RevModPhys.84.671

The yukawa suppression factors emr characteristic of

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ފdr2 þ F ðrÞr2 d2 to get to spherical coordi½ F ð rÞ þ r nates we find h dx dx ¼ ÀBðrÞdt2 þ CðrÞdr2 þ AðrÞr2 d2 ; (3.14) where 2M 3M P 2M CðrÞ ¼ À 3M P A ð rÞ ¼ 1 eÀmr ; 4 r 1 eÀmr 1 þ mr ; 4 r m 2 r2   Z dD p 1 1  T ðpÞ ; eipÁx 2 T ðpÞ À DÀ2 ð2ÞD p (3.19) R D ÀipÁx  T ðxÞ is the Fourier transform where T  ðpÞ ¼ d xe of the source. To get the retarded field, we integrate above the poles in the p0 plane. Now we specialize to D ¼ 4, and we consider as a source the point particle of mass M at the origin (3.10). For this source, the general solution (3.19) gives h ðxÞ ¼  h00 ðxÞ ¼ BðrÞ ¼ À h00 ðxÞ ¼ M Z d3 p ipx 1 M1 ; e ¼ 2MP ð2Þ3 p2 2MP 4r h0i ðxÞ ¼ 0; hij ðxÞ ¼ For later reference, we record this result in spherical spatial coordinates as well. Using the formula ½FðrÞij þ GðrÞxi xj Šdxi dxj ¼ ½FðrÞ þ r2 Gðrފdr2 þ FðrÞr2 d2 to get to spherical coordinates we find a metric of the form (3.14) with BðrÞ ¼ À (3.15) M1 ; 2MP 4r AðrÞ ¼ In the limit r ( 1=m these reduce to (3.16) Corrections are suppressed by powers of mr. For comparison, we compute the point source solution for the massless case as well. We choose the Lorenz gauge (2.49). In this gauge, the equations of motion simplify to (3.17) Taking the trace, we find hh ¼ ½2=ðD À 2ފT , and upon substituting back, we get   1  T : (3.18) hh ¼ À T À DÀ2 This equation, along with the Lorenz gauge condition (2.49), is equivalent to the original equation of motion in the Lorenz gauge. Taking @ on Eq. (3.17) and on its trace, using conservation of T and comparing, we have hð@ h À 1 @ hÞ ¼ 0, 2 so that the Lorentz condition is automatically satisfied when boundary conditions are satisfied with the property that hf ¼ 0 ) f ¼ 0 for any function f, as is the case when we impose retarded boundary conditions. We can then solve Eq. (3.17) by Fourier transforming. Rev. Mod. Phys., Vol. 84, No. 2, April–June 2012 Cð rÞ ¼ M1 ; 2MP 4r M1 : 2MP 4r (3.21) C. The vDVZ discontinuity BðrÞ ¼ À hh À 1 hh ¼ ÀT : 2 (3.20) M Z d3 p ipx 1 M1 : e ¼ 3 2 ij 2MP ð2Þ 2MP 4r ij p M 1 eÀmr 1 þ mr þ m2 r2 : 3M P 4 r m 2 r2 2M 1 ; 3MP 4r 2M 1 CðrÞ ¼ À ; 3MP 4m2 r3 M 1 : AðrÞ ¼ 3MP 4m2 r3 681 We now extract some physical predictions from the point source solution. Assume we have a test particle moving in this field, and that this test particle responds to h in the same way that a test particle in general relativity responds to the metric deviation g ¼ ð2=MP Þh . We know from the textbooks [see, for example, Chapter 7 of Carroll (2004)] that if h takes the form 2h00 =MP ¼ À2, 2hij =MP ¼ À2 c ij , h0i ¼ 0 for some functions ðrÞ and c ðrÞ, then the Newtonian potential experienced by the particle is given by ðrÞ. Furthermore, if c ðrÞ ¼ ðrÞ for some constant , called the parametrized post-Newtonian (PPN) pa...
View Full Document

This document was uploaded on 09/28/2013.

Ask a homework question - tutors are online