Alberta Thy 17-95 hep-th/9508077 Two observers calculate the trace anomaly A. Z. Capri, M. Kobayashi† and D. J. Lamb∗ Theoretical Physics Institute, Department of Physics, University of Alberta, 6 9 Edmonton, Alberta T6G 2J1, Canada 9 1 (February 1, 2008) n a J 5 2 Abstract 2 v 7 7 We adapt a calculation due to Massacand and Schmid to the coordinate 0 8 independent definition of time and vacuum given by Capri and Roy in order 0 5 tocomputethetraceanomaly foramassless scalar field inacurvedspacetime 9 / h in 1+1 dimensions. The computation which requires only a simple regulator t - p and normal ordering yields the well-known result R in a straightforward e 24π h : manner. v i X 03.70 r a Typeset using REVTEX 1 I. INTRODUCTION One of the more interesting results of the study of quantum field theory in curved spacetime is the fact that the expectation value of the trace of the stress tensor of a conformally coupled field does not vanish. It has an anomaly. This trace or conformal anomaly, as it is known, was first noticed by Capper and Duff [3] using a dimensional regularization scheme. Since then many other regularization procedures have been used and when used correctly lead to same result [4]. Unfortunately, as anyone who has ever calculated this trace anomaly knows, the computations required are rather lengthy and certainly less than illuminating. On the other hand, if one has a particle interpretation the problem can be handled more simply. This fact was first exploited by Massacand and Schmid [1]. In this paper we adapt their method to a computation in 1+1 dimensions using only the following two inputs. 1) The frame components of the stress tensor at a given point are, for two frames based at this point, related by a Lorentz transformation. 2) The vacuum expectation value of the energy momentum density (relative to a given frame) should vanish. Thus, the vacuum can have pressure, but no energy or momentum. In general there would remain the vexing question, “Which vacuum?” The answer we propose is to use the coordinate independent definition of Capri and Roy. In section II we give a brief review of this construction of the vacuum and apply the result to a calculation of the vacuum expectation value of the trace of the stress tensor in section III. Our conclusions are set out in section IV. II. COORDINATE INDEPENDENT DEFINITION OF TIME AND VACUUM In a globally hyperbolic spacetime one can choose a foliation based solely on geodesics. Thus, givenatimelike (unit)vectorN (P )atthepointP oneestablishes aframe(zweibein) µ 0 0 at P with components: 0 eµˆ0 = Nµ(P ) 0 2 eµˆ1 = pµ(P ) (2.1) 0 where pµ(P ) is a unit vector orthogonal to Nµ(P ) at P . The spacelike hypersurface (line) 0 0 0 consiting of the geodesic through P with tangent vector pµ(P ) defines the surface t = 0. 0 0 The “time” t corresponding to an arbitrary point P is the distance along a geodesic P P 1 − which intersects the line t = 0 orthogonally at some point P . The geodesic distance P P 1 0 1 − along the line t = 0 yields the space coordinate x. These geodesic normal coordinates prove to be very useful since in these coordinates the metric becomes ds2 = dt2 α2(t,x)dx2 (2.2) − where α(0,0) = 1 ∂α ∂α ∂2α ∂2α = = 0 = = (2.3) ∂t|t=0 ∂x|t=0 ∂t∂x|P0 ∂x2|P0 Also, 2 ∂2α = R (2.4) α ∂t2 where R is the curvature scalar. The field equations in these coordinates, for a massless scalar field read: 1 ∂ (√ggµν∂ )φ = 0 µ ν √g ∂2φ α˙ ∂φ α′ ∂φ 1 ∂2φ + + = 0. (2.5) ∂t2 α ∂t α3 ∂x − α2∂x2 Here, ∂α ∂α α˙ = and α′ = . (2.6) ∂t ∂x The positive frequency modes φ of this field are obtained by solving these field equations with the two initial conditions 1 1) φ (0,x) = exp( ipǫx) p > 0. (2.7) p,ǫ √4πp − 3 Here we have ǫ = +1 corresponds to left travelling waves ǫ = 1 corresponds to right travelling waves. (2.8) − and ∂φ p,ǫ 2) i = pφ . (2.9) ∂t |t=0 p,ǫ|t=0 A useful Ansatz to implement these initial conditions is: 1 φ (t,x) = exp( ipf (t,x)) (2.10) p,ǫ ǫ √4πp − where f is real. Equation (2.5) then yields that ǫ ∂f ǫ ∂f ǫ ǫ = (2.11) ∂t α ∂x The initial conditions become f (0,x) = ǫx (2.12) ǫ and near t = 0 f (t,x) t+ǫx (2.13) ǫ ≈ The quantized field is now given by ∞ Ψ(t,x) = d(ǫp) φ (t,x)a +φ∗ (t,x)a† (2.14) p,ǫ p,ǫ p,ǫ p,ǫ ǫX=±1Z0 (cid:16) (cid:17) with the vacuum defined by a 0 >= 0. (2.15) p,ǫ | These modes have been normalized such that 4 ∞ ↔ (φ ,φ ) = iǫ dx√g φ∗ (t,x) ∂ φ (t,x) p,ǫ q,ǫ p,ǫ t q,ǫ Z−∞ (cid:18) (cid:19) p+q ∞ ∂f ǫ = dxǫα exp(i(p q)f (t,x)) ǫ 4π√pq −∞ ∂t − Z p+q ∞ ∂f ǫ = dx exp(i(p q)f (t,x)) ǫ 4π√pq −∞ ∂x − Z p+q = 2πδ(p q) 4π√pq − = δ(p q) (2.16) − III. THE TRACE ANOMALY We begin with two “observers” with tangents to their world lines given by sinh(χ) Nµ(P ) = (1,0) and N¯µ(P ) = (cosh(χ), ) (3.1) 0 0 α¯ The corresponding frames are: 1 eµˆ0 = (1,0) eµˆ1 = (0, ) (3.2) α sinh(χ) cosh(χ) e¯µˆ0 = (cosh(χ), ) e¯µˆ1 = (sinh(χ), ) (3.3) α¯ α¯ Corresponding to this the metric has the two forms ds2 = dt2 α2(t,x)dx2 = dt¯2 α¯2(t¯,x¯)dx¯2 (3.4) − − We can solve for the positive frequency modes in the barred as well as in the unbarred coordinatestoobtainthecorrespondingquantizedfieldsΨ¯(t¯,x¯)andΨ(t,x). Theirrespective sets of annihilation and creation operators are (a¯ ,a¯† ) and (a ,a† ) . p,ǫ p,ǫ p,ǫ p,ǫ AtP ,thepointwithcoordinates(0,0)inbothcoordinatesystems thetwofieldscoincide. 0 Corresponding to these two fields we have their respective Fock space vacuums ¯0 > , 0 > | | defined by a¯ ¯0 >= 0 , a 0 >= 0 (3.5) p,ǫ p,ǫ | | Anybilinear expression inthefieldoperatorswhich, forphysical reasons, should havevanish- ing vacuum expectation value is defined by normal ordering with respect to its own vacuum. 5 Thus since we expect the vacuum to be the state of zero energy and momentum density we require that < ¯0 : T¯ˆ0µˆ : ¯0 >= 0 (3.6) | | and < 0 : Tˆ0µˆ : 0 >= 0, (3.7) | | where, Tαˆβˆ = eµαêνβˆT µν T¯αˆβˆ = e¯µαê¯νβˆT¯ . (3.8) µν Furthermore, since the barred and unbarred frames e¯µαˆ , eµαˆ are related by a Lorentz transformation cosh(χ) sinh(χ) βˆ Λ = (3.9) αˆ   sinh(χ) cosh(χ)   we have that at P 0 : Tαˆβˆ : = Λαˆ Λβˆ : T¯γˆδˆ : (3.10) |P0 γˆ δˆ |P0 so that in particular : Tˆ0ˆ0 : = cosh2(χ) : T¯ˆ0ˆ0 : +2cosh(χ)sinh(χ) : T¯ˆ0ˆ1 : +sinh2(χ) : T¯ˆ1ˆ1 : . (3.11) |P0 |P0 |P0 |P0 Taking the vacuum expectation value with respect to the barred vacuum of this equation, and using (3.6) we have < ¯0 : Tˆ0ˆ0 : ¯0 >= sinh2(χ) < ¯0 : T¯ˆ1ˆ1 : ¯0 > (3.12) | |P0| | |P0| Since < ¯0 : T¯ˆ0ˆ0 : ¯0 >= 0 we find that the vacuum expectation value of the trace is: | | 1 < ¯0 η : T¯αˆβˆ : ¯0 >= < ¯0 : T¯ˆ1ˆ1 : ¯0 >= < ¯0 : Tˆ0ˆ0 : ¯0 > (3.13) | αˆβˆ |P0| − | |P0| −sinh2(χ) | |P0| 6 To evaluate this expression we have to take the term : Tˆ0ˆ0 : which has been normal |P0 ordered with respect to the vacuum 0 > , rewrite it in terms of the operators (a¯ ,¯a† ) and | p,ǫ p,ǫ commute the terms so that the resulting expression is normal ordered with respect to the vacuum ¯0 > . To do this we write out the term : Tˆ0ˆ0 : explicitly. A simplification due to | |P0 the use of equation (2.11) occurs so that only time derivatives of the field operators appear. Also since Ψ(t,x) = Ψ¯(t¯,x¯) we may write ∂Ψ∂Ψ ∂Ψ¯ ∂Ψ : Tˆ0ˆ0 : = : : =: : |P0 ∂t ∂t |P0 ∂t ∂t |P0 ∂Ψ¯ ∂φ ∂φ∗ ∂Ψ¯ = d(ǫp)[ p,ǫa + p,ǫa† ] . (3.14) ∂t ∂t p,ǫ ∂t p,ǫ ∂t |P0 ǫ=±1Z X To simplify the notation we drop the , but keep in mind that these equations only apply |P0 at the point P . Also we only evaluate this expression for a fixed ǫ. Thus, 0 ∞ ∞ ∂φ¯ ∂φ¯∗ ∂φ : Tˆ0ˆ0 := d(ǫp) d(ǫq)[ ( q,ǫa¯ + q,ǫa¯† )a p,ǫ ǫ ∂t q,ǫ ∂t q,ǫ p,ǫ ∂t Z0 Z0 ∂φ∗ ∂φ¯ ∂φ¯∗ + p,ǫa† ( q,ǫa¯ + q,ǫa¯† ) (3.15) ∂t p,ǫ ∂t q,ǫ ∂t q,ǫ # The operators (a ,a† ) are related to the barred operators (a¯ ,a¯† ) by a Bogolubov trans- p,ǫ p,ǫ k,ǫ k,ǫ formation a = d(ǫq)(α a¯ +β∗ a¯† ) (3.16) k,ǫ k,q q,ǫ k,q q,ǫ Z where α = (φ ,φ¯ ) k,q k,ǫ q,ǫ β = (φ∗ ,φ¯ ) (3.17) k,q k,ǫ q,ǫ In our evaluation of the vacuum expectation value, the only term of interest is the c- number term that results from the commutator a¯q,ǫ′a¯†k,ǫ = a¯†k,ǫa¯q,ǫ′ +δǫ,ǫ′δ(k −q) (3.18) Thus, we get 7 ∂φ¯ ∂φ < ¯0 : Tˆ0ˆ0 : ¯0 >= d(ǫp)d(ǫq)[ q,ǫ p,ǫβ∗ +c.c.] (3.19) | ǫ |P0| ∂t ∂t p,q Z These terms are evaluated by replacing β by its expression (3.17) and interchanging the order of integration to first do the momentum integrals. In doing so the only regularization required is to define an integral of the form ∞ dxxexp(ixp) (3.20) Z0 This is accomplished by replacing p by p + iδ . No further regularizations are needed. Further details of such a calculation are in the appendix as well as the paper by Massacand and Schmid [1] and yield a Schwarz derivative. The final result is: 1 ∂3f¯ǫ < ¯0 : Tˆ0ˆ0 : ¯0 >= ∂x3 |P0 (3.21) | ǫ |P0| 24π ∂f¯ǫ ∂x|P0 So we only have to evaluate these terms. Now, ∂f¯ ∂f¯ ∂x¯ ∂f¯ ∂t¯ ǫ ǫ ǫ = + (3.22) ∂x ∂x¯ ∂x ∂t¯ ∂x and as initial conditions at P we have 0 ∂x¯ ∂t¯ = cosh(χ) , = sinh(χ) (3.23) ∂x|P0 ∂x|P0 Furthermore, we also have that ∂f¯ ∂f¯ ∂f¯ f¯(0,x¯) = ǫx¯ , ǫ = 1 , ǫ = ǫα¯ ǫ = ǫ (3.24) ǫ ∂t¯|P0 ∂x¯|P0 ∂t¯|P0 since α¯ = 1 . Also, as we stated earlier, |P0 ∂α¯ ∂α¯ = = 0 (3.25) ∂t¯|P0 ∂x¯|P0 ∂2α¯ ∂2α¯ = = 0 (3.26) ∂t¯∂x¯|P0 ∂x¯2|P0 and ∂2α¯ R = . (3.27) ∂t¯2 |P0 2 8 By repeatedly using the barred version of equation (2.11) , namely ∂f¯ ∂f¯ ǫ ǫ = ǫα¯ (3.28) ∂x¯ ∂t¯ as well as (3.22), (3.23) and (3.25) we find: ∂2f¯ ∂2f¯ ∂2f¯ ǫ ǫ ǫ = = = 0 (3.29) ∂t¯2 |P0 ∂t¯∂x¯|P0 ∂x¯2 |P0 as well as ∂3f¯ ∂3f¯ ∂2α¯∂f¯ ǫ ǫ ǫ = ǫ +ǫ = 0 (3.30) ∂x¯3 |P0 ∂t¯3 |P0 ∂t¯2 ∂t¯|P0 Thus we arrive at the result that ∂3f¯ ∂2α¯ R ǫ = = (3.31) ∂t¯3 |P0 −∂t¯2 |P0 −2|P0 This result now allows us to obtain that ∂3f¯ ∂3t¯ ∂3x¯ ∂2α¯∂x¯ ∂t¯ ∂f¯ ∂t¯ ∂x¯ ∂3f¯ ∂t¯ ǫ = [ +ǫα¯ +ǫ ( )2] ǫ +( +ǫ ) ǫ( )2 ∂x3 |P0 ∂x3 ∂x3 ∂t¯2 ∂x ∂x ∂t¯|P0 ∂x ∂x ∂t¯3 ∂x |P0 R ∂3t¯ ∂3x¯ = sinh3(χ)+ +ǫ (3.32) −2 ∂x3 ∂x3 To evaluate the last two terms in this expression we use the fact that (t,x) as well as (t¯,x¯) satisfy the geodesic equations, but have different initial data on the spacelike geodesic that passes through P . These initial data are: 0 dx dt = 1 , = 0 (3.33) ds|P0 ds|P0 dx¯ dt¯ = cosh(χ) , = sinh(χ) (3.34) ds|P0 ds|P0 The geodesic equations read: d2t ∂α dx = α ( )2, ds2 − ∂t ds d2x 2 ∂αdx dt 1 ∂α dx = ( )2 (3.35) ds2 −α ∂t dsds − α∂x ds d2t¯ ∂α¯ dx¯ = α¯ ( )2, ds2 − ∂t¯ ds d2x¯ 2 ∂α¯dx¯ dt¯ 1 ∂α¯ dx¯ = ( )2 (3.36) ds2 −α¯ ∂t¯dsds − α¯ ∂x¯ ds 9 By differentiating these equations as well as using (3.23) we find that ∂2t¯ ∂2x¯ = = 0 (3.37) ∂x2|P0 ∂x2|P0 and ∂3t¯ ∂2α¯ R = sinh(χ)cosh2(χ) = sinh(χ)cosh2(χ) (3.38) ∂x3|P0 −∂t¯2 |P0 −2 ∂3x¯ ∂2α¯ = 2 cosh(χ)sinh2(χ) = Rcosh(χ)sinh2(χ). (3.39) ∂x3|P0 − ∂t¯2 |P0 − Combining these results we obtain that R < ¯0 : Tˆ0ˆ0 : ¯0 >= ǫexp(ǫχ)sinh(χ) (3.40) | ǫ |P0| −48π Adding the results for both values of ǫ we obtain R < ¯0 : Tˆ0ˆ0 : ¯0 >= sinh2(χ) (3.41) | |P0| −24π Inserting this into equation (3.12) we finally obtain the vacuum expectation value of the trace of the stress-energy tensor, namely R . 24π IV. CONCLUSION For the case of a conformally coupled massless scalar field in 1+1 dimensions it is much simpler to evaluate the trace anomaly using a particle picture than to avoid this. The only regularization required is very simple, but it must be this very simple regularization that suffices to break the conformal symmetry and thus give a non-zero result for the vacuum expectation value of the trace. ACKNOWLEDGMENTS AZC would like to thank theTheoretical Physics Institute oftheUniversity of Innsbruck, as well as the Max-Planck-Inst. fuer Physik; Werner Heisenberg Inst. for their hospitality. He also acknowledges support from the Natural Sciences and Engineering Research Council of Canada (NSERC) as well as the Alexander von Humboldt Stiftung. 10

Two observers calculate the trace anomaly

Similar Two observers calculate the trace anomaly