in a straight line, without an up and down component? That's where sin lives.
It’s not that the momentum goes to 0. It starts off as 0 since the box is initially at rest. Since there are no external forces, momentum is conserved. So it must remain 0 at all times.
Yes the photon goes in a straight line… as long as the observer is not accelerating or there is no gravitational field. More specifically, we call this an inertial observer.
@@physicsalmanac there is a conundrum that I am having here. In a unit circle triangle, if one leg is 0, the other leg is 1. The two legs are energy and momentum, mc^2 is the hypotenuse (momentum is on the wrong side so it is a jacketed imaginary value). If momentum can be 0, why can't it be 1 and energy be 0? This would imply v=c. Also, isn't this contrary to everything we are being told about energy?
Also, I suppose E= ± m*c^2*sqrt(1-(v/c))^2, factoring out the c in the square root, E= ± m*c*sqrt(c^2-v^2). If we multiply everything by gamma and it is greater than 1, the speed of light also increases. Sorry, you must get a lot of this confusion.
I don’t really follow… why are you multiplying the energy by gamma? And why would that increase the speed of light?
Energy = 0 means there’s nothing there. So you can’t have non zero momentum and zero energy.
@@physicsalmanac E^2=(m*c^2)^2+(p*c)^2
Oh, I made an error in my previous post. I need to be more careful.
E^2=(m*c^2)^2+(m*v*c)^2
m*c^2 looks like a dotproduct, so I thought it might have been the hypotenuse. Solving for E^2 under this equation (instead of the other one I used).
±sqrt(c^2+v^2)*m*c
So, looking at that solution, it says at v=0 it is m*c^2. At v=c, it is m*c^2*sqrt(2).
So, I take E^2 and make it into a unit vector for:
1=(c/sqrt(c^2+v^2))^2+(v/sqrt(c^2+v^2))^2
The main reason I thought mc^2 was the hypotenuse is because of gamma.
1/sqrt(1-(v/c)^2)
Factor out the c.
c/sqrt(c^2-v^2)
cos(arctan(v/c))=c/sqrt(c^2+v^2)
cos(arctan(i*v/c))=1/sqrt(1-(v/c)^2)=c/sqrt(c^2-v^2)
I just thought that if I'm multiplying mass by gamma, maybe there should be an 'i' in the original E^2 equation. It looks like something that can be overlooked. I think if normal E^2 was in the numerator, and the 'i' version of E^2 was in the denominator, wouldn't you have tanh(2*arctan(i*v/c))?
The momentum in that energy equation is defined with respect to proper time. So p = gamma*mv. as v goes to c, E and p go to infinity. A better way to think of it is if p is very large m becomes insignificant and you can say E = pc. This is the ultra relativistic limit.
Since when is it known that photons carry not only energy but also momentum?
Since 1901 and the Nichols Radiometer experiment.
It was already known that light carried momentum since the development of Maxwells equations (mid 1800s). But this is the wave theory of light. For a single photon (particle theory of light) this was developed by Einstein earlier the same year (1905). Regarding this thought experiment, in the actual paper Einstein looked at it from an energy point of view. I modified it slightly because in his version the math and physics is a bit more advanced and requires some knowledge of special relativity.
Yo "Physics Almanac" would you like to work with me on physics?
Maybe. What do you have in mind? What do you want to work on?
@@physicsalmanac You know.... something odd - I'm tryna mix Quantum Computers and AI together
You’re working on that now?
@@physicsalmanac yup