# Why does this derivation of the Einstein Field Equations only work with the Trace-Reversed form?

The Einstein Field Equations (ignoring cosmological constant) are:

$$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = \kappa T_{\mu\nu}$$

The trace of this equation gives $$R - \frac{1}{2}4R = -R = \kappa T$$. Putting this back into the EFE, we get the "trace-reversed" form:

$$R_{\mu\nu} = \kappa (T_{\mu\nu} - \frac{1}{2}Tg_{\mu\nu})$$

I've looked at a number of sources for solving for the constant $$\kappa$$, and most use an argument that works in the trace-reversed form, but not in the "standard" form, and I can't figure out why. (An example is chapter 17 of "Gravitation" by Misner, Thorne and Wheeler.)

The argument uses these assumptions to force the EFE to match the Poisson Equation (using the $$(+---)$$ metric signature):

• Use "low-velocity limit" so the $$00$$-components of tensors dominate. In particular, $$T^{00} = \rho c^2$$.
• From geodesic equation, we get $$\Gamma^k_{00} = \frac{1}{c^2} \partial_k \Phi$$, and therefore $$R_{00} = \frac{1}{c^2} \nabla^2 \Phi$$
• Use weak field limit $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$ where $$h_{\mu\nu} << 1$$ (ends up being $$g_{00} = 1 + \frac{2\Phi}{c^2}$$)

Now, supposedly in the trace-reversed form of the EFE, in the weak field limit, we can pretend the metric $$g_{\mu\nu}$$ is equal to the flat space metric $$\eta_{\mu\nu}$$, so $$g_{00} = \eta_{00} = +1$$ instead of $$g_{00} = 1 + \frac{2\Phi}{c^2}$$. From there, solving for $$\kappa$$ is easy:

$$T = T^\mu_\mu \approx T^0_0 \approx T^{00}g_{00} = \rho c^2 (+1) = \rho c^2$$

$$R_{00} = \kappa (T_{00} - \frac{1}{2}T g_{00}) = \kappa (\rho c^2 - \frac{1}{2} \rho c^2 (+1)) = \kappa \frac{1}{2}\rho c^2$$

Assuming Poisson's Equation $$\nabla^2 \Phi = 4 \pi G \rho$$ holds in the weak field limit, we get

$$\frac{1}{c^2} \nabla^2 \Phi = \kappa \frac{1}{2}\rho c^2$$ $$\frac{1}{c^2} 4 \pi G \rho = \kappa \frac{1}{2}\rho c^2$$ $$\frac{8 \pi G}{c^4} = \kappa$$

Which is the expected constant of proportionality in the EFE. This $$g_{00} = +1$$ trick doesn't seem to work with the standard EFE form of $$R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = \kappa T_{\mu\nu}$$ (answer is off by a factor of 4), and I don't know the exact reasoning for this. Is it because $$\kappa(\frac{2\Phi}{c^2})$$ is of negligible size, so we ignore it in the trace-reversed form? That seems a strange assumption if we don't know the value of $$\kappa$$ ahead of time...

Wikipedia says this, but does not elaborate:

The trace-reversed form may be more convenient in some cases (for example, when one is interested in weak-field limit and can replace gμν in the expression on the right with the Minkowski metric without significant loss of accuracy).

• Another method is: leave $\kappa$ unspecified and obtain the Schwarzschild-Droste metric (i.e. vacuum sol with spherical symmetry), then set $\kappa$ so that predictions for orbits at large $r$ match Newtonian gravity. Commented Aug 12, 2021 at 18:41

I will use signature (+---). In the weak-field limit in which $$\mathbf g \simeq \boldsymbol \eta + \mathbf h$$, the components of the Ricci tensor are given by $$R_{\mu\nu} = -\frac{1}{2}\left(\square h_{\mu\nu} + \partial_\mu\partial_\nu h - 2\partial^\alpha \partial_{(\mu}h_{\nu)\alpha}\right)$$ Computing the trace to lowest order in $$\mathbf h$$ yields $$R = g^{\mu\nu} R_{\mu\nu} \simeq \eta^{\mu\nu}R_{\mu\nu} = -\square h + \partial^\mu\partial^\nu h_{\mu\nu}$$

Applying the condition that $$\mathbf h$$ be stationary $$(\partial_0 \mathbf h\rightarrow 0)$$, the $$(00)$$-component of the Einstein equation becomes $$R_{00} - \frac{1}{2}Rg_{00} = \kappa T_{00} \implies -\frac{1}{2}\square h_{00} + \frac{1}{2}\square h- \frac{1}{2} \partial^\mu\partial^\nu h_{\mu\nu} = \kappa T_{00}$$

On the other hand, if we assume the non-relativistic limit in which $$T^{00}$$ dominates over all of the other components of $$\mathbf T$$, the trace of the energy-momentum tensor is to lowest order given by $$T = g_{\mu\nu} T^{\mu\nu} \simeq \eta_{00} T^{00} = \rho c^2= T_{00}$$ and the $$(00)$$-component of the trace-reversed Einstein equation becomes (in 4 spacetime dimensions) $$R_{00} = \kappa(T_{00} - \frac{1}{2}T g_{00}) \rightarrow -\frac{1}{2}\square h_{00} = \kappa(T_{00} - \frac{1}{2}T_{00}) = \kappa T_{00}/2$$ $$\iff-\frac{1}{2}\square h_{00} = \frac{\kappa T_{00}}{2}$$

To see that these expressions are consistent with one another, note that the taking the trace of the full Einstein equation yields $$R - \frac{4}{2}R = \kappa T = -R = \square h - \partial^\mu\partial^\nu h_{\mu\nu}$$

The error is in assuming that $$R = g^{\mu\nu} R_{\mu\nu} \approx R_{00} = -\frac{1}{2}\square h_{00}$$. While $$\mathbf T$$ (and its trace) is dominated by its $$(00)$$-component, the same is not true of $$\mathbf R$$; indeed, $$R - R_{00} = -\frac{1}{2} \square h + \partial^\mu\partial^\nu h_{\mu\nu}$$

To see this even more explicitly, consider the case off a pressure-less fluid. In component form,

$$R_{\mu\nu} -\frac{R}{2}\pmatrix{1 & 0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1} = \kappa \pmatrix{\rho c^2 & 0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0}$$

In order for this equation to hold, each $$R_{\mu\mu}$$ must be equal to $$-R/2$$ (and hence $$R=-\kappa \rho c^2$$).

You have the wrong dimensions for a few things, and these are arising from errors that have propagated widely even in the Physics literature. Among other things, it leads to the wrong value for $$κ$$. Even the literature is confused on the matter. In it, you will see $$1/c^2$$, $$1/c^3$$ and $$1/c^4$$ as the power of $$c$$ that goes with $$κ$$.

Only $$1/c^3$$ is correct - not $$1/c^4$$!

Also: the numerical part of the coefficient, $$8π$$, is dependent on the number of dimensions, but you see that same factor carelessly lifted from the 3+1-dimensional case and used with gravity theories of different dimensions. The actual factor is a bit more involved, and your analysis already points the way to the correct value.

So, let's examine everything in greater depth. It won't matter what convention you use for the coordinates. But let's see how it works each way.

Stress Tensor Density, Not Stress Tensor!
First: the energy density $$ρc^2$$ - as the name, itself, indicates - is not the component of a tensor, but of a tensor density. In particular, $$ρc^2 = 𝔗^0_0$$, where $$𝔗^ρ_ν$$ is the stress tensor density, which is related to the stress tensor by $$\sqrt{|g|}T_{μν} = g_{μρ} 𝔗^ρ_ν$$. Fundamentally, it is a rank (1,1) tensor density, not a tensor at all. Particularly, it is the density which is involved in the transport equation $$∂_ρ 𝔗^ρ_ν = 0$$ in the zero curvature limit for the conservation of the momentum $$p_ν$$.

With the convention $$x^0 = t$$.
The dimensions are the same as for the canonical stress tensor. Thus, the diagonal components $$𝔗^ρ_ρ$$, including $$𝔗^0_0$$, have the units of action per space-time volume, which is $$(ML^2/T)/(L^3T) = M/(LT^2)$$ (for $$M$$, $$L$$, $$T$$ respectively mass, length and time). That's the dimension for energy density - and also: for pressure and momentum flux.

Second: since you suspended the $$c = 1$$ convention, the line element is $$c^2 dt^2 - dx^2 - dy^2 - dz^2$$, and $$g_{00} = c^2$$ in the zero curvature limit. You could adopt the convention of making it a proper time metric, dividing it by $$c^2$$, but that will completely alter the analysis - in ways that are even more off from what you have.

Correspondingly, $$\sqrt{|g|} = c$$ in the flat space limit (i.e. it has dimension $$L/T$$). Therefore, the de-densitized stress tensor $$T^0_0 = ρc$$ in the zero curvature limit. With the index lowered that becomes $$T_{00} = ρc^3$$. That has the dimension $$M/T^3$$.

The Ricci tensor $$R_{00}$$ has dimension $$1/T^2$$. Therefore the coefficient $$κ$$ has dimension $$T/M$$. Noting that $$G$$'s dimension is $$L^3/(MT^2)$$ and $$c$$'s is $$L/T$$, then that corresponds to the dimension of $$G/c^3$$, not $$G/c^4$$: $$κ = \frac{8πG}{c^3}.$$

With the convention $$x^0 = ct$$
You might say, "well, I'm taking $$x^0 = ct$$, not $$x^0 = t$$".

It doesn't matter. That won't actually change anything.

The line element is now $$(c dt)^2 - dx^2 - dy^2 - dz^2$$, $$g_{00} = 1$$ in the zero-curvature limit, $$g$$ is dimensionless.

However, the diagonal stress tensor density components $$𝔗^ρ_ρ$$ now have dimension action per space-time volume, which is now $$(ML^2/T)/L^4 = M/(L^2T)$$. De-densitized, this becomes $$T^ρ_ρ$$ with same dimension. Likewise with the index lowered, including $$T_{00}$$, the dimension is $$M/(L^2T)$$.

The dimension for the Ricci tensor component $$R_{00}$$ is now $$1/L^2$$. The coefficient $$κ$$, therefore, must have dimension $$T/M$$ - the same as before.

Directly From The Action Integral
The action integral $$S = \int R/(2κ) \sqrt{|g|} d^4 x$$ has dimension of action: $$ML^2/T$$. The dimension for $$\sqrt{|g|} d^4 x$$ is $$L^4$$, independent of what convention is adopted for the coordinates!

If, on the other hand, you had adopted the convention of making the metric a proper time metric, with a line element of dimension $$T^2$$, then it would have been $$T^4$$ - which would take us way far away from standard conventions, so we'll adopt the usual convention of the line element having dimension $$L^2$$.

The dimension of the Ricci component $$R_{00}$$ is that of $$1/{x^0}^2$$, while that for $$g^{00}$$ is that for $${x^0}^2/L^2$$. Thus, $$R$$ has dimension $$1/L^2$$, and $$R\sqrt{|g|}d^4x$$ has dimension $$L^2$$.

Therefore, to get an action with dimension $$ML^2/T$$, requires $$κ$$ to have dimension $$T/M$$. Therefore, $$κ = 8πG/c^3$$.

With Proper Time Line Elements
This will really screw things up. Now, $$\sqrt{|g|}d^4x$$ has dimension $$T^4$$, while $$R$$ has dimension $$1/T^2$$. Therefore, the $$R\sqrt{|g|}d^4x$$ has dimension $$T^2$$. To match that with the action requires $$κ$$ to have dimension $$T^3/(ML^2)$$, thus making $$κ = 8πG/c^5$$.

Nobody does that, and you won't see that anywhere in the literature. The only powers of $$c$$ ever seen are $$1/c^2$$ (Einstein did that), $$1/c^4$$ (Leiden's wall did that, along with everyone on the net), and $$1/c^3$$ - the correct one.

And About That $$8π$$
There's another widespread error all over the place in the literature, equally borne of carelessness. As you already see with your analysis, there is a dependency of the numerical coefficient on the dimension of space. The coefficient $$4π$$ is the area of a solid sphere in 3D. Let the surface of the solid sphere in $$m$$ spatial dimensions be $$A_m$$.

Your analysis produces a factor of $$1/(n - 2)$$ to go with $$T$$, for a space-time of dimension $$n$$. You're subtracting that from 1 to get the multiplier for $$T_{00}$$, resulting in $$(n - 3)/(n - 2)$$. Finally, you're dividing $$A_{n-1}$$ by this to get $$(n-2)A_{n-1}/(n-3)$$.

The formula is $$A_m = m π^{m/2}/(m/2)!$$, adopting the convention that $$(-1/2)! = \sqrt{π}$$. Thus $$A_3 = 3 π\sqrt{π}/(3\sqrt{π}/4) = 4π$$.

So, for space-times of dimension $$n$$, the correct numerical value (apart from multiples of $$c$$) is not $$8πG$$, but $$(n-3)A_{n-1}G_{n-1}/(n-2)$$, where $$G_m$$ is the analogue of Newton's coefficient for $$m$$ spatial dimensions.

Since $$G_{n-1}$$ has dimension $$L^{n-1}/(MT^2)$$, and $$R\sqrt{|g|}d^nx$$ has dimension $$L^{n-2}$$, then to match $$κ$$ with the action integral, we need $$κ$$ to have dimension $$L^{n-2}/(ML^2/T) = TL^{4-n}/M$$ - which is the dimension of $$G_{n-1}/c^3$$.

So, even in $$n$$ dimensions, the power of $$c$$ is still $$1/c^3$$. But the numerical part of the coefficient no longer need be $$8π$$.