# Energy-momentum for a point particle and 4-vectors

• 1effect
In summary: Sigma E)^2-c^2(\Sigma \vec{p})^2 is not a four vector and does not work to obtain a value for the invariant mass of the system, while \Sigma(E^2-p^2c^2) is a four vector and does work. Why use something that doesn't work?Since \Sigma (E^2-p^2c^2) represents the total energy of the system, and (\Sigma E)^2-c^2(\Sigma \vec{p})^2 does not represent the total momentum of the system, using (\Sigma E)^2-c^2(\Sigma

#### 1effect

...for a point particle is a 4-vector. Consequence : $$E^2-c^2(\vec{p})^2$$ is an invariant
Nevertheless, for a system of particles, the energy momentum is not a 4-vector. See here.
Hence $$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$ is not an invariant. See here

I'm assuming that $$\Sigma (E^2-p^2c^2)$$ IS invariant.

kev said:
I'm assuming that $$\Sigma (E^2-p^2c^2)$$ IS invariant.

True but the above expression doesm't represent anything meaningful.
The energy of a particle system is $$\Sigma E$$ and the resultant momentum is $$\Sigma \vec{p}$$. Neverheless, you can't make a 4-vector out of $$(\Sigma E,c \Sigma \vec{p})$$. Bummer.

1effect said:
True but the above expression doesm't represent anything meaningful.
The energy of a particle system is $$\Sigma E$$ and the resultant momentum is $$\Sigma \vec{p}$$. Neverheless, you can't make a 4-vector out of $$(\Sigma E,c \Sigma \vec{p})$$. Bummer.

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?

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kev said:
So if I sum the energies of all the individual particles in a system that does not represent the total energy of the system?

Were you summing the energies? Where? In what formula?

kev said:
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?

Sure they are.
No one argued with the above. How is your post relevant re : OP?

1effect said:
Were you summing the energies? Where? In what formula?

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?

I am trying to figure out the significance of $$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$

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.

kev said:
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?

there is no connection between your posts and the OP <shrug>

kev said:
I am trying to figure out the significance of $$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$

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.

It's explained here and in its associated links. Hope it helps you.

In the rotating rocket thread we worked out that the total energy of the rotating rocket acted like additional mass and that the initial and final energies of the rotating rocket were the same as the non-rotating rocket with additional mass equivalent to the rotational kinetic energy of the rotating rocket. All fine and dandy.

You can come to same conclusion if you have to two particles bouncing up and down in the tranverse direction of a closed box. The energy balance works out the same as accelerating a box with two non bouncing particles that have additional mass equivelent to the kinetic energy of the bouncing particles. However, if the box has a pair of particles that bouncing in the parallel direction there turns out to be a anomalous excess energy of $${m_o u^2 v^2 \over c^2 \sqrt{1-u^2/c^2}\sqrt(1-v^2/c^2)$$ This sort of gives credance to the thread about a star (a system of gas particles) requiring extra energy to accelerate it to a given velocity. I am trying to figure out how to fix this anomally. Is is something to do with centre of mass of a system of particles changing during a transformation or something to do with way energies of a system are summed?

The energy anomally come about, because when a system of particles that is initially symetrical in the rest frame becomes asymetrical in another frame. For example if 50 % of the particles are going right and the other 50% are going left in the rest frame, then in the other frame 90% might be going right at a higher velocity and 10 % going left at a lower velocity.

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OK, for what it's worth I think it is fairly easy to show that $$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$ is not invariant even for the simple case of two particles with parallel motion and different velocities.

$$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$ is not a four vector and does not work to obtain a value for the invariant mass of the system, while $$\Sigma(E^2-p^2c^2)$$ is a four vector and does work. Why use something that doesn't work?

Note: $$p = \sqrt{p_x^2+p_y^2+p_z^2}$$ for each particle.

1effect said:
...for a point particle is a 4-vector. Consequence : $$E^2-c^2(\vec{p})^2$$ is an invariant
Nevertheless, for a system of particles, the energy momentum is not a 4-vector. See here.
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 $$E^2-c^2(\vec{p})^2$$ will be invariant.

Pete

1effect said:
...for a point particle is a 4-vector. Consequence : $$E^2-c^2(\vec{p})^2$$ is an invariant
Nevertheless, for a system of particles, the energy momentum is not a 4-vector. See here.
Hence $$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$ is not an invariant. 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.

$(\Sigma \vec{p})^2$ is invariant where the constituent velocity vectors are constant. $(\Sigma p^2})$, on the otherhand, is invariant regardless (edit: with respect to an energy-momentum relation).

Regards,

Bill

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1effect said:
...for a point particle is a 4-vector. Consequence : $$E^2-c^2(\vec{p})^2$$ is an invariant
Nevertheless, for a system of particles, the energy momentum is not a 4-vector. See here.
Note that the reason why it makes sense to consider $$\Sigma E$$ and $$\Sigma \textbf{p}$$ 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 $$\left(E, \textbf{p}c \right)$$ is a "4-vector", which means that it obeys the Lorentz transform

$$E' = \gamma_u \left(E - u p_x \right)$$
$$p_x' = \gamma_u \left(p_x - \frac{u E}{c^2} \right)$$
$$p_y' = p_y$$
$$p_z' = p_z$$

(orienting the space axes such that the relative velocity between the two frames is $$\textbf{u} = (u, 0, 0)$$). We call this 4-vector the 4-momentum or the energy-momentum vector of the particle.

From this is follows that m defined by $$(mc^2)^2 = E^2-||\textbf{p}c||^2$$ is invariant (the same in all inertial frames), and we call it the invariant mass (or "rest mass") of the particle.

The question is, is $$\left(\Sigma E, \Sigma \textbf{p}c \right)$$ 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 $$\Sigma E$$ 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).

This problem doesn't matter if all the particles are moving inertially (e.g. at the beginning of an experiment before they converge on each other to interact, or at the end of an experiment after all interaction) because in that case each particle's 4-momentum remains constant over a period of time. So it doesn't matter when exactly you make each measurement. The problem arises if you try to sum the 4-momenta during interaction.

This paper, referred to by pervect in this post, seems to imply, however, that if the system of particles in question is a "closed finite" system, then in fact, the total 4-momentum is well-defined. "Closed" means there is no interaction with anything outside the system. I'm not sure whether, in this context, "finite" should be taken literally as "less than infinite" or as "non-zero volume".

I suspect that, in a closed system, if you sum the 4-momenta across any spacelike 3D-surface you'll always get the same answer due to conservation of energy-momentum, although I can't quite prove this rigorously to myself. I'd like to see a reference to this (preferably online), if anyone can supply one.

This seems to imply that, if you want to consider two macroscopic bodies (each composed of sub-particles) interacting with each other, you can't model this by considering the total 4-momentum of each of the two bodies (because each body is not a closed system). Is that correct?

The way to tackle these issues when 4-momentum won't work appears to be to use the stress-energy tensor (effectively a 4 x 4 matrix that is compatible with Lorentz transformations), but my understanding is that this models objects as a continuous density distribution instead of a collection of discrete particles.

1effect said:
Hence $$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$ is not an invariant. 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.

kev #8 said:
I am trying to figure out the significance of $$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$
In those circumstances where the total 4-vector $$\left(\Sigma E, \Sigma \textbf{p}c \right)$$ is well defined (a "closed finite" system, see my last post), it represents $$(Mc^2)^2$$, where M is the invariant mass ("rest mass") of the whole system.

kev said:
OK, for what it's worth I think it is fairly easy to show that $$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$ is not invariant even for the simple case of two particles with parallel motion and different velocities.

$$(\Sigma E)^2-c^2(\Sigma \vec{p})^2$$ is not a four vector and does not work to obtain a value for the invariant mass of the system, while $$\Sigma(E^2-p^2c^2)$$ is a four vector and does work. Why use something that doesn't work?
Neither $$(\Sigma E)^2-c^2(\Sigma \textbf{p})^2$$ nor $$\Sigma(E^2-||\textbf{p}||^2c^2)$$ 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).

1effect said:
kev said:
I'm assuming that $$\Sigma (E^2-p^2c^2)$$ IS invariant.
True but the above expression doesm't represent anything meaningful.
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.

DrGreg said:
Note that the reason why it makes sense to consider $$\Sigma E$$ and $$\Sigma \textbf{p}$$ 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 $$\left(E, \textbf{p}c \right)$$ is a "4-vector", which means that it obeys the Lorentz transform

$$E' = \gamma_u \left(E - u p_x \right)$$
$$p_x' = \gamma_u \left(p_x - \frac{u E}{c^2} \right)$$
$$p_y' = p_y$$
$$p_z' = p_z$$

You can't use the above transformation in the case of a multiparticle system because you can't arrange for $$u$$ 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 $$p_y$$ and $$p_z$$.

The question is, is $$\left(\Sigma E, \Sigma \textbf{p}c \right)$$ 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 $$\Sigma E$$ 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).

Yes, you solved the issue!

DaleSpam said:
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.

It is equal to the sum of the squares of the invariant masses of the individual particles. making up the system.

DrGreg said:
In those circumstances where the total 4-vector $$\left(\Sigma E, \Sigma \textbf{p}c \right)$$ is well defined (a "closed finite" system, see my last post), it represents $$(Mc^2)^2$$, where M is the invariant mass ("rest mass") of the whole system.

Neither $$(\Sigma E)^2-c^2(\Sigma \textbf{p})^2$$ nor $$\Sigma(E^2-||\textbf{p}||^2c^2)$$ 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).

First, can you define a "closed finite" system?
Would gas particles in a thermally insulated pressure tight box or the gravitationally bound particles of a star count as one?

I have found out something interesting about the formula $$(\Sigma E)^2-c^2(\Sigma \textbf{p})^2$$

If we take 2 particles with mass m/2 each going in opposite directions with speed u then then the output of $$(\Sigma E)^2-c^2(\Sigma \textbf{p})^2$$ is $$\left({mc^2 \over \sqrt{1-u^2/c^2} \right)}^2$$

This value turns out to be invariant for any reference frame and represents the square of the "rest energy" of the system of particles in the centre of mass frame, (but not the square of the rest mass of the particles)

I am at this point using an open system as I have not allowed for the particles bouncing and reversing direction. (That is another story ;)

[EDIT] I have checked that $$(\Sigma E)^2-c^2(\Sigma \textbf{p})^2$$ is invariant for particles moving transverse and parallel to any reference frame. The simplest way to add up the whole system would seem to be to take the x, y and z components of all the particle momentums and then sum all parallel components. eg:

Total invariant rest energy = $$(\Sigma E_x)^2-c^2(\Sigma \vec{p_x})^2+(\Sigma E_y)^2-c^2(\Sigma \vec{p_y})^2+(\Sigma E_z)^2-c^2(\Sigma \vec{p_z})^2$$

= $$(\Sigma E)^2-c^2\left((\Sigma \vec{p_x})^2+(\Sigma \vec{p_y})^2+(\Sigma \vec{p_z})^2\right)$$

It is worth noticing that the total momentum of the particles in the centre of mass rest frame is zero (by definition) which simplifies things a lot.

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1effect said:
You can't use the above transformation in the case of a multiparticle system because you can't arrange for $$u$$ 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 $$p_y$$ and $$p_z$$.
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.)

DaleSpam said:
1effect said:
kev said:
I'm assuming that $$\Sigma (E^2-p^2c^2)$$ IS invariant.
True but the above expression doesm't represent anything meaningful.
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.
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 $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$.

kev said:
First, can you define a "closed finite" system?
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 $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ 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. $$\Sigma E / c^2$$.

DrGreg said:
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 $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ 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. $$\Sigma E / c^2$$.

Hold the press... I think I might have made an error in the calculations in post #21. I am checking them. :(

[EDIT] Ah..It's OK. The problem was with the spreadsheet I was checking it with. The algebra and formula seems OK for the limited conditions of an unclosed system (ie no changes of direction for the particles bouncing off walls etc)

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There is no doubt that the invariance holds, as pam said, this is a standard approach.
The particles might even interfere with a wall or something, as long as this wall is included in the sum. You may even let the particles disintegrate or create new particles from kinetic energy in a collision, it is exactly this expression that is always conserved. The rest mass of the system.

Ich said:
There is no doubt that the invariance holds, as pam said, this is a standard approach.
The particles might even interfere with a wall or something, as long as this wall is included in the sum. You may even let the particles disintegrate or create new particles from kinetic energy in a collision, it is exactly this expression that is always conserved. The rest mass of the system.

I can show a closed system where $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ is NOT invariant.

Also bear in mind that we have two distinct definitions of rest mass being used in this thread.

1) $$m_oc^2$$

2) $${m_oc^2 \over \sqrt{1-v^2/c^2}}$$

$$\Sigma (E^2-c^2||\textbf{p}||^2)$$ has definition 1 as the invariant quantity and

$$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ has definition 2 as the invariant quantity which includes the thermal kinetic energy in the centre of mass rest frame as part of the rest mass.

kev said:
I can show a closed system where $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ is NOT invariant.
Show it.

Also bear in mind that we have two distinct definitions of rest mass being used in this thread.

1) $$m_oc^2$$

2) $${m_oc^2 \over \sqrt{1-v^2/c^2}}$$
The second one is energy, not rest mass. It's difficult enough when people confuse different meanings of "mass", don't introduce a second meaning of "rest mass".
$$\Sigma (E^2-c^2||\textbf{p}||^2)$$ has definition 1 as the invariant quantity and

$$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ has definition 2 as the invariant quantity which includes the thermal kinetic energy in the centre of mass rest frame as part of the rest mass.
Definition 2 is not invariant.

The first equation is the sum of the rest masses of the constituends. A sum of scalars, invariant, but not conserved.
The second equation is the norm of a sum of four vectors, also a scalar, and the definition of the rest mass of a system. Invariant and always conserved.
In flat spacetime, of course.

DrGreg said:
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 $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$.
Are you sure about that? That goes against what I understood about the conservation of the 4-momentum (of course I am not a physicist). I understood that the sum of the 4-momenta was conserved even across particle decay.

IIRC, they had example problems that ran something like this: An electron at rest is anhilated by a positron moving at .6 c in the x direction. One of the two resulting photons travels only along the y axis. Find the 4-momenta of the two photons using units where c=1 and mass of an electron = 1.

So the 4 momenta are as follows:
electron (1, 0, 0, 0) -> mass = 1
positron (1.25, 0.75, 0, 0) -> mass = 1
system (2.25, 0.75, 0, 0) -> mass = sqrt(4.5)

photon1 (e1, 0, y1, 0, 0)
photon2 (e2, x2, y2, z2)
system (e1+e2, x2, y1+y2, z2)

So you have 6 unknowns, you get four from the conservation of 4-momentum for the system system before = system after, you get one from the norm of photon1 = 0 and one from the norm of photon2 = 0.

photon1 (1, 0, 1, 0) -> mass = 0
photon2 (1.25, .75, -1, 0) -> mass = 0
system (2.25, 0.75, 0, 0) -> mass = sqrt(4.5)

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kev said:
Also bear in mind that we have two distinct definitions of rest mass being used in this thread.

1) $$m_oc^2$$

2) $${m_oc^2 \over \sqrt{1-v^2/c^2}}$$

Ich said:
The second one is energy, not rest mass. It's difficult enough when people confuse different meanings of "mass", don't introduce a second meaning of "rest mass".

That is why I called it "rest energy" in earlier posts.

kev said:
$$\Sigma (E^2-c^2||\textbf{p}||^2)$$ has definition 1 as the invariant quantity and

$$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ has definition 2 as the invariant quantity which includes the thermal kinetic energy in the centre of mass rest frame as part of the rest mass.
Ich said:
Definition 2 is not invariant.
The first equation is the sum of the rest masses of the constituends. A sum of scalars, invariant, but not conserved.
The second equation is the norm of a sum of four vectors, also a scalar, and the definition of the rest mass of a system. Invariant and always conserved.
In flat spacetime, of course.

Case study A: Two particles of mass $m_o$ moving in opposite directions with speed u parallel to the x axis. (Open system - No collisions)

In frame S at rest with centre of mass of the two particles

$$\Sigma (E^2-||\textbf{p}||^2) = \left({m_o^2 \over (1-u^2)} -{m_o^2u^2\over (1-u^2)} \right) +\left({m_o^2 \over (1-u^2)} -{(-m_ou)^2\over (1-u^2)} \right) = 2m_o^2$$

$$(\Sigma E)^2-||\Sigma \textbf{p}||^2 = \left({m_o\over \sqrt{1-u^2}} +{m_o\over \sqrt{1-u^2}} \right)^2 -\left({m_o u\over \sqrt(1-u^2)} +{(-m_ou)\over \sqrt{1-u^2}} \right)^2 = {2m_o^2 \over (1-u^2)}$$

In frame S' moving at velocity v relative to frame S parallel to the x axis.

$$u ' = {v+u \over (1+uv)}$$ and $$-u ' = {v-u \over (1-uv)}$$

$$\Sigma (E^2-||\textbf{p}||^2) = \left({m_o^2 (1+uv)^2\over (1-u^2) (1-v^2) } -{m_o^2(v+u)^2\over (1-u^2)(1-v^2) } \right) +\left({m_o^2 (1-vu)\over (1-u^2)(1-v^2) } -{m_o^2(v-u)^2\over (1-u^2)(1-v^2) } \right) = 2m_o^2$$

$$(\Sigma E)^2-||\Sigma \textbf{p}||^2 = \left({m_o(1+uv)\over \sqrt{1-u^2} \sqrt{1-v^2}} +{m_o(1-uv)\over \sqrt{1-u^2}\sqrt{1-v^2}} \right)^2 -\left({m_o (v+u)\over \sqrt{1-u^2}\sqrt{1-v^2}} +{m_o(v-u)\over \sqrt{1-u^2}\sqrt{1-v^2}} \right)^2 = {2m_o^2 \over (1-u^2)}$$

kev said:
I can show a closed system where $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ is NOT invariant.
Ich said:
Show it.

Case study B: 2*N particles of mass $m_o$ moving with speed u parallel to the x-axis in a closed box.

In the rest frame (S) 50% of the particles are moving in the positive x direction and the other 50% are moving in the negative direction. The calculations for invariant mass/ energy in frame S are the same as in case A.

When we move to frame S' we notice that the forward moving particles take longer to catch up with front face of the box than it takes to return to the rear face. Taking relativistic velocity addition into account it turns out that at anyone time..

$${N(1+uv) }$$ particles are moving forward and

$${N(1-uv) }$$ particles are moving backward

.. so the masses and energies are weighted by this factor in the different directions creating an asymmetric situation.

$$\Sigma (E^2-||\textbf{p}||^2) = N(1+uv)\left({m_o^2 (1+uv)^2\over (1-u^2) (1-v^2) } -{m_o^2(v+u)^2\over (1-u^2)(1-v^2) } \right) + N(1-uv)\left({m_o^2 (1-vu)\over (1-u^2)(1-v^2) } -{m_o^2(v-u)^2\over (1-u^2)(1-v^2) } \right) = 2Nm_o^2$$

$$(\Sigma E)^2-||\Sigma \textbf{p}||^2 = \left({Nm_o(1+uv)^2\over \sqrt{1-u^2} \sqrt{1-v^2}} +{Nm_o(1-uv)^2\over \sqrt{1-u^2}\sqrt{1-v^2}} \right)^2 -\left({Nm_o (v+u)(1+vu)\over \sqrt{1-u^2}\sqrt{1-v^2}} +{Nm_o(v-u)(1-vu)\over \sqrt{1-u^2}\sqrt{1-v^2}} \right)^2 = {2Nm_o^2 (1-u^4v^2)\over (1-u^2)}$$

so if I have calculated correctly $$(\Sigma E)^2-||\Sigma \textbf{p}||^2$$ is not invariant for particles in a box.

P.S. The rest energy of the particles in the box is invariant using $$(\Sigma E)^2-||\Sigma \textbf{p}||^2$$ if the moving observer moved along the y axis, transverse to the motion of the particles. I hope I have made a mistake at the last step (case B) as this asymmetry seems very unsatisfactory.

DrGreg said:
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 did (unknowingly). By using the transforms that affect only $$p_x$$ and leave $$p_y$$ and $$p_z$$ unchanged.
I believe that kev just posted something that proves that $$(\Sigma E)^2-c^2(\Sigma\vec{p})^2$$ is not an invariant via a counterexample.

For quite a while I have pointed out that ,in the case of two particles $$E_1E_2-c^2 \vec{p_1} \vec{p_2}$$ is not an invariant.

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1effect said:
You did (unknowingly). By using the transforms that affect only $$p_x$$ and leave $$p_y$$ and $$p_z$$ unchanged.
I believe that kev just posted something that proves that $$(\Sigma E)^2-c^2(\Sigma\vec{p})^2$$ is not an invariant via a counterexample.

For quite a while I have pointed out that ,in the case of two particles $$E_1E_2-c^2 \vec{p_1} \vec{p_2}$$ is not an invariant.

Dr Greg is right that the y and z components of momentum do not change under transformation if the frame is moving in the x direction and it is correct mathematically and convenient to align one axis of the momentum components with the motion. Doing it any other way is just making life hard for yourself.

$$p_x ' = {m_o (v+u_x)\over \sqrt{1-u_x^2}\sqrt{1-v^2}} = { p_x\over \sqrt{1-v^2}}+{m_o v\over \sqrt{1-u_x^2}\sqrt{1-v^2}}$$

$$p_y ' = {m_o u_y\sqrt{1-v^2}\over \sqrt{1-u_y^2}\sqrt{1-v^2}} = p_y$$

$$p_z ' = {m_o u_z\sqrt{1-v^2}\over \sqrt{1-u_z^2}\sqrt{1-v^2}} = p_z$$

Where $$u_x, u_y, u_y$$ are the velocity components of the particles in the centre of mass rest frame S.

Note that caution is required when abbreviating the $${ {m_o \over \sqrt{1-u_x^2}}$$ term in the $$p_x '$$ expression to $$E_o$$ because it does not include any energy associated with any initial momentum in the y and z directions.

It seems that it is fair to say that $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ is invariant most of the time, but care is required in its use. It includes thermal energy and any change in temperature during the transformation will provide unreliable results. Remember that at the moment there is not even widespread agreement of how temperature transforms. My counter example shows that $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ breaks down when there is motion of a cyclic nature such as spin or vibration in the system. An example would be a gas with non monatomic molecules spinning and thermal kinetic motion. In a particle collider the formula is probably OK as most of the motion in a collision is linear and non cyclic.

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kev said:
so if I have calculated correctly $$(\Sigma E)^2-||\Sigma \textbf{p}||^2$$ is not invariant for particles in a box.

As I said earlier:
Ich said:
The particles might even interfere with a wall or something, as long as this wall is included in the sum.

If you do not include the wall, you run into the simultaneity problems pervect mentioned in the last thread.

double posting

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kev said:
Dr Greg is right that the y and z components of momentum do not change under transformation if the frame is moving in the x direction and it is correct mathematically and convenient to align one axis of the momentum components with the motion. Doing it any other way is just making life hard for yourself.

$$p_x ' = {m_o (v+u_x)\over \sqrt{1-u_x^2}\sqrt{1-v^2}} = { p_x\over \sqrt{1-v^2}}+{m_o v\over \sqrt{1-u_x^2}\sqrt{1-v^2}}$$

$$p_y ' = {m_o u_y\sqrt{1-v^2}\over \sqrt{1-u_y^2}\sqrt{1-v^2}} = p_y$$

$$p_z ' = {m_o u_z\sqrt{1-v^2}\over \sqrt{1-u_z^2}\sqrt{1-v^2}} = p_z$$

Where $$u_x, u_y, u_y$$ are the velocity components of the particles in the centre of mass rest frame S.

You seem to be missing the point. While you can surely align $$u$$ with the momentum of one particle, all the other momenta will not be necessarily aligned with $$u$$.

It seems that it is fair to say that $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ is invariant most of the time, but care is required in its use. .

This is also incorrect. You can easily try it for 2 particles , take $$\vec{v_1}=-\vec{v_2}$$ and look at $$E_1E_2-\vec{p_1} \vec{p_2}$$
You will quickly see that the expression is not invariant.