thermodynamics - Difference between kinetic energy "relativistic" mass and heat mass

I have been told in several answers that the term "relativistic mass" is no longer considered adequate wrt. the mass/energy increase in kinetic energy.

Yet, I read that 1 kg of gold's mass increases by 10^-14 kg if its temperature is raised by a couple of degrees, and that the very "invariant mass" of the proton is actually only 1% proper mass and the rest is energy in the form of virtual photons/ gluon soup.

I am baffled; can you please clarify when something qualifies for the attribute of "mass"? Can you explain why energy bound as heat in gold is more "mass" then energy bound as kinetic energy in motion? Former "relativistic" mass is sensitive to gravity as is "heat mass" in gold; it determines a larger trajectory at LHC, just as a proton does (an electron with 1836 "relativistic" masses has the same distance from the center of LHC as a proton, doesn't it?) So, what properties of mass has kinetic energy mass lacking in order to qualify for that title?

Answer

Suppose you start with your (stationary) 1kg block of gold. If you raise its temperature you have to add energy to it, and that means it's different after you've raised its temperature. For example you could shine an infrared lamp on it, in which case you've added the energy from the IR lamp. The mass changes because if you add an energy $E$ the mass goes up by $E/c^2$.

However suppose instead of heating the gold you go shooting off in a rocket at $0.999c$. In your rest frame the gold is now moving at $-0.999c$ so in your rest frame the momentum of the gold is given by the relativistic momentum formula:

$$ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$

where $m$ is the invariant mass i.e. 1 kg.

The old idea of relativistic mass came from writing the momentum as:

$$ p = \frac{m}{\sqrt{1 - \frac{v^2}{c^2}}} v = m_r\,v $$

where $m_r$ is defined as the relativistic mass:

$$ m_r = \frac{m}{\sqrt{1 - \frac{v^2}{c^2}}} $$

But you need to be clear that the relativistic mass is a computational device and it does not mean the mass is changing. Nothing happens to the gold. If I am sitting by the gold while you speed off in your spaceship I will see nothing happen to it.

This is the key point. When we heat the gold all observers, no matter what their speed, will agree that the gold has changed because we all see it absorbing the light from the IR lamp.

The invariant mass is the parameter $m$ in the relativistic energy momentum equation:

$$ E^2 = p^2c^2 + m^2c^4 $$

This energy $E$ is one component of the four-momentum and the four-momentum is a coordinate independent object. So any observer, moving at any speed in any way, can start with the four-momentum transformed into their coordinates and calculate the invariant mass $m$. And all such observers will end up with the same value for $m$.