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Conserved quantities in the Doran Metric? 
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#1
Mar712, 04:17 AM

P: 90

I've been doing some amateur simulations of particle trajectories in a few wellknown metrics (using GNU Octave and Maxima), and in the case of the Schwarzschild and GullstrandPainleve metrics I have the ability to check my results using freely available equations for conservation of energy and angular momentum.
In the case of the Doran metric however, the geodesic equations are far messier(!), but I believe I now have a "correct" simulation. I would like to check this also, but have not seen any reference to equations for conserved quantities for this metric. Has anyone here looked into this, or does anyone know any useful links, or are there any computer algebra wizards that can help? 


#2
Mar912, 08:02 AM

P: 95

What's the Doran metric? If it doesn't have (m)any symmetries, don't hold your breath for conserved quantities. If it does, work them out yourself: the dot product of a killing vector with the fourvelocity is conserved along a geodesic.



#3
Mar912, 12:47 PM

Physics
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PF Gold
P: 6,161

By the "Doran metric", do you mean the Doran chart on Kerr spacetime? This chart is described, for example, by Matt Visser in this article (Section 7):
http://arxiv.org/abs/0706.0622 If that is what you mean, then the conserved quantities are dictated, as Sam Gralla notes, by the symmetries of the spacetime. These can be read off, roughly speaking, by looking at the metric and seeing which coordinates it doesn't depend on. (I say "roughly speaking" because you need to be using the right set of coordinates for this to work. Strictly speaking, the symmetries are captured by the Killing vectors of the spacetime; the "right" set of coordinates is one where as many coordinates as possible match up with Killing vectors.) 


#4
Mar912, 03:15 PM

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P: 7,634

Conserved quantities in the Doran Metric?
Some remarks:
If a metric has explicit time symmetry, i.e. t does not appear in the metric, then the vector with components <1,0,0,0>, which we will call [itex]\xi^a[/itex] is a Killing vector of that metric. And we can write the dot product [itex]E = g_{ab} \, \xi^a \, u^b[/itex] = constant along any geodesic, [itex]u^b[/itex] being the tangent vector of that geodesic at any point, the same [itex]u^b[/itex] that appears in the geodesic equation. To confirm this (including to prevent any errors by confusing vectors with one forms or other mistakes in interpreting the notation, or typos in my post) you should be able to confirm that one of the geodesic equations is equivalent to [itex] d E/ d \tau = 0[/itex], where you expand the total derivative (dE/d \tau), which is taken to be along the geodesic curve which is parameterized by [itex]\tau[/itex] as usual using the chain rule. Example: for the Schwarzschild metric we can write , for some geodesic parameterized by [itex]\tau[/itex] so that we have the geodesic curve [itex]t(\tau), r(\tau), \theta(\tau), \phi(\tau)[/itex], and the tangent vector [itex]u^a[/itex] of this curve with components [itex] u^t = dt/d\tau, u^{r} = dr/d \tau, u^{\theta} = d \theta / d \tau, u^{\phi} = d \phi / d \tau [/itex] [tex] E(\tau) = (12m/r) u^t = \left(1  \frac{2m} {r(\tau)} \right) \left( \frac{ d \, t(\tau) } { d \, \tau } \right) [/tex] and we can expand using the chain rule [tex] dE / d\tau = (12m/r) ) \frac{d^2 t}{d \tau^2} \frac{2m}{r^2} \left( \frac{d \, r } {d \tau} \right) \left( \frac{d \, t} {d \tau} \right) = 0 [/tex] Which with some algebra this can be seen to be identical to one of the standard geodesic equations for the Schwarzschild metric. Thus we see that one of the geodesic equations is equivalent to a statement that E is constant along a geodesic curve. Furthermore, if you convert to a coordinate chart that doesn't have explicit time symmetry, the tensor equations above will still be true, but the Killing vector will have different components. And because the Killing vector transforms like a vector, you can use the standard vector transformation laws to find the components in the new coordinate chart  so you can convert the Kerr Killing vectors to the Doran chart, for instance. 


#5
Mar1012, 05:46 AM

P: 90

OK thanks for the replies so far, for information my main link for the metric is arxiv.org/pdf/grqc/0411060 (sorry but the whole link is too messy). You might be able to imagine that the Christoffel symbols (34 nonunique) are hideous to work with (at least the output from Maxima, which is very poorly factorized) so I don't fancy my chances at much of the maths on my own, hence my question here. I don't have GRTensor and I just hoped that it would do a better job . . . .
Pervect: I will have to take my time to absorb your comments fully, I think you have given me some good information, thank you. 


#6
Mar1012, 10:37 PM

Physics
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PF Gold
P: 6,161




#7
Mar1112, 01:36 AM

P: 611

'A new form of the Kerr solution' Chris Doran http://lanl.arxiv.org/abs/grqc/9910099 'PainleveGullstrand Coordinates for the Kerr Solution' Jose Natario http://arxiv.org/abs/0805.0206v2 


#8
Mar1112, 09:17 AM

P: 90




#9
Mar1212, 01:07 PM

P: 90

[EDIT] OK I see both now, np ;) 


#10
Mar1312, 12:59 PM

P: 90

I am also now intrigued at the possibility of replacing two huge geodesic equations with much simpler expressions, Maxima permitting ;) Still working my way through the last part of your post (quoted). As it happens I need to implement some control over coordinates in order to set initial conditions in a consistent way. 


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