Is $E=mc^2$ strictly just a conversion formula?

Question

$E=mc^2$
Does not this imply that there is no energy without mass, or is it strictly a conversion formula?

Answer 1

That equation allows mass units to be converted to energy units. In this sense, it is a unit conversion formula - but it is also much more than just that.

It tells us that energy and mass can actually be transformed back and forth into each other. For example, in a reaction in which a large amount of energy is released, the products of the reaction will weigh less than the reactants did before the reaction occurred. Conversely, if we store a large amount of energy in a system, that system will weigh more than it did before the energy was added to it.

It also means that in a system that is gravitationally bound together because of the mass of its constituent parts, that binding energy behaves just as if it had mass itself - which then adds to the mass of the system. If the starting mass is big enough, which squeezes the system together more strongly, which makes it act like its mass was increased, and so forth - then this effect will create a gravity-driven runaway condition which ends with the whole mass collapsing into a black hole.

This means that the "unit conversion formula" $E = mc^2$ also predicts black holes.

Answer 2

This is the case for objects without momentum, the full Energy-momentum formula is $$E^2=(pc)^2+(mc^2)^2$$ From this we can see that a particle without mass still can have energy (a photon)

Answer 3

It appears that you are new to the subject just like I am. I heard a statement some time ago which helped me understand the stuff easily even though I knew practically nothing about it. The statement was

"The traditional mass in Newtonian Physics is itself in a way composed of energy."

Meaning what we call as matter - electrons, protons , neutrons are itself composed of energy. Hence the formula, in this way, tells us that if we measure a mass of say "m" units of an object,then this mass is composed of energy of "$mc^2$" units.

This energy is called rest-mass energy , which is same for any reference frame.

Does not this imply that there is no energy without mass

When you are saying the above statement you are assuming that the rest mass energy of our object is the only energy possessed by the object. However there may be some other form of energy in our object, say Kinetic Energy. So even when mass is zero (photon), it still may possess momentum and hence energy.

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Conclusion:

The equation $E=mc^2$ gives only the rest-mass energy relation of our object. This is not the complete description of our object's energy. Hence saying that "energy implies mass" is invalid but "mass implies energy" is valid.

Answer 4

So, this appears to be true but it requires a very particular way of thinking about mass. Most of the people who are saying that it's false do not share that way of thinking about mass, that is why they say it is false.

In the way that it is true, you have to think of mass as an aggregate property of a system, the resistance of that system to acceleration. And you similarly have to be very careful with energy, as potential energy has an arbitrary zero and thus total energy is not well-defined as a scalar. But if you do this carefully, one of Einstein's big calculations was that when you heat up the gas in a box the box indeed has more inertia.

It turns out this also applies to a box with photons in it, even though photons are “massless,” for which there is a shortcut: uniform acceleration is the same as gravitational redshift which has the well-known formula for wavelength, $$ \frac{\Delta \lambda}{\lambda}=\frac{ g~\Delta y}{c^2}.$$ The momentum carried by a photon is $p=h/\lambda$ and some calculus gives $\Delta p= -h\Delta\lambda/\lambda^2$ for its momentum loss as it ascends, so if it bounces back and forth between two mirrors held at height $\Delta y$ apart it delivers a net momentum of $2|\Delta p|$ downwards (doubled due to reflection) over a time $\Delta t=2\Delta y/c$ equivalent to a force $h g /(c\lambda)$ downwards, but $hc/\lambda=hf=E$ is the energy of the photon and so there is an added downward force the same as if there were any additional $E/c^2$ of mass, due to gravitational redshift. Since gravitation and acceleration are equivalent, this means that any box containing a photon gas has an extra component of how it resists that acceleration due to the energy in the photon gas.

So yes, energy does in fact function as a sort of inertia for a system in a very general sense. However, I should point out that the fact that massless particles can sum up together with mirrors to create a more-massive system immediately implies that systems do not compose by adding masses in general in relativity. In other words you have a formula like

$$M_\text{system}=\sqrt{\left(\sum_{k=1}^N E_k/c^2\right)^2-\left(\sum_{k=1}^N \mathbf p_k/c\right)^2},$$

And this idea that $M_\text{sys} = \sum_k E_k/c^2$ requires that you are looking at a system with no net momentum $\sum_{k}\mathbf p_k=0$, and if you build a system from subsystems those subsystems had better also have no net momentum to get your mass addition formula to work right. (I think. This argument is very rusty in my memory.)