Problems with Mannheim s conformal gravity program

Transcription

1 Poblems with Mannheim s confomal gavity pogam Abstact We show that Mannheim s confomal gavity pogam, whose potential has a tem popotional to 1/ and anothe tem popotional to, does not educe to Newtonian gavity at shot distances, unless one assumes undesiable singulaities of the mass density of the poton. Theefoe, despite the claim that it successfully explains galaxy otation cuves, unless one assumes the singulaities, it seems to be falsified by numeous Cavendish-type expeiments pefomed at laboatoies on Eath whose wok have not found any deviations fom Newton s theoy. Moeove, it can be shown that as long as the total mass of the poton is positive, Mannheim s confomal gavity pogam leads to negative linea potential, which is poblematic fom the point of view of fitting galaxy otation cuves, which necessaily equies positive linea potential. 1 The Weyl tenso This section closely follows Intoducing Einstein s Relativity by Ray D Inveno. Weyl tenso C abcd is defined in n 3 dimensions by The C abcd = R abcd + 1 n 2 (g 1 adr cb +g bc R da g ac R db g bc R ca )+ (n 1)(n 2) (g acg db g ad g cb )R (1) It is easy to check that the Weyl tenso has the following same symmeties as the Riemann tenso (Poblem 1.): C abcd = C abdc = C bacd = C cdab (2) C abcd + C adbc + C acdb = (3) Howeve, it has an additional symmety the Riemann tenso doesn t have: C a bad = (4) (Poblem 2. Check this.) Thus, we say the Weyl tenso is tace-fee as the tace vanishes. Theefoe, one can think of the Weyl tenso as the tace-fee component of the Riemann tenso. 1

2 Now, conside the following tansfomation called confomal tansfomation. g ab (x) e 2φ(x) g ab (x) (5) Unde this tansfomation, the angle is peseved. (Poblem 3. Check this. Hint: use cos θ = a b a and expess the dot poduct and the magnitude using metic.) That is the b eason why confomal is tanslated into dengjiao in Chinese which means same angle. The Weyl tenso has a vey special popety. Unde the confomal tansfomation (5), the Weyl tenso is invaiant: C a bcd C a bcd (6) Fo this eason, the Weyl tenso is also called the confomal tenso. 2 Mannheim s confomal gavity Instead of Einstein-Hilbet action, in Mannheim s confomal gavity pogam, we have the following action. S = α g d 4 x gc λµνκ C λµνκ = 2α g d 4 x g[r µν R µν 1 3 R2 ] (7) whee C λµνκ is the confomal Weyl tenso, and α g is a puely dimensionless coefficient. By adding this to the action of matte and vaying it with espect to the metic, one can obtain the confomal gavity vesion of the Einstein equation. Poblem 4. Show that the above action is invaiant unde confomal tansfomation using (6). (Hint: Fist show that g e 8φ g.) 3 The metic solution in confomal gavity The following deivation closely follows Mannheim and Kazanas s in Ref. [3]. [See in paticula Eqs. (9), (13), (14) and (16) in thei pape.] When thee is a spheical symmety in the distibution of the mass, one can wite the metic as follows: ds 2 = B()dt 2 + d2 B() + 2 dω 2 (8) Plugging this into the confomal gavity vesion of the Einstein equation, and assuming that all the matte is inside the adius, Mannheim and Kazanas obtain 2

3 B( > ) = 1 2β + γ (9) 4 B() = f() (1) whee The solution is given by 2β = 1 6 γ = 1 2 f() without any appoximation whatsoeve. d 4 f( ) (11) d 2 f( ) (12) 3 4α g B() (T T ) (13) 4 Non-Newtonian potential If we ignoe T in the above equation because it is small, set T we get: = ρ, and use B() 1, 2β = 1 ( 1 d 4 ρ) (14) 8α g Compae this with the Newtonian case, which is the following: 2β = 2G c 2 d 4π 2 ρ = 2GM total c 2 (15) Thus, unlike in the Newtonian case, we see that in Mannheim s confomal gavity, the gavitational attaction depends not only on the total mass, but also on the mass distibution. Theefoe, if two spheically symmetic objects with the same mass but diffeent density distibutions yield the same stength of gavitational foces, then confomal gavity is toublesome. On the othe hand, if they yield diffeent stengths of gavitational foce, in pecisely the manne that confomal gavity pedicts, then confomal gavity will be veified. Notice that the diffeence of the gavitational foce would be big; it would be in leading ode, not in next-to-leading ode. Fo example, if the mass of two objects is the same, but the fist one s size is double that of the second one, the fome will exet quaduple the amount of gavitational foce. Confomal gavity seems toublesome, as many Cavendish-type expeiments have been pefomed, and none of them has detected that gavity depends on the density distibution [4]. We intoduce Mannheim and Kazanas s cicumvention of this dilemma in the next section. 3

4 5 The wong sign of the linea potential tem Mannheim compaes the gavitational potential in Newtonian gavity and confomal gavity in Ref. [1]. He consides the case in which all the matte is inside the egion ( < R), and the mass distibution only depends on (i.e., spheically symmetic). In the case of Newtonian gavity, the potential is given by The solution is given by φ( < R) = 1 2 φ( ) = g( ) (16) φ( ) = 1 4π φ( > R) = 1 d 3 g( ) R d 2 g( ) (17) d 2 g( ) (18) R In the case of Mannheim s confomal gavity, we have d g( ) (19) 4 φ( ) = h( ) (2) The solution is given by φ( ) = 1 8π d 3 h( ) (21) φ( < R) = 1 6 φ( > R) = 1 6 d 4 h( ) 1 2 R R d 4 h( ) 2 d 3 h( ) 2 Then, Mannheim notes that the following h(): h( < R) = γc 2 N i=1 δ( i ) 2 yields the following gavitational potential: 3βc2 2 φ( > R) = Nβc2 i=1 R d 2 h( ) (22) d 2 h( ) 2 6 R N ] [ ] [ 2 2 δ( i ) Nγc2 2 d h( ) Thus, by teating odinay matte as being composed of point-like mass singulaities epesenting potons and neutons, Mannheim ecoves a potential fo a spheically symmetic object that is independent of its intenal mass distibution and that matches Newton s law (with an additional tem popotional to ). (23) (24) (25) 4

5 Howeve, a close look at the last tem of Eq. (22) shows that this cicumvention will not wok. Notice that h is the mass density up to a cetain positive coefficient. Theefoe, h 2 d (26) should be equal to the total mass of the poton divided by the positive coefficient. This is obvious fom the following elementay fomula: M = 4π 2 ρd (27) Theefoe, the last tem of Eq. (22) yields the negative linea gavitational potential. Howeve, we know that we want a positive linea gavitational potential to fit the galaxy otation cuve; what we need is not exta epulsion but exta attaction. Theefoe, Mannheim s confomal gavity pogam seems poblematic, unless someone can come up with an agument that α g in Eq. (7) can take a negative value. 1 We want to note that Mannheim s confomal gavity pogam, with which we have dealt in this pape, should not be confused with Andeson-Babou-Foste-Muchadha confomal gavity [5]. 6 Mannheim s citicism About two yeas afte my pape Poblems with Mannheim s confomal gavity pogam was published, Mannheim uploaded his own pape to the axiv citicizing my pape. He noted T µ µ =, which should be satisfied fo a confomal theoy, is not satisfied fo my appoximation T ρ and T. Let me fist explain why the tace of the enegy momentum must vanish in a confomal theoy (i.e. a theoy which has a confomal symmety which means that the action is invaiant unde confomal tansfomation). Let s wite the total action of confomal theoy as the sum of the gavitational action and the matte action as follows: S = d 4 x g(l g + L m ) (28) Now, fo a small vaiation of the metic, the action is invaiant as follows: ( δ( = δs = d 4 glg ) x δg ab δg ab + δ( ) glm ) δg ab δg ab (29) 1 It is unlikely that α g is negative. Fom the action, one can calculate the enegy density (i.e. Hamiltonian density) and it has to have a cetain minimum. If α g is negative, the enegy would have no minimum and the field would continue to fall down to the minimum enegy foeve. 5

6 Given this, notice that fom (5) an infinitesimal confomal tansfomation can be witten as follows: δg ab (x) = 2φ(x)g ab (x) (3) If we plug this into (29), the fist tem must be zeo since the gavitational action of confomal theoy doesn t change unde confomal tansfomation. Theefoe, the second tem must be zeo as well. Recall now, T ab 2 g δ( gl m ) δg ab (31) Plugging this in, we get: = d 4 x gφ(x)g ab (x)t ab (x) (32) Since this has to be tue fo any φ(x), we conclude Ta a =. To make the tace of the enegy-momentum tenso vanish, Mannheim assumes that the Higgs field is distibuted within galaxies in such a manne that it compensates the tace of the enegy-momentum tenso of odinay matte (i.e. stas, etc.). Howeve, it is also vey difficult to popose a mechanism that would distibute the Higgs field in such a manne. It is simply unlikely. Refeences [1] P. D. Mannheim, Altenatives to dak matte and dak enegy, Pog. Pat. Nucl. Phys. 56, 34 (26) [asto-ph/55266]. [2] P. D. Mannheim, Making the Case fo Confomal Gavity, Found. Phys. 42, 388 (212) [axiv: [hep-th]]. [3] P. D. Mannheim and D. Kazanas, Newtonian limit of confomal gavity and the lack of necessity of the second ode Poisson equation, Gen. Rel. Gav. 26, 337 (1994). [4] G. T. Gillies, The Newtonian gavitational constant: ecent measuements and elated studies, Rept. Pog. Phys. 6, 151 (1997). [5] E. Andeson, J. Babou, B. Foste and N. O Muchadha, Scale invaiant gavity: Geometodynamics, Class. Quant. Gav. 2, 1571 (23) [g-qc/21122]. 6