# Gravity without force of gravity. What does it mean?

As we know, the gravitation force in General Relativity is the semblant phenomenon. Let's look at this phenomenon in more detail.

## A brief preface

From the Landau-Lifshitz "Volume 2. The Classical Theory of fields" (LL2)(Problem 1 in §88), we can see the equation of motion (EOM) and an expression for gravitation force (we put $c=1$):

EOM(also known as $\frac{dp}{dt} =F$ or $ma = F$) : $$\sqrt{1-v^2} \frac{d}{ds}\frac{v^{\alpha}}{\sqrt{1-v^2}} + \lambda^{\alpha}_{\beta\gamma}\frac{mv^{\beta}v^{\gamma}}{\sqrt{1-v^2}} = f^{\alpha}$$

Gravitation force (also known as $F = G\frac{mM}{r^2}$): $$f_{\alpha} = \frac{m}{\sqrt{1-v^2}}\left[ -\frac{\partial \ln\sqrt{h}}{\partial x^{\alpha}} + \sqrt{h} \left(\frac{\partial g_{\beta}}{\partial x^{\alpha}} - \frac{\partial g_{\alpha}}{\partial x^{\beta}}\right){v^{\beta}}\right]$$ where $h = g_{00}$, $g_{\alpha} = -\frac{g_{0\alpha}}{g_{00}}$.

As one can see, the origin of gravitation force in GR is mostly due to the curvature of time, because it depend on component of metrics $g_{00}$. And this force in LL2 expressed in 3-vector form, which looks similar to Lorentz force: $$\label{3Force} \vec{F} = \frac{m}{\sqrt{1-v^2}} \left( -\vec\nabla\ln\sqrt{h} + \left[\vec{v}\times\left[ \sqrt{h}\vec\nabla\times\vec{g}\right] \right] \right),$$

Now consider an EOM. The RHS of EOM consists of two terms. The second term contains spatial Cristoffel symbols $\lambda^{\alpha}_{\beta\gamma}$ which is also depend on metrics.

# Attempts to ask a question

Lets consider some kind of spherical-symmetric metrics (now we are not interested in the distribution of matter that caused it):

$$ds^2 = dt^2 - \frac{dr^2}{1-\frac{2M}{r}} - r^2(d\theta^2 + \sin^2\theta d\phi^2).$$

Nonzero components of $\Gamma^i_{jk}$: \begin{align} \Gamma^r_{rr} &= -\frac{M}{r^2}\frac{1}{1-\frac{2M}{r}},\\ \Gamma^r_{\theta\theta} &= 2M - r\\ \Gamma^r_{\phi\phi} &= (2M - r)\sin^2\theta\\ \Gamma^{\theta}_{r\theta} &= \Gamma^{\phi}_{r\phi} = \frac1r\\ \Gamma^{\theta}_{\phi\phi} &= -\sin\theta\cos\theta \\ \Gamma^{\phi}_{\theta\phi} &= \cot\theta, \end{align} $\Gamma\text{'s} = \lambda\text{'s}$. The Riemann Tensor nonzero components: \begin{align} R_{r\theta r\theta} & = \frac{M}{r}\frac{1}{1-\frac{2M}{r}} \\ R_{r\phi r\phi} & = \frac{M}{r}\frac{\sin^2\theta}{1-\frac{2M}{r}} \\ R_{\theta\phi\theta\phi} &= - 2Mr\sin^2\theta \end{align} This metrics is only spattialy curved.

Scalar curvature $R = 0$.

According to gravitation force expression, we conclude there is no force $f_{\alpha} = 0$ in such metrics. Then (in case $\theta =\frac\pi 2$, $\frac{d^2\theta}{ds^2} = 0$) the EOM has the simple form:

\begin{align} \frac{d^2r}{ds^2} &= \frac{M}{r^2}\frac{1}{1-\frac{2M}{r}}\frac{(v^r)^2}{1-v^2} + (r-2M)\frac{(v^{\phi})^2}{1-v^2}\\ \frac{d^2\phi}{ds^2} &= \frac1r \frac{v^r v^{\phi}}{1-v^2}. \end{align}

Due to nonzero values of $\lambda^{\alpha}_{\beta\gamma}$ in EOM, the geodesics are not straight lines in such metrics. If particle has zero initial velocity, then will be no force of attraction to the origin.

But, if the particle has the radial velocity, then it will be attract to the origin. Thus, we have the gravity without force of gravity.

How to correctly judge the reality of gravity, by the existence of force $f_{\alpha}$ or by the geodesics deviation? What is the meaning of the force of gravity in LL2?

• The $g_{00}$ term is not zero in your spherically symmetric metric there since there is a term proportional to $dt^2$ which comes from the term $g_{00}dx^0 dx^0$ – Triatticus Apr 7 '18 at 6:07
• @Triatticus ops! I mean $g_{00} = 1$. Thanks. – Sergio Apr 7 '18 at 6:12
• Have you worked out $\Gamma^r_{tt}$ for your metric? I can't do the calculation right now, but I bet it isn't zero. That means moving in the time direction will accelerate you in the $r$ direction. – John Rennie Apr 7 '18 at 6:45
• @JohnRennie Yes I do it in MapleSoft. $\Gamma^r_{tt} = 0$. – Sergio Apr 7 '18 at 7:03
• Why are you setting $g_{00}$? The Schwarzschild metric satisfies $g_{00}=1-\frac{2M}{r}$. – J.G. Apr 7 '18 at 7:49

The geodesic equation tells us:

$$\frac{d^2x^\alpha}{d\tau^2} = -\Gamma^\alpha{}_{\mu\nu}u^\mu u^\nu$$

The reason we often say the acceleration is due to the curvature in time is because in everyday life the four-velocity is dominated by the time component i.e. $u^t \gg u^r, u^\theta, u^\phi$. That means to a first approximation we can ignore the terms in $u^tu^a$ and $u^au^b$ ($a$ = $r$, $\theta$ or $\phi$) and consider only the term in $u^tu^t$. Then the only significant Christoffel symbols are $\Gamma^a{}_{tt}$.

But you've chosen a metric for which (in the coordinates being used) $\Gamma^a{}_{tt}=0$. That means you are quite correct that a stationary object will not accelerate in space. Only objects moving in the spatial coordinates will accelerate in space.

This isn't that unusual. The Ellis wormhole has exactly the same property. If you're interested I discuss this in my answer to How do spatial curvature and temporal curvature differ? (actually that is close to a duplicate of this question).

So I guess the answer to your question is that the gravitational acceleration is due to the curvature in time in most circumstances but not all. You have managed to choose one of the exceptional cases.

One last comment, be careful about statements like:

This metric is only spatially curved

Remember that the Christoffel symbols are not tensors and we can always choose a coordinate system that makes them all zero i.e. the normal coordinates. In your coordinates it's certainly true that the curvature is spatial, but that's just down to your choice of coordinates and it wouldn't be true of a different coordinate choice.

• As far as I understand, this type of metrics does not cause the initially immobile body to move to $r = 0$, but causes tidal forces in moving bodies. If the body immobile , no forces act on it. – Sergio Apr 7 '18 at 9:27
• @Sergio correct! :-) – John Rennie Apr 7 '18 at 9:34
• Hi John, minor nitpick: I believe you mean "Riemann normal co-ordinates" in your last paragraph. – WetSavannaAnimal Apr 7 '18 at 13:54
• @WetSavannaAnimalakaRodVance I did mean Fermi normal coordinates though actually just normal coordinates is fine. – John Rennie Apr 7 '18 at 14:38
• @WetSavannaAnimalakaRodVance well we were talking about geodesic motion :-) – John Rennie Apr 7 '18 at 15:30

The geodesic deviation equation is $\ddot{x}^\mu=-\Gamma^\mu_{\nu\rho}\dot{x}^\nu\dot{x}^\rho$, where dots denote derivatives with respect to proper time. Speeds $\ll c$ give $\partial_t\approx\partial_\tau$ and $\dot{x}^j\ll \dot{x}^0$ so $\partial_t^2 x^i\approx -\Gamma^i_{00}$. This recovers Newtonian gravity, and expresses its acceleration in terms of $\Gamma^i_{00}$. You might want to say that's "time curvature", but then there is that pesky spatial index.

Update: in the special case $\Gamma^i_{00}=0$, the approximation becomes $\partial_t^2 x^i\approx -\Gamma^i_{jk}\beta^j\beta^k$.

• Sergio's point is that $\Gamma^r{}_{tt} = 0$, which indeed it is - I've just checked using Mathematica. – John Rennie Apr 7 '18 at 7:57
• I think you've missed the point. Sergio has not incorrectly written down the Schwarzschild metric. He is asking what happens if the metric is what he has written down, whatever stress-energy tensor is needed to produce this geometry. – John Rennie Apr 7 '18 at 8:09
• @JohnRennie I've edited my answer. – J.G. Apr 7 '18 at 8:13