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Nordstrom's theory of a particle moving in the presence of a scalar field $\varphi (x)$ is given by $$ S = -m\int e^{\varphi (x)}\sqrt{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \lambda}\frac{dx^{\beta}}{d \lambda}}d\lambda , $$ where $\lambda$ is the parametrization of the worldline of the particle, ignoring the free field part $\int \eta_{\alpha \beta}\partial^{\alpha}\varphi \partial^{\beta} \varphi d^{4}x$.

How does one derive the equations of motion in terms of the parameter $$d\tau = \sqrt{\eta_{\alpha \beta}\frac{dx^{\alpha}}{d \lambda}\frac{dx^{\beta}}{d \lambda}} d\lambda. $$ $u^{\alpha} = \frac{dx^{\alpha}}{d\tau} \Rightarrow u_{\alpha}u_{\beta}\eta^{\alpha \beta} = 1$?

My attempt:

$$ \delta S = 0 \Rightarrow \int \left( \frac{\partial (e^{\varphi } \sqrt{...})}{\partial x^{\alpha}}\delta x^{\alpha} +\frac{\partial (e^{\varphi } \sqrt{...})}{\partial \left( \frac{d x^{\alpha}}{d \lambda } \right)}\frac{d}{d\lambda} \delta x^{\alpha} \right)d\lambda = |d\tau = \sqrt{...}d\lambda | = $$ $$ = \int \left(\sqrt{...}e^{\varphi}\partial_{\alpha}\varphi - \frac{d}{d \lambda} \left(\frac{d x_{\alpha}}{d\tau}e^{\varphi}\right) \right)\delta x^{\alpha}d \lambda = $$ $$ =\int \left( \partial_{\alpha}\varphi - \frac{d^{2}x_{\alpha}}{d \tau^{2}} - \frac{dx_{\alpha}}{d\tau} \frac{d \varphi }{d\tau }\right) \delta x^{\alpha} e^{\varphi} \sqrt{...}d\lambda = $$ $$ =\int \left( \partial_{\alpha}\varphi - \frac{d u_{\alpha}}{d \tau} - u_{\alpha} u_{\beta} \partial^{\beta} \varphi \right)\delta x^{\alpha} e^{\varphi} \sqrt{...}d\lambda \Rightarrow $$ $$ \partial_{\alpha}\varphi - \frac{d u_{\alpha}}{d \tau} - u_{\alpha} u_{\beta} \partial^{\beta} \varphi = 0 \Rightarrow \partial_{\alpha} \varphi = e^{-\varphi}\frac{d }{d \tau}\left( e^{\varphi } u_{\alpha}\right). $$ Unfortunately, this equation doesn't look like the equation from Wikipedia, $$ \frac{d (\varphi u_{\alpha})}{d \tau} = -\partial_{\alpha } \varphi. $$ I can explain the part of differences by renaming the function, $e^{\varphi } \to \varphi $, in the expression for action (then my equation reduces to the form $ \partial_{\alpha} \varphi = \frac{d }{d \tau}\left( \varphi u_{\alpha}\right)$), but I can't explain why my equation has the wrong sign.

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    $\begingroup$ The minus sign comes from the definition of $\eta$. You are using particle physics convention and I guess wikipedia uses the opposite one. A simple way to check is to dot $u^{\alpha}$ to your equation. And you see your equation with this sign and mostly negative metric convention is correct. $\endgroup$ – anecdote Jan 2 '14 at 0:53
  • $\begingroup$ @anecdote : in my equation there aren't convolutions like the corresponding one in Klein-Gordon equation (which can be written in form $(\square \pm m^2)\psi = 0$ in dependence of metric convention), so I don't see how the metric convention can change the sign in my case. Excuse me, but can you describe your statement in details? $\endgroup$ – Andrew McAddams Jan 3 '14 at 0:41
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    $\begingroup$ In the definition of $d\tau$, you use $\sqrt{\eta ...}$ instead of $\sqrt{-\eta...}$. In the variation $\delta(\sqrt{-\eta...})$, the opposite convention will differ by a minus sign. $\endgroup$ – anecdote Jan 3 '14 at 0:53

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