# I Question about deriving E=mc^2

1. Mar 13, 2017

### JohnnyGui

I have a question about the velocity of a mass that is being accelerated from its resting position, for deriving its energy according to $E=mc^2$

The energy needed for accelerating a mass is: $E = F \cdot s = m \cdot a \cdot s$. Where s is the distance over which the mass is moved.
Now, I understand that with each small increase in s, the mass would increase according to
$\frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}=m$. So I should integrate
$$\int_{0}^{d} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot a \cdot ds$$
Since the relativistic mass is dependent on the velocity we could write the distance $ds$ and $a$ in terms of velocity. Regarding the acceleration $a$, we know that $a \cdot t$ will give the final velocity after time $t$. So $a \cdot t = v_{final}$. I’ll explain later why I’m emphasizing on this velocity being the final velocity.

The distance $s$ is, as known, calculated by $\frac{1}{2}at^2$. Combining this with $a \cdot t = v_{final}$ would give: $\frac{1}{2}v_{final}\cdot t$. Substituting these formulas for $a$ and $ds$ in the integration formula would give:
$$\int_{0}^{v} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot \frac{v_{final}}{t} \cdot \frac{1}{2} \cdot dv_{final} \cdot t$$
Which, after simplifying gives:
$$\int_{0}^{v} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot v_{final} \cdot \frac{1}{2} \cdot dv_{final}$$
I googled this and it seems to be the correct formula that needs to be integrated. However, I can’t see how this is possible.
Here’s why; I think the $v_{final}$ that you’d calculate from the acceleration and distance is different from the velocity $v$ in the Lorentz formula part of the integration. If a mass is being accelerated over a small distance $ds$, then the mass increases based on the mean velocity in that small distance $ds$ and not based on the final velocity at the moment the mass finishes that small distance $ds$. The $v$ in the Lorentz formula should be the velocity at half of $ds$, so that $v = \frac{1}{2}v_{final}$. How can the $v$ in the Lorentz term and the $v_{final}$ still be considered the same?

Last edited: Mar 13, 2017
2. Mar 13, 2017

### Staff: Mentor

You've already gone wrong by trying to use $E = mc^2$ for an object that is not at rest.

Only in the non-relativistic limit, where $a \cdot s << c^2$.

You've gone wrong again by trying to use "relativistic mass" here.

This is also only true in the non-relativistic limit.

Only in the non-relativistic limit.

If you're really trying to do a correct relativistic calculation, you need to use the correct relativistic formulas. Assuming that the fixed, known quantities are the acceleration $a$ (and you also have to decide whether that is proper acceleration or coordinate acceleration--I recommend the former) and the distance $s$ (assumed to be the distance in the inertial frame in which the object starts out at rest, but you should specify that too), you can use the relativistic rocket equations to find out the other quantities. See here:

http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

Note that Baez does not give an explicit formula for energy, but he does give formulas for the relativistic $\gamma$ factor, which amounts to the same thing (since the energy of an object with rest mass $m$ is just $\gamma m c^2$).

3. Mar 13, 2017

### JohnnyGui

Apologies, I wasn't clear enough on this. I did mean the needed energy to accelerate a mass that was at rest at the start. I'll try and see what I can pick up from the link.

Here's a link that I've found where they integrate $\int_0^d F \cdot ds$ just as I did. How come they were able to derive $E=mc^2$ from that? If it's possible that way, doesn't my question in my OP then stand?

4. Mar 13, 2017

### Staff: Mentor

The mean speed in any short interval $\Delta{t}$ starting at time $t$ is not $v(t)/2$, it is $v(t)+\frac{a\Delta{t}}{2}$. In the limit as $\Delta{t}$ approaches zero (which is basically what you're doing when you replace $\Delta{t}$ with $dt$ and integrate) the $\Delta{t}$ term disappears.

5. Mar 13, 2017

### Staff: Mentor

No, they don't. They don't use $F = ma$. They use $F = d(mv) / dt = m_0 d(\gamma v) / dt$ (I've rephrased what they actually did somewhat to make the difference between theirs and yours clear). So they are using relativistically correct formulas and you are not.

6. Mar 14, 2017

### vanhees71

The most clear way to derive what's energy and momentum is to use Nother's theorem to translation invariance of Minkowski space. It turns out that the four-vector of energy and momentum is given by
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} = m \gamma \frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} = \frac{m}{\sqrt{1-\vec{v}^2/c^2}} \begin{pmatrix} c \\ \vec{v} \end{pmatrix}=\begin{pmatrix} E/c \\ \vec{p} \end{pmatrix}.$$
Note that $m=\text{const}$ is the invariant mass of the particle, which can also be expressed as the "on-shell condtion" for the four-momentum,
$$p_{\mu} p^{\mu}=m^2 c^2.$$
Now you have, for $\vec{v}^2 \ll c^2$
$$E=m \gamma c^2= m c^2 + \frac{m}{2} \vec{v}^2 + \mathcal{O}(v^4/c^4).$$
This shows that in order to get a four-vector for energy and momentum, you have to include the "rest energy" $E_0=mc^2$ in the total energy of the particle. That's how "$E=m c^2$" comes into play in the most simple case of poin-particle mechanics.

Last edited: Mar 14, 2017
7. Mar 14, 2017

### JohnnyGui

I have noticed something when using your mentioned formulation of the mean speed $\overline{v} =v(t)+\frac{a\cdot{dt}}{2}$ which leads to the correct formulation of my integration for giving $E=mc^2$. I'm not sure if this is coincidence or not but here it goes;

If a mass is being accelerated over a small distance $ds$, then in that distance one could give the following formulas regarding the mean velocity $\overline{v}$, the initial velocity $v_i$ (which is your mentioned $v(t)$) and the increase in velocity $dv$:

$\overline{v} = v_i + \frac{a \cdot dt}{2}$
$dv = a \cdot dt$
So that: $ds = v_idt + \frac{a \cdot dt^2}{2}$

If these formulas are correct, one could substitute the $ds$ in the integration $\int_{0}^{d} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot a \cdot ds$ given in my OP which will give:
$$\int_{0}^{d} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot a \cdot (v_i \cdot dt + \frac{a \cdot dt^2}{2}) = \int_{0}^{d} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot a \cdot dt (v_i + \frac{a \cdot dt}{2})$$
Since $v_i + \frac{a \cdot dt}{2} = \overline{v}$ and $a \cdot dt = dv$ as discussed, this means that:
$$\int_{0}^{v} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot dv \cdot \overline{v}$$
This is now indeed shows the mean velocity $\overline{v}$ just as I considered the $v$ term in the Lorentz formula to be the mean velocity at which a mass increases. I can see that this link agrees with this integration. Is this correct by coincidence or am I doing it the right way here?

I might be missing something here but, doesn't the part $dv / dt$ in $F = d(mv) / dt = m_0 d(\gamma v) / dt$ equal $a$ so that $F = d(mv) / dt = m_0 \cdot d\gamma \cdot a$? Also, please see my above calculation as reply to @Nugatory that does seem to agree with the link I've given.

Last edited: Mar 14, 2017
8. Mar 14, 2017

### Staff: Mentor

If $a$ is interpreted as coordinate acceleration, then $a = dv/dt$ is correct--but then $F = m a$ is not correct, because the time derivative taken is of $\gamma v$, not $v$.

If $a$ is interpreted as proper acceleration, then no, $a = dv/dt$ is not correct (it is only an approximation valid at low speeds).

9. Mar 14, 2017

### JohnnyGui

But then how come my calculation in my previous post #7 agrees with the formula in the link? Have I concluded it the right way or is this merely coincidence?

10. Mar 14, 2017

### Staff: Mentor

I don't think so, since the link never uses "the mean value of v" as an integration variable.

I'm also not entirely sure the derivation at that link is correct anyway (not so much as a matter of math but as a matter of physics).

Since your "mean value of v" is not a correct integration variable, I would say it is coincidence.

11. Mar 14, 2017

### JohnnyGui

It's me who called it the mean value of velocity just to show that it's the same $v$ as the $v$ in the Lorentz transformation for mass. Perhaps the link already considered them to be the same $v$ values without having to mention that they're the mean velocities over a distance $ds$.

Isn't the $v$ in the Lorentz transformation the mean velocity of a mass that accelerates over a distance $ds$? If it's the final velocity after finishing $ds$ or the initival velocity before traveling $ds$ then it can't give the correct mass increase, right?

12. Mar 14, 2017

### Staff: Mentor

No. The Lorentz transformation does not describe a physical process of an object changing speed. It describes the mathematical process of switching from one inertial coordinate system to another.

13. Mar 15, 2017

### vanhees71

I lost a bit track of what the problem in this discussion is. What I can see is that there is some cofusion about how to derive the equivalent of what's called "work-energy theorem" in non-relativistic physics. In fact it's pretty simple, starting with manifest covariant formulations of the equations of motion. For a massive particle one uses the proper time as the "natural" scalar parameter to describe the trajectory of the particle under the influence of external fields. The equation of motion then takes the simple manifest covariant form
$$\frac{\mathrm{d}}{\mathrm{d} \tau} p^{\mu}=K^{\mu}, \quad p^{\mu}=m \frac{\mathrm{d}}{\mathrm{d} \tau} x^{\mu}.$$
Of course, these are not four independent equations since the definition of four-momentum and proper time implies that the particles obey the on-shell condition
$$p_{\mu} p^{\mu} =m^2 c^2=\text{const}$$
along the worldline of the particle, and this implies
$$\frac{\mathrm{d}}{\mathrm{d} \tau} (p^{\mu} p_{\mu})=2 p_{\mu} \frac{\mathrm{d}}{\mathrm{d} \tau} p^{\mu}=2 p_{\mu} K^{\mu}=0,$$
i.e., the Minkowski force $K^{\mu}$ (which in general is a function of position and momentum) must be Minkowski-perpendicular to the momentum. This constraint means that you simply solve the three spatial equations of motion
$$\frac{\mathrm{d}}{\mathrm{d} \tau} \vec{p}=\vec{K},$$
and the constraint guarantees the compatibility of the temporal component of the equation. Now one can easily derive the "work-energy theorem" in the usual way: Multiply the equation of motion with $\mathrm{d} \vec{p}/\mathrm{d} \tau$,
$$\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} \tau} \vec{p}^2=\vec{p} \cdot \vec{K}=p^0 K^0=\frac{E}{c} K^0.$$

14. Mar 15, 2017

### JohnnyGui

I understand. But if you'd want to integrate the Lorentz transformation over a distance $ds$ because a mass accelerates in that distance $ds$, then which velocity $v$ of that mass during that $ds$ would you have to choose to fill in the Lorentz transformation?

Apologies, but I have a hard time understanding your way of deriving the equation. I'm not familiar with that level of physics I think.
My problem is that, given the integration from this link :

I was not convinced that the two marked $v$'s are the same $v$'s. I thought that the $v$ in the numerator, which you calculate from acceleration over a distance $ds$, would be the final velocity at the moment a mass finishes that distance $ds$. While I think that for integrating a Lorentz transformation for an accelerating mass, the $v$ in the denominator would have to be the mean velocity of a mass during that distance $ds$.
That is until @Nugatory showed me in his post #4 that I was formulating the $v$ in the numerator the wrong way, to which I realised and concluded in my post #7 that they're both indeed the same mean velocities.

However, PeterDonis explained that my conclusion in my post #7 is probably a coincidence since they don't use the mean velocities over $ds$ in that integration and that the derivation in the given link is not entirely physically correct. That's where I'm at, at the moment.

15. Mar 15, 2017

### vanhees71

I see. Forget this link. This doesn't make sense for usual relativistic forces since they usually depend on both position and momentum (or velocity) of the particle.

16. Mar 15, 2017

### Staff: Mentor

The thing you've been integrating is not the Lorentz transform. The Lorentz transforms are something different (and much useful and important).

17. Mar 15, 2017

### Staff: Mentor

That doesn't make sense. The Lorentz transformation is not something you integrate over a distance. It transforms the coordinates of events from one chart to another. That's all it does.

18. Mar 15, 2017

### JohnnyGui

This might have nothing to do with the equation $E=mc^2$ but how would one then calculate the total relativistic mass of a proper accelerating object over a distance $s$?

19. Mar 15, 2017

### Staff: Mentor

A better term than "relativistic mass" would be "total energy in the original rest frame". You can do the entire calculation in that frame (the original rest frame); there is no need to Lorentz transform anything. (And you need to keep to a single frame anyway for the distance $s$ to be meaningful, since distance is frame-dependent.)

How you do the calculation depends on what you want to take as given to start with. The approach that I prefer is to take the basic spacetime geometry as given, i.e., that we have a flat spacetime metric, or, equivalently, that the interval between any two events in spacetime, given an inertial coordinate chart, is $ds^2 = - dt^2 + dx^2 + dy^2 + dz^2$. The other thing we need to take as given is that the proper acceleration of an object is the path curvature of its worldline. From those two things it is straightforward to derive the relativistic rocket equations in the article I linked to in post #2, which are for the case of constant proper acceleration, i.e., constant path curvature. Those equations give us the $\gamma$ factor as a function of distance $s$ (which is given the symbol $d$ in the article).

The only other thing we need is that total energy is given by $E = m_0 \gamma$, where $m_0$ is the rest mass. If we are taking spacetime geometry as given, as above, the simplest way to get that is the way vanhees71 described in post #6, since that way is basically just applying the properties of the spacetime geometry (basically time translation and space translation symmetry) to obtain formulas for energy and momentum.

20. Mar 15, 2017

### Mister T

Note that $\gamma$ is the Lorentz factor. I see no use of the Lorentz transformations in this derivation. Perhaps "Lorentz factor" is what you mean? If so, that's likely a source of confusion in this thread.

Let's look at the first part of that expression.

$$K=\int_{0}^{s}\frac{d\big(\gamma m_ov\big)}{dt}ds.$$

The $s$ that's the upper limit on the integral is the final value of the position, whereas the $s$ in $ds$ is a variable over which the integration takes place. It's common for mathematicians to warn against this notational sloppiness, but physicists largely ignore that warning. Note, though, that it gets you in trouble when the change of variables is carried out.

$$K=\int_{0}^{v}vd\bigg(\frac{m_ov}{\sqrt{1-(v/c)^2}}\bigg).$$

That is, $v$ is both the upper limit on the integral (the final velocity) and the variable over which you are integrating.

Note that this derivation is not as general as one might think. It applies only to particles. And secondly, only to particles with mass. When we use the phrase "a mass" it's not clear we're talking about a particle. A composite object might also be called "a mass" and these objects are capable of changing their internal energy. Thus an applied force might change not only their kinetic energy. It might also change their internal energy.

In the newtonian approximation the work-energy principle is a perfectly valid dynamical relation, but it's not a valid thermodynamic relation. That is, it's valid only for particle-like objects that have no internal energy.

21. Mar 15, 2017

### Staff: Mentor

That symbol can be used to denote the $\gamma$ that appears in the Lorentz transformation. But it can also be used to denote the function $\gamma = 1 / \sqrt{1 - v^2 / c^2}$, where $v$ is the velocity, possibly changing with time, of an object in a particular inertial frame. No Lorentz transformation is required to use $\gamma$ in this second sense.

22. Mar 18, 2017

### JohnnyGui

My bad, I indeed meant the integration of the Lorentz factor!

Just to make sure, could you please elaborate how you got $v$ and $dv$ from $ds$?

Is this the cause why integrating the Lorentz factor over a distance $d$ to get the total increased relativistic mass in terms of total energy is a wrong approximation for getting $E=mc^2$?

23. Mar 18, 2017

### Mister T

I didn't! I just read the link you posted and followed what they did. $dv$ was introduced using the integration by parts technique and $v$ was introduced in the definition of $F$.

No. All you're doing is integrating $\gamma m_o$ where $m_o$ is a constant. $\gamma$ is a function of $v$ so all you're doing is integrating that function with respect to $v$.