The vDVZ Discontinuity and Vainshtein Limit

The linear theory of a massive graviton does not approach linearized general relativity smoothly as its mass tends to zero. Its helicity-0 polarization retains a finite coupling to the trace of a conserved source: the van Dam–Veltman–Zakharov (vDVZ) discontinuity. Nonlinear derivative interactions can suppress that polarization inside a source-dependent Vainshtein radius, but this statement must be checked on a complete solution and within a controlled effective field theory.

Required background. Variational and Symplectic Geometry supplies the field/source variational language, and Weak-Field Gravity supplies conserved-source exchange and linear metric potentials.

Helpful background. Einstein Equations fixes the general-relativistic comparison; Gravitational Actions and Boundary Terms supplies the action principle; Formulations of General Relativity supplies constraint intuition; and Gravity Beyond General Relativity places screening among modified-gravity mechanisms.

The discontinuity in one exchange amplitude

Take two conserved sources, $\partial_\mu T^{\mu\nu}=\partial_\mu T'^{\mu\nu}=0$. Terms in the propagator proportional to momentum vanish against these sources. Up to an overall coupling and the propagator denominator, Fierz–Pauli exchange is therefore

$$\mathcal A_m(k)\propto \frac{T_{\mu\nu}T'^{\mu\nu}-\frac13TT'}{k^2+m^2},$$

whereas a genuinely massless graviton gives

$$\mathcal A_0(k)\propto \frac{T_{\mu\nu}T'^{\mu\nu}-\frac12TT'}{k^2}.$$

The trace weights are $-1/3$ and $-1/2$, respectively. Sending $m\to0$ changes the denominator continuously but not that finite difference. The propagators and source amplitudes are derived in de Rham 2014, § 2.2.3, eqs. (2.49)–(2.66).

The mismatch is the helicity-0 exchange. A traceless source cannot excite it, but ordinary nonrelativistic matter has $T\ne0$. If the coupling is normalized to reproduce the Newtonian force between two slow sources, the response to a traceless electromagnetic source is then different from general relativity. The discontinuity is not the claim that every finite-mass nonlinear solution disagrees with observation; it says that linear perturbation theory cannot be used uniformly as $m\to0$ near a source.

The nonlinear scale

Use the reduced Planck convention $M_{\mathrm{Pl}}^2=(8\pi G)^{-1}$ and a source of physical mass $M$, with Schwarzschild radius $r_s=2GM$. For ghost-free massive gravity, the most important weak-mass interactions survive in the decoupling limit

$$m\to0, \qquad M_{\mathrm{Pl}}\to\infty, \qquad \Lambda_3=(m^2M_{\mathrm{Pl}})^{1/3}\ \text{fixed}.$$

A cubic Galileon model captures the basic radial scaling,

$$\mathcal L_\pi=-\frac12(\partial\pi)^2 -\frac{1}{4\Lambda_3^3}(\partial\pi)^2\Box\pi +\frac{\pi T}{M_{\mathrm{Pl}}}.$$

For a static point source $T=-M\delta^{(3)}(\boldsymbol r)$, define $u(r)=\pi'(r)/r$. One radial integration reduces the equation of motion to the algebraic equation

$$u+\frac{u^2}{\Lambda_3^3} =\frac{M}{4\pi M_{\mathrm{Pl}}r^3}.$$

The two terms become comparable at

$$r_V=\frac{1}{\Lambda_3} \left(\frac{M}{4\pi M_{\mathrm{Pl}}}\right)^{1/3} =\left(\frac{r_s}{m^2}\right)^{1/3},$$

where the equality follows from the stated Planck and Schwarzschild-radius conventions for this cubic model. The radius fixes the order-one transition scale; demanding exact equality of the two radial terms would place the transition at $2^{-1/3}r_V$. Other scalar normalizations and interactions can change that numerical factor. This is a source-dependent radius, not a new universal interaction range.

The branch that decays at infinity is explicit:

$$u(r)=\frac{\Lambda_3^3}{2} \left[\sqrt{1+4(r_V/r)^3}-1\right].$$

The other quadratic root approaches $-\Lambda_3^3$ at infinity and does not satisfy the same asymptotically trivial scalar condition. Selecting the root is part of specifying the physical problem. For a finite source, the exterior formula applies only outside its surface and must be matched to a regular interior.

Screening inside the Vainshtein radius

For $r\gg r_V$, the linear term dominates, so $u\propto r^{-3}$ and $\pi'\propto r^{-2}$: the scalar force is comparable in scaling to the Newtonian force. For $r\ll r_V$, the quadratic term dominates,

$$u\sim\Lambda_3^3\left(\frac{r_V}{r}\right)^{3/2}, \qquad \frac{F_\pi}{F_N}\sim\left(\frac{r}{r_V}\right)^{3/2}\ll1.$$

Thus the scalar is present but its response is screened. The exact radial equation, its branches, and the $3/2$ force-suppression exponent are given in de Rham 2014, § 10.1.2, eqs. (10.11)–(10.20). Quartic and quintic interactions change the detailed power laws and can change perturbation propagation, so the cubic result should not be silently transferred to every dRGT parameter choice.

The same background that makes the nonlinear term large also enlarges the kinetic matrix $Z^{\mu\nu}$ for fluctuations. Canonically normalized perturbations then couple more weakly to matter. Screening is therefore kinetic, not the disappearance of the helicity-0 mode.

What must be checked beyond the scaling argument

The decoupling-limit solution is a local asymptotic description. A physical claim additionally requires a branch that is regular across the source surface or horizon, matches the desired boundary condition at infinity, and remains continuously connected through $r\sim r_V$. Time-dependent and nonspherical sources need not obey the static cubic scaling.

Effective-field-theory control is a separate question. Large $\partial^2\pi/\Lambda_3^3$ can be resummed while higher-derivative operators remain small only in an appropriate hierarchy; one must check the derivatives of the background and of fluctuations, not just the field amplitude. Moreover, $\Lambda_3$ is a strong-coupling scale and need not equal the ultraviolet cutoff, as emphasized in de Rham 2014, §§ 10.2 and 10.5, pp. 100–105. A screened classical profile is consequently not, by itself, evidence for ultraviolet completeness, causal propagation, or global stability.

Exercises

1. See the vDVZ mismatch explicitly. Let both slow sources have only $T_{00}=\rho$ and $T'_{00}=\rho'$. Compare the massive and massless numerator, then rescale the overall massive-exchange normalization so that the two Newtonian amplitudes agree. What happens for a traceless probe?

Solution

With the $(-,+,+,+)$ convention, $T=T^\mu{}_{\mu}\simeq-\rho$, while $T_{\mu\nu}T'^{\mu\nu}=\rho\rho'$. The massive numerator is $(1-1/3)\rho\rho'=2\rho\rho'/3$; the massless numerator is $(1-1/2)\rho\rho'=\rho\rho'/2$. Matching Newtonian exchange therefore multiplies the overall massive amplitude normalization by $(1/2)/(2/3)=3/4$. For a traceless probe, the trace term vanishes in both theories, so this rescaling leaves the massive response at $3/4$ of the general-relativistic value. This finite mismatch survives $m\to0$ in linear theory.

2. Recover the Vainshtein exponent. Use the algebraic radial equation to find the inner scaling of $\pi'$ and its force relative to $F_N\propto r^{-2}$.

Solution

At $r\ll r_V$, $u^2/\Lambda_3^3\simeq M/(4\pi M_{\mathrm{Pl}}r^3)$, so $u\propto r^{-3/2}$ and $\pi'=ru\propto r^{-1/2}$. Dividing by a Newtonian force proportional to $r^{-2}$ gives $F_\pi/F_N\propto r^{3/2}$. Matching at $r=r_V$ fixes the dimensionless form $(r/r_V)^{3/2}$ up to the normalization of the scalar coupling.

References

• de Rham, Claudia. “Massive Gravity.” Living Reviews in Relativity 17 (2014): 7. doi:10.12942/lrr-2014-7; Open PDF.