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Wikipedia says that Nordstrom theory with equations of motion of the test particle $$\tag{1} \frac{d (\varphi u_{\alpha})}{d \tau} = \partial_{\alpha} \varphi $$ and field equation $$\tag{2} \varphi \square \varphi = 4 \pi \rho $$ is equal to the conditions $$\tag{3} C_{\alpha \beta \gamma \delta} = 0, \quad R = 24 \pi T, \quad T = g^{\mu \nu} T_{\mu \nu}, $$ so the metrics $$ g_{\alpha \beta} = e^{2\psi}\eta_{\alpha \beta}, \quad \varphi = e^{\psi} $$ with condition $R = 24 \pi T$ represents Nordstrom theory.

I have some questions about it.

If the statement about the equivalence of Nordstrom theory and $(3)$ is correct, I will get from it geodesic and field equations which are equal to $(1), (2)$ respectively.

First, let's get the geodesic equation. By using expression $$ \Gamma_{\alpha \beta \gamma} = (g_{\alpha \beta}\partial_{\gamma}\psi + g_{\alpha \gamma}\partial_{\beta}\psi - g_{\beta \gamma}\partial_{\alpha}\psi ) $$ I got $$ \frac{d u_{\alpha}}{d \tau} + \Gamma_{\alpha \beta \gamma}u^{\beta} u^{\gamma} = \frac{d u^{\alpha}}{d \tau} + \left(g_{\alpha \beta}\partial_{\gamma}\psi + g_{\alpha \gamma}\partial_{\beta}\psi - g_{\beta \gamma}\partial_{\alpha}\psi \right) u^{\beta}u^{\gamma} $$ using $\left|g_{\beta \gamma} u^{\beta} u^{\gamma} = 1\right|$ $$\tag{4} \frac{d u_{\alpha}}{d \tau} + 2 u_{\alpha}u^{\beta}\partial_{\beta} \psi - \partial_{\alpha}\psi = 0 \quad\Longrightarrow\quad \frac{d}{d \tau}(\varphi^{2} u_{\alpha}) = \frac{1}{2}\partial_{\alpha} \varphi^{2} $$ which isn't equal to $(1)$.

Second, let's get the field equation.

By using $R = 24 \pi T$ I got (in terms of $\psi$) \begin{align} e^{-2\psi}\left(-6\square \psi - 6 \partial_{\lambda} \psi \partial^{\lambda} \psi \right) &= -6\frac{ \square \varphi}{\varphi^{3}} \\ &= 24 \pi T \end{align} then using $T\approx\rho$ we get $$\tag{5} 24\pi T = 24 \pi \rho, $$ which also isn't equal to $(2)$ (I used the non-relativistic limit of stress-energy tensor of macroscopic body $T_{\mu \nu} = (p + \rho)u_{\mu} u_{\nu} - pg_{\mu \nu}$).

Where is the problem?

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No, your computation for $R=-6\varphi^{-3}\Box\varphi$ is quite correct (it even shows up as such in the Wikipedia article). – Alex Nelson Jan 3 '14 at 3:17
@alexNelson : so what is the consensus? Where is the mistake? Even the equation of motion isn't equal to corresponding one from "usual" Nordsrtom theory. – Andrew McAddams Jan 3 '14 at 11:53
You should not mix the (confusing) notations of Wikipedia. In the first part, the gravitational field/potential is $\phi$, while, in the second part (from Fokker..), it it $\phi^2$. Your relation $(4)$ is exact and can be differently obtained from the Euler-Lagrange equations for the Lagrangian $L= \phi(x)^2 \eta_{ab} u^au^b$, where $u^a = \dfrac{dx^a}{d\tau}$. The relation $R=-6\varphi^{-3}\Box\varphi$ is correct too. However, the relations between $R$ and $T$ are not completely clear for me, because we have the supplementary $\rho$ term.... – Trimok Jan 3 '14 at 12:14
I mean that, in this Wiki article, the origin of the relation $R = 24 \pi T$ is not completely clear for me (does it come from a Lagrangian formulation for the field, and which one ?, and this should imply $\rho$...) – Trimok Jan 3 '14 at 12:43
I think (or, at least, I have some vague recollection) that MTW's Gravitation uses these coefficients. It may be the Wiki article is conflating MTW's treatment of Norstrom gravity with another source's treatment of it... – Alex Nelson Jan 3 '14 at 16:10

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