# Is this relativistic mass?

I have seen in a lot of places in here clearly stating that relativistic mass is outdated, that we can make do just fine with the concept of invariant mass,etc. But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object. This confuses me. Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases? Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?

But I've also seen people saying that a hotter object is heavier than a colder object. Where they say that the internal energy of the constituent atoms contribute to the mass of the object.

Yes, and this is not in contradiction with the convention of invariant mass. Mass is defined by the identity $$m^2=E^2-p^2$$ (in units where $$c=1$$), which implies that it isn't additive. So say I have two electrons, each with mass $$m$$. If one is moving to the right at $$0.9c$$, and the other is moving to the left at $$-0.9c$$, then the mass of the whole system is greater than $$2m$$. However, each electron individually still has mass $$m$$.

• Okay thanks and does relaivistic mass add linearly? Commented Apr 9, 2019 at 17:21
• No, that's what I mean by saying that it isn't additive.
– user4552
Commented Apr 11, 2019 at 13:42
• But you said that for invariant mass. I asked if it was the same case for relativistic mass. Commented Apr 12, 2019 at 8:07

Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?

Outdated does not mean wrong. It means confusing, since we have much better tools to study the microcosm of atoms, since it was realized that special relativity in the motion of particles is completely and cleanly described by defining the relativistic four vectors, which obey vector equations.

Here is the energy momentum four vector:

$$\vec{p}=\begin{bmatrix}E\\p_xc\\p_yc\\p_zc\end{bmatrix}=\begin{bmatrix}E\\\vec{p}c\end{bmatrix}$$

$$\sqrt{P\cdot P}=\sqrt{E^2-(pc)^2}=m_0c^2$$

and to the right the definition of the invariant mass:

The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.

As with the everyday length of vectors, lengths are not addivive, one has to use vector addition, in the special relativity case as defined on the right.

Doesn't an increase in internal energy mean an increase in the constituent atom's velocity?

Note what you said:

Doesn't relativistic mass imply that I should observe your mass to increase as your velocity increases?

bold mine.

When you hold a solid, is the solid moving with respect to your observation? The statement holds mathematically for each individual electron and atom with respect to the other, but statistically there is no motion that an external observer can measure. The four vector formalism simplifies this. The addition of all the four vectors in a solid will give the total four vector whose length is the mass you can measure in the laboratory. Hotter items have higher momenta and the total addition of four vectors will give higher invariant mass for a hot object than it has when cold.