Exploring the Mass-Energy Equation

[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
y²where
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?

 

[Janus]

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.

 

[Bill_K]

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?

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.

 

[DrStupid]

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.

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

$p = m \cdot v$

This leads to

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

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

$u' = - u = u - v$

leads to

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

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

$dE = m \cdot v \cdot dv$

The integration leads to

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

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

$dE = dm \cdot c^2$

Now Integration leads to

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

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

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

the inertial mass is additive:

$m = \sum\limits_i {m_i }$

Therefore my assumption above has been correct.

 

[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?

 

[Spacie]

... 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...

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.

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?

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.

 

[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.

 

[kmarinas86]

A more fundamental derivation can be found here:

http://www.adamauton.com/warp/emc2.html

The Derivation of E=mc2

Perhaps the most famous equation of all time is E = mc2. The equation is a direct result of the theory of special relativity, but what does it mean and how did Einstein find it? In short, the equation describes how energy and mass are related. Einstein used a brilliant thought experiment to arrive at this equation, which we will briefly review here.

First of all, let us consider a particle of light, also known as a photon. One of the interesting properties of photons is that the have momentum and yet have no mass. This was established in the 1850s by James Clerk Maxwell. However, if we recall our basic physics, we know that momentum is made up of two components: mass and velocity. How can a photon have momentum and yet not have a mass? Einstein’s great insight was that the energy of a photon must be equivalent to a quantity of mass and hence could be related to the momentum.

Einstein’s thought experiment runs as follows. First, imagine a stationary box floating in deep space. Inside the box, a photon is emitted and travels from the left towards the right. Since the momentum of the system must be conserved, the box must recoils to the left as the photon is emitted. At some later time, the photon collides with the other side of the box, transferring all of its momentum to the box. The total momentum of the system is conserved, so the impact causes the box to stop moving.

Unfortunately, there is a problem. Since no external forces are acting on this system, the centre of mass must stay in the same location. However, the box has moved. How can the movement of the box be reconciled with the centre of mass of the system remaining fixed?

Einstein resolved this apparent contradiction by proposing that there must be a ‘mass equivalent’ to the energy of the photon. In other words, the energy of the photon must be equivalent to a mass moving from left to right in the box. Furthermore, the mass must be large enough so that the system centre of mass remains stationary.

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.

 

[DrStupid]

A more fundamental derivation can be found here:
http://www.adamauton.com/warp/emc2.html

After reading Einstein's original paper [http://www.physik.uni-augsburg.de/annalen/history/einstein-papers/1905_18_639-641.pdf" ] 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.

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

How do you get the first equation without Lorentz transformation?

 

[kmarinas86]

Experiments, that's how.

 

[DrStupid]

Experiments, that's how.

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

(1) E=m·c²

 

[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.

 

[dm4b]

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

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?

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".

 

[DrStupid]

p=E/c for photons was discovered first

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.

 

[kmarinas86]

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.

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.

 

[DrStupid]

There is no corresponding source for the exact point I made.

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