Is c^2 = m/E a Valid Representation of Stable Elemental States?

[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..

 

[AdrianTheRock]

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 = m_ev​

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

m_e = 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 m_e varies between observers as it depends directly on how fast the particle is traveling in each one's own frame of reference.

 

[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.