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 Exercise [18.22] 
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Joined: 03 Jun 2010, 15:18
Posts: 136
Post Exercise [18.22]
Attached my solution.

Exercise_18_22.pdf [65.67 KiB]
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29 Aug 2010, 14:06

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [18.22]
I'd like to add something that Roberto didn't mention, but that's important in comparing the relativistic and non-relativistic cases, and in reaching a definite conclusion.

In exercise 18.17 we were able to invoke Newton's 3rd law to show that internal forces come in pairs that cancel out when summed together, so in the absence of external forces we have \frac{d}{dt}\left(\sum \mathbf{p}}\right) = 0. However, the "equal and opposite" part of Newton's 3rd law implicitly means "equal and opposite at any given instant". It thus involves a notion of simultaneity that's inapplicable in the relativistic case: In fact we would need to introduce the notion of a field (e.g. the electromagnetic field, etc) in order to correctly handle interactions between distant particles in special relativity... so see the next chapter!

Without fields, our solution can only cover impulse forces that occur at single points in spacetime, when two structureless point particles collide at that point. That is, the changes in \mathbf{p} for each of the two colliding particles sum to zero; total \mathbf{p} is conserved in the collision, as is total relativistic p.

Mass is just the zero'th component of p, so in such a single-point collision it also is preserved, and thus \sum\left(\frac{dm}{dt}\boldsymbol{x}\right) is also zero: The changes in m cancel between the two colliding particles, and they occur at the same point \boldsymbol{x}.

Note that the conservation of \overline{\mathbf{p}} (which here denotes the total, rather than the average value) and the vanishing of \sum\left(\frac{dm}{dt}\boldsymbol{x}\right) together imply the conservation of Roberto's \overline{\boldsymbol{N}} (this is the reverse direction of the implication Roberto states - but in fact both directions hold).

We can thus conclude that in the relativistic case, for a collection of structureless point particles that interact only via collisions (and that do not collide with any other, external particles not in the collection), the center of mass does indeed move in a uniform straight line. (Also, as Roberto notes, it's postion and velocity actually depend on the reference frame one calculates it in; it is not a physical, frame-independent property).


Also, Roberto...

What were these subtleties in defining the \={M}^{\kappa\lambda} tensor that you refer to in your 2nd footnote (for those of us who don't have Rindler's "Introduction to Special Relativity" on hand)?

Did it have anything to do with the fact that tensors (as they've been discussed in the book so far) exist in the tangent (and/or cotangent) space of each spacetime point, and that the M^{\kappa\lambda} tensors for the different particles are therefore (apparently) defined at different spacetime points?

We are able to sum them because Minkowski space is completely flat, so tensors can be transferred from any one point to any other in a path-independent manner - and indeed without even changing the values of their components, if the usual (t,x,y,z) coordinate system is used. Thus there's effectively a single vector (or tensor) space for 4-vectors (and tensors) in Minkowski space, and we can thus (for example) add the 4-momenta belonging to spatially separated particles.

I'm not sure what happens when we go to General Rel. and curved spacetime... maybe quantities like "center of mass" and "total (angular) momentum" are no longer well-defined? (Except of course for small masses in small, locally flat neighborhoods - which reduces to the Special Rel. case).

If there's some other subtlety there, I've missed it... so please let us know!

10 Jan 2012, 16:01

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [18.22]
The subtlety is due to the fact that total 4-angular momentum needs that the individual 4-angular momenta, to be summed over, are evaluated simultaneously, i.e. at the same coordinate time t, as I mentioned in my solution to the exercise referring to total N, that is the spatial part of total 4 angular momentum.

For simplicity, let analyse a system made of just two particles, P1 and P2.
A first observer, in coordinate frame F, evaluates the total 4-angular momentum M summing the 4-angular momentum of first particle, M1, measured at space-time point (x1, y1, z1, t1) and of second particle, M2, measured at space-time point (x2, y2, z2, t2) ; this has to be done simultaneously, i.e. at time t1=t2=t
For a second observer point of view, in coordinate frame F', the first observer summed up the 4-angular momentum of first particle measured at space-time point (x1', y1', z1', t1') and of second particle at space-time point (x2', y2', z2', t2'); but now t1' is different from t2'; therefore the result is not, for the second observer, a total 4-angular momentum, because the momenta of the two particle are measured at different times.
This means that total 4-angular momentum is not (in general) Lorentz-invariant, i.e. is not a tensor, i.e. simply makes no sense.
However, the definition makes sense if interactions between the particles occurs only when the two particles are at the same space-time point (and we may call these local interaction collisions).
When there is no interaction, evaluating individual momenta at different times is not a problem, because they don't change with time, by angular momentum conservation.
When there is an interaction, occurring at the same space-time point, the two momenta can be measured just after collision simoultaneously in all frames.
Then with this restriction that total 4-angular momentum is Lorentz invariant.
You wrote that this restriction is needed to have total angular momentum conservation; actually it is needed also to have total angular momentum Lorentz invariance.
You may find more details at http://panda.unm.edu/Courses/Finley/P49 ... angmom.pdf

I found this result quite amazing: if you just assume that:
a) special relativity is correct, that is obviously very reasonable

b) total 4-angular momentum exists (i.e. you can define it as a 4-tensor), that is also very reasonable, taking into account the relation between angular momentum and spin (see chapter 22 of RTR)

then you may deduce that all interactions that may exist in nature occur via collisions among particles.
This is very elegant, because would imply that all interactions are due to a single and simple physical mechanism: collision , i.e. interaction occurring when the particles have the same spatial position at same time, and that have instantaneous duration.
In particular, it would be possible to think that any field (e.g. electromagnetic field) is made by particles, that colliding with other particles causes the interaction. But this is the basic principle of quantum field theory (see chapter 26 of RTR)!

12 Jan 2012, 12:15

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [18.22]
Ah, of course! Doh!!! I should've noticed that!

...Just because the flattness of Minkowski space allows you to unambiguously add tensors defined at different points in the space, it doesn't mean that the result of that operation will still be a tensor! (In general; this case is an exception, as noted).


But how do you conclude that Special Relativity plus the existence of a 4- angular momentum tensor implies "that all interactions that may exist in nature occur via collisions among particles"?

I haven't gotten to chapter 22 yet, so perhaps there's an explanation there, but as far as I had understood it, both concepts are compatible with a picture of nature that includes point particles interacting with classical fields (which unlike quantum fields, can't be interpreted as being made of colliding particles). Or is it not possible to define an angular momentum tensor for a classical field?

...I would've reached a somewhat weaker conclusion, which is that any interaction that affects momentum must be local: A collision in the case where two particles interact, and a localised disturbance in the field when a field interacts with a particle (which of course would then propagate outwards through the field from the point of interaction).

Obviously, if you are considering a quantum picture of nature where everything is particles, then the implication is that the only possible kind of interaction is an (instantaneous) collision. (Where the case of 2+ particles colliding and sticking together, or a single particle suddenly splitting into two or more should also count as a "collision" for our purposes).

13 Jan 2012, 11:39

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [18.22]
The discussion is becoming complex, so I posted my reply in the forum "Exercise discussion" http://roadtoreality.info/viewtopic.php?f=5&t=1942&p=2851#p2851.

13 Jan 2012, 17:48
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