# Reconciling classical mechanics with special relativity - energy lost as heat in an inelastic collision

If a particle of rest mass $$m_0$$ travelling with a velocity $$u$$ collides with a stationary particle also of rest mass $$m_0$$ and then coalesces to form a new particle with invariant rest mass $$M_0$$ travelling at velocity $$v$$ parallel to $$u$$, then special relativity would give following:

$$\gamma(u)m_0c^2+m_0c^2=\gamma(v)M_0c^2$$

This is because energy is always conserved in this way in special relativity. Now, $$M_0$$ cannot be less than $$2m_0$$, because this would mean there was an increase in kinetic energy, which is impossible in a collision of this sort. $$M_0$$ can, however, be greater than $$2m_0$$ if the collision is inelastic (i.e. kinetic energy is not conserved).

In classical mechanics, an inelastic collision means that energy has been lost to the surroundings as heat or sound. Energy wastage as sound is an everyday experience and an example of energy loss as heat is given here (https://www.youtube.com/watch?v=I4f87E8Z0JA).

However, this seems to me like a contradiction between the reality predicted by the two theories. According to classical mechanics the rest mass stays the same and energy is wasted as heat or sound. In special relativity, the lost kinetic energy does not leave the system as such, but rather makes the coalesced particle heavier. If energy is lost as heat and makes the particle heavier, then we seem to have gained energy from somewhere, which would violate energy conservation laws.

How do we reconcile these two together?

• - Momentum conservation is an additional constraint here, you should include it. - Energy is not lost to the environment unless you change your equations too include this. The final particle can become hot. - I suggest to drop the subscript "0" everywhere. Mass is always rest mass. Mar 20, 2019 at 18:03
• Just a comment on nomenclature: many physicists lump special relativity into "classical mechanics" these days. Which is both strange for a historical perspective but fairly natural because the tooling is the same. But it does leave us with a problem expressing what is meant here as "Newtonian mechanics" won't do either (it would often be understood to exclude variational mechanics). Mar 21, 2019 at 0:32
• "According to classical mechanics the rest mass stays the same and energy is wasted as heat or sound." This is too restrictive an understanding of the process. In inelastic collisions some of the energy previous in bulk kinetic channels goes into other channels. Heat and sound, sure (and thermal energy is really the channel of last resort), but also into latched elastic energy in one or both bodies. Or converted into electricity is piezoelectric processes or ... The big problem with using energy conservation is always tracking down all the contributions. Mar 21, 2019 at 0:39

"In classical mechanics, an inelastic collision means that energy has been lost to the surroundings as heat or sound"

This is where your difficulty arises. Immediately after the collision (which is when the right hand side of your special relativity equation applies) no energy has been lost to the surroundings. Instead, energy has been transferred from kinetic to internal energy (of random vibration of atoms) within the colliding bodies.

This increase in internal energy corresponds to an increase in rest mass of the bodies, as your equation predicts.

It's much easier to see this if we deal with the collision (between identical bodies) in the centre-of-mass frame, the frame in which the vector sum of momenta of the bodies is zero before and after the collision. In this case, using $$m_i$$ and $$m_f$$ for initial and final (rest) masses of the bodies, $$\gamma (u) m_i=\gamma (v) m_f$$

For example, for $$u< and $$v<, if we multiply this equation by $$c^2$$ and expand the $$\gamma$$ factors binomially, discarding terms beyond $$u^2/c^2$$, we get $$(c^2+\tfrac12 u^2)m_i=(c^2+\tfrac12 v^2)m_f\ \ \ \ \text{that is}\ \ \ \ \ (m_f-m_i)c^2=\tfrac12 m_i u^2 -\tfrac12 m_f v^2$$

But energy conservation tells us that the increase in internal energy is $$U_f-U_i=\tfrac12 m_i u^2 -\tfrac12 m_f v^2\ \ \ \ \ \text{thus}\ \ \ \ \ U_f-U_i=m_f c^2-m_i c^2$$

After a while the extra internal energy is indeed lost in the form of heat to the surroundings, and the bodies regain their former masses.

• So in classical mechanics we focus only on the 'net result' once the bodies have regained their former masses, because the initial change in mass was negligible anyway? I like this answer, but could you please provide a reference to a source that says the same thing? Thanks. Mar 25, 2019 at 9:07
• (a) In pre-relativity mechanics, the bodies' total mass, Σ𝑚, doesn't change in a collision (interaction), though mass may be transferred from one body to another. Total energy (kinetic plus internal) is also conserved. Butt the conservation of total energy is not related to the conservation of mass. The one doesn't imply the other in pre-relativity Physics. (b) I recommend Taylor/Wheeler: 𝑆𝑝𝑎𝑐𝑒𝑡𝑖𝑚𝑒𝑃ℎ𝑦𝑠𝑖𝑐𝑠, as an excellent introduction to the concepts of Special Relativity, including mass and energy. Mar 28, 2019 at 17:29

The problem is that if you are talking about inelastic collisions in SR you normally talk about colliding particles, while in mechanics you talk about macroscopic objects. So in both cases you have inelastic kinematics, but the reasons are completely different. The reason for kinetic energy being transferred into mass, is that either the particle has an inner structure/inner degrees of freedom which allows for excitations like an electron in an atom being excited into another energy level, or one collides two fundamental particles (no inner structure) and gets out another two of different mass or a bound state. While the bigger mass of the emerging fundamental particles follows from their interaction with the Higgs potential, the bigger mass of a bound/excited system just comes from the additional potential energy. On this scale talking about heat and sound doesn't make any sense.

So if you are crashing together two cars at relativistic speeds, you will have energy losses through heat and sound that can't be accounted for without thinking about the microscopical structure of the cars. For example heat is just kinetic energy being transferred into the inner degrees of freedom of the car (the atoms in molecules are "wiggling harder") and the car gets heavier.