Is mass the only store of energy

[Daniel Wareham]

After thinking about E=MC^2, I thought that mass is the only store of energy. I have found many “stores of energy” that are actually just increase in mass. For example kinetic energy, it’s not a store of energy by itself, it’s actually because of special relativity that it has more mass. When something is hotter it doesn’t store energy by temperature, it has energy because the particles are vibrating faster and so have more kinetic energy, therefore having more mass due to special relativity. I just want to ask is it true for all stores of energy that it’s just a increase in mass somehow.

 

[Ibix]

Here, you are using mass in the sense of "relativistic mass", which turns out to be a synonym for "total energy". Taken the way you are meaning mass, you are correct.

However, for the last several decades (at least) physicists have used "mass" to mean "invariant mass" (aka "rest mass"). In this sense, you are incorrect - kinetic energy, for example, does not necessarily contribute to invariant mass, and light has energy but not mass.

So yes and no, in short. Probably more "no" than "yes", in modern terminology, I'm afraid.

 

[sysprog]

This seems to me to be definitional -- if you pre-define all changes in energy levels of any kind as changes in mass, then mass change is the only required determinant of energy change.

 

[Dale]

For example kinetic energy, it’s not a store of energy by itself, it’s actually because of special relativity that it has more mass.

The mass you are referring to is relativistic mass, which is as you point out just another name for total energy. Because this makes it a completely redundant concept, as @Ibix says, this concept of mass has been discarded by professional physicists for decades. Now, the term “mass” refers to the invariant mass which is not equal to energy but rather $m^2 c^2 =E^2/c^2-p^2$

 

[sysprog]

In @Dale's equation, $p$ is the kinetic part of the energy of the rest-mass-possessing object.

$E=mc^2 \pm\omega$ is valid for rest-massive objects when $p=0$, and (now disregarding the positive or negative remaining infinitesimal) $E^2 = m^2c^4 + p^2c^2$, or $E=\sqrt{(mc^2)^2+(pc)^2}$, is valid for any object, rest-massive or not, moving or at rest, while for 'zero' rest-mass particles (photons), $E=pc$.

It's insightful, but perhaps not always useful, to recognize the inter-conversibility of energy and mass.

 

[Dale]

In @Dale's equation, p is the kinetic part of the energy of the rest-mass-possessing object

Just to be clear, $p$ is momentum, not any kind of energy. I believe that you understand that, but the way you wrote it seemed confusing to me.

Obviously, any object with momentum also has kinetic energy, but momentum is not part of energy. (Although they are both parts of the four-momentum)

 

[sysprog]

Just to be clear, $p$ is momentum, not any kind of energy. I believe that you understand that, but the way you wrote it seemed confusing to me.

Obviously, any object with momentum also has kinetic energy, but momentum is not part of energy. (Although they are both parts of the four-momentum)

I agree that momentum, per se is not energy; it's a vector quantity that requires a direction in order to be fully described. I understand that $p$ is momentum, but given that momentum is a composite that includes mass ($p=mv$), and that I was trying to elucidate the distinction between mass in se, i.e mass qua mass, not mass in toto, i.e. mass inclusive of its kinetic (or other extrinsic) energy, and given that the ratio of kinetic energy of two objects having the same momentum is inversely proportional to their static masses, I used the term 'momentum' not quite as strictly in my reply as you did in yours.

 

[Ibix]

($p=mv$)

$p=\gamma mv$ in this context.

 

[sysprog]

$p=\gamma mv$ in this context.

Fair enough, @Ibix -- that's true -- I used the non-relativistic form because I sought by that parenthetical expression in that sentence only to indicate that momentum includes mass.

 

[Ibix]

The problem with $p=mv$ in this context is that, besides being the correct Newtonian expression, it is the correct relativistic expression if you interpret $m$ as relativistic mass. Which we're trying to tell the OP not to do...