RevModPhys.84.671

# The yukawa suppression factors emr characteristic of

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Unformatted text preview: Þdr2 þ F ðrÞr2 d2 to get to spherical coordi½ F ð rÞ þ r nates we ﬁnd h dx dx ¼ ÀBðrÞdt2 þ CðrÞdr2 þ AðrÞr2 d2 ; (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 ð2ÞD 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 ð2Þ3 p2 2MP 4r h0i ðxÞ ¼ 0; hij ðxÞ ¼ For later reference, we record this result in spherical spatial coordinates as well. Using the formula ½FðrÞij þ GðrÞxi xj dxi dxj ¼ ½FðrÞ þ r2 GðrÞdr2 þ FðrÞr2 d2 to get to spherical coordinates we ﬁnd a metric of the form (3.14) with BðrÞ ¼ À (3.15) M1 ; 2MP 4r 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 2ÞT , 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 4r M1 : 2MP 4r (3.21) C. The vDVZ discontinuity BðrÞ ¼ À hh À 1 hh ¼ ÀT : 2 (3.20) M Z d3 p ipx 1 M1 : e ¼ 3 2 ij 2MP ð2Þ 2MP 4r ij p M 1 eÀmr 1 þ mr þ m2 r2 : 3M P 4 r m 2 r2 2M 1 ; 3MP 4r 2M 1 CðrÞ ¼ À ; 3MP 4m2 r3 M 1 : AðrÞ ¼ 3MP 4m2 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...
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## This document was uploaded on 09/28/2013.

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