Boyer-Lindquist Cooridinates - incorrect sign on cross term?

[m4r35n357]

OK, I realize I am probably on a hiding to nothing here, but please bear with me for now, I have tested my calculations so carefully that I am totally convinced that I have found an error in two published accounts of the BL line element . . .

The first is Visser, equation (57) http://arxiv.org/abs/0706.0622
The second is a document by someone (possibly) with initials "ELR" Chapter 7: The Kerr Metric – Rotation and Rotating Black Holes: pisces.as.utexas.edu/GenRel/KerrMetric.pdf

Basically, in both cases the sign of the dt dPhi cross term is given as negative, but I can only get sensible answers with a positive term (changing no other signs in the line element).

In my defence, I should explain that I am doing a full 4D geodesic simulation in GNU Octave, using BL coordinates to set initial conditions by specifying L and E, noting the 4-velocity norm is exactly -1.0, then converting to Doran coordinates for the integration, noting absolute rock-solid constancy of conserved energy, momentum, and of course the 4-velocity norm is absolutely stable at -1.0 at all times.

To reiterate, this all happens with my positive signed cross term, if I change it back to the (supposedly correct) negative value, all the 4-velocity norms are off, and of course the trajectory is completely different.

So, is anyone here able to check the validity of the two references above? I can of course give more information as required, but I think this is plenty for now.

 

[m4r35n357]

Pardon typo in title, I can't see how to edit it :(

 

[Mentz114]

I've scanned the Visser paper but I can't determine if a >= 0. This could be relevant because the cross term is the only one that depends on a and not a².

Sounds like some very interesting stuff you're doing.

 

[m4r35n357]

I've scanned the Visser paper but I can't determine if a >= 0. This could be relevant because the cross term is the only one that depends on a and not a².

Sounds like some very interesting stuff you're doing.

Heh, thanks for the kind words, I've put a lot of effort into understanding this ;) As for full disclosure, I'm actually just an unemployed amateur, trying to keep my mind occupied. The most challenging part has been generating the Christoffel symbols for the Doran metric in wxMaxima, then factorizing the resulting mess by hand to generate derivatives for the integration. Sorting out the initial conditions for non-equatorial orbits was quite challenging too. You can probably understand how I much have relied on all the various checks to preserve my confidence in the results!

Now I'm happy with the calculations, I need to think of a way of encapsulating them into something grander and more visual than Octave plots . . .

 

[Mentz114]

I can understand the problems you have with the Maxima simplifications. It's not too clever sometimes.

Have you tried checking if the metric in (57) is a vacuum ? I doubt if it is with both signs of the cross term. I could do this but not until the weekend.

 

[m4r35n357]

I can understand the problems you have with the Maxima simplifications. It's not too clever sometimes.

Have you tried checking if the metric in (57) is a vacuum ? I doubt if it is with both signs of the cross term. I could do this but not until the weekend.

Quite, I wish I could tell it that sin²x+cos²x=1 ;)
As to the vacuum question, I have taken it that the metric is a vacuum solution, particularly bearing in mind section 2 - "No Birkhoff Theorem", in which he says that the Kerr metric in general is ONLY applicable to black holes, and not a general rotating mass. I have to accept that for the time being because he lost me at the word "quadrupole".
On reflection I think the point you made about the ambiguity in definition of the sign of a is key here, and on that basis I am happy to assume that my work has resolved it (for me) ;)
I forgot to mention that I am also comparing results with GROrbits http://stuleja.org/grorbits/ for the equatorial case, to which I also agree with good accuracy (6 significant figures or better).

 

[PeterDonis]

Basically, in both cases the sign of the dt dPhi cross term is given as negative, but I can only get sensible answers with a positive term (changing no other signs in the line element).

The sign of the cross term in an actual computation also depends on the sign of the angular momentum a (or J), relative to the direction of the t time coordinate and the phi angular coordinate; see below. Also, by "the sign of the dt dphi cross term is negative", I assume you mean "the same sign as the dt^2 term", or "the opposite sign as the dphi^2 term", since it's really the relative signs of the terms that matters.

I've scanned the Visser paper but I can't determine if a >= 0.

The sign of a (or J) should depend on which way the hole is rotating, relative to the direction of the t time coordinate and the phi angular coordinate. A positive a means that the sense of rotation, in the future direction of time, is in the direction of positive phi. With the normal "right hand rule" definition of angular coordinates, this means the hole rotates counterclockwise when viewed from above the "North Pole" (the same as the Earth does).

A good way to cross check the sign conventions is to compute the 4-velocity of a "zero angular momentum observer" or ZAMO, i.e, of a worldline that is orthogonal to a surface of constant time, and find that it has a nonzero phi component (i.e., a ZAMO is not static but "rotates with the hole" at an angular velocity that depends on radius). Here's a quick computation (I'm simplifying to the equatorial plane, theta = pi/2, since the metric looks simpler there, but the key result generalizes): we write the line element (with dr = dtheta = 0) in the equatorial plane as:

$$ds^2 = - \left( 1 - \frac{2m}{r} \right) dt^{2} - \frac{4ma}{r} dt d\phi + \left( r^{2} + a^{2} + \frac{2ma}{r} \right) d\phi^{2} = g_{tt} dt^{2} + 2 g_{t\phi} dt d\phi + g_{\phi\phi} d\phi^{2}$$

So we have three nonzero terms of interest in the metric, $g_{tt}$, $g_{t\phi}$, and $g_{\phi\phi}$. We consider the 4-velocity of a ZAMO, $u^{a} = u^{t} \partial_{t} + u^{\phi} \partial_{\phi}$, and compute its inner product with any vector we like in a surface of constant t, the simplest one being $e^{\phi} \partial_\phi$, where $e^{\phi} = 1$. The condition for orthogonality is then inner product = 0, or (nonzero terms only):

$$0 = g_{t\phi} u^{t} e^{\phi} + g_{\phi\phi} u^{\phi} e^{\phi} = - \frac{2ma}{r} u^{t} + \left( r^{2} + a^{2} + \frac{2ma}{r} \right) u^{\phi}$$

This has the obvious solution:

$$u^{\phi} = \frac{\frac{2ma}{r}}{r^{2} + a^{2} + \frac{2ma}{r}} u^{t}$$

Note that the sign of $u^{\phi}$ depends on the sign of a, so the sign of a defines the "sense of rotation" of the hole, as above.

But note also that this solution depends on the signs of the two metric coefficients, $g_{t\phi}$ and $g_{\phi \phi}$, being opposite; if they were the same sign there would be no solution at all for positive a, which is obviously wrong.

So I think the signs given in the papers you linked to are correct.

 

[m4r35n357]

So I think the signs given in the papers you linked to are correct.

Thanks for your detailed reply, Peter. It will take me some time to digest . . .

 

[PeterDonis]

But note also that this solution depends on the signs of the two metric coefficients, $g_{t\phi}$ and $g_{\phi \phi}$, being opposite; if they were the same sign there would be no solution at all for positive a

Actually, I mis-stated this; there would be a solution, but it would have the sign of $u^{\phi}$ reversed relative to the sign of a, which is wrong given the sign conventions for a as discussed.

 

[Mentz114]

Quite, I wish I could tell it that sin²x+cos²x=1 ;)

Try trigsimp( expr), that will do it.

On reflection I think the point you made about the ambiguity in definition of the sign of a is key here, and on that basis I am happy to assume that my work has resolved it (for me) ;)
I forgot to mention that I am also comparing results with GROrbits http://stuleja.org/grorbits/ for the equatorial case, to which I also agree with good accuracy (6 significant figures or better).

Yes, PeterDonis has it spot on.

With agreement that good I'd be confident.

 

[m4r35n357]

Try trigsimp( expr), that will do it.

Yes, PeterDonis has it spot on.

With agreement that good I'd be confident.

I do have trigsimp: true set globally; it would be a PITA to apply it to each element of a tensor, but I'll look into it anyway, cheers.

BTW I'm working through Peter's explanation, getting there slowly, but I do seem to be getting the right answers for the wrong reasons ;) Still, the pressure's off, I can sort this one out at my leisure.

Thanks guys!

 

[Mentz114]

I do have trigsimp: true set globally; it would be a PITA to apply it to each element of a tensor, but I'll look into it anyway, cheers.

BTW I'm working through Peter's explanation, getting there slowly, but I do seem to be getting the right answers for the wrong reasons ;) Still, the pressure's off, I can sort this one out at my leisure.

Thanks guys!

I calculated the Ricci tensor for the metric in eqn(57) of Vissers paper for both +a and -a.
They are both vacuums (vacua ?). Of course I could have made a mistake.

m4r35n357, this is the Maxima script, perhaps you could check I haven't mistranscribed the metric into matrix form. Changing the sign of the off-diagonal term still gives a null Ricci tensor.

/********************************
Kerr spacetime in Boyer-Lindquist coords

***************************/
kill(all);
load(ctensor);
ratflag:true;
ratricci:true;
ratriemann:true;
ct_coords:[t,r,theta,phi];
lg:matrix([(2*m*r)/(a^2*cos(theta)^2+r^2)-1,0,0,(-2*a*m*r*sin(theta)^2)/(a^2*cos(theta)^2+r^2)],[0,(a^2*cos(theta)^2+r^2)/(r^2-2*m*r+a^2),0,0],[0,0,a^2*cos(theta)^2+r^2,0],[(-2*a*m*r*sin(theta)^2)/(a^2*cos(theta)^2+r^2),0,0,sin(theta)^2*((2*a^2*m*r*sin(theta)^2)/(a^2*cos(theta)^2+r^2)+r^2+a^2)])$
ug:trigsimp(invert(lg));
 
ricci(false);
Ric:zeromatrix(4,4)$
for _a thru 4 do for _b thru 4 do Ric[_a,_b]:trigsimp(ric[_a,_b]);
Ric;

 

[m4r35n357]

m4r35n357, this is the Maxima script, perhaps you could check I haven't mistranscribed the metric into matrix form. Changing the sign of the off-diagonal term still gives a null Ricci tensor.

Yes, as far as my eyes can tell you are using the same metric tensor as me. Not sure exactly what your script does, although it does gives a (zero) ricci tensor for ±a where my script just grinds away until I bail out of it . . . (I use cmetric where you invert the matrix, that's the bit I don't get, but I've never really wanted to see the output for any kerr-based metric!)

 

[Mentz114]

Yes, as far as my eyes can tell you are using the same metric tensor as me. Not sure exactly what your script does, although it does gives a (zero) ricci tensor for ±a where my script just grinds away until I bail out of it . . . (I use cmetric where you invert the matrix, that's the bit I don't get, but I've never really wanted to see the output for any kerr-based metric!)

Well, if Maxima is calculating the Ricci correctly, it shows that the metric is valid for ±a. Which was your original question. I reckon if you use -a in the metric and flip a sign somewhere in your calculation, you'll get a good simulation.I'm surprised your script hangs. When my machine hits the ricci(false) line it pauses for about a second. So it's doing a heavy calculation.

cmetric() is necessary if cframe_flag is true, but otherwise just giving lg and ug is enough.

If you want to talk about Maxima, I'm open to private messages (PMs).

 

[julian]

According to Hartle's book "Gravity An introduction to Einstein's general relativity" the d phi dt term has the same sign as the dt^2 term. I've done a detailed derivation and got the same answer as Hartle (my calculation was based on Alder's book where he gets the sign wrong!). You can find my calculation at: http://www.ianbay.comeze.com/, Volume I, chapter 3.

 

[m4r35n357]

I'm surprised your script hangs. When my machine hits the ricci(false) line it pauses for about a second. So it's doing a heavy calculation.

cmetric() is necessary if cframe_flag is true, but otherwise just giving lg and ug is enough.

If you want to talk about Maxima, I'm open to private messages (PMs).

OK cheers for the offer, but I used the term "grinds" rather than "hangs" - it's not locked up, it's genuinely producing (and parsing) output far too copious for me to care about; to me, one second is not a heavy calculation ;)
I'm going to have another go at resolving this today, rest assured if I find out the problem I'll report back here!

 

[m4r35n357]

According to Hartle's book "Gravity An introduction to Einstein's general relativity" the d phi dt term has the same sign as the dt^2 term. I've done a detailed derivation and got the same answer as Hartle (my calculation was based on Alder's book where he gets the sign wrong!). You can find my calculation at: http://www.ianbay.comeze.com/, Volume I, chapter 3.

I did wonder about that, except in this case the dt² term is the sum of two terms with opposite signs, which sort of obscures the point a bit in my eyes . . . eg. what is the sign of -(b-a), bearing in mind it can be written as (a-b)?
I haven't yet got my head around signs of cross terms for non-diagonal metrics, particularly WRT changes in metric signature (ie. I'm not intentionally messing about with signs). In fact the only reliable way I can tell the metric signature is via the sign of the 4-velocity norm. In my particular case, I am using a -1 metric (Boyer-Lindquist) to set initial conditions, and a +1 metric (Doran) to do the simulation!
Anyway, the sooner I resolve this, the sooner I can stop waffling . . .