# Just some basic question on e=mc2

1. Mar 1, 2012

### jeremynull

I know that any time energy is generated, you can use the equation in the form of E = mc^2.

Also, that energy becoming matter can be described as m = E/(c^2).

But, going as those are, is it also assumable that you could say c^2 = m/E? In other words, would the speed of light in a perfect vacuum represent all the ratios of mass to energy that each element can exist, up to a certain point in natural occurrence? What I mean is, we have stable points at which an element is represented on the periodic table. So if the equation holds, and mass isn't being converted to energy or visa-versa, wouldn't this form of the equation represent its resting state, or in other words, each element's stable state?

I was just wondering, because I haven't seen anybody talk about it much or relate to the equation in that way..

2. Mar 2, 2012

You have to be a bit careful about terminology here, because the word "mass" is most commonly used to describe the mass which a particle would be measured to have when stationary. This "rest mass" is sometimes denoted m0, and probably the more important equation when considering particles' masses and energies is

E2 = p2c2 + m02c4

This relates the total energy E of a particle to its rest mass m0 and (the magnitude of) its momentum p, thus incorporating its kinetic energy as well. When the particle is stationary, the first term on the right vanishes and it reduces to E=m0c2. But in the case of a "massless" particle such as a photon, m0 = 0 and the equation then reduces to another well-known formula

E = pc​

If you measure the velocity of a particle of energy E and momentum p, you will indeed find that, in terms of the Newtonian formula

p = mev​

the "effective" mass me of the particle appears to be larger than m0 - in fact

me = E/c2 = m0γ​

where

γ = 1/√(1 - v2/c2)​

It's important not to get these two quantities confused. The more fundamental and important is m0 - this is a Lorentz invariant, which means that observers in all frames of reference will agree on its value if they measure it, whereas me varies between observers as it depends directly on how fast the particle is travelling in each one's own frame of reference.

3. Mar 2, 2012

### jeremynull

Oh - I see now. The equation for energy is just a conversion equation, so the way I arranged it in the latter doesn't actually mean anything.

And then so the equation that I was windering about was the one that all speeds or frames of reference would agree on the object as having the same value.

Thanks - I'm not a physics guy, so this had me confused for a while.