# Thread: How did Einstein connect mass and the speed of light?

1. Order of Kilopi
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## How did Einstein connect mass and the speed of light?

Is there a laymans explanation, that could vaguely answer that?  Reply With Quote

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It's easy and hard.

1) Assume the speed of light is constant, this is the definition of the theory of special relativity.
2) Assume that the center of mass of a system doesn't suddenly start moving. This is the conservation of momentum. If you pick a reference frame where the center of mass is stationary, it stays stationary even if the system explodes.

Done. All you need now is a system with mass and light. The rest as they say is algebra. That's the hard part because you need to work in two reference frames, which brings out the Lorentz transform. Not a newbie friendly exercise.

The system I heard of is a lightbulb in a box. The lightbulb emits a photon. But photons have momentum, so the box must recoil while the photon is travelling. Once absorbed by the wall of the box, the box comes to rest again. So tell me, how does a center of mass which can't move suddenly shift over? It doesn't! Because the energy carried by the photon is equivalent to carrying mass. Sure the box moved, but the center of mass didn't. Tada!

When you do the math, you get E2 = m2c4 + p2c2, where p is momentum.

Let E=A, mc2=B and pc=C . The equation become A2 = B2 + C2. And when C=0, then A=B (or E=mc2). Do you recognize it as a right angle triangle? This means the energy from mass and the energy from momentum 'add at a right angle' instead of just plain adding up. Which means they are independent of each other. In other words, changing momentum doesn't change mass and vice versa.
Last edited by ShinAce; 2017-Jan-28 at 12:41 AM.  Reply With Quote

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Ok, thanks.........bit beyond me I suppose, but nice to see the link process.  Reply With Quote

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2,144 Originally Posted by Frog march Ok, thanks.........bit beyond me I suppose, but nice to see the link process.
Which part is beyond you?

Here's a longer description of the same derivation:
Last edited by ShinAce; 2017-Jan-28 at 11:31 AM.  Reply With Quote

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That's a remarkably cute and simple derivation that if a system contains some localized energy E within it, and that E moves as a photon to somewhere else in the system, then to understand the behavior of the center of mass of the system, we need to imagine that the energy E is associated with a mass E/c2. Since the center of gravity is obeying that rule, it suggests that gravity will also, i.e., that a localized E produces a gravitational mass E/c2. It also suggests that mass and energy are in some sense equivalent, so it's a very powerful gedankenexperiment!  Reply With Quote

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1,814 Originally Posted by Frog march Ok, thanks.........bit beyond me I suppose, but nice to see the link process.
Frog March, let's match a few things up here. Take the first few words of your first post and match them up with the first few words of ShinAce's post.
You asked for a layman's explanation and ShinAce mentioned, "It's easy and hard." What is easy is every thing he mentioned after that. What is hard and complicated is a layman's view.  Reply With Quote

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1,519 Originally Posted by ShinAce It's easy and hard.

1) Assume the speed of light is constant, this is the definition of the theory of special relativity.
2) Assume that the center of mass of a system doesn't suddenly start moving. This is the conservation of momentum. If you pick a reference frame where the center of mass is stationary, it stays stationary even if the system explodes.

Done. All you need now is a system with mass and light. The rest as they say is algebra. That's the hard part because you need to work in two reference frames, which brings out the Lorentz transform. Not a newbie friendly exercise.

The system I heard of is a lightbulb in a box. The lightbulb emits a photon. But photons have momentum, so the box must recoil while the photon is travelling. Once absorbed by the wall of the box, the box comes to rest again. So tell me, how does a center of mass which can't move suddenly shift over? It doesn't! Because the energy carried by the photon is equivalent to carrying mass. Sure the box moved, but the center of mass didn't. Tada!

When you do the math, you get E2 = m2c4 + p2c2, where p is momentum.

Let E=A, mc2=B and pc=C . The equation become A2 = B2 + C2. And when C=0, then A=B (or E=mc2). Do you recognize it as a right angle triangle? This means the energy from mass and the energy from momentum 'add at a right angle' instead of just plain adding up. Which means they are independent of each other. In other words, changing momentum doesn't change mass and vice versa.

That's a very accessible explanation  Reply With Quote

8. Originally Posted by ShinAce The system I heard of is a lightbulb in a box. The lightbulb emits a photon. But photons have momentum, so the box must recoil while the photon is travelling. Once absorbed by the wall of the box, the box comes to rest again. So tell me, how does a center of mass which can't move suddenly shift over? It doesn't! Because the energy carried by the photon is equivalent to carrying mass. Sure the box moved, but the center of mass didn't. Tada!
I hope I'm close to the "Tada", but I could use a little help getting there.

Here is the picture I have. Let's just us a single atom for our light source centered in the box. It spits a photon causing the atom to shift very slightly yet in accord with m1v1=m2v2, momentum conservation intact. Then the photon when absorbed by the wall will cause the wall to move slightly.

Is the wall's mass part of the story, or is it anchored solidly so that the c.g. story is only that of the atom and the photon? I would guess not since an absorbed photon (by the wall) would remove our photon and we would be left with a moving atom, and a moving c.g. So the box must be part of the story, too. Ok, I guess that makes sense even if the photon bounces and causes 2x the wall's movement, and even if the reflected photon happens to get reabsorbed by the atom, causing it to move quicker. Things would be more dynamic but the c.g. would be the same if we are looking (reference frame) outside the box. Or am I "outside the box"?  Reply With Quote

9. Yes, the box is included in this. And for the original thought experiment, the light source in the center is anchored to the box. So when the photon is emitted, it's not just an atom in the middle that recoils, it's the whole box. And yes, for this to work, you have to assume that not only does the photon have an equivalent energy while it's moving, that mass ends up wherever the photon is eventually absorbed.  Reply With Quote

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To be clear, you only have to assume the energy moves from place to place. That there is an associated movement of effective mass is a conclusion of the gedankenexperiment, based in the idea that the c.o.g. cannot move. So in this light, we can say that energy is "equivalent to mass," and this arises as a necessary conclusion rather than an input assumption.

Indeed, a useful way to think of the gedankexperiment is a general way to derive an effective mass associated with any way to move energy. So let's say you had no concept of the mass of a particle, you only have a concept of the macroscopic mass M of a rigid box, and you have the concept of the velocity v and momentum p associated with carrying some energy E from one side of the box of length L to the other. You can always attribute an effective mass to that energy carrier, simply by figuring out how far the box moves (to keep total momentum zero), and saying the c.o.g. of the whole system must not move when the enegy moves a distance L (all done in the limit that the energy is very small). That gives you an inferred effective mass m that should be associated with the energy that moved from place to place via:
m = Md/L
where M is the mass of the box and d is how far it moves. But from conservation of momentum, we know that Md/t = p where t=L/v, so d = pL/Mv, and so:
m = p/v
to lowest order. What this means is, as soon as we define p as mass times velocity for the macroscopic box, we immediately get the need for a mass concept that applies to any element of the system that moves at v and carries momentum p-- even photons! For photons, we know that p=E/c, so that mass concept obeys m = E/c2, in the limit of small E. In other words, the concept of "relativistic mass" comes entirely from conservation of momentum and the definition of the center of mass.
Last edited by Ken G; 2017-Feb-02 at 05:41 PM.  Reply With Quote

11. Originally Posted by Grey Yes, the box is included in this. And for the original thought experiment, the light source in the center is anchored to the box. So when the photon is emitted, it's not just an atom in the middle that recoils, it's the whole box.
Ok, that would be the simpler -- einfacher gedankenexperiment -- approach. I don't think it matters as far as a steady c.g., but it sure helps to restrict the emitter.

And yes, for this to work, you have to assume that not only does the photon have an equivalent energy while it's moving, that mass ends up wherever the photon is eventually absorbed.
I think this is simple enough, but in case I'm missing something, whatever atom absorbs the photon will normally bump an electron to a higher energy level, so can we use either mass or energy for the wall's increase, right? However, wouldn't the momentum exchange still be the better thing to watch if are other eye is on the c.g.?  Reply With Quote

12. Originally Posted by Ken G To be clear, you only have to assume the energy moves from place to place. That there is an associated movement of effective mass is a conclusion of the gedankenexperiment, based in the idea that the c.o.g. cannot move. So in this light, we can say that energy is "equivalent to mass," and this arises as a necessary conclusion rather than an input assumption.

Indeed, a useful way to think of the gedankexperiment is a general way to derive an effective mass associated with any way to move energy. So let's say you had no concept of the mass of a particle, you only have a concept of the macroscopic mass M of a rigid box, and you have the concept of the velocity v and momentum p associated with carrying some energy E from one side of the box of length L to the other. You can always attribute an effective mass to that energy carrier, simply by figuring out how far the box moves (to keep total momentum zero), and saying the c.o.g. of the whole system must not move when the enegy moves a distance L (all done in the limit that the energy is very small). That gives you an inferred effective mass m that should be associated with the energy that moved from place to place via:
m = Md/L
where M is the mass of the box and d is how far it moves. But from conservation of momentum, we know that Md/t = p where t=L/v, so d = pL/Mv, and so:
m = p/v
to lowest order. What this means is, as soon as we define p as mass times velocity for the macroscopic box, we immediately get the need for a mass concept that applies to any element of the system that moves at v and carries momentum p-- even photons! For photons, we know that p=E/c, so that mass concept obeys m = E/c2, in the limit of small E. In other words, the concept of "relativistic mass" comes entirely from conservation of momentum and the definition of the center of mass.
Ah ha! "That there's" a good explanation. I read Brian May's E=mc^2 book and all of sudden, late in the book, E=mc^2. I had missed the explanation expecting something different, I suppose.

Now I understand Grey's post a little better. I expected something far more cryptic.

Thanks to both of you.  Reply With Quote

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