Energy, Mass and the... Speed of Light?!

[Trekker4747]

Einstein's famou mass-energy equivalence formula says that if you multiply the mass of an object by the speed of light squared you get how much energy would be released if that object were to be converted into energy.

So I'm wondering.... What does the speed of light have to do with it?

 

[Jadzia]

There's something you can do in physics called quantitative analysis I think it's called, where you look at equations in terms of their units.

Energy is measured in newton-metres.
mass is measured in kilograms.
velocity is measured in metres per second.

Combining these into an equation, there are limitations to how they can combine because the units have to match up on both sides of the equation.

So qualitatively, it's the same as Kinetic Energy = (1/2) mass . velocity^2
albeit for the factor (1/2)

mass-energy must use absolute velocity = speed of light. While the half is lost somewhere in the theory.

 

[Lindley]

Just a guess, but could the half getting lost have anything to do with antiparticles?

 

[Trekker4747]

Qualitatively, it's like Kinetic Energy = (1/2) mass . velocity^2

But mass-energy uses absolute velocity = speed of light. The half is lost somewhere in the theory.

Yes, but my question is what c has to do with a direct energy-mass conversion.

Kinetic energy makes sense. You've more energy the greater your velocity.

But we're not talking about "mass moves and the the speed of light and then hits something" we're talking about "mass converts to energy."

 

[Jadzia]

Just a guess, but could the half getting lost have anything to do with antiparticles?

I can't remember now but I think it's just lost in the integration, because this isn't newtonian mechanics. The equations are bent, so the constants are different... Not about antiparticles. But I might be wrong.

 

[Asbo Zaprudder]

Assuming conservation of energy and momentum, Special Relativity theory yields the following equation for total energy of an object with rest mass m0 and velocity v relative to the observer's frame of reference (derived by considering the collision of two objects as seen from each of their frames of reference):

E= gamma * m0 * c^2

where gamma is the Lorentz factor:

gamma = 1 / sqrt(1 - (v/c)^2)

The series expansion of gamma is:

1 + ((v/c)^2)/2 + ((v/c)^4)*3/8 + ...

Multiplying out gives:

E = (m0 * c^2) + (m0 * v^2)/2 + ...

That can be interpreted as:

E = rest mass (no velocity term) + kinetic energy

For v << c (non-relativistic) case, we get the Newtonian approximation:

Kinetic energy ~ (m0 * v^2)/2

The term (gamma * m0) when written as m, is known as the relativistic mass, so yielding the familiar:

E = m * c^2

 

[Jadzia]

I think trekker wants to see in more direct visual terms, why it is c and not another number.

Is there a visual explanation involving the familiar kinetic energy equation, where mass has a 'rest' energy = mc^2, perhaps since it moves at constant rate 'c' through space-time, or something like that?

 

[Lindley]

You may as well ask why c is what it is.

 

[Asbo Zaprudder]

Sorry, I haven't come across a pictorial explanation although that doesn't mean that there isn't one. The term basically drops out of the equations as a result of requiring momentum and energy to be conserved. The equations imply that energy and mass are equivalent, and that the conversion factor between their units of measurement is c^2. As you mention, in our rest frame, coordinate time and proper time are the same, and I've heard is said that we are effectively travelling into the future at the speed of light. I'm not sure that this helps develop the argument, however.

ETA: I just checked the derivation, and the c^2 factor come from integrating the longitudinal force(gamma^3)*m0*dv/dt when deriving the relativistic form of the kinetic energy -- (gamma-1)*m0*c^2. This reduces to the Newtonian form when v << c, and can be written as (gamma*m0*c^2) - (m0*c^2). The second term is identified as the rest mass energy and the first as the total energy.

 

[hyzmarca]

It's simple, all particles with zero rest mass propagate at the speed to light in all reference frames.

However, particles with zero rest mass have momentum and inertia. This is why solar sails work, among other things. The formula for momentum is p=mv. A particle with a mass of 0 would not have any momentum. Yet, these massless particles do have momentum. Which means that they have mass. They're massless, meaning that they don't have mass, but they do have mass. And it's this big WTF moment that leads us into the concept of relativistic mass and mass-energy equivalence.

The formula E/(c^2) = m is, perhaps, more useful in understanding the principle than the more traditional formulation is.

We know that photons have momentum because we can measure it. We know that photons have energy because we can measure it. And we know that photons are massless.

When using relativistic mass as (m), momentum remains p=mv, just as in Newtonian physics. The momentum of a photon is, thus, is p=m(xc) where x is the relative vector of the photon (since the speed of light is a dimensionless constant, and velocity is a vector).

From there, we can get the formula p/v = m and p/(xc) = m

We know, from measurement, that p = x(the plank constant)/(the wavelength of the photon) and that this equals x(E/c)

So that give us m = x(E/c)/(xc) = xE/x(c^2) = E/c^2.

In other words, the relativistic mass of a photon is equal to it's energy divided by the speed of light squared.

And of course, the energy of the photon equals it's relativistic mass times the speed of light squared. The famous E=mc^2.

At this point, let's say you take a massive object and convert it to photons. E=mc^2 applies here.
There are several ways to do it. Nuclear fission is one, though this is an inefficient and incomplete process, no where near total conversion. Matter-antimatter annihilation is better, probably. No matter how you accomplish it, the energy released by transforming a massive object into massless particles is equal to the lost mass times the speed of light squared.
This is the case because the mass lost in such a conversion will be equal to the relativistic mass of the resulting photons.

 

[Asbo Zaprudder]

Erm, gamma is infinity when v = c, it's not 1.

After De Broglie, the momentum is p=h*(frequency/c), or simply p=E/c as E=h*frequency (Planck). This yields E=m*c^2 if the effective momentum for a photon is taken to be p = m*c, where m is the equivalent mass due to energy, not rest mass (which is zero). In other words, we've assumed what we set out to derive. The argument is tautological.

It doesn't necessarily follow that this equation applies to particles with a non-zero rest mass. Their momentum is (gamma * m0 * v). The equation relating momentum and energy for them is E^2 = (p*c)^2 + (m0*c^2)^2.

 

[TIN_MAN]

Just a guess, but could the half getting lost have anything to do with antiparticles?

I, of course, would say yes, but ultimately, I think it has to with anti-photons? In conventional physics the photon is considered to be its own anti-particle, but in scalar electrodynamics there is complete time symetry, so the photon travels forward in time, and the anti-photon travels backwards in time (and is properly a tachyon). All electromagnetic radiation is composed of photon/anti-photon pairs, or when seen as a wave, is a longitudinal scalar wave pair. One half travels from the past to the future, the other half travels from the future to the past. When they "meet" at the point of 'observation' a particle (photon) is seen to manifest. An extremely rapid succession of such 'conjugations' gives us the measure we know as 'C'. This is an oversimplification, there is much more to it than this, of course, but this'll give you something new to chew on?

 

[hyzmarca]

Erm, gamma is infinity when v = c, it's not 1.

Thank you, I missed that. It would then be more correct to say that the Lorentz factor doesn't apply when dealing with massless particles, because they are massless. I'll amend that.

My mistake was confusing invariant mass for relativistic mass, which calculating momentum using relativistic mass as m, p=mv remains the correct formula. The Lorentz factor only works if invariant mass is used, and doesn't work for massless particles due the 0/0 problem.

I don't think that the tautological nature of the explanation is a problem here, because a tautological explanation is what was called for.