# Doubt about proof that E=mc²

1. May 6, 2010

### pc2-brazil

Good night,

I've read the following proof that E=mc², through work and relativistic mass.
The expression for work is:
$$W=\int Fds$$
$$W=\int \frac{d(mv)}{dt}vdt=\int d(mv)v=\int (vdm+mdv)v=\int v^2 dm+mvdv$$
The expression for relativistic mass is:
$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$,
where m is the mass with velocity v and m0 is the rest mass.
When I differentiate the expression for relativistic mass, I find that:
$$c^2 dm=v^2 dm+mvdv$$
The second term (v2dm+mvdv) is the expression which is inside the integral of the work above. I can substitute it in that expression, thus obtaining the work with respect to the mass:
$$W=\int c^2 dm$$
When the work is performed in an object initially at rest, the velocity varies from 0 to v and the mass varies from m0 (rest mass) to m (mass with velocity v). Solving the above integral from m0 to m:
$$W=\int_{m_0}^m c^2 dm=mc^2-m_0 c^2$$
As the work is a variation in kinetic energy, one concludes that m0c² is the rest energy and E = mc² = γm0c², where γ is the Lorentz factor is the relativistic energy for any velocity v.

Now, here is my doubt:
The situation I'm using to prove E = mc² is a situation in which work is being performed by a force; thus, the object is accelerating. But Special Relativity is only valid for inertial frames of reference...
Then, why is this proof correct?

Last edited: May 6, 2010
2. May 6, 2010

### TMFKAN64

Objects may accelerate within an inertial frame of reference... it is the frame itself that is not accelerating.

3. May 6, 2010

### starthaus

Not true, there are extensions to accelerated motion (see my blog for detail)

4. May 6, 2010

### Integral

Staff Emeritus
I thought that the proof of E=mc2 was in the yield of a nuclear power plant or of an A bomb.

What you are doing is not a proof, it is a derivation.

5. May 6, 2010

### Fredrik

Staff Emeritus
The last statement isn't true. I would define SR as the set of theories of matter and interactions in Minkowski spacetime. Inertial frames are just convenient coordinate systems on Minkowski spacetime. To go beyond SR, you would have to consider a different spacetime, not just a less convenient coordinate system.

If you for some reason would like to change coordinates to an "accelerating" coordinate system, you would have to use some other function than a Lorentz transformation. That's all. It would still be SR, because we're not talking about changing the topology or metric of spacetime. And it's in exactly this sense that inertial frames are more "convenient": As long as you only use inertial frames, you can use a Lorentz transformation every time you want to change coordinates.

You're not changing coordinates anywhere in that calculation, so you don't have to worry about what function you would use for that.

You might be interested in my version of the calculation you posted. Note in particular the comments about definitions.

6. May 6, 2010

### Dickfore

There is no such thing as relativistic mass. There is only one mass - the mass of the object as measured when it is in rest. The formula $E = m c^{2}$ signifies that a body of mass m contains energy E.

7. May 6, 2010

### Fredrik

Staff Emeritus
I wouldn't say that there's "no such thing", but I do consider the term "relativistic mass" useless and obsolete.

8. May 6, 2010

### Dickfore

Please provide an experiment that measures the "relativistic mass" of an object.

9. May 6, 2010

### Fredrik

Staff Emeritus
Any experiment that measures the kinetic energy would do. But you knew that already, so why do you ask?

10. May 6, 2010

### Dickfore

ORLY? Then how would you calculate the relativistic mass if you knew the kinetic energy?

11. May 6, 2010

### Fredrik

Staff Emeritus
Are you serious? If K is the kinetic energy, then the relativistic mass is K/c2+m, where m is the rest mass. It's even simpler if the experiment measures the total energy. If the total energy is E, then the relativistic mass is E/c2.

By the way, I didn't mean to edit my posts after you replied. You just answered extremely fast.

12. May 6, 2010

### Dickfore

But, this means you already use a formula which he tries to derive. How did you derive that?

13. May 7, 2010

### Fredrik

Staff Emeritus
What does that have to do with anything? You were asking about how to measure relativistic mass, so I answered that. Now you're moving the goalpost. You can click the link in #5 if you're interested in my comments about "derivations". I have to go to bed so I won't be posting any more replies for a while.

"Relativistic mass" may be a useless term, but to say that it doesn't exist because it's the total energy divided by c² is like saying that the number 2 doesn't exist because it's =1+1.

14. May 7, 2010

### Dickfore

So, giving random formulas is a way to measure things now? I didn't know that.

Cool story bro. It seems you're doing it.

No, it is nothing like that. Why do you play with math? This is not Mathematics. This is Physics. A physical quantity is defined in one of two ways: operationally or analytically. Analytically means you connect it to already well defined physical quantities. However, in this way you can construct an arbitrary "physical quantity". To give an example, I define Dickfore's mass as:

$$M_{\textup{Dickfore}} = M_{p} \ln\left(\frac{E}{M_{p} c^{2}}\right),$$

where $M_{p}$ is some "fundamental constant" with the dimension of mass. You can "measure" Dickfore's mass if you know the "total energy" of the body E (let's not even ask how you can measure this). But, this ambiguity is redundant.

So, is your definition of relativistic mass.

15. May 7, 2010

### Al68

That proof does use an inertial reference frame. The object, not the reference frame used, is accelerating.

The "v" in the above proof refers to the velocity of the accelerated object relative to an inertial reference frame.

16. May 7, 2010

### Dmitry67

even the notion of 'relativistic mass' is obsolete,
most of the mass of proton and neutron is relativitic :)
as well as most of the mass of stars, planets etc

different observers dont agree on the rest mass of the body
if one looks at proton at whole, while another looks at it as bound system of quarks (and the third does not even know that it consists of quarks).
How can the value of an objective physical property depend on one's KNOWLEDGE and INTERPRETATION?

17. May 7, 2010

### Fredrik

Staff Emeritus
There's nothing random about anything I said, and when I gave you the answer to what you had asked, you acted as if the information I gave you was inadequate. If the correct answer to the question you asked doesn't satisfy you, then there's something wrong with the question, not with the answer. And how exactly did I move the goalpost?

That's what I've been saying! You need to stop acting as if you're proving me wrong when you're just saying essentially the same thing that I've been saying.

18. May 7, 2010

### Dickfore

Whatever. I don't plan to quarrel over the Internet.

Last edited by a moderator: May 7, 2010
19. May 7, 2010

### Fredrik

Staff Emeritus
You've made it clear that you're only here to quarrel. Your behavior yesterday (moving the goalposts, and acting as if you had just proved me wrong after saying what I've been saying all along) was just as inappropriate as the direct insult.

20. May 7, 2010

### Dickfore

Dear pc2-brazil. It is advisable to refrain from using the concept of relativistic mass. The only mass that should be used is the so called rest mass. In fact, your derivation starts out from two assumptions, namely, that momentum is

$$\mathbf{p} = m \mathbf{v}$$

where

$$m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

is the relativistic mass. There is no need for this disassembling of the formula for momentum in relativistic dynamics. Instead. The momentum of a particle with mass $m$ (I made the substitution $m_{0} \rightarrow m$) moving with velocity $\mathbf{v}$ is:

$$\mathbf{p} = \frac{m \mathbf{v}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

If you are familiar with the Hamilton's formulation of mechanics, you can convince yourself that this momentum is derivable from a Lagrangian for a free particle:

$$L = - m c^{2} \sqrt{1 - \frac{v^{2}}{c^{2}}}$$

as

$$\mathbf{p} = \frac{\partial L}{\partial \mathbf{v}}$$

This is the only Lagrangian (up to an always present ambiguity up to a total time derivative) that satisfies the following conditions:
1. leaves the Hamilton's action

$$S = \int {L dt}$$

invariant under Lorentz transformations;

2. leaves space-time homogeneous;

3. leaves space isotropic;

4. reduces to the classical Lagrange function for a free particle when $v/c << 1$.

From here, you can also derive the energy of a point particle (material point):

$$\mathcal{E} = \mathbf{v} \cdot \frac{\partial L}{\partial \mathbf{v}} - L = \frac{m c^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$$

One can show that the energy and momentum derived above behave like a 4-vector $P^{\mu} = (\mathcal{E}/c, \mathbf{p})$. The magnitude of this 4-vector is a Lorentz invariant:

$$P_{\mu} P^{\mu} = \left(\frac{\mathcal{E}}{c}\right)^{2} - \mathbf{p}^{2} = (m c)^{2}$$

In the reference frame where the particle is at rest, the 4-momentum has components $P_{0}^{0} = \mathcal{E}_{0}/c, P_{0}^{i} = 0$, and, by the invariance stated above:

$$\mathcal{E}_{0} = m c^{2}$$

This is called the rest energy of the particle because it is the energy of a particle at rest.

21. May 7, 2010

### starthaus

This is a nice approach. I have put together a different solution (see the latest attachment in my blog).

22. May 8, 2010

### yuiop

Not being a calculus expert, I did not get this step at first, so just incase there is anyone else in the world like me that not is not a calculus whizz, here is how it breaks down:

Assume

The expression for relativistic mass is:

$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}} }$$.

Differentiate the above with respect to velocity:

$$d\frac{m}{dv} = \frac{m_0 v}{c^2(1-\frac{v^2}{c^2})^{3/2}}= \frac{m_0 v}{c^2(1-\frac{v^2}{c^2}) \sqrt{1-\frac{v^2}{c^2}} }}$$.

Substitute the original expression for relativistic mass back into the equation:

$$d\frac{m}{dv} = \frac{m v}{c^2(1-\frac{v^2}{c^2})} = \frac{m v}{(c^2-v^2)}$$.

Rearrange:

$$dm (c^2-v^2) = m v dv$$,

$$c^2 dm - v^2 dm = m v dv$$,

$$c^2 dm = v^2 dm + m v dv$$.

Seems to work OK.

As others have mentioned, this is an invalid concern. SR is perfectly capable of handling acceleration.

Last edited: May 8, 2010
23. May 8, 2010

### pc2-brazil

Thank you for posting this. I am not a calculus expert either, but I didn't include this step because I was afraid the text of the topic became too long.

24. May 8, 2010

### yuiop

$$\Delta W = m_0\int_{v_0}^{v_1} \gamma^3 v dv = m_0 \int_{v_0}^{v_1} \frac{v dv}{(1-\frac{v^2}{c^2})^{3/2}}$$

and not:

$$\Delta W = m_0\int_{v_0}^{v_1} \gamma^3 v dv = m_0 \int_{v_0}^{v_1} \frac{v dv}{\sqrt{1-\frac{v^2}{c^2}}}$$

25. May 8, 2010

### starthaus

It's an obvious typo. The solution is correct.