E=mc^2 Explained: Momentum and c

In summary, the conversation revolved around a proof of E = mc^2 involving the equation E = Pc. The question raised was how this equation could make sense if energy is typically calculated as force times distance, and where the c in this equation comes from. The conversation then delved into discussing the physical assumptions being made and the errors that may have been made in the proof. The concept of the fulcrum in the proof was clarified and it was determined that the center of mass of the system is what is actually moving, not a physical fulcrum. The use of E = Pc for a photon was also questioned, and it was suggested that it may be an empirical result from quantum mechanics. It was also mentioned that this equation
  • #1
guss
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0
In this proof of E = mc^2: http://www.youtube.com/watch?v=g1_Z7UZWmVQ

At around 5:40, he says that E = Pc. I understand that this means that energy is equal to the momentum of the system (in this case 2mv) multiplied by c. However, this doesn't make sense to me. If energy is force times distance, where does the c in this equation come from?

Thanks.
 
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  • #2
guss said:
In this proof of E = mc^2:

At around 5:40, he says that E = Pc. I understand that this means that energy is equal to the momentum of the system (in this case 2mv) multiplied by c. However, this doesn't make sense to me. If energy is force times distance, where does the c in this equation come from?

Thanks.

Are you just asking how they both have the same units? Force times distance has the same units as mass * distance^2 / time^2, since force has units of mass * acceleration = mass * distance / time^2. And pc has units of momentum * velocity while momentum itself has units of mass * velocity, so that means pc has units of mass * velocity^2 = mass * distance^2 / time^2. So you can see from this that pc has the same units as energy.

BTW, E=pc only works for particles with zero rest mass (like photons), for particles with some nonzero rest mass m the full equation is E2 = m2c4 + p2c2.
 
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  • #3
JesseM said:
BTW, E=pc only works for particles with zero rest mass (like photons), for particles with some nonzero rest mass m the full equation is E2 = m2c4 + p2c2.

That's what I'm asking. How can he use E = Pc, if he is talking about massive radiation? You may need to watch the first few minutes of the video to understand what I'm talking about.
 
  • #4
guss said:
That's what I'm asking. How can he use E = Pc, if he is talking about massive radiation? You may need to watch the first few minutes of the video to understand what I'm talking about.
I looked at the video, I don't really understand the physical assumptions being made, perhaps the book he got it from ('Instant Physics' by Tony Rothman) would spell it out more...I gather the idea is that the fulcrum of the seesaw moves when the photon transfers energy from one block to the other, but I don't quite understand why that happens, as opposed to the fulcrum staying at the same point and the seesaw just becoming out-of-balance. Also, I guess v is supposed to be the velocity at which the fulcrum movies, since he says the distance dx by which it move is equal to vt, and he says t=2x/c, the time for the light to pass from one block to the other. Anyway, I would guess that E is supposed to be the energy transferred by the photon from one block to the other, which causes one block to increase in mass by dm and the other to decrease in mass by the same amount...that would explain why he sets E=pc if p is supposed to be the momentum of the photon. But then in that case I don't quite understand p=2mv, what object's momentum is this supposed to be the equation for, and why is the velocity of the fulcrum relevant to it?
 
  • #6
Bill_K said:
This guy is completely wrong.

I think he just left a bunch of stuff out of the explanation. The way he explains it is crap, but I think I like what he is getting at.

Can anyone help solve this puzzle? I'm really interested in the solution/his errors.
 
  • #7
Yeah this is ridiculous. He's assuming
[tex]E^2=m^2c^4+p^2c^2[/tex]
To begin with, which is how he gets E=pc. But if you're assuming this, you just neglect the momentum of the system and immediately arrive at E=mc^2. No fulcrums needed.
 
  • #8
I think I figured it out--I was taking the "seesaw" drawing too literally, it's not an actual physical fulcrum that moves, but just the center of mass of the two blocks (not including the photon). If the center of mass is initially at rest before the photon is emitted, then in order to satisfy conservation of moment, after the photon is emitted the center of mass must move in the opposite direction with equal and opposite momentum, and then come to rest again once the photon is absorbed after a time of t=2x/c. So v is the velocity of the center of mass while it's moving, and since the mass of the two blocks is 2m, their momentum is 2mv, which must be equal and opposite to the photon's momentum of pc.

edit: though as Nabeshim says, seems like cheating to use E=pc for a photon, unless they're assuming this is an empirical result from quantum mechanics.
 
  • #9
JesseM said:
I think I figured it out--I was taking the "seesaw" drawing too literally, it's not an actual physical fulcrum that moves, but just the center of mass of the two blocks (not including the photon). If the center of mass is initially at rest before the photon is emitted, then in order to satisfy conservation of moment, after the photon is emitted the center of mass must move in the opposite direction with equal and opposite momentum, and then come to rest again once the photon is absorbed after a time of t=2x/c. So v is the velocity of the center of mass while it's moving, and since the mass of the two blocks is 2m, their momentum is 2mv, which must be equal and opposite to the photon's momentum of pc.

edit: though as Nabeshim says, seems like cheating to use E=pc for a photon, unless they're assuming this is an empirical result from quantum mechanics.

Ah, I see.

But, wouldn't the center of mass move in the same direction as the photon is moving, because the mass is shifting with the photon?

I believe E = Pc can be derived from Maxwell's equations (I could be wrong). That should definitely be included in this proof, though. Maybe we can discuss that when we have the main derivation solved.
 
  • #10
guss said:
Ah, I see.

But, wouldn't the center of mass move in the same direction as the photon is moving, because the mass is shifting with the photon?
I was just talking about the center of mass of the blocks alone, note the part where I said "the center of mass of the two blocks (not including the photon)". And by conservation of momentum, if the center of mass of the whole system was at rest before the photon was emitted, then after the photon is emitted the center of mass of the whole system (blocks + photon) must still be at rest since no outside force is acting on this system, which means the momentum of the photon must be canceled out by the equal and opposite momentum of the center of mass of the two blocks.
 
  • #11
guss said:
I believe E = Pc can be derived from Maxwell's equations (I could be wrong).

I don't think so. Classical electromagnetism has nothing to say about individual quanta of electromagnetism -- i.e photons. You can describe the momentum density of an electromagnetic wave in terms of its constituent E and B fields, but I don't think you can get anything about something we'd call a photon.
 
  • #12
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  • #14
Well, this page from mathpages.com (always pretty reliable as far as I've seen) says:
From Maxwell’s equations, and more fundamentally from the invariant space-time interval, we know that the energy E and momentum p of a massless particle (e.g., a photon) are related by pc = E.

...

This derivation is sometimes presented as the entirety of Einstein’s argument in the 1906 paper, but if this were so, it would be a somewhat weak argument, since it would apply only to the energy of light (whose inertia is already entailed by the electromagnetic relation E = pc), rather than to all forms of energy. However, Einstein’s paper actually describes a complete cyclical process, consisting first of the emission and absorption of the light pulse at opposite ends of a tube, and then the conveyance of that energy in any form whatsoever back to the emitting end of the tube by some arbitrary mechanism.
So it sounds like Einstein actually used E=pc in his own derivation of E=mc^2 in a 1906 paper, which is probably the one whose title is translated http://www.natscience.com/Uwe/Forum.aspx/relativity/11202/Einstein-s-1906-paper as "The principle of conservation of the centre of gravity movement and the inertia of energy"
Annalen der Physik, 1906, 20(8), pp.627-633.
 
  • #15
Nabeshin said:
This derivation assumes E=mc^2 ... not sure how effective that would be in this context...

Yes. I think a non-cheating derivation must go along the lines of considering the effect of a plane electromagnetic wave on a charged particle.
 
  • #16
I can't vouch for its correctness but Gonzolo gave a derivation of E=pc from Maxwell's laws in [post=297597]this post[/post].
 
  • #17
Classical electrodynamics does not have the notion of "photon", but one can show that the energy density of the electromagnetic field equals the magnitude of the momentum density times c.
 
  • #18
Here is a derivation of E=pc that doesn't assume anything about quantum mechanics or relativity: http://www.lightandmatter.com/html_books/0sn/ch11/ch11.html See "Momentum of light waves." (For a pdf version without the occasional math glitches, see http://www.lightandmatter.com/area1sn.html .)
 
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  • #19
One last thing, at the beginning of the video he says that the mass is transferred via radiation. He later goes on to say that the radiation is transferred at the speed of light. How is this possible? If the radiation has mass, wouldn't it not be able to travel at the speed of light?
 
  • #20
Radiation has no rest mass. It has energy, which one can think of heuristically as relativistic mass. The energy of radiation can be transformed into particles with kinetic energy and non-zero rest mass.
 
  • #21
Here is a geometric vector based derivation. Sorry for the lengthy post, but I wanted to present it in sufficient detail for those visitors to the forum who know some math and vector analysis but don't particularly know special relativity. [Edit: looking over my derivation, I really should have used m0 to make clear the reference to rest mass]
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  • #22
Thanks, I have it all figured out now. I think I'm going to make an E = mc^2 proof video on youtube, all of the other ones are crap. I'll make a thread when I'm done so people can point out any mistakes.
 

FAQ: E=mc^2 Explained: Momentum and c

What does the equation E=mc^2 mean?

The equation E=mc^2 represents the relationship between energy (E), mass (m), and the speed of light (c). It states that the energy of an object is equal to its mass multiplied by the speed of light squared.

How does the equation E=mc^2 relate to momentum and the speed of light?

E=mc^2 shows that energy and mass are directly proportional, and the speed of light is a constant that relates the two. This equation also demonstrates that any object with mass has energy, even if it is at rest.

What is the significance of the speed of light in the equation E=mc^2?

The speed of light (c) is a fundamental constant in the universe, and it represents the maximum speed at which anything with mass can travel. In the equation, it serves as a conversion factor between mass and energy.

Can you provide an example of how E=mc^2 is used in real life?

One example is nuclear energy, where mass is converted into energy through the process of nuclear fission. Another example is in particle accelerators, where particles are accelerated to near the speed of light, increasing their energy.

Is E=mc^2 the full equation for energy and mass?

No, E=mc^2 is a simplified version of the full equation, which includes the momentum of an object. The full equation is E^2=(mc^2)^2+(pc)^2, where p represents momentum. This equation is used in situations where an object's momentum cannot be neglected.

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