# E=mc^2 derivation

1. Feb 23, 2005

### villiami

I know this is basic, but how is E=mc^c derived.

Thanks Heaps

2. Feb 23, 2005

### dextercioby

Take the free particle Lagrangian and compute:
$$W=\vec{v}\cdot\frac{\partial L}{\partial\vec{v}}-L$$

Daniel.

3. Feb 23, 2005

### da_willem

4. Feb 23, 2005

### cepheid

Staff Emeritus
I notice that you, yourself, da willem, answered the question three times: once in each thread. If only that kind of persistence were rewarded somehow!

5. Feb 23, 2005

### villiami

What do you mean?

6. Feb 23, 2005

### villiami

Sorry, now I see

7. Feb 23, 2005

### dextercioby

Those links DO NOT GIVE PROOF TO E=mc^{2}...Both noncovariant and covariant lagrangian formulations of free relativistic particle give the proof.

Daniel.

8. Feb 23, 2005

### scilover89

Can you explain in detail? I found myself difficult to understand the statement.

9. Feb 23, 2005

### dextercioby

There's a lotta say,really.U need to build a lagrangian which would give the correct dynamics (motion equations) and be a relativistic/Lorentz invariant/scalar.
I think this stuff is described in electrodynamics books.In order to discuss relativistic particle in interaction with a classical EM field,u need to analyze the free particle first.I don't remember seing this issue in Jackson,nor in Goldstein,but i'm sure it has to be somewhere,just have the interest to look for.

Daniel.

10. Feb 23, 2005

### dextercioby

Okay,here's an elementary proof,i'd call it HS level,using differential calculus.
Assume for simplicity only one space-component.So the whole discussion would involve scalars.

Newton's second law:

$$\frac{d(m_{rel}v)}{dt}=F$$(1)

Multiply with dt:

$$d(m_{rel}v)=Fdt$$ (2)

Substitute dt with:

$$dt=\frac{dx}{v}$$(3)

$$d(m_{rel}v)=F\frac{dx}{v}$$(4)

Define differential work:

$$\delta L=Fdx$$(5)

Use the Leibniz theorem in differential form:

$$dE=\delta L\Rightarrow dE=Fdx$$(6)

Rewrite (4) in terms of the differential of energy:

$$v d(m_{rel}v)=dE$$ (7)

Everybody knows that:

$$m_{rel}=\gamma m_{0}=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$ (8)

Expand (7):

$$v^{2}dm_{rel}+m_{rel}vdv=dW$$(9)

Differentiate (8):

$$dm_{rel}=\frac{m_{0}\gamma v dv}{c^{2}-v^{2}}$$(10)

Plug (10) in (9) and factor:

$$m_{0}\gamma v dv(\frac{v^{2}}{c^{2}-v^{2}}+1)=dE$$(11)

Therefore:

$$m_{0}\gamma v dv \frac{c^{2}}{c^{2}-v^{2}}=dE$$(12)

Or:

$$c^{2}(\frac{m_{0}\gamma v dv}{c^{2}-v^{2}})=dE$$(13)

Taking into account (10),one finally finds the diferential form of Einstein's formula:

$$c^{2}dm_{rel}=dE$$(14)

Integrating with corresponding limits (zero relativistic mass,zero energy),one finds:

$$m_{rel}c^{2}=E$$

Daniel.

EDIT:THAT is a proof... It took me 10 minute to cook. :yuck: Though the Lagrangian approach is simply PERFECT.

Last edited by a moderator: May 3, 2014
11. Feb 23, 2005

### Andrew Mason

I like the conceptual 'light box' approach that daWillem uses, which was originally proposed by Einstein, I believe. However, it is somewhat inaccurate because it assumes instantaneous transmission of forces through the box.

The box part really is not needed. All you need are two masses. Conceptually, it goes like this:

Two equal masses, m1 and m2 (=m) at co-ordinates -d,0 and d,0 (origin at centre of mass). Since light has energy E = hf and momentum E/c (= hf/c), when a photon leaves m1, m1 recoils with momentum E/c. When the photon is stopped by m2, m2 takes on momentum E/c = mv (so v=E/cm). In time t=2d/c, m1 moves distance s=vt = E2d/mc^2. At time t after m1 begins moving, m2 receives momentum E/c.

Now you can see where there is a problem. Unless some mass is transferred, the centre of mass has moved!! Newton's third law takes care of this where the masses are not separated by a distance as the changes in motion occur at the same instant. But when the momentum change is provided by light, there will be a shift in the centre of mass unless light carries mass with it.

How much mass does it have to carry with it? Work it out: In order to conserve the centre of mass at time t, m1(-d) + m2d = 0 = m1'(-d1') + m2'd. Since d1'=d+s = d+E2d/mc^2, we have:

$$(m + \Delta m)(d) = (m-\Delta m)(d + E2d/mc^2)$$

$$md + \Delta md = md - \Delta md + mE2d/mc^2 - \Delta mE2d/mc^2)$$ or:

$$2\Delta md + \Delta mE2d/mc^2 = 2Ed/c^2$$

Ignoring the negligible 1/mc^2 term (this actually disappears if you take into account the $\gamma$ factor but we can avoid the math because we can see that $m>>\Delta m$):

$$\Delta m = E/c^2$$ or:

$$E = \Delta mc^2$$

The explanation from Einstein's "m = L/c^2" paper in 1905 is also quite understandable. The original translation is here:
http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

AM