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Deriving E=mc² · Acedra
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Deriving E=mc²
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0:00
So, the goal of today is indeed to go back to the first paper where this equation, E equals mc squared, has been presented, and to explain how in that original presentation that equation was found.
0:18
Now, the paper in which this has been presented was published in September 1905, and it had the title, Is die Trägheit eines Körpers von seinem Energieinhalt abhängig?
0:33
Which in English means, does the inertia of a body depend on its energy content? And the word inertia there you have to interpret in current language as mass. Does the mass of a body depend on its energy content?
0:50
And the paper is really a short one. This is the entire paper. Three pages.
0:58
In this presentation, we will go through that paper and follow the reasoning that has been developed there. But before coming to that, I want to sketch why that paper was possible at all. Because it is a follow-up paper of a paper that Einstein had published a few months earlier, in which he had developed special relativity. And it is a paper that he had published a few months earlier, in which he had developed special relativity.
1:25
And this paper is a consequence of special relativity. And in order to understand how that follows, we have to have a little understanding of what special relativity was about.
1:39
So, special relativity resolved a conflict that people had realized was going on between the two major theoretical frameworks of that time, the classical mechanics developed by Newton, and the electromagnetism, which was where the theoretical foundations had been constructed by Maxwell. So, classical mechanics made it possible to understand and predict everything about the motion of the planets and the stars.
2:10
And electromagnetism, that was the theoretical backbone of the electric revolution of the 19th century. So, two very important theoretical frameworks. But they did not agree with each other. They were in conflict with each other. And there is a very simple experiment that can illustrate that conflict. So, let's imagine that you have an electrical charge, Q, a ball, an electrically charged ball with charge Q, and a bar magnet. And you hold these next to each other.
2:53
You can ask that question to a first-year student in physics or engineering. You can ask, what will happen? And the student will answer, nothing will happen. Because a static electric charge and a magnetic field, they do not interact. So, that's right, nothing will happen. But now imagine you observe, that phenomenon, from within a train that passes this scene. Now, the observer on the train is as good an observer as the person who holds these two objects in his hands.
3:38
So, you can do the physics from within that moving train. What does the person in the train see? Well, that person sees this magnet and charge passing from the other direction. Something like this. But now, this is not a static phenomenon anymore. For this observer, what this observer actually sees is a moving charge. So, a current and a magnetic field. And a current and a magnetic field, they do interact. So, we have a conflict here. For one observer, nothing happens.
4:19
And for the observer in the train, they should interact and something should happen. Obviously, only one of the two can be right. So, which one is right? Special relativity was meant to solve that problem. And the way in which it was solved is that the transformation from the axis system for the observer at rest to the observer in the train is not the simple Galilean transformation. As was used in classical mechanics up to now, up to that time. It was the Lorentz transformation.
4:58
Which mixes time and space. And which is responsible for funny effects like the length contraction, the time dilatation. So, size, length and time, they do depend on the speed of the observer. Now, these Lorentz transformations, they also mix electric and magnetic fields in this way. So, the electric field that is observed by the person in the train is a combination of the electric field that the static observer observes plus a term that depends on the magnetic field for the static observer.
5:40
And the same for the magnetic field seen from the train. That's a mixture of the static magnetic field and the static electric field. So, the magnetic field that the observer in the train sees will be constructed in such a way that it counteracts the tendency of the magnet to move. And the result will be that they don't move at all. So, both observers will see the same. There is no conflict. But you have to go through the Lorentz transformations to see that there is no conflict.
6:16
So, all of that was published a few months before this paper came. So, let's work out the reasoning in that paper. We will observe a system, and I will tell you which kind of systems that can be, a system that will be in an excited state. And what could that system be? It could be, for instance, a cup of hot water. Cold water has less energy content than hot water. So, the hot water is the system in the excited state. And the excited system can de-excite,
7:01
can go to the lower state by emitting infrared radiation. Or, our excited system could be an excited atom that can de-excite by emitting visible radiation. Like in these sodium lamps that are used to light the streets. Or, the excited system can be an excited nucleus that can de-excite by emitting gamma radiation, in radioactive decay. So, three examples of excited systems that are observed by a static observer. So, there is a person standing next to the hot water and looking at the hot water.
7:45
So, that system is in its excited state, indicated here by the 1. And the energy of that system, we call it E. That's the energy of the excitation. So, the ground state of the system has an energy 0. My cold water has an energy 0. My excited water has an energy E.
8:11
That system will emit radiation, infrared visible gamma with a certain wavelength, like in this water picture. And so, we can represent that by the electromagnetic spectrum. If this is hot water, it will emit this infrared radiation. The atoms will emit visible lights. The nuclei will emit gamma rays. We will make, a bit of bookkeeping, we will take track of the internal energy of the system, the kinetic energy of the system, and the total energy of the system,
8:56
where the total energy is just the sum of the internal energy and the kinetic energy. And the static observer will tell, well, my internal energy, that's E. The system is E above the ground state. The system is at rest, so it has no kinetic energy. So, the sum of these two is E.
9:21
Now, we will move to a different observer, an observer that is in a spaceship, that travels at a particular speed, V. And we will look at a situation where the system has come back to its ground state. So, the system is here in that zero state. What is now the energy bookkeeping? The internal energy is zero. And, seen from the observer in the spaceship, that system is moving at a speed V, the same speed as the spaceship has. Therefore, the kinetic energy of that system is mv squared over two.
10:04
And the total energy, is the sum of these two, is mv squared over two.
10:11
What we will examine now, is what is the difference between the total energy seen by the static observer for the excited system, and the total energy seen by the moving observer for the de-excited system. You may wonder, why are you interested in that property? And, okay, I agree, it's a weird question. But, I hope you can agree with me, that it is a question that, in principle, can be answered. It's a legitimate question to ask. So, we will try to find that difference.
10:53
How do we do that? Well, we will work in two steps. We will first let the static observer observe the decay of the system. And then, we will look at the de-excited system by the moving observer. And in these two steps, we will examine what is the change in total energy. If the system has decayed, its internal energy is now zero. For the static observer, it's still at rest, so kinetic energy is zero, and therefore, the total energy is zero. The change in total energy between these two situations
11:35
is the total energy in the final situation, minus the total energy in the original situation, so zero minus E. The change in total energy is minus E.
11:50
Then, we make the step from the static to the moving observer. We have already all the numbers. So, what is the change in total energy? Total energy in the final situation, minus the one in the initial situation, that's mv squared. So, the energy difference between this situation and this situation is the sum of these two. Fair enough, but we can also calculate this in a different way. We can first bring a change to the moving observer, who sees the excited system,
12:32
and we can then have the decay observed by the moving observer. So, let's calculate these energy differences too. What is the internal energy seen by the moving observer for the excited system? That's still the same E. Now, that system has a kinetic energy, because it is moving from the point of view of the moving observer, and therefore the total energy is E plus mv squared over two. The change in total energy in this step is this expression minus this expression, so mv squared over two.
13:16
And in this second step, vertical, the energy difference between mv squared over two minus E plus mv squared over two, that's minus E. Of course, whichever route you take, that must be the same energy difference, because the two reasonings, they give you the energy difference between this and this situation. So, we have here everything summarized. I add here a label classical, because so far we just applied Newtonian reasoning. Newton would have agreed with that, with everything we said here.
13:57
And so, in this classical picture, these two parts should give the same energy difference. They should give the same energy difference in every reasoning. So, I write that down. The first route and the second route must give the same energy difference. And we had calculated these energy differences. That was, at the left-hand side, our first reasoning, the second reasoning at the right-hand side. And obviously, these two expressions are exactly the same. That's trivial.
14:31
But this was for a classical derivation. So, let's now do this in a relativistic derivation. What will change here? Not much. There will be one step where it changes, and that is observing the decay, observing the emitted photon. Now, characteristic for the photon is its velocity. Wavelength, or frequency. And, well, I have here a sine function that illustrates the wavelength of one photon. And I draw it to draw your attention to the time axis. So, this would be a sine with a wavelength
15:22
of just one on this. Well, let me say it correctly. The sine function reaches the same value after one second of time has passed. So, the frequency is one. But time, in a relativistic picture, depends on the observer. If the observer moves, time can run more slowly. So, the same photon observed by an observer where time runs slower could have a frequency that is maybe twice as large, because there are two full periods in one second. So, the observed frequency will depend on the speed
16:14
of the observer by this time effect in special relativity. And, thanks to his first paper of a few months before, Einstein could calculate how this change was, and the result is that the change in energy here, which is the energy that is carried away by the photon, that this change in energy is this value. We will not derive that here. We take that as a given. That is a result that can be found in the previous paper. So, if the observer would not be moving, v equals zero,
16:55
then this is just minus e, just as we had here. But, if the observer is moving, then the frequency is a bit different. Now, we will be paranoic a bit, and we will give a few of the symbols we have used so far, we will give it, well, two different meanings, just to be on the safe side. So, I will call the mass of the system in the excited state, n1, so this energy difference here will be m1 v squared over 2. The kinetic energy seen by this moving observer will be m1 v squared over 2,
17:39
and so that appears also here in the total energy. And the mass of the de-excited system I will call m0. So, everywhere in the lower half of the slide you see m0, in the upper half of the slide you see m1. I know, of course, why we are doing that, because I know the result of this story. But, in case this would not be needed, we will see it in our later derivations that these two symbols will become equal again. So, it is innocent so far. We don't play any trick, we are super-cautious,
18:19
but we don't introduce something that could drive our reasoning into a bias, in a biased way. Good. We will play the same game as before. We will write the energy difference seen in the first way, so sum of these two terms, and that should be identical to the energy difference observed in the second way. And now we have these expressions. So, we have this factor here due to observing the photon by the moving observer. I can simplify this a bit. The e goes away at both sides.
19:09
The v squared is everywhere present and can be divided away. And the factor 1 half is everywhere and can be divided away. And the result of this expression is that the mass of the expression of the excited system minus the mass of the de-excited system is the internal energy of the excited system divided by c squared. So, here we see that giving these two names to these two masses was necessary, because if these two masses would have been identical, then that's only possible
19:49
if the system has an energy zero, so if the system stays forever in its ground state. So, you are forced by the formalism of special relativity to introduce these two different values for the masses. And if you now write that expression in its common form, and you call this mass difference delta m, then you have e equals delta m times c squared. So, now we can answer the question that Einstein asked in the title of that paper. Does the mass of an object depend on its energy content?
20:28
Is the mass of an excited system different from the mass of the system in its ground state? And the answer is yes, and the mass difference is given by this expression.
20:42
In the paper that is formulated in the following way, so the sentence where e equals m c squared is introduced for the first time in the green sentence. Gibt dein Körper die Energie L in Form von Strahlung ab, so verkleinert sich seine Masse um L über v². So, in English, this equation has the following immediate consequence. If a body gives off the energy e in the form of radiation, then its mass diminishes by e over c squared.
21:20
And if you understand that, you will understand this kind of cartoon that you may find sometimes on the web, Einstein stating I didn't write e as m c squared, I wrote if a body gives off the energy L in the form of radiation, its mass diminishes by L over c squared. Which is indeed two different ways of saying the same thing. But he never wrote in the first paper e equals m c squared. He had different notation and it's only a few years later that the traditional notation got into use.
21:58
Now you may wonder is this just theoretical playing or does this have actual consequences? Well, of course this expression has actual consequences and let's see how large they are. So let's take the practical example of one kilogram of water which we keep just at well, a little a marginal fraction above zero Celsius, so we will do it in Centigrade, not in Kelvin, so it's still liquid water. One kilogram of liquid water and we compare it with the same amount of water that is boiling.
22:42
According to what we just derived, there should be a different mass for the boiling water than for the cold water and that mass difference is 3.8 nanograms. So very, very small of course nobody has experimentally detected that before somebody made the derivation. This would always have been within the error bar of the mass measurements.
23:12
If you do it with atoms then it becomes a bit different. So let's take one kilogram of hydrogen atoms and compare that with the same amount of hydrogen atoms where the electron and the proton are fully dissociated. So these are highly excited hydrogen atoms. You have a plasma of protons and electrons. So that is the high energy state. That high energy state will be 14.5 micrograms more massive than the low energy state. 1000 times more but still a very small quantity.
23:54
It becomes more measurable when you go to radioactive decay. So let's take one kilogram of barium, krypton and neutrons, which are the fission products of uranium-235 and compare that to the uranium-235 that was needed to produce this one kilogram of fission products. In that case it's 0.79 grams which is the mass difference. That is already sizeable. And the largest effect, which is a bit hypothetical, would be the one if you take one kilogram of protons, which are made of
24:38
two up quarks and one down quark and you would disentangle every proton into isolated free quarks. Which is impossible. You have a phenomenon of color confinement that inhibits this. But if you would have the same amount of free quarks then the difference would be more than 100 kilograms. Because the strong force between the quarks that strong force is so strong that it represents an enormous amount of mass. If you let free quarks interact and form a proton, an enormous amount of mass is
25:20
converted into the energy of that interaction.
25:26
So you see that it is an expression with sometimes small effects, but in some cases at least experimentally detectable effects. So I leave you with this summary slide. I hope that now you know where that expression has been presented the first time, in which context it played. And I'm sure you will have questions. I'm also sure that I probably will not be able to answer these questions, because this is not really my field. But at least I could have a try.
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