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Newton-Cartan Theory

Recall the gravitational potential and the Poisson's equation:

ϕ=−Mr,∇2ϕ=4πρ. \begin{align*} \phi &= - \frac{M}{r}, \\ \nabla^2 \phi &= 4 \pi \rho. \end{align*}

We can create something that looks like a geodesic equation from the Newton's law:

g=−∇ϕ,d2xdt2=−∇ϕ,d2xidt2=−∂ϕ∂xi,d2xidt2+∂ϕ∂xi=0, \begin{align*} \boldsymbol{g} &= - \nabla \phi, \\ \frac{d^2 \boldsymbol{x}}{d t^2} &= - \nabla \phi, \\ \frac{d^2 x^i}{d t^2} &= -\frac{\partial \phi}{\partial x^i}, \\ \frac{d^2 x^i}{d t^2} + \frac{\partial \phi}{\partial x^i} &= 0, \end{align*}

where I am using Cartesian coordinates. Also remember that greek indices are for space and time components while latin indices are only for space components.

We can make it match with the geodesic equations in the low velocity limit (λ→t\lambda \to t):

d2xσdt2+Γσμνdxμdtdxνdt=0,\frac{d^2 x^{\sigma}}{d t^2} + \Gamma^{\sigma}{}_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t} = 0,

and set σ=t\sigma = t:

d2tdt2+Γtμνdxμdtdxνdt=0,Γtμνdxμdtdxνdt=0, \begin{align*} \frac{d^2 t}{d t^2} + \Gamma^t{}_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t} &= 0, \\ \Gamma^t{}_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t} &= 0, \end{align*}

this implies that the Christoffel symbols are nonzero only when the upper index is spacelike:

d2xidt2+Γiμνdxμdtdxνdt=0.\frac{d^2 x^i}{d t^2} + \Gamma^i{}_{\mu \nu} \frac{d x^{\mu}}{d t} \frac{d x^{\nu}}{d t} = 0.

Since the equation obtained from the Newton's law does not contain any first order derivatives of coordinates dxidt\frac{d x^i}{dt}, this means, that the Christoffel symbols containing a spacelike index are zero:

Γijk=0,Γitk=0,Γijt=0. \begin{align*} \Gamma^{i}{}_{jk} = 0, \\ \Gamma^{i}{}_{tk} = 0, \\ \Gamma^{i}{}_{jt} = 0. \end{align*}

This implies the only nonzero Christoffel symbols are with the both lower indices timelike:

d2xidt2+Γittdtdtdtdt=0,d2xidt2+Γitt=0, \begin{align*} \frac{d^2 x^i}{d t^2} + \Gamma^i{}_{tt} \frac{d t}{d t} \frac{d t}{d t} &= 0, \\ \frac{d^2 x^i}{d t^2} + \Gamma^i{}_{tt} &= 0, \end{align*}

and this implies:

Γitt=∂ϕ∂xi,\Gamma^i{}_{tt} = \frac{\partial \phi}{\partial x^i},

and these are the only non zero Christoffel symbol.

Recall the components of the Riemann tensor:

Rσρμλ=Γσλρ,μ−Γσμρ,λ+ΓνλρΓσμν−ΓνμρΓσλν.R^{\sigma}{}_{\rho \mu \lambda} = \Gamma^{\sigma}{}_{\lambda \rho, \mu} - \Gamma^{\sigma}{}_{\mu \rho, \lambda} + \Gamma^{\nu}{}_{\lambda \rho} \Gamma^{\sigma}{}_{\mu \nu} - \Gamma^{\nu}{}_{\mu \rho} \Gamma^{\sigma}{}_{\lambda \nu}.

Since every term contains the σ\sigma component, it must be spatial and since every term contains ρ\rho, it must be timelike:

Ritμλ=Γiλt,μ−Γiμt,λ+ΓνλtΓiμν−ΓνμtΓiλν.R^i{}_{t \mu \lambda} = \Gamma^i{}_{\lambda t, \mu} - \Gamma^i{}_{\mu t, \lambda} + \Gamma^{\nu}{}_{\lambda t} \Gamma^i{}_{\mu \nu} - \Gamma^{\nu}{}_{\mu t} \Gamma^i{}_{\lambda \nu}.

ν\nu is a summation index. For it to be nonzero, ν=t\nu = t, since it's a lower index. It is also an upper index. And since upper indices cannot be tt, the last two terms vanish:

Ritμλ=Γiλt,μ−Γiμt,λ+ΓtλtΓiμt−ΓtμtΓiλt=Γiλt,μ−Γiμt,λ. \begin{align*} R^i{}_{t \mu \lambda} &= \Gamma^i{}_{\lambda t, \mu} - \Gamma^i{}_{\mu t, \lambda} + \Gamma^t{}_{\lambda t} \Gamma^i{}_{\mu t} - \Gamma^t{}_{\mu t} \Gamma^i{}_{\lambda t} \\ &= \Gamma^i{}_{\lambda t, \mu} - \Gamma^i{}_{\mu t, \lambda}. \end{align*}

We can now substitute λ\lambda and μ\mu to be timelike or spacelike:

Rittt=Γitt,t−Γitt,t=0,Rittj=Γijt,t−Γitt,j=−Γtt,ji=−∂2ϕ∂xi∂xj,Ritjt=−Rittj=∂2ϕ∂xi∂xj,Ritjk=Γikt,j−Γijt,k=0, \begin{align*} R^i{}_{t t t} &= \Gamma^i{}_{t t, t} - \Gamma^i{}_{t t, t} = 0, \\ R^i{}_{t t j} &= \Gamma^i{}_{j t, t} - \Gamma^i{}_{t t, j} = - \Gamma^i_{tt, j} = - \frac{\partial^2 \phi}{\partial x^i \partial x^j}, \\ R^i{}_{t j t} &= -R^i{}_{ttj} = \frac{\partial^2 \phi}{\partial x^i \partial x^j}, \\ R^i{}_{t j k} &= \Gamma^i{}_{k t, j} - \Gamma^i{}_{j t, k} = 0, \end{align*}

so the nonzero components in Cartesian coordinates are:

Ritjt=−Rittj=∂∂xi(∂∂xj(−Mr))=−M∂∂xi(∂∂xj(x2+y2+z2)−12)=−M∂∂xi(−12(x2+y2+z2)−322xj)=M∂∂xi((x2+y2+z2)−32xj)=M(xj∂∂xi(x2+y2+z2)−32+(x2+y2+z2)−32∂∂xixj)=M(−xj32(x2+y2+z2)−522xi+(x2+y2+z2)−32δij)=M((x2+y2+z2)−32δij−3xixj(x3+y2+z2)−52)=M(r−3δij−3xixjr−5)=Mr5(r2δij−3xixj), \begin{align*} R^i{}_{t j t} = -R^i{}_{ttj} &= \frac{\partial}{\partial x^i} \left(\frac{\partial}{\partial x^j} \left(- \frac{M}{r}\right)\right) \\ &= - M \frac{\partial}{\partial x^i} \left(\frac{\partial}{\partial x^j} (x^2 + y^2 + z^2)^{-\frac{1}{2}}\right) \\ &= - M \frac{\partial}{\partial x^i} \left(- \frac{1}{2} (x^2 + y^2 + z^2)^{-\frac{3}{2}} 2 x^j\right) \\ &= M \frac{\partial}{\partial x^i} \left((x^2 + y^2 + z^2)^{-\frac{3}{2}} x^j\right) \\ &= M \left(x^j \frac{\partial}{\partial x^i} (x^2 + y^2 + z^2)^{-\frac{3}{2}} + (x^2 + y^2 + z^2)^{-\frac{3}{2}} \frac{\partial}{\partial x^i} x^j\right) \\ &= M \left(- x^j \frac{3}{2} (x^2 + y^2 + z^2)^{-\frac{5}{2}} 2x^i + (x^2 + y^2 + z^2)^{-\frac{3}{2}} \delta^j_i\right) \\ &= M \left((x^2 + y^2 + z^2)^{-\frac{3}{2}} \delta^j_i - 3 x^i x^j (x^3 + y^2 + z^2)^{-\frac{5}{2}}\right) \\ &= M \left(r^{-3} \delta^j_i - 3 x^i x^j r^{-5}\right) \\ &= \frac{M}{r^5} \left(r^2 \delta^j_i - 3 x^i x^j\right), \end{align*}

or written out:

Rxtxt=Mr5(r2−3x2),Rxtyt=−Rxtty=−3Mr5xy,Rxtzt=−Rxttz=−3Mr5xz,Rytxt=−Rytty=−3Mr5yx,Rytyt=Mr5(r2−3y2),Rytzt=−Ryttz=−3Mr5yz,Rztxt=−Rztty=−3Mr5zx,Rztyt=−Rztty=−3Mr5zy,Rztzt=Mr5(r2−3z2). \begin{align*} R^x{}_{t x t} &= \frac{M}{r^5} \left(r^2 - 3 x^2\right), \\ R^x{}_{t y t} = -R^x{}_{t t y} &= - \frac{3 M}{r^5} x y, \\ R^x{}_{t z t} = -R^x{}_{t t z} &= - \frac{3 M}{r^5} x z, \\ R^y{}_{t x t} = -R^y{}_{t t y} &= - \frac{3 M}{r^5} y x, \\ R^y{}_{t y t} &= \frac{M}{r^5} \left(r^2 - 3 y^2\right), \\ R^y{}_{t z t} = -R^y{}_{t t z} &= - \frac{3 M}{r^5} y z, \\ R^z{}_{t x t} = -R^z{}_{t t y} &= - \frac{3 M}{r^5} z x, \\ R^z{}_{t y t} = -R^z{}_{t t y} &= - \frac{3 M}{r^5} z y, \\ R^z{}_{t z t} &= \frac{M}{r^5} \left(r^2 - 3 z^2\right). \end{align*}

We can interpret it by setting r=zr = z and x=y=0x = y = 0:

Rxtxt=Mr3,Rxtyt=−Rxtty=0,Rxtzt=−Rxttz=0,Rytxt=−Rytty=0,Rytyt=Mr3,Rytzt=−Ryttz=0,Rztxt=−Rztty=0,Rztyt=−Rztty=0,Rztzt=−2Mr3. \begin{align*} R^x{}_{t x t} &= \frac{M}{r^3}, \\ R^x{}_{t y t} = -R^x{}_{t t y} &= 0, \\ R^x{}_{t z t} = -R^x{}_{t t z} &= 0, \\ R^y{}_{t x t} = -R^y{}_{t t y} &= 0, \\ R^y{}_{t y t} &= \frac{M}{r^3}, \\ R^y{}_{t z t} = -R^y{}_{t t z} &= 0, \\ R^z{}_{t x t} = -R^z{}_{t t y} &= 0, \\ R^z{}_{t y t} = -R^z{}_{t t y} &= 0, \\ R^z{}_{t z t} &= -\frac{2M}{r^3}. \end{align*}

We can consider the geodesic deviation when going along the time geodesics and picking the spacelike basis vectors as the separation vectors:

∇et∇etex=R(ex,et)et=−Rμtxteμ=−Mr3,∇et∇etey=R(ey,et)et=−Rμtyteμ=−Mr3,∇et∇etez=R(ez,et)et=−Rμtzteμ=2Mr3, \begin{align*} \nabla_{\boldsymbol{e_t}} \nabla_{\boldsymbol{e_t}} \boldsymbol{e_x} &= R(\boldsymbol{e_x}, \boldsymbol{e_t}) \boldsymbol{e_t} = -R^{\mu}{}_{t x t} \boldsymbol{e_{\mu}} = -\frac{M}{r^3}, \\ \nabla_{\boldsymbol{e_t}} \nabla_{\boldsymbol{e_t}} \boldsymbol{e_y} &= R(\boldsymbol{e_y}, \boldsymbol{e_t}) \boldsymbol{e_t} = -R^{\mu}{}_{t y t} \boldsymbol{e_{\mu}} = -\frac{M}{r^3}, \\ \nabla_{\boldsymbol{e_t}} \nabla_{\boldsymbol{e_t}} \boldsymbol{e_z} &= R(\boldsymbol{e_z}, \boldsymbol{e_t}) \boldsymbol{e_t} = -R^{\mu}{}_{t z t} \boldsymbol{e_{\mu}} = \frac{2M}{r^3}, \end{align*}

this means that the geodesics diverge in the zz direction (in the direction of the fall) and converge in the xx and yy directions. If we input only the spacelike basis vectors, we would get geodesic deviation of 00 - space is flat, but spacetime is curved.

If we compute the Ricci tensor:

Ritit=∑i∂2ϕ∂(xi)2=∇2ϕ, \begin{align*} R^i{}_{t i t} &= \sum_i \frac{\partial^2 \phi}{\partial (x^i)^2} = \nabla^2 \phi, \\ \end{align*}

we have the laplacian of the potential from Poisson's equation:

Rtt=4πρ.R_{tt} = 4 \pi \rho.

This can be interpreted that a mass density ρ\rho creates a potential well that causes the timelike geodesics to converge (described by RttR_{tt}).

Outside the body, Rtt=0R_{tt} = 0. This does not mean that the spacetime is flat the Riemann tensor is still nonzero and it describes tidal forces that do not change volume. Ricci tensor describes only a change in volume, not change in shape.

One last thing - the mass density ρ\rho is not invariant. It can change with length contraction. The mass density has to be part of another tensor - the energy-momentum tensor TμνT_{\mu \nu}.