I see tidal gravity mentioned in the literature sometimes, and sometimes people even say something like that is the “real gravity”. I am confused about the significance of tidal gravity, and what role it plays in curvature. Please correct me if I have any misconceptions, and I will try to explain my confusion.

Tidal gravity, by my understanding, is the difference in gravity between two points. As the strength of gravity falls as the inverse square of the distance from a spherical source, tidal in this sense indicates a strictly geometrical attribute. What I mean is that there is no difference between this effect and any other geometrical inverse square law, such as for the intensity of light as emitted by a point source. It is due to geometry. Individually, the radiated photons travel to infinity, but the overall intensity (flux) falls off as the inverse square of the distance, due to equal amounts of light passing through increasingly larger surface patch areas.

As for what curvature means, I like how it is presented in the Feynman lectures “curvature expressed in terms of the excess radius is proportional to the mass inside a sphere”. They also give the second law of gravitation as ” How objects move if there are only gravitational forces—namely, that objects move so that their proper time between two end conditions is a maximum.”

So it seems to me that tidal gravity is only kind of indirectly related to curvature. The spherical nature of the gravitational source of bodies such as planets and stars make the distribution of matter (energy) unevenly distributed. This (uneven density distribution of energy) gives rise to the curvature, or excess radius. The fact that the bodies are spheres (or point sources as viewed from a larger perspective) is also what gives rise to the inverse square law.

We could consider the situation for a gravitational body which is an infinite plane. In this case the gravitational strength would not fall as the inverse square of the distance. It seems to me that below the surface there would still be an excess radius, and therefore curvature would exist. But above the surface there are no tidal effects. Gravity is still attractive above the surface, but is space-time there curved?


I could be wrong, but from what I think I understand about tidal forces, there are three cases that I can think of. Please correct me if I am incorrect.

Gravitational tidal force 1: (The inverse square law) This is the decrease in gravitational strength as a body travels further from the center of an (idealized) spherical gravitational source.

Gravitational tidal force 2: (density distribution) These tidal forces are due to the uneven density distribution of mass/energy in the gravitational source. If the mass in the center of the earth is more dense than the mass closer to the surface, this will be a second kind of tidal force. Mountains, valleys, and in general any deviation from a perfectly spherical shape contribute to these tidal forces.

Gravitation tidal force 3: (non-linearity, or the gravity of gravity) I am a bit fuzzy on this, but it seems to me how this works is that above and beyond the direct gravitation due to mass/energy, there is additional gravitation due to the fact that that gravitation itself is a form of energy.

Errors in my assessment of tidal forces notwithstanding, my question is what is the relationship between gravitational tidal forces and curvature.

When I brought up the scenario of an infinite plane gravitational source, I did so because I thought (and still do think) that the relationship between tidal forces and curvature might be more easily understood if we could eliminate gravitational tidal force #1 above. With an infinite plane gravitational source the strength of gravity does not fall off as a function of distance from the surface.


  • $\begingroup$ This question is not a duplicate of the question you referenced. The question you referenced is really three questions in one. The first and third have to do with a particle in free fall in an inertial frame of reference, and gravity as a fictitious force experienced in a non-inertial reference frame. The second question does regard curvature, but he is asking if curvature is only required in order to explain tidal forces. My question differs in that I am trying to understand the difference between gravitational tidal forces and curvature, if there are any. $\endgroup$
    – pigdog627
    Aug 19, 2015 at 12:05
  • $\begingroup$ point taken, I should have deleted the comment when I saw the other answer, apologies for that $\endgroup$
    – user81619
    Aug 19, 2015 at 12:09
  • $\begingroup$ No problem. I tried to respond to your other comment, but the comment size limit is ridiculously low. Is there another way (that is not frowned upon). $\endgroup$
    – pigdog627
    Aug 19, 2015 at 12:17
  • $\begingroup$ Nope, they even stopped a Nobel prize winner trying that on this site, if I have my facts right:). My comments are those of a newbie, every second answer on this site is a jaw-dropping revelation to me , so no response necessary. thanks $\endgroup$
    – user81619
    Aug 19, 2015 at 12:23
  • $\begingroup$ Actually, ignoring this particular question, you can go into chat, I have never done it, but I think if either side has a high enough rep, that's one way to get things sorted out. $\endgroup$
    – user81619
    Aug 19, 2015 at 12:27

1 Answer 1


I'm going to answer the question of your title, and also address the curious statement that "tidal gravity=real gravity".

Let's begin with your statement:

Tidal gravity, by my understanding, is the difference in gravity between two points.

You're very much on the right track here. When people talk of "tidal effects" and "tidal gravity" when not describing the actual phenomenon of ocean (or other fluid) tides on a planet, these are, I believe vague statements that to me mean something like "a gravity problem involving a spacetime neighborhood big enough that special relativity - the relativity of flat, Minkowski spacetime, cannot accurately describe the physics and events there". This would also explain the "real gravity" phrase: i.e. gravity for which the tools of general relativity are needed.

The curvature tensor indeed tells us when we are dealing with this "real", non-special-relativity-described gravity and when we are not, i.e. when we could handle the problem with special relativity. There is a mathematical construction that explicitly spells this delineation out very clearly: it is called the Riemann normal co-ordinate system for a neighborhood about a given event $\mathscr{P}_0$ in spacetime and these are explained in more detail below my main answer. Riemann normal co-ordinates annul the connexion co-efficients (the Christoffel Symbols) at the point in question, so they make a small neighborhood of $\mathscr{P}_0$ look exactly like Minkowski spacetime. The fact that such a neighborhood and co-ordinates exist can be thought of as how the Equivalence Principle enters GR: a momentarily co-moving inertial frame always exists at $\mathscr{P}_0$ and physics follows special relativity if we use the Minkowskian co-ordinates of this frame and we don't stray too far from $\mathscr{P}_0$. How far? The curvature tensor determines how far. As long as our proper distance/ time from $\mathscr{P}_0$ is much less than the reciprocals of the curvature tensor's components, special relativity is good to go. Informally, "tidal effects are negligible" as long as our spacetime neighborhood is much smaller than the reciprocals of all the curvature components. In this sense, curvature and the need to consider "tidal effects" are intimately related.

This all sounds sophisticated, but it is really nothing more than the situation shown below. We have a "curved" manifold $\mathcal{M}$ - think of the 2-sphere as a concrete example. The geometry of the manifold $\mathcal{M}$ - for example, the sums of angles inside triangles, how long parallel straight lines can be whilst remaining equidistant (how well Euclid's parallel postulate is fulfilled) - is approximately the same as the geometry of the tangent plane $\mathcal{T}$ as long as we as we stay within a distance $s$ from our initial point $\mathscr{P}_0$ such that $s\,\kappa \ll 1$, where $\kappa$ is the sectional curvature.

Tangent Space Approximation

Afterword: Construction of Riemann Normal Co-ordinates and the Role of the Curvature Tensor in Defining the Extent of a Minkowskian Approximation

Take the target event $\mathscr{P}_0$ in spacetime and, for every tangent vector $\vec{v}$ to spacetime at this point, you work out the spacetime geodesic through $\mathscr{P}_0$ with the tangent given by $\vec{v}$, so we end up with a path $\mathscr{P}(\vec{v},\,\tau)$ with $\mathscr{P}(\vec{v},\,0)=\mathscr{P}_0$ parameterized by the arclength (proper time/distance) $\tau$ along it. We follow the path for 1 time / length unit to arrive at $\mathscr{P}(\vec{v},\,1)$. By varying $\vec{v}$ both in direction and magnitude, we can indeed describe any point in the neighborhood as the point defined by $\mathscr{P}(\vec{v},\,1)$, as long as the neighborhood is small enough. "Small enough" here means a definite, nonzero extent and we're not yet talking first order approximations. Small enough simply means that none of geodesics will crash into singularities or anything untoward, so that our co-ordinates are well defined. So we can define nearby points by tangent vectors (technically what we are doing is called "exponentiating" the tangent vector $\vec{v}$ out to the point $\mathscr{P}(\vec{v},\,1)$).

Now choose a basis $\{\vec{e}_j\}_{j=0}^3$ for the tangent space, so that $\vec{v} = \vec{e}_j\,v^j$. The Riemann normal co-ordinates of $\mathscr{P}(\vec{v},\,1)$ are then the components $v^j$ of the tangent vector $\vec{v}$ which "exponentiates" to the point $\mathscr{P}(\vec{v},\,1)$. They have the following dramatic properties:

$$\Gamma(\mathscr{P}_0) = 0\text{ (all Christoffel symbols knocked out at }\mathscr{P}_0\text{)}$$ $$\Gamma^a{}_{b\,c,\,j}(\mathscr{P}_0) \stackrel{def}{=}\frac{\partial}{\partial v^j}\Gamma^a{}_{b\,c}(\mathscr{P}_0) = -\frac{1}{3}\left(R^a{}_{b\,c\,j}+R^a{}_{c\,b\,j}\right)$$

where the $R^a{}_{b\,c\,j}$ and like components are components of the curvature tensor. So the $\Gamma$'s are annuled at $\mathscr{P}_0$ and therefore the covariant derivative there is simply the same partial derivative one would use with flat Minkowski spacetime. The second line works out the derivative of the Christoffel symbols, i.e. defines "how far" you can wander from $\mathscr{P}_0$ before the Christoffel symbols become appreciable and things don't look like flat spacetime anymore. As you can see, the curvature components directly define this distance.

See Misner Thorne and Wheeler "Gravitation", section 11.6 for more details.

  • $\begingroup$ +1 Put all your GR answers together, they would make a good concise guide, seriously. $\endgroup$
    – user81619
    Aug 17, 2015 at 15:14
  • $\begingroup$ Thank you for the reference. I am a bit new to the tensor calculus of GR, so I need a little more time to study this, but even if it is discouraged around here I just have to say, what a fantastic answer! $\endgroup$
    – pigdog627
    Aug 19, 2015 at 12:03
  • $\begingroup$ @pigdog627 Thanks. Keep the diagram in mind. Also I was forgetting that another interpretation of the curvature tensor is through the geodesic deviation equation. Neighboring, initially parallel geodesics track alongside one another: they stay parallel in flat space but they wobble relative to one another (change their separation) and the acceleration of this separation change is directly related to the curvature tensor. So, yes, a small cloud of matter falling in a gravitational field squeezes and swells orthogonal to its motion as it ... $\endgroup$ Aug 19, 2015 at 12:12
  • $\begingroup$ @pigdog627 ... falls, so this is rather like the tidal bulge phenomenon. PS Your username reminds me of this wonderful scene: 1:20 especially youtube.com/watch?v=A8yjNbcKkNY $\endgroup$ Aug 19, 2015 at 12:17
  • $\begingroup$ @WestSavannaAnimal LOL! Yes, that's where it's from. Nice link. $\endgroup$
    – pigdog627
    Aug 20, 2015 at 22:25

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