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Equivalence principle for test fields

+ 4 like - 0 dislike
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I originally posted it here: http://physics.stackexchange.com/questions/183041/equivalence-principle-for-test-fields In order to get a more professional answer, I paste it in the following.

My question is very simple. We all know that, for a test particle(classical) in a gravitational field, the motion is only determined by the geodesic lines(let's forget about the initial conditions for now), and has no dependence on the "structure" of the particle, such as spin(in the classical sense), charge, etc..

But let's now consider the motion of a test field, scalar, spinor, vector or tensor, under gravity. I know how to describe this kind of motion using the corresponding field equations, Klein-Gordon equation, covariant Dirac equation, or master equations accounting for any masslesss fields. Of course, the results do depend on the spin(quantum). But is there a way to think about this in a "equivalence principle" sense, which can satisfactorily account for the difference due to spin?

Thank you in advance!

asked May 12, 2015 in Theoretical Physics by ruifeng14 (65 points) [ no revision ]
recategorized May 12, 2015 by Dilaton

2 Answers

+ 5 like - 0 dislike

All wave equations on a curved background can be solved in the eikonal approximation if the typical curvature length of the space-time is much larger than the wave-length of motion of the particle. If the equations are minimally coupled to the background, i.e. only through the d'Alembertian, the eikonal equation is to leading order equivalent to the equation of a geodesic of a corresponding particle. Obviously, if the typical curvature length is comparable to the wavelength, interference comes to play and the solutions may strongly disagree with a "geodesic streamlining" of the wave-packet.

But for example spinning matter is never minimally coupled to the background. A spinning classical particle actually interacts with the curved background in a non-trivial way and the interaction can be obtained by taking a zero-volume limit of a classical extended rigid body to yield the so-called Mathisson Papatreou equations
$$\frac{\rm d}{{\rm d}\tau} (m u^\mu - u_\sigma \dot{S}^{\mu \sigma}) = - \frac{1}{2} R^\mu_{\;\; \nu \sigma \rho} u^\nu S^{\rho \sigma}$$
$$\dot{S}^{\mu \nu} = 2 u^{[ \mu} \dot{S}^{\nu] \rho}u_\rho$$ where $S^{\mu \nu}$ represents the internal angular momentum of the particle (there are some supplementary technicalities which need not be discussed here). I.e. on a "pole-dipole" classical particle level, there is additional spin-curvature interaction. This is also true for a spinor field on a curved background where the square of the Dirac operator on a curved background yields $$(\square + m^2 + R/4) \psi = 0$$ That is, on any level there is non-minimal interaction between the curvature and spin. I have never investigated the connection between these two; feel free to dig, the keyword would probably be Frenkel and his 1926 theory of the spinning electron.

On the other hand, for example for a massless scalar it is possible to have a non-minimal coupling to curvature $$(\square+\xi R) \phi = 0$$ where $\xi$ can be a non-zero constant. A rather popular choice is the conformal coupling $\xi=1/6$ which under a conformal rescaling $g \to \tilde{g} = \Omega g$ allows to obtain solutions of the new equation from the original ones as $\tilde{\phi} = \Omega^{-1} \phi$. To state the point of this paragraph differently, the translational invariance of Minkowski allowed us to determine the respective classical equations unambiguously but this is no longer true on a curved background, here the field equation may sometimes not reflect the classical geodesic -- even if no spin is involved.

answered May 12, 2015 by Void (1,505 points) [ no revision ]

@Void Thank you very much. Very clear and insightful answer. As the square of the covariant Dirac equation for electrons is the same as that for a scalar field with coupling 1/4 with the Ricci scalar. Should then be a correspondence between the two, like in the greybody factors, or normal modes, say in black holes? Thanks again!

Note that the cases you mention are mostly vacuum space-times, i.e. $R=0$ cases where the scalar is "wave-equivalent" to $\psi$ for any $\xi$. A spinor wave passing without interaction through matter would be rather contrived and interaction would be complicated (or hard to even formulate) so only a non-zero cosmological constant $R=4 \Lambda$ is admissible.

In the case of the scalar field one is solving $\phi = \exp(iS(x)), k_\mu = \partial_\mu S(x)$ $k^2 = R/4 = \Lambda$ (maybe the signs are somewhat different depending on the convention). But in the spinor case it is impossible to do in general $\psi = \psi_0 \exp(iS(x))$ because of an additional condition with respect to the constant spinor amplitude $\psi_0$: $k_\mu \gamma^\mu \psi_0 = 0$ which will not be possible to satisfy in the general case. I cannot recall any explicit comparison of the $\xi=1/4$ with spinors but even in the $R=0$ case the spinor-solution tends to be rather different (both technically and physically) from the scalar one so I would not expect overly strong analogies.

@Void Thanks again! I really learnt a lot.

+ 0 like - 0 dislike

As commonly used, a test particle is assumed to have no internal structure, so that it follows the geodesic by definition, and has no back reaction to the field.

Gravitation in the presence of another field must be discussed in terms of a joint field theory, and if the latter is to be diffeomorphism invariant (as one wants to have it), this joint field theory contains proper interactions that work in both directions. The interactions are different for spin 0 and spin 1/2, hence there is no such equivalence principle.

answered May 12, 2015 by Arnold Neumaier (12,355 points) [ no revision ]

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