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

1. Mar 6, 2008

### 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

2. Mar 6, 2008

### yuiop

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

3. Mar 7, 2008

### 1effect

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.

4. Mar 7, 2008

### yuiop

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?

Last edited: Mar 7, 2008
5. Mar 7, 2008

### 1effect

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

6. Mar 7, 2008

### 1effect

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

7. Mar 7, 2008

### yuiop

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?

8. Mar 7, 2008

### yuiop

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.

9. Mar 7, 2008

### 1effect

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

10. Mar 7, 2008

### 1effect

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

11. Mar 7, 2008

### yuiop

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.

Last edited: Mar 7, 2008
12. Mar 7, 2008

### yuiop

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.

13. Mar 7, 2008

### pmb_phy

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

14. Mar 7, 2008

### pam

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.

15. Mar 7, 2008

### Antenna Guy

$(\Sigma \vec{p})^2$ 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

Last edited: Mar 7, 2008
16. Mar 7, 2008

### DrGreg

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.

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.

17. Mar 7, 2008

### DrGreg

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).

18. Mar 7, 2008

### Staff: Mentor

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.

19. Mar 7, 2008

### 1effect

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 $$p_z$$.

Yes, you solved the issue!

20. Mar 7, 2008

### 1effect

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