I'm assuming that [tex]\Sigma (E^2-p^2c^2)[/tex] IS invariant.
So the sum the energies of all the individual particles in a system does not represent the total energy of the system?True but the above expression doesm't represent anything meaningful.
The energy of a particle system is [tex] \Sigma E[/tex] and the resultant momentum is [tex] \Sigma \vec{p} [/tex]. Neverheless, you can't make a 4-vector out of [tex] (\Sigma E,c \Sigma \vec{p}) [/tex]. Bummer.
Were you summing the energies? Where? In what formula?So if I sum the energies of all the individual particles in a system that does not represent the total energy of the system?
Sure they are.So the sum the energies of all the individual particles in a system does not represent the total energy of the system?
.... and the sum of the individual momentums of all the individual particles in a system does not represent the total momentum of the system?
It's hypothetical. For example how can you be sure of the total mass of a system of particles, without knowing the exact velocity of evry particle in the system?Were you summing the energies? Where? In what formula?
there is no connection between your posts and the OP <shrug>It's hypothetical. For example how can you be sure of the total mass of a system of particles, without knowing the exact velocity of evry particle in the system?
It's explained here and in its associated links. Hope it helps you.I am trying to figure out the significance of [tex](\Sigma E)^2-c^2(\Sigma \vec{p})^2[/tex]
Where do we use it? Why do we need it? It's related to a problem I am working on and so I really am interested.
You have neglected to state under what circumstances that is true. If the system is a closed system and you're speaking about the total energy E and the total momentum P of the system (including the energy and momentum of fields if present) then, in general, the quantity [tex]E^2-c^2(\vec{p})^2[/tex] will be invariant....for a point particle is a 4-vector. Consequence : [tex]E^2-c^2(\vec{p})^2[/tex] is an invariant
Nevertheless, for a system of particles, the energy momentum is not a 4-vector. See here.
The second expression is standard, and in most textbooks. It is called the "invariant mass squared" of a collection of free particles, and has been used to find most of the resonances in the PDG. I don't have to look at a website that denies it.
Note that the reason why it makes sense to consider [tex]\Sigma E[/tex] and [tex]\Sigma \textbf{p}[/tex] is because of conservation of energy and momentum. We want to replace our multi-particle system with a single particle which behaves like the system, as far as possible....for a point particle is a 4-vector. Consequence : [tex]E^2-c^2(\vec{p})^2[/tex] is an invariant
Nevertheless, for a system of particles, the energy momentum is not a 4-vector. See here.
Well the above seems to suggest it is invariant in the case of a "closed finite" system. The only reason for it not being invariant would be if the summation is over different particle 4-vectors for different observers. A sum of 4-vectors is always a 4-vector, and so its "length" is invariant.Hence [tex](\Sigma E)^2-c^2(\Sigma \vec{p})^2[/tex] is not an invariant. See here
In those circumstances where the total 4-vector [tex] \left(\Sigma E, \Sigma \textbf{p}c \right) [/tex] is well defined (a "closed finite" system, see my last post), it represents [tex](Mc^2)^2[/tex], where M is the invariant mass ("rest mass") of the whole system.I am trying to figure out the significance of [tex](\Sigma E)^2-c^2(\Sigma \vec{p})^2[/tex]
Neither [tex](\Sigma E)^2-c^2(\Sigma \textbf{p})^2[/tex] nor [tex]\Sigma(E^2-||\textbf{p}||^2c^2)[/tex] are 4-vectors. They are scalars. But the first is a square-norm of a 4-vector (proportional to square mass), and the second is a sum of square-norms of 4-vectors (proportional to a sum of square masses).OK, for what it's worth I think it is fairly easy to show that [tex](\Sigma E)^2-c^2(\Sigma \vec{p})^2[/tex] is not invariant even for the simple case of two particles with parallel motion and different velocities.
[tex](\Sigma E)^2-c^2(\Sigma \vec{p})^2[/tex] is not a four vector and does not work to obtain a value for the invariant mass of the system, while [tex]\Sigma(E^2-p^2c^2)[/tex] is a four vector and does work. Why use something that doesn't work?
I disagree most emphatically. The above expression is the mass of the system, or (if you prefer) the rest energy of the system. Not only is it frame invariant but it is conserved. I cannot think of a more useful physical concept than something that is both frame invariant and conserved.True but the above expression doesm't represent anything meaningful.I'm assuming that [tex]\Sigma (E^2-p^2c^2)[/tex] IS invariant.
You can't use the above transformation in the case of a multiparticle system because you can't arrange for [tex]u[/tex] to align with more than one momentum (if they all have different directions).In other words, you need the generic transformation , the one that changes bot [tex]p_y[/tex] and [tex]p_z[/tex].Note that the reason why it makes sense to consider [tex]\Sigma E[/tex] and [tex]\Sigma \textbf{p}[/tex] is because of conservation of energy and momentum. We want to replace our multi-particle system with a single particle which behaves like the system, as far as possible.
For each particle, the four dimensional vector [tex]\left(E, \textbf{p}c \right) [/tex] is a "4-vector", which means that it obeys the Lorentz transform
[tex] E' = \gamma_u \left(E - u p_x \right) [/tex]
[tex] p_x' = \gamma_u \left(p_x - \frac{u E}{c^2} \right) [/tex]
[tex] p_y' = p_y [/tex]
[tex] p_z' = p_z [/tex]
Yes, you solved the issue!The question is, is [tex] \left(\Sigma E, \Sigma \textbf{p}c \right) [/tex] a 4-vector, i.e. does it also obey the Lorentz transform? You might think the answer is "yes", but we have to ask, what does [tex] \Sigma E [/tex] really mean? It tells us what to sum, but it doesn't say when to measure all the individual energies. In Newtonian theory, we measure them simultaneously. But in relativity, "simultaneity" is a frame-dependent concept. Sure, you can add up all the 4-momenta simultaneously relative to some observer and get a 4-vector, but if you add up all the 4-momenta simultaneously relative to another observer you might get a different 4-vector (because you are adding a different set of 4-vectors).
It is equal to the sum of the squares of the invariant masses of the individual particles. making up the system.I disagree most emphatically. The above expression is the mass of the system, or (if you prefer) the rest energy of the system. Not only is it frame invariant but it is conserved. I cannot think of a more useful physical concept than something that is both frame invariant and conserved.
In those circumstances where the total 4-vector [tex] \left(\Sigma E, \Sigma \textbf{p}c \right) [/tex] is well defined (a "closed finite" system, see my last post), it represents [tex](Mc^2)^2[/tex], where M is the invariant mass ("rest mass") of the whole system.
Neither [tex](\Sigma E)^2-c^2(\Sigma \textbf{p})^2[/tex] nor [tex]\Sigma(E^2-||\textbf{p}||^2c^2)[/tex] are 4-vectors. They are scalars. But the first is a square-norm of a 4-vector (proportional to square mass), and the second is a sum of square-norms of 4-vectors (proportional to a sum of square masses).
I made no assumptions here about the relative directions of u and p. u does not align with the momentum, it is the relative velocity between an arbitary pair of frames. All I've done is align the spatial-coordinate system with u. So, for a given pair of aligned frames, the equations I quoted here are valid for all particles. (You can work with the unaligned version of the Lorentz transform if you wish, but there's no need.)You can't use the above transformation in the case of a multiparticle system because you can't arrange for [tex]u[/tex] to align with more than one momentum (if they all have different directions).In other words, you need the generic transformation , the one that changes bot [tex]p_y[/tex] and [tex]p_z[/tex].
The sums of squares of the invariant masses of the particles is indeed invariant. It is conserved if the particles never merge or decay, but not otherwise. But it definitely is not the square of the system's invariant mass, which is (for a closed system) proportional to [tex](\Sigma E)^2-c^2||\Sigma \textbf{p}||^2[/tex].I disagree most emphatically. The above expression is the mass of the system, or (if you prefer) the rest energy of the system. Not only is it frame invariant but it is conserved. I cannot think of a more useful physical concept than something that is both frame invariant and conserved.True but the above expression doesm't represent anything meaningful.I'm assuming that [tex]\Sigma (E^2-p^2c^2)[/tex] IS invariant.
I'm quoting this paper as I said in this post, where I gave my interpretation, but I'm not 100% sure. I believe that "closed" means your particles cannot be interacting with anything outside the system (it doesn't mean they have to be "sealed in a box").First, can you define a "closed finite" system?
Hold the press... I think I might have made an error in the calculations in post #21. I am checking them. :(I'm quoting this paper as I said in this post, where I gave my interpretation, but I'm not 100% sure. I believe that "closed" means your particles cannot be interacting with anything outside the system (it doesn't mean they have to be "sealed in a box").
Your calculations with [tex](\Sigma E)^2-c^2||\Sigma \textbf{p}||^2[/tex] are confirming that it does represent invariant-mass-squared (rescaled into square-energy units), in the circumstances you have tested. Of course, in the centre-of-mass frame, invariant mass is the same as "relativistic mass", i.e. [tex]\Sigma E / c^2 [/tex].