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Exercise [19.14] b

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Exercise [19.14] b
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Roberto Joined: 03 Jun 2010, 15:18 Posts: 136	Exercise [19.14] b Attached my solution. The last part (conservation low for a discrete system of particles) gives the same result obtained by Laura (http://sites.google.com/site/vascoprat/rtr-solutions ); but Laura demonstrated it (if I understood correctly) only for special relativity (Minkowski space ) while I think the exercise required (again, if I understood it correctly) a demonstration valid also when there is gravity. Attachments: Exercise_19_14.pdf [60.78 KiB] Downloaded 167 times
07 Sep 2010, 14:03

deant Joined: 12 Jul 2010, 07:44 Posts: 154	Re: Exercise [19.14] b I had basically the same solution as Roberto for the last part, except I put it in (what I feel) is a more explicit conservation-law form. $\nabla_a L^a = 0$ is a conservation law expressed in differential form, as applicable to differentiable tensor fields. (It also implies an integral form applicable to compact 4-volumes). In the case of a set of point particles, however, we are dealing with a sum of discrete quantities, rather than fields, so there's no exactly analagous differential or integral conservation laws. Instead, we can define a "time" parameter, $\tau$ , by dividing spacetime into a continous stack of 3-space slices, parameterized by $\tau$ . (Like slicing up a salami; it doesn't matter exectly how we do this - any stack of slices will do, so long as they are everywhere spacelike, and the stack covers all spacetime without any gaps). Then we can phrase our conservation law as saying that the conserved quantity is "constant over time", i.e. $\frac{\mathrm{d}}{\mathrm{d}\tau}(conserved\ quantity) = 0$ . In particular, since particle trajectories are always timelike, they're never parallel to a slice, so $\tau$ can always be used to parameterize the particle trajectories, i.e. the position and momentum of the $i^{th}$ particle can be denoted by $x_i(\tau)^a$ and $p_i(\tau)^a$ (where the is the particle index, and is a tensor index). Then the conservation law () that we're after is: $\frac{\mathrm{d}}{\mathrm{d}\tau} \sum\limits_i p_i(\tau)^a\kappa_a = 0$ Collision points are single spacetime points, so $\displaystyle{\kappa_a}$ is constant and since total (summed) 4-momentum is conserved, so is $\displaystyle{\sum p_i(\tau)^a\kappa_a}$ for the particles involved in the collision at that point. It remains to show that $\frac{\mathrm{d}}{\mathrm{d}\tau}(p^a\kappa_a)$ is zero along the particle trajectories between collision points. Since $\tau$ parametrizes the particle trajectories, then along these trajectories: $\frac{\mathrm{d}}{\mathrm{d}\tau} \propto p(\tau)^b\nabla_b<br />\;\;\;\;\mathrm{i.e.}\;\;\;\;<br />\frac{\mathrm{d}}{\mathrm{d}\tau} \equiv S(\tau)p(\tau)^b\nabla_b$ for some positive scalar function . Then: $\frac{\mathrm{d}}{\mathrm{d}\tau} \sum\limits_i p_i(\tau)^a\kappa_a =<br />\sum\limits_i S_i(\tau)p_i(\tau)^b\nabla_b\left(p_i(\tau)^a\kappa_a\right)$ but $p^b\nabla_b(p^a\kappa_a) = \kappa_a(p^b\nabla_b p^a) + p^a p^b \nabla_b\kappa_a$ And we note that $p^b\nabla_b p^a = 0$ along particle trajectories because they are geodesics (see p.304), and $p^a p^b \nabla_b\kappa_a = p^a p^b \nabla_{(a}\kappa_{b)} = 0$ because $\kappa$ is a Killing vector field, satisfying $\nabla_{(a}\kappa_{b)} = 0$ . So along particle trajectories the summed term in () equates to zero as required, and we have thus demonstrated that (*) holds everywhere (both at collision points and along the particle trajectories between collisions). This holds true for any parameterization of time $\tau$ and for any spacetime that has a Killing vector field $\kappa$ on it, whether it's flat (i.e. Minkowski space) or not.
02 Apr 2012, 12:12

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