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Exercise [18.22]
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Joined: 03 Jun 2010, 15:18
Posts: 136
Exercise [18.22]

Actually, the deduction is mine: in the following of the book I found no additional discussion of this issue. So maybe it is not correct; but I can't see any flaw.

The basic point is that definition of total 4-angular momentum MS of a system S requires that interactions are local (i.e. are "collisions") and between particles belonging to S. To allow for local interactions between a classical field and the particles, you should include the field into the system S, i.e. add also the 4-angular momentum due to the field.
And for a classical field (not made of quanta) you cannot define a 4-angular momentum, but only a 4-angular momentum density; calling m(x,y,z,t) this density, the 4-angular momentum associated to a 4-volume dV at point (x,y,z,t) is M=m dV.
And you cannot add a 4-angular momentum (a set of 16 numbers) to a 4-angular momentum density ( a set of 16 function of space time).
Therefore the field, to interact, has to have the same point-like nature of the particles, i.e. has to be made of quanta.

Indeed, one could say: OK, maybe assuming existence of total 4-angular momentum is a too much restrictive hypothesis. Just assume that a total 4-angular momentum density exists. In this case, summation cause no trouble because it is always made at same 4-point.
Then one could define the 4-angular momentum density of a point-like particle as a Dirac delta.
But this would mean to consider point-like particles just as the very localised special case (Dirac delta) of a not localised class of objects. And this not localised object is just a field. So the hypothesis that total 4-angular momentum density exist (and not total 4-angular momentum) led us to the opposite conclusion: particles are fields (possibly as localised as Dirac deltas).
And this is quantum mechanics.

So a generalised version of the deduction is that (assuming existence of total 4-angular momentum or of its density) point-line particle can only interact with point-like particles (so a field to interact with a particle must be made of quanta) and fields can only interact with fields (so a particle to interact with a field must be a field herself). Mixed interactions (particle with fields) are forbidden. And again, this is the duality field-quanta of QFT.

What I found amazing is that this result can be deduced in a purely classical context, from special relativity.

13 Jan 2012, 17:43

Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [18.22]
I don't agree with that, Roberto.

I don't see any problem with having particles and fields interact, and calculating total (conserved) angular momentum as

(sum of particle Ms) + (integral of field M)

The situation is analagous to the one for mass/energy - also a conserved quantity - where total energy = (mass/energy of particles) + (integral of energy stored in field).

[In the non-relativistic case it's (kinetic energy of particles) + (integral of potential energy stored in the field)].

I agree that there's something awkward about mixing discrete sums with integrals, but that just reflects the fact that in this model, fields and particles are two disparate kinds of entity. Just because it's awkward, doesn't mean it's incorrect!

It might help to note that when particles collide, the change of momentum is instantaneous; whereas when a particle interacts with a field, the change of momentum is gradual. This means that if between the times t and t+dt a particle's momentum changes by an amount dp due to its interaction with a field, then this causes a localised change in the field of maximum extent c.dt around the particle's position at t. This local "bump" in the field will contribute a momentum of -dp to the total integrated field momentum. So no dirac delta calculations are actually necessary. (Though I know, that's not quite why you introduced them earlier).

13 Jan 2012, 19:34

Joined: 03 Jun 2010, 15:18
Posts: 136
Re: Exercise [18.22]
Yes, you are indeed right.
I missed to take into account the (obvious!) fact that when also the field (or fields) 4-angular momentum (i.e. density integrated over the 3-volume) is summed over, the total 4-angular momentum is conserved, because the system now includes all particles and fields, i.e. everything that may interact. This solves the problem of simultaneity that (without taking into account the angular momentum of field) caused to miss Lorentz invariance.

So I think we can deduce two things from the fact that total angular momentum M is Lorentz-invariant:
1) all interactions affecting angular momentum (particle/particle or particle/field) must be local, as you already pointed out in a previous post;
2) because M is Lorentz-invariant only for systems having no interaction, that may cause changes to angular momentum, with external objects (i.e. system closed with respect to angular momentum), and for those system M is conserved, Lorentz invariance of M implies its conservation.

15 Jan 2012, 11:39

Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [18.22]
I'd also note that the entire discussion applies equally well to linear momentum, too.

(In fact, since M can be defined in terms of X^P, you might consider linear momentum to be the more fundamental property... at least until you get to considering particles with intrinsic spin, anyway).

-------------------------

Roberto, I think you've got the direction of implication backwards in (2):

Local conservation of some tensor quantity T implies that the sum or integral of T across any spacelike slice is constant, given a particular reference frame. So this sum or integral depends only on the choice of reference frame, not on the choice of spacelike slice to sum/integrate over. A change of frame applies a Lorentz transformation to T at every point, and (since it's a linear transformation) therefore transforms the sum/integral the same way. So this sum or intergral of T behaves like a tensor, albeit one not anchored to any specific point in Minkowski space. (But that's OK since Minkowski space is flat, so you can define a single "tensor space" that maps identically onto the tensor space at each/any individual spacetime point).

What you seem to be contending in (2) is the reverse of this: Given a sum or integral of some tensor quantity S over a particular spacelike slice (e.g. t=const) relative to a given reference frame, if we know that this sum or integral behaves like a tensor under changes of frame, can we therefore conclude that S is locally conserved?

Possibly that's true, but it's a fairly strong statement, and it's by no means obvious - you'd definitely have to prove it!

Nothing in either our discussion r.e. exercise 18.22 so far, or in the paper you linked to earlier (http://panda.unm.edu/Courses/Finley/P495/TermPapers/relangmom.pdf) demonstrates this, or implies it, in any way that I can see.

15 Jan 2012, 15:41

Joined: 03 Jun 2010, 15:18
Posts: 136
Re: Exercise [18.22]
No, you misunderstood my statement (2): I mentioned conservation of M, total angular momentum, that is global conservation, not local conservation.
Restating (2) in your terms it is:
Given a sum or integral S of some tensor quantity T over a spacelike slice (e.g.t=const ) relative to a given reference frame, if we know that this sum or integral S behaves like a tensor under changes of frame for ervery t , can we therefore conclude that S is globally conserved (i.e. is not a function of t).

In the frame F' we have S'(t'); performing a Lorentz transformation to frame F we obtain S''(k(t+w x)), where k is the usual relativistic factor and, just to keep notation simple, I assumed that transformation from F to F' is just a Lorenz boost with speed w.
Computing the same quantity in frame F we obtain S(t). Lorentz invariance implies the identity
S''(k(t-w x))=S(t).
This may happen only if there is no dependency on the argument t, i.e. if S(t)=S.
In the physical case we are discussing, S is total angular momentum M and the relationship is total angular momentum conservation.

17 Jan 2012, 09:43

Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [18.22]
Ah, in that case, I'd agree with you!

We have:

globally conserved locally conserved tensor quantity globally conserved

------------------------------------------

By the way, you appear to be using the term "Lorentz invariant" to mean "behaves like a tensor", which is not correct. A quantity is Lorentz invariant if its value at each spacetime point is unchanged by a Lorentz transformation. Scalar fields are Lorentz invariant, but vectors, 1-forms, and general tensors ARE NOT, since their component values change. For example, a vector field pointing in the "+t" direction no longer points that way after a boost. See this link for the correct definition.

Total angular momentum M is most certainly NOT Lorentz invariant!

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While I followed your proof quite well, allow me to present a slightly different version, that might be clearer to some readers:

Let S(t) be a "tensor-like quantity" defined as the sum or integral of the tensor T over all space (that is, over all of x,y,z, with t held constant), as calculated in a given reference frame F.

By construction, S is invariant w.r.t. purely spatial translations in F, i.e.: S(t,x,y,z) S(t)

Now, furthermore suppose that S is indeed a tensor: That is to say, any Lorentz transformation that takes F to any other reference frame F', when applied to the components of S at any spacetime point, yields the same result as calculating S directly at that point in the frame F'.

Since this latter calculation yeilds a tensor field that's invariant w.r.t. purely spatial translations in F', we can conclude that S (as calculated in frame F) must be invariant w.r.t. any spacetime translation that is purely spatial in some reference frame F'.

In particular, the translations and in F both represent purely spatial translations in two other (correctly chosen) reference frames, respectively. Individually each translation must leave the value of S unchanged, and composing them yeilds the translation which therefore must also. Hence:

S(t) = S() for all

This can only be satisfied when S is constant for all t (as calculated in F); i.e. S is "conserved" and does not change over time.

Such a "conserved" S is therefore a tensor quantity that is constant across all spacetime, with component values that depend on the frame F in which it is calculated (as usual for a tensor).

Note that in earlier discussion, we described the conservation of S over time by (equivalently) saying that T is "globally conserved".

------------------------------------------

Note that we could construct an analogous proof (corresponding to a slightly different definition of "globally conserved") if the sum or integral S were to be defined over some particular (frame-relative) set of non-intersecting spacelike slices, where a "slice" is defined so as to bisect all of spacetime into +t and -t regions, and where for each frame, this set covers all space. (i.e. As an alternative to using the flat surfaces t=const). All that is required for the proof is that any point in spacetime be reachable from any other, by traversing the union of these slices (as collected from all reference frames). This works since S must be constant over any such slice, irrespective of reference frame. (Where we define S(t,x,y,z) calculated in reference frame F as the value of the sum or integral of T across the unique F-relative slice that the point (t,x,y,z) happens to lie on).

Note that saying that T is "globally conserved" in this manner is not as strong a statement as saying that the value of S in a given frame F will be the same when computed over any arbitrary spacelike slice. This latter property is implied by local conservation of T, however.

------------------------------------------

We have demonstrated that: "S is a tensor" "T is globally conserved"

It remains an open question (to me) as to whether either of the (successively) stronger implications hold:

"S is a tensor" "the value of S in a given frame F will be the same when computed over any arbitrary spacelike slice"

or

"S is a tensor" "T is locally conserved"

"the value of S in a given frame F will be the same when computed over any arbitrary spacelike slice" "T is locally conserved"

...Anybody have any thoughts on this?

17 Jan 2012, 21:59

Joined: 03 Jun 2010, 15:18
Posts: 136
Re: Exercise [18.22]
The proposition
"S is a tensor" implies "T is locally conserved"
can be restated, in more pysical-oriented terms, as
"S is globally conserved " implies "T is locally conserved"
I doubt that a formal proof may be found, without adding some additional hypothesis.

Let's take the physical example of electric charge, that it is locally (and therefore also globally) conserved. In this case S is the electric charge Q (so it is a scalar) and T is the 4-vector which components are (j,q), the current density vector j and the charge density q.
Global conservation means that the total charge Q of a closed system (e.g. the entire space) is constant in time.
Local conservation is the continuity equation, which physical interpretation is that "decrease of q at a point P1 occurs because there is a current j that transports away some charge from there; and the charge is transported to a near point P2 where q increases of the same amount".
In other words, locality implies that Q conservation is realised by a transport of it from a place to another, by a current; and only this transport (not creation nor destruction) may occur.

Let's now try to think to global conservation without local conservation.
If there is a charge conservation without current (i.e. global without local conservation), then the physical picture is that an amount of charge disappear from a place and the same amount of charge appears at a different place. There are two possible explanations for that:
a) it is the same charge, that moves instantaneously from one point to another; this will imply that this charge moves at infinite speed, in particular faster than light;
b) there two independent, but coordinated, phenomena: some charge is destroyed (by some unknown mechanism) here and some other charge is created (by some other unknown mechanism) there; the two mechanism are related in some way, so that the the amount of destroyed charge is exactly equal to the amount of created charge.
The first explanation violates special relativity, so we can immediately discard it.
The second explanation, despite it may look quite absurd, is more difficult to be disputed. Creation and destruction are not absurd at all: classical heat equation is essentially a continuity equation with an extra "heat source" term (that is a an "heat creation") that is used to represent e.g. heating by Joule effect. What looks "absurd" is that creation and destruction needs to be coordinated; i.e. it is needed a communication channel, allowing the creation and destruction mechanism to exchange information about the amount of charge to be created and destroyed; and this communication has to be done faster than light, so again violating special relativity. But there are quantum phenomena (quantum entanglement, chapter 23 of RTR) that show a very similar non-local behaviour, and have a lots of subtleties and open questions. So I can't see any strong reason to discard explanation b), without adding some new hypothesis.

18 Jan 2012, 17:22

Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [18.22]
Yes, but suppose (by some mechanism) some charge was created at point A, while at exactly the same time (by my watch) an equal amount of charge was destroyed at point B. According to me, then, there's conservation.

But to an observer in another frame, the two events are no longer simultaneous, so for the time duration between them, charge is NOT conserved!

I wasn't really thinking about physical mechanisms here; more about the mathematical question: Can you define a tensor field T that fails to satisfy the continuity equation, but that nevertheless yields the same constant value when integrated over every spacelike plane through Minkowski space?

The requirement that the value be the same for every such plane imposes a very strong constraint on T. Intuitively, it seems to me that this might be enough to guarantee that the continuity equation is satisfied, but I'm not certain of that... and right now neither a proof, nor a simple counterexample comes to mind.

...I think I've gotten a bit off the topic of the exercise, though!

18 Jan 2012, 18:10
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