# Table of contents

Gravitational redshift in Einstein and Jordan frames

- 1. Gravitational redshift in General Relativity
- 2. Gravitational redshift in $f(R)$: the Jordan frame
- 3. Gravitational redshift in $f(R)$: the Einstein frame
- 4. Explanation for cosmological redshift in a static frame
- 5. Discussion

# Gravitational redshift in Einstein and Jordan frames

This problem came about trying to understand gravitational redshift in $f(R)$ theories in both frames (i.e. the physical expanation directly in each frame independently) and is a good example that highlights the difference in interpretation of observables in the two frames. To kick this off lets start off by making two contradictory arguments for what the gravitational redshift in $f(R)$ is:

- 1. We can work in the Jordan frame. Gravity is modified, but photons and matter follow trajectories of the space-time metric $\tilde{g}_{\mu\nu}$. The usual derivation for gravitational redshift makes no reference to the gravity theory we use so it should apply and give that the gravitational redshift is given by the $\tilde{g}_{00}$ component of the metric. The modifications of gravity would enter in that the relation between what $\tilde{g}_{00}$ is given a matter distribution.
- 2. We can make a conformal transform to the Einstein frame where the metric is $g_{\mu\nu}$ which is related via $\tilde{g}_{\mu\nu} = A^2 g_{\mu\nu}$ where $A$ is the conformal factor. In this frame the gravity theory is GR, but non-relativitstic particles would now feel a fifth-force in addition to the usual gravitational force. The photons on the other hand is not affected since the electromagnetic action is conformally invariant. Since gravity is GR and photons move as they would in GR we would naively expect that the gravitational redshift would be given by $g_{00}$. Thus we seem to get a different result than in the Jordan frame.

So this leads us to the question, which frame is correct? Is it the one where the geodesic equation takes the standard form or is it the frame where the equivalence principle applies for the (pure) gravity sector? The answer is that both are equaly correct and gives the same result as long as we take into account one important thing that we ignored above: the spacetime dependence of inertial masses in the Einstein frame. Lets go through and try to understand this in more detail.

## Gravitational redshift in General Relativity

First a rough derivation of the usual expression for the gravitational redshift: consider for simplicity a time-independent metric for which $\xi^\mu = (1,0,0,0)$ is a Killing vector. Let $p^\mu$ be the 4-momentum of a photon and consider a stationary observer ($u^i = 0$) with 4-velocity $u^\mu$. From $g_{\mu\nu}u^\mu u^\nu = -1$ we have $u^0 = (-g_{00})^{-1/2}$. From the observed energy of a photon $E = -g_{\mu\nu}u^\mu p^\nu$ we find that $$\xi^\mu p^\nu g_{\mu\nu} = \frac{E}{(-g_{00})^{-1/2}}$$ and since the left hand side is a constant of motion (see e.g. Hartle) $$ \left(\frac{E}{(-g_{00})^{-1/2}}\right)_{\rm obs} = \left(\frac{E}{(-g_{00})^{-1/2}}\right)_{\rm source}$$ Finally we have $E = h/\lambda$ and from the definition of redshift, $z \equiv \frac{\lambda_1}{\lambda_2}-1$ , we find that the gravitational redshift is given by $$z = \sqrt{\frac{g_{00}^{\rm source}}{g_{00}^{\rm obs}}} - 1 \simeq \Psi_{\rm source} - \Psi_{\rm obs}$$ Importantly this only depends on having a metric theory and the geodesic equation for photons - the exact form of the laws of gravity plays no role except in what $\Psi,\Phi$ really is. In the Newtonian gauge we have $\Psi = \Phi = \Phi_N$ where $\nabla^2 \Phi_N = 4\pi G \rho$ so the gravitational redshift is directly determined by the mass distribution via the usual Poisson equation.

## Gravitational redshift in $f(R)$: the Jordan frame

Lets consider $f(R)$ whose action is $S = \int \sqrt{-\tilde{g}}\frac{f(\tilde{R})}{16\pi G} + S_{\rm matter}(\tilde{g}_{\mu\nu})$. The Newtonian gauge metric $\tilde{g}_{\mu\nu}$ is given by $$ds^2 = -(1+2\Psi)dt^2 + (1-2\Phi)dx^2$$ and we can generally write (see reviews of $f(R)$) $\Psi = \Phi_N + \phi$ and $\Phi = \Phi_N - \phi$ where $\phi$ encompass the modifications of gravity. The sum of the two potentials is the lensing potential $\frac{\Psi + \Phi}{2} = \Phi_N$ and takes the same form as in GR as expected from the fact that the lightcone is preserved under a conformal transformation, but what about the energy of the photons (and thereby the redshift)? In this Jordan frame the geodesic equation for $\tilde{g}_{\mu\nu}$ takes the same form as in GR. Thus the derivation for the gravitational redshift is the same as in GR giving rise to $$z = \frac{E_{\rm source}}{E_{\rm obs}} - 1 = \sqrt{\frac{\tilde{g}_{00}^{\rm obs}}{\tilde{g}_{00}^{\rm source} }} - 1 \simeq \Psi_{\rm obs} - \Psi_{\rm source}$$ The modifications of gravity enter through the modified expressions for $\Psi$ (which differs from $\Phi_N$). In the extreme regime (no screening at all) we have $\phi = \Phi_N/3$ and we get the result that if we have exactly the same mass-distribution as in GR the gravitational redshift prediction is $4/3$ larger in $f(R)$. See Wojtak et al for the first observation of this effect from cluster stacks (which also compares this maximum $f(R)$ prediction to the data) and Zhao, Peacock & Li for pointing out some additional effects (like the transverse Doppler effect) that should also be taken into account when studying this effect in real data. See Gronke et al for the effect on the cosmological gravitational redshift signal from clusters from just having a different matter distribution due to modifications of gravity (they have used the wrong Einstein frame expression for the gravitational redshift so its not showing the full signal, but for screened regions - which we expect the clusters used in Wojtek et al. to be for $f(R)$ model parameters in agreement with observations - the results are nevertheless good).

## Gravitational redshift in $f(R)$: the Einstein frame

Now lets look at this in the Einstein frame where gravity is GR and we have a fifth-force. The action in this frame is (schematically) $S = \int \sqrt{-g}\frac{R}{16\pi G} + S_{\rm matter}(\tilde{g}_{\mu\nu}) + S_{\rm fifth-force~dynamics}$. The Newtonian metric $g_{\mu\nu}$ is here $$ds^2 = -(1+2\Phi_N)dt^2 + (1-2\Phi_N)dx^2$$ The fifth-force comes from the fact that matter couple to $\tilde{g}_{\mu\nu} = A^2g_{\mu\nu}$ instead of just $g_{\mu\nu}$ (and $A \simeq 1+ 2\phi$). The EM action is conformally invariant so photons should not feel the presence of this force so in this frame gravity and photons behaves just like in GR (if there are no matter present). It is therefore very reasonable to think that the same derivation that applies for GR for the gravitational redshift would apply just as well leading us to the conclusion $$z = \frac{E_{\rm source}}{E_{\rm obs}} - 1 = \sqrt{\frac{g_{00}^{\rm obs}}{g_{00}^{\rm source} }} - 1 \simeq \Phi_N^{\rm obs} - \Phi_N^{\rm source} \not= \Psi_{\rm obs} - \Psi_{\rm source}$$ which is in contradiction with what we got in the Jordan frame. Here the gravitational redshift is exactly the same as in GR given the same mass distribution. How do we explain this? To understand this its useful to see what happens to the matter fields when we go from one frame to the other. For example if we consider a massive scalar field with mass-term $\frac{1}{2}m_0^2\varphi^2$ in the Jordan-frame then in the Einstein frame this becomes $\frac{1}{2}A^2m_0^2\varphi^2$ thus the mass is now space-time dependent $$m = m_0 A$$ where $A \simeq 1 + 2\phi$ varies from place to place. The mass of the electron is what sets the energy levels in atoms thus in this frame the frequency of a photon that is emmitted from an atom (and later absorbed when it is observed) is proportional to the conformal factor. And in general any non-gravitational mass-scale will get rescaled in the same way. We can now see the (subtle) error in the argument. We forgot about the implict assumption that is present when we talk about the energy of a photon that moves from the source to the observer. We have assumed that the masses of the fundamental particles are constant across space. This is indeed true in the Jordan frame (but there the effective gravitational constant would instead change from place to place), but in the Einstein frame they do change. Thus we need to take this into account when computing the gravitational redshift and the correct formula is therefore $$z = \frac{A^{-1}_{\rm source}E_{\rm source}}{A^{-1}_{\rm obs}E_{\rm obs}} - 1 = \sqrt{\frac{A^2g_{00}^{\rm obs}}{A^2g_{00}^{\rm source} }} - 1 = \sqrt{\frac{\tilde{g}_{00}^{\rm obs}}{\tilde{g}_{00}^{\rm source} }} - 1$$ which is exactly the same result as in the Jordan frame. Note that the variation of the masses is not an observable effect (as all matter has the same spacetime dependence), its rather a matter of interpretation (we either have constant masses and a varying effective gravitational constant (Planck mass), or we have varying masses and a constant Planck mass and any real observable is in effect a dimensionless number - e.g. a ratio of mass scales which is the same in both frames) and if we do the calculation consistently in both frames we get the same result for observables. To make this point more clear lets try to apply this same type of argument to a much more familar situation: cosmological redshift in General Relativity.

## Explaining cosmological redshift in a static frame

We start with GR and the usual Friedmann-LemaĆ®re-Robertson-Walker metric $$ds^2 = a^2(-d\eta^2 + dx^2)$$ and as we all know photons are redshifted $z = \frac{a_0}{a} - 1$ as they propagate through our the Universe. We can here also change frame by performing a conformal transformation $g_{\mu\nu} = a^2 \tilde{g}_{\mu\nu}$ with the scalefactor being the conformal factor to get that the metric reduces to Minkowski $\tilde{g}_{\mu\nu} = \eta_{\mu\nu}$. In this frame the Universe is now static! By the same argument as above the photons are not affected by the conformal transformation so how do we explain cosmological redshift? The interpretion here is the same as above: the masses of particles and thereby the frequency of the photons emitted and absorbed depend on the scale-factor so if we take this into account we do get that photons change their wavelengths in exactly the same way. The physial interpretation of the redshift is completely different in the two frames, but the most important thing - predictions for what we would observe - are exactly the same.

## Discussion

We have seen that the two frames gives the same result as long as we do the calculation properly in both frames, but the physical interpretation is quite different. The discussion here applies not just to $f(R)$, but to the whole range of scalar-tensor theories that have a conformal coupling to matter (i.e. theories where matter fields "feel" an effective metric $A^2(\phi)g_{\mu\nu}$).

Having different physical interpretations is not uncommon in GR. A classic example (closely related to what we have talked about here, but unrelated to the variation of inertial masses) is astronomical redshifts which can often be interpreted as as pure gravitational effect due to the expansion of space or as a pure Doppler effect (See e.g. Astronomical Redshifts and the Expansion of Space by Nick Kaiser and A diatribe on expanding space by John Peacock for a nice discussion regarding this point).

What is the most "fundamental" frame then? There is no consensus on this, it depends on your taste and what physical principle you hold dearest. The pros for the Jordan frame is that gravity is purely geometrical (and consequently we have the usual geodesic equation) and that the inertial masses are truly constant. The cons is that the equivalence principle is broken and the effective gravitational constant varies from place to place. The cons for the Einstein frame is that we have an additional fifth-force and particle masses are not constant across space. The pros is that gravity is given by good old General Relativity so the equivalence principle is true for the pure gravity sector (violations in the Jordan frame is a consequence of the presence of the fifth-force in the Einstein frame) and Newtons constant is just Newtons constant. The con is that we have an additional fifth-force and particle masses are not constant. Instead of seeing this as a problem I like to look at this in the same way as I view the "gauge problem" in both QFT and GR: its not a problem, but an opportunity as some calculations or interpretations will be much easier in one frame than the other (e.g. local tests of gravity are argubly most simple to compute and intuitively more clear in the Einstein frame while cosmology predictions are arguably easiest in the Jordan frame). I therefore take the view that a frame is "fundamental" if its convenient to think of it as "fundamental" for the problem at hand.

I should mention that all of the disussion in this note applies to purely classical phenomena. For quantum phenomena its not obvious that this equivalence always holds (you can find conflicting views regarding this in the litterature). Another situation where its also not perfectly clear (meaning I expect it to hold, but I have not seen a clear demonstration of this) is for systems whose gravitational binding energy is significant.

For a much more detailed discussion about the (classical) equivalence of the two frames (from which the example in the previous section is taken from) see the very nice paper Conformal equivalence in classical gravity: the example of "veiled" General Relativity by Nathalie Deruelle and Misao Sasaki (and thanks to Kazuya Koyama for pointing this out to me and to John Peacock, Pedro Ferreira and Torsten Bringmann for discussion regarding this problem).