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You seem confused by how conservation works, and intent on misunderstanding it. All I can say is that momentum conservation is Lorentz invariant in my system,

You seem confused by how conservation works, and intent on misunderstanding it. All I can say is that momentum conservation is Lorentz invariant in my system, and hence cannot depend on choice of reference frame. The conserved quantity is not invariant, and conserved quantities do not need to be Lorentz invariant, and fact almost always are not. If this confuses you, sorry, but I cannot follow your reasoning anymore.

pdonis 10y ago
> All I can say is that momentum conservation is Lorentz invariant in my system If by "my system" you mean your hypothetical theory where energy is not conserved but 3-momentum is, I strongly suspect you have not done the math. See below. > The conserved quantity is not invariant, and conserved quantities do not need to buslye Lorentz invariant, and fact almost always are not. If this confuses you, sorry, but I cannot follow your reasoning anymore. I understand what you are saying here, but it does not refute what I was saying. You are saying that, for example, in standard relativity, energy is conserved--in a given frame, it stays the same through any series of events--but it is not frame invariant; changing frames changes the energy. I agree with that. What I am saying is something different: in your hypothetical relativistic theory in which energy is not conserved, then 3-momentum cannot be conserved in any frame other than one particular one (which in my example was the center of mass frame). I am not saying the numerical value of the 3-momentum has to be the same in all frames. I am saying that in all frames but one, the numerical value of the 3-momentum, as evaluated in that frame, is different after an inelastic collision than before, i.e., 3-momentum is not conserved in that frame. Here is the math backing up that assertion. Suppose 3-momentum is conserved in the center of mass frame. In that frame the 3-momentum is zero. We assume that our system consists of just two billiard balls, each with the same rest mass m. Before the collision, one ball has speed v and the other has speed -v (we can restrict ourselves to one spatial direction). After the collision, one ball has speed w and the other has speed -w, where w < v (because the collision is inelastic). So 3-momentum is zero before and after the collision in this frame, and energy is not conserved--it is smaller after the collision. Now transform to any other frame. Call the relative velocity of the Lorentz transformation u. Then, in this frame, the speeds of the two balls before the collision are (I am using units where c = 1) v+' = (u + v)/(1 + uv) v-' = (u - v)/(1 - uv) and the speeds after the collision are w+' = (u + w)/(1 + uw) w-' = (u - w)/(1 - uw) Now we evaluate the 3-momentum in this frame. Before the collision, it is m ( v+'/sqrt(1 - (v+')^2) + v-'/sqrt(1 - (v-')^2) ) After the collision, it is m ( w+'/sqrt(1 - (w+')^2) + w-'/sqrt(1 - (w-')^2) ) Substituting and straightforward algebra simplifies these to before = 2mu/sqrt[(1-u^2)(1-v^2)] after = 2mu/sqrt[(1-u^2)(1-w^2)] These are obviously not equal, hence 3-momentum is not conserved in this frame. (In standard relativity, where energy is conserved, 3-momentum would be conserved as well; as I mentioned before, the simplest way for that to happen would be for the two billiard balls to heat up, increasing their rest mass. More complicated ways would involve other particles, or the billiard balls emitting radiation, or something like that.)
- pdonis 10y ago
  Just to clean up one item: I did assume in the above that the rest mass of the billiard balls was unchanged in the inelastic collision. But we can drop that assumption and still prove that, if energy is not conserved in the center of mass frame, 3-momentum cannot be conserved in any frame other than the center of mass frame. Suppose the rest mass of the balls after the collision is M (instead of m). Then we have for the 3-momentum before (P0') and after (P1') the collision, in a frame with velocity u relative to the center of mass frame: P0' = 2mu/sqrt[(1-u^2)(1-v^2)] P1' = 2Mu/sqrt[(1-u^2)(1-w^2)] But we can simplify this by writing down the total energy before (E0) and after (E1) the collision, in the center of mass frame: E0 = 2m/sqrt(1-v^2) E1 = 2M/sqrt(1-w^2) So we can see that P0' = E0 (u/sqrt(1-u^2)) P1' = E1 (u/sqrt(1-u^2)) Hence, if E0 > E1, we must also have P0' > P1. In other words, the only reason we happen to have P0 = P1 in the center of mass frame is that being in that frame is equivalent to having u = 0 in the above formulas. (It is also straightforward to show that if energy is not conserved in the center of mass frame, it is not conserved in any frame. So the Lorentz invariance of the two conservation laws, energy and momentum, cannot be separated--they are inseparably linked.)

pdonis 10y ago

> All I can say is that momentum conservation is Lorentz invariant in my system If by "my system" you mean your hypothetical theory where energy is not conserved but 3-momentum is, I strongly suspect you have not done the math. See below. > The conserved quantity is not invariant, and conserved quantities do not need to buslye Lorentz invariant, and fact almost always are not. If this confuses you, sorry, but I cannot follow your reasoning anymore. I understand what you are saying here, but it does not refute what I was saying. You are saying that, for example, in standard relativity, energy is conserved--in a given frame, it stays the same through any series of events--but it is not frame invariant; changing frames changes the energy. I agree with that. What I am saying is something different: in your hypothetical relativistic theory in which energy is not conserved, then 3-momentum cannot be conserved in any frame other than one particular one (which in my example was the center of mass frame). I am not saying the numerical value of the 3-momentum has to be the same in all frames. I am saying that in all frames but one, the numerical value of the 3-momentum, as evaluated in that frame, is different after an inelastic collision than before, i.e., 3-momentum is not conserved in that frame. Here is the math backing up that assertion. Suppose 3-momentum is conserved in the center of mass frame. In that frame the 3-momentum is zero. We assume that our system consists of just two billiard balls, each with the same rest mass m. Before the collision, one ball has speed v and the other has speed -v (we can restrict ourselves to one spatial direction). After the collision, one ball has speed w and the other has speed -w, where w < v (because the collision is inelastic). So 3-momentum is zero before and after the collision in this frame, and energy is not conserved--it is smaller after the collision. Now transform to any other frame. Call the relative velocity of the Lorentz transformation u. Then, in this frame, the speeds of the two balls before the collision are (I am using units where c = 1) v+' = (u + v)/(1 + uv) v-' = (u - v)/(1 - uv) and the speeds after the collision are w+' = (u + w)/(1 + uw) w-' = (u - w)/(1 - uw) Now we evaluate the 3-momentum in this frame. Before the collision, it is m ( v+'/sqrt(1 - (v+')^2) + v-'/sqrt(1 - (v-')^2) ) After the collision, it is m ( w+'/sqrt(1 - (w+')^2) + w-'/sqrt(1 - (w-')^2) ) Substituting and straightforward algebra simplifies these to before = 2mu/sqrt[(1-u^2)(1-v^2)] after = 2mu/sqrt[(1-u^2)(1-w^2)] These are obviously not equal, hence 3-momentum is not conserved in this frame. (In standard relativity, where energy is conserved, 3-momentum would be conserved as well; as I mentioned before, the simplest way for that to happen would be for the two billiard balls to heat up, increasing their rest mass. More complicated ways would involve other particles, or the billiard balls emitting radiation, or something like that.)

pdonis 10y ago
Just to clean up one item: I did assume in the above that the rest mass of the billiard balls was unchanged in the inelastic collision. But we can drop that assumption and still prove that, if energy is not conserved in the center of mass frame, 3-momentum cannot be conserved in any frame other than the center of mass frame. Suppose the rest mass of the balls after the collision is M (instead of m). Then we have for the 3-momentum before (P0') and after (P1') the collision, in a frame with velocity u relative to the center of mass frame: P0' = 2mu/sqrt[(1-u^2)(1-v^2)] P1' = 2Mu/sqrt[(1-u^2)(1-w^2)] But we can simplify this by writing down the total energy before (E0) and after (E1) the collision, in the center of mass frame: E0 = 2m/sqrt(1-v^2) E1 = 2M/sqrt(1-w^2) So we can see that P0' = E0 (u/sqrt(1-u^2)) P1' = E1 (u/sqrt(1-u^2)) Hence, if E0 > E1, we must also have P0' > P1. In other words, the only reason we happen to have P0 = P1 in the center of mass frame is that being in that frame is equivalent to having u = 0 in the above formulas. (It is also straightforward to show that if energy is not conserved in the center of mass frame, it is not conserved in any frame. So the Lorentz invariance of the two conservation laws, energy and momentum, cannot be separated--they are inseparably linked.)