Why in a collision between particles is the four-momentum conserved within a frame of reference but not invariant between frames of reference?
[Physics] Invariance and conservation
conservation-lawscovarianceinvariantsmomentumreference frames
Related Solutions
I understand that the inner product of two 4-vectors is conserved under the Lorentz transformations
Yes, $p_1.p_2$ is a Lorentz invariant
So that the absolute value of the four momentum is the same in any reference frame.
It is not correct to speak about the "absolute value" of a (quadri)vector. Which is conserved in a Lorentz transformation is $p^2 = (p^o)^2 - \vec p^2$
This is what I (most likely mistakenly) thought was meant by the conservation of momentum.
No, conservation of momentum is a completely different thing. Ultimately, you have some theory describing fields and interactions, describing by an action which is invariant by some symmetries. If the action is invariant by space and time translations, then there is a conserved quantity which is momentum/energy.
I don't understand why equations such as P 1 =P 2 +P 3 (P i are 4-momentum vectors for different particles in a collision for example) should hold, within a reference frame. I've been told that you can't just add four velocities together on collision of particles, so why should you be able to do this with the momentum vectors?
If the theory action is invariant by space/time translations, then the momentum/energy is conserved, so the total momentum/energy of the initial particles is the same as the total momentum/energy of the final particles :
$$(p_\textrm{tot})_\textrm{in}^\mu = (p_\textrm{tot})_\textrm{out}^\mu\tag{1}$$
If there are several initial particles, they are considered as independent (the global state is the tensor product of the states of the initial particles). The independence means that you have :
$$(p_\textrm{tot})_\textrm{in}^\mu = \sum_i p_i^\mu\tag{2}$$ where the sum is about all the initial particles. A similar equation holds for the final particles.
The reason your expectations of kinetic energy loss are violated is because you picked a non-inertial reference frame.
In fact, you didn't notice, but based on your assumptions, momentum isn't even conserved.
Let's look more closely at your first equation:
$$m_e * 0 + m_i v_i = m_e * 0 + m_f v_f$$
Let's say the ball's mass is 5 kg. If the ball is falling down towards the ground, the initial velocity before the collision should be negative (assuming we have adopted a coordinate system where "up" is positive). Let's say the ball was moving at 10 m/s. Here is our equation so far:
$$m_e * 0 + (5 kg) (-10 m/s) = m_e * 0 + (5 kg) v_f$$
Let's simplify, given your assumption that the Earth doesn't move:
$$(5 kg) (-10 m/s) = (5 kg) v_f$$ Which gives us
$$ v_f = -10 m/s $$
Whaaa? The final velocity is negative? That means . . . after this "collision", the ball is still falling downward at the same speed! If we assume, as you did, that the velocity was positive, then the momentum of the system can't be conserved! Clearly, something is wrong here.
Here's the trouble: You began in the reference frame of the Earth. Once the ball hit the Earth, the Earth's reference frame is no longer inertial! You either have to introduce a fictitious force on the ball to account for the fact that the Earth accelerated (a tiny bit, yes, but an important tiny bit), or you need to find an inertial reference frame.
So, Let's pick the frame of reference in which the Earth starts out at rest. When the Earth moves, we'll let our frame of reference stay where it is, so as to allow it to remain inertial:
$$m_e * 0 + (5 kg) (-10 m/s) = m_e * v_{e, f} + (5 kg) v_f$$ $$-50 kg m/s = m_e * v_{e, f} + (5 kg) v_f$$
NOW we can introduce a coefficient of elasticity and demand that energy be either conserved or not, and find the final velocity of the Earth and the ball as they rebound from each other.
Note that although $v_{e, f}$ will be incredibly tiny, due to the enormous mass of the Earth $m_e$, the final momentum of the Earth will be non-negligible - in fact, the final momentum of the Earth must be comparable to the final momentum of the ball in order for momentum to be conserved!
So, to sum up: You picked a non-inertial reference frame, and didn't account for it, so your reliance on the laws of physics was betrayed.
Best Answer
Conservation and invariance are fundamentally different things. Conservation means "doesn't change with respect to time". While invariance means "doesn't change with respect to Lorentz transformations". Components of four-momentum transform like vector components and are thus NOT invariant under Lorentz Transformations. But that doesn't prevent them from being conserved. Suppose the four-momentum is conserved in one frame. If you switch to a different frame, the four-momentum components will all be different, but the conservation is preserved.