Is the Proof of E=mc² Valid for Accelerating Objects in Special Relativity?

[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?

 

[TMFKAN64]

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

 

[starthaus]

But Special Relativity is only valid for inertial frames of reference.

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

 

[Integral]

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.

 

[Fredrik]

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...

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.

 

[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.

 

[Fredrik]

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

 

[Dickfore]

Yeah, I agree. I wouldn't say that there's "no such thing", but I consider the term useless and obsolete.

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

 

[Fredrik]

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

 

[Dickfore]

Any experiment that measures the kinetic energy would do (if we know the rest mass already). But you knew that already, so why do you ask?

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

 

[Fredrik]

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.

 

[Dickfore]

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.

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

 

[Fredrik]

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.

 

[Dickfore]

What does that have to do with anything? You were asking about how to measure relativistic mass, so I answered that.

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

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

"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.

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:

<br /> M_{\textup{Dickfore}} = M_{p} \ln\left(\frac{E}{M_{p} c^{2}}\right),<br />

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.

 

[Al68]

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?

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.

 

[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 don't 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?

 

[Fredrik]

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.

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?

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.

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.

 

[Dickfore]

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.

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

 

[Fredrik]

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.

 

[Dickfore]

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?

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

<br /> \mathbf{p} = m \mathbf{v}<br />

where

<br /> m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}<br />

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:

<br /> \mathbf{p} = \frac{m \mathbf{v}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}<br />

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:

<br /> L = - m c^{2} \sqrt{1 - \frac{v^{2}}{c^{2}}}<br />

as

<br /> \mathbf{p} = \frac{\partial L}{\partial \mathbf{v}}<br />

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

<br /> S = \int {L dt}<br />

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 &lt;&lt; 1.

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

<br /> \mathcal{E} = \mathbf{v} \cdot \frac{\partial L}{\partial \mathbf{v}} - L = \frac{m c^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}<br />

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:

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

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:

<br /> \mathcal{E}_{0} = m c^{2}<br />

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

 

[starthaus]

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

<br /> \mathbf{p} = m \mathbf{v}<br />

where

<br /> m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}<br />

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:

<br /> \mathbf{p} = \frac{m \mathbf{v}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}<br />

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:

<br /> L = - m c^{2} \sqrt{1 - \frac{v^{2}}{c^{2}}}<br />

as

<br /> \mathbf{p} = \frac{\partial L}{\partial \mathbf{v}}<br />

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

<br /> S = \int {L dt}<br />

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 &lt;&lt; 1.

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

<br /> \mathcal{E} = \mathbf{v} \cdot \frac{\partial L}{\partial \mathbf{v}} - L = \frac{m c^{2}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}<br />

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:

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

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:

<br /> \mathcal{E}_{0} = m c^{2}<br />

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

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

 

[yuiop]

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

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}} }}<br />.

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})}<br /> = \frac{m v}{(c^2-v^2)}<br />.

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.

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...

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

 

[pc2-brazil]

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}} }}<br />.

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})}<br /> = \frac{m v}{(c^2-v^2)}<br />.

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.

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.

 

[yuiop]

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

Step (5) of your solution has an error. It should read:

\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}}}

 

[starthaus]

Step (5) of your solution has an error. It should read:

\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}}}

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

 

[starthaus]

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.

The problems with the solution are:

1. It is very limitted, it works for the unidimensional case only.
2. It uses the antiquated concept of relativistic mass.

Either Dickfore's solution or mine do not suffer from any limitations.

 

[pc2-brazil]

starthaus: Your derivation was very helpful. Unfortunately, I can't make use of Dickfore's
approach for now, because I still don't know Hamiltonian and Lagrangian mechanics.
Since I have a very superficial knowledge of vector calculus, it took me a while to understand some things. I would like to make 2 observations:
1) In the part following part:
\vec{v}\frac{d\vec{v}}{dt} = \frac{1}{2}\frac{d(\vec{v}^2)}{dt}=\frac{1}{2}\frac{d(v^2)}{dt}=v\frac{dv}{dt}
I see you are doing the dot product, which comes from the work definition. I would like to
ask if I can use the following reasoning in order to arrive at the same result: dot product between two vectors a and b is \vec{a}\cdot\vec{b}=ab\cos{\theta}; if they have the same direction and orientation,the angle is 0, and cosine of 0 is 1; so, the dot product is the product ab. Since \vec{v} and \frac{d\vec{v}}{dt} necessarily have the same direction and orientation, the dot product between them will be the product between their magnitudes: v\frac{dv}{dt}.
2) In the part where you say "(3) becomes after some simple manipulation:" (Step 5), I've tried to do the same manipulation that you did to arrive at that \gamma^3v³/c² dv/dt, so I would like to expose it here:
The expression after the last equal sign in Step 3 contains the following member:
\frac{d\gamma}{dt}v^2 (1)
The manipulation involves only that member; so, I will work with this member only; then, it can be put back into the expression in Step 3. Now I will calculate the derivative of gamma, d\gamma/dt, by using the chain rule:
\frac{d\gamma}{dt} = \frac{d}{dt}\left (1 - \frac{v^2}{c^2}\right )^{-\frac{1}{2}} = -\frac{1}{2}\left (1 - \frac{v^2}{c^2}\right )^{-\frac{3}{2}}\left (\frac{d}{dt}\left(1 - \frac{v^2}{c^2}\right)\right ) = -\frac{1}{2}\left (1 - \frac{v^2}{c^2}\right )^{-\frac{3}{2}}\left (-\frac{1}{c^2}\frac{d(v^2)}{dt}\right )
Now, since \left (1 - \frac{v^2}{c^2}\right )^{-\frac{3}{2}}=\gamma^3 and, by the product rule, \frac{d(v^2)}{dt} = v\frac{d(v)}{dt} + v\frac{d(v)}{dt} = 2v\frac{dv}{dt}, the derivative of gamma becomes:
\frac{d\gamma}{dt} = -\frac{1}{2}\gamma^3\left (-\frac{2v}{c^2}\frac{dv}{dt}\right ) = \frac{\gamma^3v}{c^2}\frac{dv}{dt}
Substituting it into (1), that member becomes:
\frac{\gamma^3v^3}{c^2}\frac{dv}{dt}
which can be put back into the expression of Step 3, obtaining the first expression in Step 5.

Thank you in advance.

 

[starthaus]

Since \vec{v} and \frac{d\vec{v}}{dt} necessarily have the same direction and orientation,

No, they do not. Think about rotation as a clear counter-example.
The derivation that I am using does not assume anything about the directions of \vec{v} and \frac{d\vec{v}}{dt}

 

[pc2-brazil]

No, they do not. Think about rotation as a clear counter-example.
The derivation that I am using does not assume anything about the directions of \vec{v} and \frac{d\vec{v}}{dt}

OK, I was forgetting about centripetal acceleration.

 

[starthaus]

2) In the part where you say "(3) becomes after some simple manipulation:" (Step 5), I've tried to do the same manipulation that you did to arrive at that \gamma^3v³/c² dv/dt, so I would like to expose it here:
The expression after the last equal sign in Step 3 contains the following member:
\frac{d\gamma}{dt}v^2 (1)
The manipulation involves only that member; so, I will work with this member only; then, it can be put back into the expression in Step 3. Now I will calculate the derivative of gamma, d\gamma/dt, by using the chain rule:
\frac{d\gamma}{dt} = \frac{d}{dt}\left (1 - \frac{v^2}{c^2}\right )^{-\frac{1}{2}} = -\frac{1}{2}\left (1 - \frac{v^2}{c^2}\right )^{-\frac{3}{2}}\left (\frac{d}{dt}\left(1 - \frac{v^2}{c^2}\right)\right ) = -\frac{1}{2}\left (1 - \frac{v^2}{c^2}\right )^{-\frac{3}{2}}\left (-\frac{1}{c^2}\frac{d(v^2)}{dt}\right )
Now, since \left (1 - \frac{v^2}{c^2}\right )^{-\frac{3}{2}}=\gamma^3 and, by the product rule, \frac{d(v^2)}{dt} = v\frac{d(v)}{dt} + v\frac{d(v)}{dt} = 2v\frac{dv}{dt}, the derivative of gamma becomes:
\frac{d\gamma}{dt} = -\frac{1}{2}\gamma^3\left (-\frac{2v}{c^2}\frac{dv}{dt}\right ) = \frac{\gamma^3v}{c^2}\frac{dv}{dt}
Substituting it into (1), that member becomes:
\frac{\gamma^3v^3}{c^2}\frac{dv}{dt}
which can be put back into the expression of Step 3, obtaining the first expression in Step 5.

Thank you in advance.

Correct.

 

[starthaus]

OK, I was forgetting about centripetal acceleration.

Vector calculus is a very powerful tool in obtaining the most general solutions. This is why I mentioned that the solution I provided is much more general than the one you found. Where did you find iit? What book?

 

[pc2-brazil]

Vector calculus is a very powerful tool in obtaining the most general solutions. This is why I mentioned that the solution I provided is much more general than the one you found. Where did you find iit? What book?

Actually, I found it in an online video: http://www.youtube.com/user/matmania1#p/c/7E2FA65CB78B54B0/14/4cS5qvNJJmA". This video is in Portuguese.
My knowledge is somewhat fragmented, since I'm not following a course (just so you have an idea, I've learned the substitution method in integration for the first time today in order to understand a particular step in your derivation). I'm learning from what the Internet has to offer. This forum is a very helpful resource.

 

[starthaus]

Actually, I found it in an online video: http://www.youtube.com/user/matmania1#p/c/7E2FA65CB78B54B0/14/4cS5qvNJJmA". This video is in Portuguese.
My knowledge is somewhat fragmented, since I'm not following a course. I'm learning from what the Internet has to offer. This forum is a very helpful resource.

If I may suggest an internet source, find C.Moller's book on relativity (The Theory of Relativity). It is very good, a model of rigor and it is...free. I think that it is one of the best, most complete books written on the subject. It covers both SR and GR.

 

[pc2-brazil]

If I may suggest an internet source, find C.Moller's book on relativity (The Theory of Relativity). It is very good, a model of rigor and it is...free. I think that it is one of the best, most complete books written on the subject. It covers both SR and GR.

Thank you for the suggestion, I will have a look.
By the way: you've answered to my reply so fast that you may haven't noticed the edition I made.

 

[Dickfore]

Also this reference:

http://www.fourmilab.ch/etexts/einstein/E_mc2/www/"

 

[pc2-brazil]

Also this reference:

http://www.fourmilab.ch/etexts/einstein/E_mc2/www/"

Thank you too.