Conservation of generalized energies in Newtonian and Relativistic Systems

Question

I was considering the following problem:

In a closed system, it is assumed that mass, momentum and energy is conserved. If we label the total mass of the System $M(i)$ at a time i, the total Momentum as $ M(i)V(i) $ and the energy as $ \frac{1}{2} M(i)V(i)^2$, then the following holds true

For any time $i,j$ in the domain of the system (ie while the system remains closed)

$$ M(i) = M(j)$$ $$M(i)V(i) = M(j)V(j)$$ $$\frac{1}{2}M(i)V(i)^2 = \frac{1}{2}M(j)V(j)^2$$

Now I was interested in other conserved quantities. Particularly

$$ \frac{1}{6} M(i)V(i)^3$$

My reason for asking is that the sequence above is simply the integration of Mass (assumed to be constant w.r.t velocity) with respect to velocity.

Under what conditions is this conserved? Is it actually conserved in real life?

Answer 1

The pattern does not continue. In fact it doesn't always hold as written. In particular:

• $\sum_i m_i = \sum_j m_j$ only if rest mass is conserved, which it is not when, say, you have particles and antiparticles being created by and annihilating into photons.
• $\sum_i m_i \vec{v}_i = \sum_j m_j \vec{v}_j$ only if all momentum in the system is of the form $\mathrm{(mass)} \times \mathrm{(velocity)}$. This is not the case if anything is moving relativistically.
• $\sum_i m_i \vec{v}_i^2 = \sum_j m_j \vec{v}_j^2$ follows from the first condition together with the assumptions that there is no energy transfer between kinetic energy and any other form of energy, and again that nothing is moving relativistically.

Noether's theorem indeed assures us total momentum and total energy are conserved. In fact, they are combined into the 4-momentum with components $p^\mu$ in relativity. The timelike component is $p^t = \gamma mc$ for massive particles, but it is $h\nu/c$ for photons. The spacelike components are $p^k = \gamma mv^k$ for massive particles, but again this formula does not work for massless particles.

Since $\gamma = 1 + \mathcal{O}(v^2/c^2)$, collections of non-relativistic particles conserve the three components $\sum_i p_i^k \approx \sum_i m_i v_i^k$. Similarly, such particles will conserve $\sum_i p_i^t \approx \sum_i m_i c$, meaning $\sum_i m_i$ is approximately conserved. If (and only if!) kinetic energy is independently conserved for a collection of non-relativistic particles, we can subtract conserved mass from conserved total energy to find the formula for conserved energy: $\sum_i (p_i^t-m_ic) = \sum_i (\gamma_i - 1) m_ic$. Since $\gamma = 1 + (1/2) (v/c)^2 + \mathcal{O}(v^4/c^4)$, the non-relativistic approximation to the conserved kinetic energy is proportional to $\sum_i (1/2) m_i v_i^2$.

Four spacetime dimensions give us four conserved quantities, plus the restatement of the assumption that rest mass is conserved. There's nothing else to do. In particular, note that these formulas are not obtained from integration.