How Can I Understand Slowly Rotating Perturbations of the Whittaker Metric?

  • Thread starter Thread starter MstoneHall
  • Start date Start date
  • Tags Tags
    Metric
MstoneHall
Messages
2
Reaction score
0
I am interested in considering slowly rotating perturbations of the Whittaker metric. Unfortunately I have very little knowledge of this metric. Could someone help me with the exact form of this metric and maybe where to get more material? I understand this may be out of place here but...
 
Physics news on Phys.org
http://www.journals.royalsoc.ac.uk/openurl.asp?genre=article&id=doi:10.1098/rspa.1968.0133

J. M. Whittaker,
An Interior Solution in General Relativity
Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences (1934-1990)
ISSN: 0080-4630
Issue: Volume 306, Number 1484 / July 30, 1968

has links to other papers that reference it.

http://www.jstor.org/view/00804630/di002761/00p00437/0 [via JSTOR]


http://adsabs.harvard.edu/abs/1968RSPSA.306...1W
lists additional citations.
 
Last edited by a moderator:
Wont' say!

Well, a moment with Google yields the paper http://arxiv.org/abs/gr-qc/9910001, from which it seems that the so-called Whittaker metrics are obtained from the coframe (first covector field timelike, rest spacelike, all pairwise orthogonal)
\sigma^1 = -\sqrt{1+2 \, h(r,\theta)} \, A(r) \, dt
\sigma^2 = \sqrt{1+2 \, m(r,\theta)} \, B(r) \, dr
\sigma^3 = \sqrt{1 + 2 \, k(r,\theta)} \, C(r) \, d\theta
\sigma^4 = \sin(\theta) \, \left( d\theta + (\Omega - \omega(r)) \, dt) \right)
via
g = -\sigma^1 \otimes \sigma^1 + \sigma^2 \otimes \sigma^2<br /> + \sigma^3 \otimes \sigma^3 + \sigma^4 \otimes \sigma^4
I believe the motivation is Hartle's approximation scheme for slowly rotating axisymmetric fluid balls in gtr, you can look up the citations they give and see if any of the earlier papers clarify this.

(While I was composing this reply, robphy posted some relevant information. Rob, I don't have on-line access to that journal here--- do you know if this is an exact fluid solution or, as I guess, an approximation valid to say second order in a rotation parameter?)

Picking up the counting theme, the metric functions here comprise three functions of two variables and four functions of one variable, and all are independent of time. One can ask: how many metric functions of so many variables are required to represent an arbitrary stationary axisymmetric Lorentzian spacetime? An arbitrary stationary axisymmetric fluid solution?
 
Last edited:
Much thanks Chris and robphy. Your explanations and links have been of tremendous help.
 
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
ASSUMPTIONS 1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both. 2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals. 3. The speed of light is anisotropic. METHOD 1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A. 2. Clock B receives the signal from A at time t[B2] and...
So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right. Couls someone point me in the right direction? "What have you tried?" Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason. I thought it would be a bit of a challenge so I made a derivation or...
Back
Top