# I'm wondering why the value that converts mass to energy happens to be lightspeed

1. Sep 21, 2011

### Jarwulf

Alright I've asked before but this is still bugging me.

given that the mass energy equation is this

energy=mass(lightspeed)2

xz2=x(lightspeed(z/y) )2
y2

where
x=arbitrary unit for mass
y =arbitrary unit for time
z=arbitrary unit for length

I'm wondering why the value that converts mass to energy just happens to be lightspeed and not some other arbitrary/nonarbitrary value.

I don't see how the speed of light in a vacuum can 'cause' the rest energy of a unit of mass or vice versa if you get my drift?

Last edited: Sep 21, 2011
2. Sep 21, 2011

### Janus

Staff Emeritus
Re: e=mc2

Its a consequence of the the fact that c is an invariant speed (everyone measures it to have the same value relative to themselves.)

Once you establish that such a speed exists, it follows that this speed becomes the speed limit for the universe, (And thereby making it the only invariant speed.) and further consideration leads to the conclusion that mass and energy are related to each other by this speed.

3. Sep 23, 2011

### Bill_K

Re: e=mc2

Are you asking why E = mc2? The main reason is it's confirmed experimentally, but one might also ask why is it plausible. Can a good theoretical argument be constructed for it?

If that's the question, here's what I would say. Relativistic mechanics must be consistent with classical mechanics for small v, and we know that for small velocities E = ½ mv2. Secondly you want mechanical quantities to behave simply when a Lorentz transformation is applied. Under a Lorentz transformation t = γ t0, where t is the time in an arbitrary rest frame and t0 is the proper time. So a reasonable guess to make is that E behaves the same way that time behaves, namely E = γ E0 where E0 is the energy in the rest frame.

Now γ = 1/(1-v2/c2) and for small v this is γ ≈ 1 + ½ v2/c2. So

E = γ E0 ≈ E0 + ½ v2/c2E0. You want this to be E0 + ½ mv2, so to get the second terms to agree you must have E0 = mc2.

4. Sep 23, 2011

### DrStupid

Re: e=mc2

It's better not to guess but to calculate:

The change of energy is

$dE = F \cdot ds$

Newton says

$F = \frac{{dp}}{{dt}}$

and

$p = m \cdot v$

$dE = m \cdot v \cdot dv + v^2 \cdot dm$

Now we need an expression for the inertial mass m. If there is a dependence from velocity it should be the same for all bodies. Therefore I start with

$m\left( v \right): = m_0 \cdot f\left( v \right)$

where f(v) is a function of velocity independent from the body and from the frame of reference and m0 is the mass of the body in rest. So we already know

$f\left( 0 \right) = 1$

Due to isotropy f must also be symmetric

$f\left( { - v} \right) = f\left( v \right)$

Now let’s assume a body A with the velocity v and a body B at rest. Both bodies should have the same rest mass m0. The product C of a perfectly inelastic collision shall have the rest mass M0 and the velocity u. The conservation of momentum leads to

$p = m_0 \cdot f\left( v \right) \cdot v = M_0 \cdot f\left( u \right) \cdot u$

At this point I have to make a reasonable assumption (I will check it later): If energy is conserved the mass of C shall be the sum of the masses of A and B:

$M_0 \cdot f\left( u \right) = m_0 \cdot f\left( v \right) + m_0 \cdot f\left( 0 \right)$

This results in

$f\left( v \right) = \frac{u}{{v - u}}$

To get the velocity u I change to a frame of reference moving with the velocity v. Now body B moves with -v and body A is at rest. As the situation is symmetric the velocity of C is

$u' = - u$

The next steps depends on the transformation. Galilei transformation

$u' = - u = u - v$

$f\left( v \right) = 1$

Therefore in classical mechanics inertial mass is independent from the frame of reference and the change of Energy is reduced to

$dE = m \cdot v \cdot dv$

$E = E_0 + {\textstyle{1 \over 2}}m_0 \cdot v^2$

In SRT Galilei transformation is replaced by Lorentz transformation

$u' = - u = \frac{{u - v}}{{1 - \frac{{u \cdot v}}{{c^2 }}}}$

Everything else remains unchanged. This leads to

$m\left( v \right) = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$

Although this is often called "relativistic mass" it is still the good old inertial mass as used in Newton’s definition of momentum. With

$dm = \frac{{m_0 \cdot v \cdot dv}}{{c^2 \cdot \sqrt {1 - \frac{{v^2 }}{{c^2 }}} ^3 }} = \frac{{m \cdot v \cdot dv}}{{c^2 - v^2 }}$

The change of Energy is

$dE = dm \cdot c^2$

$E = E_0 + \left( {m - m_0 } \right) \cdot c^2$

With Einstein’s equation for rest mass and rest energy we get an expression for inertial mass and total energy:

$E = m \cdot c^2$

Due to the additivity of energy

$E = \sum\limits_i {E_i } = \sum\limits_i {\left( {m_i \cdot c^2 } \right) = \left( {\sum\limits_i {m_i } } \right)} \cdot c^2 = m \cdot c^2$

$m = \sum\limits_i {m_i }$

Therefore my assumption above has been correct.

5. Sep 24, 2011

### PhilDSP

In deference to the most fascinating posts already given, maybe it's worth pointing out that it is not the speed of light involved but the speed of light squared. Physically $c$ represents the distance covered by a photon or an EM wave in vacuum in one second. One possible interpretation of $c^2$ is that it represents the spatial area in a plane swept out by 2 planar waves each orthogonal to the other in one second, though there are certainly other possible interpretations.

In the equation $E = mc^2$ mass is represented in only one spatial dimension, is it not? (Because mass is modeled as being centered at a single spatial point) But doesn't $c^2$ seem to imply that 2 spatial dimensions are necessary to represent what is happening with energy transported at the speed of light in this case when it encounters mass or is a component of mass?

Last edited: Sep 24, 2011
6. Sep 25, 2011

### Spacie

I too get inspired by this interpretation. To me it says that a force is a distortion in the fabric of space. I envision it straightforward: empty space is flat and its "curvature" is non-existent. A curved space literally describes how that a formerly flat side of a... cube? ...becomes curved like a sail in the wind. Then mass acts on the geometry of space like wind that inflates the sail.

Hmm... In my understanding of the geometry involved, the mass/gravity is truly a 5-dimensional event, while EM forces act in 4-dimensions. So, yeah, I guess that's about the same as you saying that, if mass is represented by one spatial dimension, then its effect on the geometry of space should involve an additional spatial dimension. Yes, I totally agree with you. I wonder what the local gurus have to say.

7. Sep 25, 2011

### PhilDSP

Just to make sure that no confusion is generated by what I posted: gravity, GR and curved space do not need to be invoked to describe or explain the relationship $E = mc^2$. It is associated strictly with SR. Three orthogonal spatial dimensions seem to be quite enough to model the movement of energy (with respect to time) in this case. Mass here has more the meaning of being a localized collection of energies than a measure of how attractive the object is to other objects.

Last edited: Sep 25, 2011
8. Sep 25, 2011

### kmarinas86

A more fundamental derivation can be found here:

There are less than 12 steps in this very simple derivation, and it has nothing at all to do with Lorentz transformations.

This derivation assumes that all the energy in a mass comes from fundamental force particles each with a momentum equal to E/c, where E is the energy of each force particle and c is a constant. If this is not the case for all fundamental force particles in a mass, then E = mc^2 is not true.

Last edited: Sep 25, 2011
9. Sep 26, 2011

### DrStupid

After reading Einstein's original paper [http://www.physik.uni-augsburg.de/annalen/history/einstein-papers/1905_18_639-641.pdf" [Broken]] I can not agree. As far as I understand it, Einstein didn't derived the equivalence of the so called "relativistic mass" and total energy but of rest mass and rest energy of a body and his derivation has almost nothing to do with the derivation in your link above.

How do you get the first equation without Lorentz transformation?

Last edited by a moderator: May 5, 2017
10. Sep 27, 2011

### kmarinas86

Experiments, that's how.

11. Sep 27, 2011

### DrStupid

Then here is my single-step derivation of mass-energy equivalence:

(1) E=m·c²

12. Sep 28, 2011

### kmarinas86

That may work for matter-antimatter creation/annihilation, but p=E/c for photons was discovered first, and it can be measured with higher accuracy.

13. Sep 28, 2011

### dm4b

Im sure you will get plenty of replies with people saying matter of factly that because experiment tells us so and this calculation shows this and that, and as a result we totally understand why this is the case.

I disagree.

Perhaps, you are like me and think that this hints at something deeper about the Universe, or reality, that we don't quite yet understand.

Having studied both GR and SR, I was still left with this sneaking suspicion.

So, I think the honest answer is, "we don't FULLY know the answer to your question yet".

14. Sep 28, 2011

### DrStupid

Please refer to a corresponding source.

By the way: Starting from p=E/c you only need the momentum p=m·c of the photon (according to Newton's definition) to get E=m·c². There would be no need for any thought experiments or complicated derivations.

15. Sep 29, 2011

### kmarinas86

Arthur Schuster conceived of antimatter-matter annihilation in a letter he wrote to Nature in 1898. Obviously, the value of the energy involved was not experimentally tested until antimatter could be generated. Positrons were experimentally found by Carl Anderson in 1932, over 15 years after Einstein's development of General Relativity. Planck won the Nobel prize on his study of quantum theory, including his model of blackbody radiation (where p=E/c), in 1918.

There is no corresponding source for the exact point I made. You will just have to connect the dots.

16. Sep 29, 2011

### DrStupid

I'm not surprised. Einstein wouldn't write three pages where a single sentence is sufficient. He wasn't that stupid.