Understanding Einstein's E=mc^2 Theory

[taylrl3]

How does E=mc^2 ??

Hey,

After looking on the internet I am having trouble finding some explanation of how E=mc^2. I know what it means and how to use it but how did Einstein arrive at this conclusion? Don't worry if the answer is a bit complex I can handle some meaty maths :-)

Thanks!
:-)

 

[ZealScience]

Just substituting transformation of mass into ∫F·ds could derive it. Basically, transformation of mass is implying that when energy increases, mass increases. Thus there is certain equivalence.

 

[Trifis]

Take a look at the relativistic kinetic energy here: http://en.wikipedia.org/wiki/Kinetic_energy ;)

 

[crocque]

Everyone else was thinking explosions. All of a sudden he though, "what about an implosion... hmm". It's about the "meaty math" but then again, it's not somethimes.

 

[SamRoss]

Why not go straight to the source?

This is Einstein's original paper on relativity http://wikilivres.info/wiki/On_the_...._Dynamics_of_the_Slowly_Accelerated_Electron

This is the paper he wrote soon after that where he proposed E=mc^2 http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

With regard to the first paper, the part that you really need to look at is section 10 where Einstein develops the formula for kinetic energy.

I would also recommend looking at a short thread I started which will help you compare the two papers and the E=mc^2 proof. https://www.physicsforums.com/showthread.php?t=501738

 

[Superstring]

Find the relativistic kinetic energy by finding the work done on a particle:

$$\Delta E_k = W = \int\mathbf{F}\cdot d\mathbf{s} = \int \frac{d\boldsymbol{p}}{dt}\cdot d\mathbf{s}=\int \mathbf{v}\cdot d\mathbf{p} = \int \mathbf{v}\cdot d(\frac{m\mathbf{v}}{\sqrt{1-v^2/c^2}})$$

After you solve that integral (and assuming the particle you're doing work on starts from rest), you should get:

$$E_k = \frac{mc^2}{\sqrt{1-v^2/c^2}} +K$$

where K is a constant of integration.

Now take the case where the particle is at rest (v=0). The kinetic energy in this case is taken to be zero:

$$0 = mc^2 +K$$

or:

$$K=-mc^2$$


Now we have the formula for the kinetic energy of a particle:

$$E_k = \frac{mc^2}{\sqrt{1-v^2/c^2}} - mc^2$$


Einstein realized that the second term meant that a particle has energy, even at rest, by virtue of the fact that it has mass. He called mc² the "rest energy." He also realized that the total energy of a body can thought of as the sum of the rest energy and the kinetic energy, or:

$$E = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$

To reinforce the concept of rest energy, when v=0 in the above equation you get:

$$E_0 = mc^2$$

 

[SamRoss]

Now we have the formula for the kinetic energy of a particle:

$$E_k = \frac{mc^2}{\sqrt{1-v^2/c^2}} - mc^2$$


Einstein realized that the second term meant that a particle has energy, even at rest, by virtue of the fact that it has mass. He called mc² the "rest energy." He also realized that the total energy of a body can thought of as the sum of the rest energy and the kinetic energy, or:

$$E = \frac{mc^2}{\sqrt{1-v^2/c^2}}$$

I could be wrong, but I'm not convinced that the relativistic kinetic energy equation that you gave (although it is correct) predicts the existence of rest energy. When we plug v=0 into the equation we get 0 (as we should, since there is clearly no kinetic energy when v=0). I'm pretty sure the formula for total energy that you gave (also correct) comes from the addition of the famous rest energy equation E=mc^2 (which Einstein published after he published the kinetic energy equation) to the kinetic energy equation.

In other words, I agree with everything you said except that rest energy comes from the second term of the kinetic energy equation. It comes from a similar-looking yet different equation.

 

[Drakkith]

I could be wrong, but I'm not convinced that the relativistic kinetic energy equation that you gave (although it is correct) predicts the existence of rest energy. When we plug v=0 into the equation we get 0 (as we should, since there is clearly no kinetic energy when v=0). I'm pretty sure the formula for total energy that you gave (also correct) comes from the addition of the famous rest energy equation E=mc^2 (which Einstein published after he published the kinetic energy equation) to the kinetic energy equation.

In other words, I agree with everything you said except that rest energy comes from the second term of the kinetic energy equation. It comes from a similar-looking yet different equation.

Wouldn't V=0 mean that you divide the top by 1?

 

[SamRoss]

Yes, divide the first term by 1, then subtract the second term. We're talking about the kinetic energy equation, remember. Not the total energy equation.

 

[Drakkith]

Yes, divide the first term by 1, then subtract the second term. We're talking about the kinetic energy equation, remember. Not the total energy equation.

Yeah, so the kinetic energy is equal to 0 since you would have mc^2 - mc^2. While I don't know what Einstein was thinking exactly, it seems logical to say that mc^2 is the rest energy of a particle. At least to me. But don't take my opinion too seriously.

 

[bcrowell]

Find the relativistic kinetic energy by finding the work done on a particle:

$$\Delta E_k = W = \int\mathbf{F}\cdot d\mathbf{s} = \int \frac{d\boldsymbol{p}}{dt}\cdot d\mathbf{s}=\int \mathbf{v}\cdot d\mathbf{p} = \int \mathbf{v}\cdot d(\frac{m\mathbf{v}}{\sqrt{1-v^2/c^2}})$$

When I've seen this derivation, the point that I have never seen addressed at all is why it is valid to assume that $W=\int F ds$ holds without modification in relativity.

 

[SamRoss]

Is it possible that it leading to the same answer as other derivations for kinetic energy could actually be considered a derivation of the fact that W=$\int$Fds holds in relativity? So you would still need another proof for kinetic energy but then you could go on and say "and here's this nice corollary".

 

[Trifis]

When I've sees this derivation, the point that I have never seen addressed at all is why it is valid to assume that $W=\int F ds$ holds without modification in relativity.

This is just the definition of work. I don't think there is a reason to be motified, The same happens with all the other physical definitions. Besides the components of the line integral are already relativistic.

 

[SamRoss]

Yeah, so the kinetic energy is equal to 0 since you would have mc^2 - mc^2. While I don't know what Einstein was thinking exactly, it seems logical to say that mc^2 is the rest energy of a particle. At least to me. But don't take my opinion too seriously.

If the kinetic energy is zero when the particle is at rest, that tells you that you need to look somewhere else to find the rest energy. The rest energy being mc^2 is not something that is just "logical". It comes from somewhere, namely this line of reasoning by Einstein http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

 

[bcrowell]

Is it possible that it leading to the same answer as other derivations for kinetic energy could actually be considered a derivation of the fact that W=$\int$Fds holds in relativity? So you would still need another proof for kinetic energy but then you could go on and say "and here's this nice corollary".

Sure, that makes sense. But many people, including Einstein in his 1905 paper on SR, use it to *establish* the relativistic equation for KE.

This is just the definition of work. I don't think there is a reason to be motified, The same happens with all the other physical definitions. Besides the components of the line integral are already relativistic.

Personally, I consider the definition of work to be a transfer of energy by a macroscopic force. But regardless of whether you take that to be the definition or $W=\int F ds$ to be the definition, these derivations are assuming that $W=\int F ds$ equals the transfer of energy. That equality isn't just a matter of definition.

 

[Drakkith]

If the kinetic energy is zero when the particle is at rest, that tells you that you need to look somewhere else to find the rest energy. The rest energy being mc^2 is not something that is just "logical". It comes from somewhere, namely this line of reasoning by Einstein http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

I really don't follow if your saying that the kinetic energy equation came before the absolute energy equation, as you did earlier.

 

[Orion1]

The work done on a body is equal to the amount of force integrated over the distance in which it moves, that work turns into the body's kinetic energy:
$$W = \int_{x_0}^{x_1} F \; dx = K$$
$$F \rightarrow |x_0 \rightarrow x_1|$$
However, force changes momentum:
$$F = \frac{dp}{dt}$$
$$K = \int \frac{dp}{dt} \; dx = \int \frac{dx}{dt} \; dp = \int v \; dp$$
And momentum depends on mass, which also changes with velocity:
$$p = \gamma m_0 v = \frac{m_0 v}{\left(1 - \frac{v^2}{c^2} \right)^{\frac{1}{2}}}$$
$$\frac{dp}{dv} = \frac{m_0}{\left(1 - \frac{v^2}{c^2} \right)^{\frac{3}{2}}}$$
$$dp = \frac{m_0}{\left(1 - \frac{v^2}{c^2} \right)^{\frac{3}{2}}} \; dv$$
Kinetic energy may be expressed as an integral over changing velocity from rest until the force stops pushing:
$$K = m_0 \int_{0}^{v_1} \frac{v}{\left(1 - \frac{v^2}{c^2} \right)^{\frac{3}{2}}} \; dv = \frac{m_0 c^2}{\left(1 - \frac{v^2}{c^2} \right)^{\frac{1}{2}}} - m_0c^2$$
The result appears more complicated than it really is, at low velocity:
$$K = \frac{mv^2}{2}$$
However, at higher velocity, the curve for increasing kinetic energy resembles the curve for increasing mass:
$$K = \frac{m_0 c^2}{\left(1 - \frac{v^2}{c^2} \right)^{\frac{1}{2}}} - m_0 c^2 = \gamma m_0 c^2 - m_0 c^2 = mc^2 - m_0c^2$$
The equation demonstrates that at any velocity, the kinetic energy is equal to the total change in mass times $c^2$
$$K = (m - m_0) c^2$$
And since m_0 is equal to the body mass at rest, the energy at rest is called the Rest-mass Energy:
$$E_0 = m_0 c^2$$
Adding the kinetic energy to the Rest-mass Energy gives the total body energy:
$$E_t = K + E_0 = (m - m_0) c^2 + m_0 c^2 = mc^2$$
$$\boxed{E_t = mc^2}$$

 

[SamRoss]

I really don't follow if your saying that the kinetic energy equation came before the absolute energy equation, as you did earlier.

In Einstein's original paper on relativity, "On the Electrodynamics of Moving Bodies", he gave a derivation of the kinetic energy equation. In a paper he published a little later, the one I linked to my last post, he gave the rest energy equation, E(rest)=mc^2. Adding these two together gives total energy.

 

[Drakkith]

In Einstein's original paper on relativity, "On the Electrodynamics of Moving Bodies", he gave a derivation of the kinetic energy equation. In a paper he published a little later, the one I linked to my last post, he gave the rest energy equation, E(rest)=mc^2. Adding these two together gives total energy.

Ok, so you are saying that the paper you linked was the original KE equation, and the E=MC^2 equations came later, meaning that it is unlikely that Einstein developed the idea of MC^2 being the rest energy from that equation?

 

[Trifis]

The way that Einstein did it is demonstrated in Orion's post. He first calculated the Kinetic Energy and then he added the rest energy to find the whole one. After all it was not so hard in comparison with the first derivation. Just check what happens if v=0.

 

[SamRoss]

Ok, so you are saying that the paper you linked was the original KE equation, and the E=MC^2 equations came later, meaning that it is unlikely that Einstein developed the idea of MC^2 being the rest energy from that equation?

Unlikely that he developed it from the kinetic energy equation, right. The kinetic energy equation simply doesn't tell you anything about a particle's energy when it is at rest.

 

[Orion1]

And the relativistic mass and the relativistic kinetic energy are related by the formula:
$$E_k = m c^2 - m_0 c^2$$
Einstein wanted to omit the unnatural second term on the right-hand side, whose only purpose is to make the energy at rest zero, and to declare that the particle has a total energy which obeys:
$$E = m c^2$$
which is a sum of the rest energy m_0 c^2 and the kinetic energy. This total energy is mathematically more elegant, and fits better with the momentum in relativity. But to come to this conclusion, Einstein needed to think carefully about collisions. This expression for the energy implied that matter at rest has a huge amount of energy, and it is not clear whether this energy is physically real, or just a mathematical artifact with no physical meaning.

In a collision process where all the rest-masses are the same at the beginning as at the end, either expression for the energy is conserved. The two expressions only differ by a constant which is the same at the beginning and at the end of the collision. Still, by analyzing the situation where particles are thrown off a heavy central particle, it is easy to see that the inertia of the central particle is reduced by the total energy emitted. This allowed Einstein to conclude that the inertia of a heavy particle is increased or diminished according to the energy it absorbs or emits.

Reference:
http://en.wikipedia.org/wiki/Mass–energy_equivalence#Mass.E2.80.93velocity_relationship"