# LOG#017. Momenergy (II).

Imagine you own a set of n different particle species, and you make them to collide. In general, even in classical mechanics, you can get some particles loose their identities and become “new particles”. Suppose you make m different “new” species of particles. The collision can be represented as the following chemical reaction:

$\displaystyle{\boxed{\sum_{i=1}^{n}A_i \rightarrow \sum_{j=1}^{m}B_j}}$

or

$\boxed{A_1+\cdots +A_n \rightarrow B_1+\cdots +B_m}$

Now, we will study what happens in this type of collsions in the classical mechanics realm and in the SR framework.

Let us begin with Classical Mechanics. In absence of external forces, Newton’s second law implies that the momentum must be conserved, i.e., momentum does not change in time if there is no applied force. Therefore, the momentum is the same before and after the collision:

$\boxed{\mathbf{p}_{\mbox{before}}=\mathbf{p}_{\mbox{after}}}$

or equivalently

$\boxed{\mathbf{p}_{A_1}+\cdots +\mathbf{p}_{A_n}=\mathbf{p}_{B_1}+\cdots +\mathbf{p}_{B_m}}$

Classical momentum is $\mathbf{p}=m\mathbf{v}$, so we get

$\boxed{m_{A_1}\mathbf{v}_{A_1}+... +_{A_n}\mathbf{v}_{A_n}=m_{B_1}\mathbf{v}_{B_1}+...+m_{B_m}\mathbf{v}_{B_m}}$

This equation holds in an inertial frame S. If we consider other inertial frame S’ moving at constant speed $\mathbf{V}$ with respect to S, conservation of momentum provides:

$\boxed{m_{A_1}\mathbf{v}^{'}_{A_1}+\cdots +m_{A_n}\mathbf{v}^{'}_{A_n}=m_{B_1}\mathbf{v}^{'}_{B_1}+\cdots +m_{B_m}\mathbf{v}^{'}_{B_m}}$

where

$\mathbf{v'}=\mathbf{v}-\mathbf{V}$

Substracting the last two boxed equations yields:

$m_{A_1}(\mathbf{v}_{A_1}-\mathbf{v}^{'}_{A_1})+...+m_{A_n}(\mathbf{v}_{A_n}-\mathbf{v}^{'}_{A_n})=m_{B_1}(\mathbf{v}_{B_1}-\mathbf{v}^{'}_{B_1})+...+m_{B_m}(\mathbf{v}_{B_1}-\mathbf{v}^{'}_{B_m})$

i.e.,

$\left(m_{A_1}+\cdots +m_{A_1}\right)\mathbf{V}= \left(m_{B_1}+\cdots +m_{B_m}\right)\mathbf{V}$

Therefore, the total inertial mass must be conserverd:

$\boxed{\left(m_{A_1}+\cdots +m_{A_1}\right) =\left(m_{B_1}+\cdots +m_{B_m}\right)\leftrightarrow m_{\mbox{before}}^{\mbox{total}}=m_{\mbox{after}}^{\mbox{total}}}$

Conclusion: TOTAL (INERTIAL) MASS IS CONSERVED!

Do you recognize this fact/law? Yes: Newton’s second law plus the no force condition provide the celebrated Lavoisier’s law of Chemistry (the conservation of mass) if we consider the transformation between different inertial frames in Classical Mechanics. Lavoisier’s law remained exact until the discovery of radioactive substances. Special Relativity, thanks to the equivalence between mass and energy, could explain the mysterious origin of the radioactivity that Classical Mechanics could not understand.

Let us go back to the relativistic domain. The relativistic momentum is:

$\mathbf{P}=M\mathbf{v}=m\gamma \mathbf{v}$

Again the condition of no external forces provide the conservation of the relativistic momentum $\mathbb{P}$:

$\boxed{\mathbb{P}_{\mbox{before}}=\mathbb{P}_{\mbox{\mbox{after}}}}$

Firstly, we will focus on the spatial part of the last four dimensional equation. It yields:

$\boxed{\mathbf{P}_{\mbox{before}}=\mathbf{P}_{\mbox{after}}}$

or

$\boxed{\left( \sum_i \gamma (v_{A_i})m_{A_i} \mathbf{v}_{A_i}\right)_{\mbox{before}}=\left( \sum_j \gamma (v_{B_j})m_{B_j} \mathbf{v}_{B_j}\right)_{\mbox{after}}}$

It can be written shortly as a feynmanity:

$\boxed{\Delta \left(\sum \gamma (v) m \mathbf{v} \right)=0= \left(\sum_j \gamma (v_{B_j})m_{B_j} \mathbf{v}_{B_j} - \sum_i \gamma (v_{A_i})m_{A_i} \mathbf{v}_{A_i} \right)}$

This last equation is valid in certain inertial frame S. In other inertial frame S’, the same conservation law should hold:

$\boxed{\Delta \left(\sum \gamma (v') m \mathbf{v}' \right)=0}$

From previous logs, we do know that the relativistic transformation between velocities of different frames is:

$\mathbf{v}'=-\mathbf{V}\biguplus \mathbf{v}=\dfrac{\mathbf{v}_\parallel+\gamma^{-1}(V)\mathbf{v}_\perp -\mathbf{V}}{1-\dfrac{\mathbf{V}\cdot \mathbf{v}}{c^2}}$

Using the identity:

$\gamma (v')=\gamma (V)\gamma (v)\left(1-\dfrac{\mathbf{V}\cdot \mathbf{v}}{c^2}\right)$

we deduce

$\boxed{\Delta \left(\Sigma \gamma (v')m\mathbf{v}' \right)=\Delta \left(\Sigma \gamma (v)m\mathbf{v}\right)-\gamma (V)\Delta \left(\Sigma \gamma (v)m\right)\mathbf{V}+\left(\gamma (V)-1\right)\Delta \left(\Sigma \gamma (v)m\mathbf{v}_\parallel\right)}$

Due to the conservation of the relativistic momentum, and specially its parallel projection into $\mathbf{V}$ as well, the lefthanded side and the first and third terms in the righthanded side vanish, so we are left with:

$\gamma (V)\Delta \left(\Sigma \gamma (v)m\right)\mathbf{V}=0$

If the relative velocity is not equal to zero, the momentum in the S’-frame will be conserved if and only if:

$\boxed{\Delta \left(\Sigma \gamma (v)m\right)=0\leftrightarrow \left( \Sigma \gamma (v)m\right)_{\mbox{before}}=\left( \Sigma \gamma (v)m\right)_{\mbox{after}}}$

But this last equation means that the relativistic mass must be conserved in order to conserve the full momenergy. Multiplying by $c^2$ we get indeed the conservation of the total relativistic mass during the collision! Mathematically speaking:

$\boxed{\Delta Mc^2=0\leftrightarrow E_{\mbox{before}}=E_{\mbox{after}}}\leftrightarrow \boxed{\left( \Sigma \gamma (v)mc^2\right)_{\mbox{before}}=\left( \Sigma \gamma (v)mc^2\right)_{\mbox{after}}}$

In the case of our reaction above, we obtain from these last relationship:

$\boxed{\gamma (v_{A_1})m_{A_1}c^2+\cdots+\gamma (v_{A_n})m_{A_n}c^2=\gamma (v_{B_1})m_{B_1}c^2+\cdots+\gamma (v_{B_m})m_{B_m}c^2}$

This equation can be also written using the relativistic mass and relativistic kinetic energy variations as follows:

$\boxed{\left(\Delta \sum m\right)c^2+\Delta \sum K =0}$

In the low velocity limit, this last equation provides:

$(m_{A_1}+...+m_{A_n})c^2+\dfrac{1}{2}(m_{A_1}v_{A_1}^2+...+m_{A_n}v_{A_n}^2)=(m_{B_1}+...+m_{B_m})c^2++\dfrac{1}{2}(m_{B_1}v_{B_1}^2+...+m_{B_m}v_{B_m}^2)$

We have to be careful to apply this last equation since in the approximation we used, the kinetic energies have small numeric values compared with the rest masses. Generally, this equation is more conveniently recasted into:

$\boxed{\Delta \sum m c^2+\Delta\sum \dfrac{1}{2}mv^2=0}$

Remark: In the special case that the initial and final objects match, we get the conservation of the rest mass (a.k.a. Lavoisier’s law) as a trivial identity AND the commonly known kinetic conservation law. They are “emergent” from SR in certain limit. When we say that the initial and the final objects match, we mean that they are completely identical, i.e., a complete identity. If, for instance, in the collision some body gets some heat (i.e., it increases its temperature) or if some body change its inner rotational state, or if some body is “excited”,…Or generally, if the inner energetic content “changes” in any of the colliding objects, the internal energy changes, therefore its inert mass too, and the collision is not elastic.