# Is acceleration caused by curvature or space or time or both?

I'm trying to get a hold of the idea of gravity in general relativity and spacetime. I've seen plenty of demonstrations of the rubber mat analogy to describe gravity and spacetime curvature. Is this curvature a deformation of 3 dimensional space or a deformation of time, or a combination of the two?

If it's a curvature of 3D space, then acceleration makes sense that if you compress the distance you are trying to go, but time is still linear and uncompressed, then you will cover that distance more quickly.

If time is curved/deformed/compressed, then... my brain can't wrap around that very well.

Sorry for my naivety!

• Ever heard of Gravitational Time Dilation? – Schrödinger's Cat Nov 24 '14 at 19:22
• Gravity is curvature of Spacetime (Don't talk Space and Time separately).. – Schrödinger's Cat Nov 24 '14 at 19:23
• Only seen time dilation mentioned, haven't gotten to researching it just yet. Hmm, spacetime is confusing. I can think about 3D space fine, it can be broken down into it's components and events can be mentioned in relation to a component, but spacetime... More research I guess. – pbcrazy Nov 24 '14 at 19:26
• This might help you: physics.stackexchange.com/questions/111900/… – Schrödinger's Cat Nov 24 '14 at 19:29

In a certain sense (regime) acceleration is caused by the curvature of time more than the curvature of space. Actually, the curvature is of the spacetime so that, making rigid distinctions has no much sense. However, if you consider the motion of a particle free falling in a region of spacetime, the equation of its story is the geodesical one:

$$\frac{d^2x^{\mu}}{d\tau^2}= - \Gamma^\mu_{\alpha \beta}\frac{dx^{\alpha}}{d\tau}\frac{dx^{\beta}}{d\tau}\:.$$ Everything here is described in a coordinate frame $x^0=ct, x^1,x^2,x^3$ where the metric is approximatively the flat one $g_{\mu\nu} = \eta_{\mu\nu}+ h_{\mu\nu}$. It is possible to prove that under physically admissible approximations (weak fields $|h_{\mu\nu}|<< 1$, field almost stationary'', velocities small with respect to $c$, etc...) the written equation can be approximated with

$$\frac{d^2x^{i}}{dt^2} = \frac{c^2}{2} \frac{\partial h_{00}}{\partial x^i}\quad i=1,2,3$$ so that, the Newtonian gravitational potential which is the cause of the acceleration in the Newtonian picture, can be approximated by $$\varphi(t,\vec{x})= - \frac{c^2}{2} h_{00}(t,\vec{x})\:.\tag{1}$$ recovering the Newtonian equation of motion of a particle in a gravitational field $$\frac{d^2\vec{x}}{dt^2} = - \nabla \varphi(t,\vec{x})$$ (1) is the same identification used to compute the redshift to correct, for instance, the GPS instruments.

You see that, in this semiclassical approximation, where it makes still sense to think in terms of classical acceleration (instead of four-acceleration), what really matters is the deviation of the the temporal component of the metric from the flat metric $\eta_{00}=-1$, the other components play a negligible role.

I'm hardly a GR expert, so if you want a more technical analysis I'm sure others will be able to give you one. However, the answer to your apparent questions is fairly straight forward.

It is not the curvature of space or the curvature of time that causes accelerations, it is the curvature of space-time. We live in a four dimensional universe (ignoring possible String Theory implications) which includes the familiar three spatial dimensions, as well as one time dimension. Together, these dimensions form the four dimensional space-time.

If you're having a hard time visualizing the "curving" of time, try not to over-think it too much. Just be aware that in the vicinity of very massive objects, the passage of time is altered in the vicinity. Although I hesitate to put another "Interstellar" reference on Physics.SE, the fact that the rate of the passage of time in the vicinity of the black hole "Gargantua" is slower than the rate of the passage of time at Earth does demonstrate this truth. (Although of course some artistic license was taken).

• The word is Spacetime, not Space-time.. – Schrödinger's Cat Nov 24 '14 at 19:30
• I've heard it both ways – Sean Nov 24 '14 at 19:31
• @SachinShekhar It really doesn't matter; different conventions are used. – HDE 226868 Nov 24 '14 at 19:45
• It would be nice if the Stack Exchange organization could get Google to accept the spelling of spacetime as one word, like physicists or their students (the largest group using the term) usually do: The wavy red lines scattered thru notes are distracting for those with short attention spans, like myself, and the rarer space-time version confusingly suggests space MINUS time. In the U.S., maybe there'd be some ADA (Americans with Disabilities Act) potential here. – Edouard Jul 20 '19 at 19:26

In Relativity (General and Special) there is no unique way to divide spacetime into space and time. Different observers, using different coordinate systems, will disagree about whether a four vector is just a displacement in time or just a displacement in space. So to this extent your question can't be answered because it's an artificial distinction.

A related question would be whether we can choose a coordinate system in which the curvature is just in time, or just in space. I asked a question along these lines in What makes a coordinate curved?. The answers are possibly a bit too deep for this discussion, but they boil down to that's a silly question :-)

But I'd like to pick up on Valter's last point because it's an interesting one. We have a tendancy to set $c = 1$ when writing down the metric, so we get nice symmetric looking equations like:

$$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$

But written in the units we use for everyday observations the metric is really:

$$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$$

where $c \approx 3 \times 10^8$m/s. So a displacement in time of 1 second contributes a factor of $3 \times 10^8$ more to the line element than does a displacement of 1 metre. What this means is that when considering weak gravitational fields, like the Earth's gravity, we get a good approximation by just ignoring the spatial curvature and only considering the curvature in the time coordinate.

This makes sense because experience tells us that thrown objects move in curves (parabolae) but also that space isn't obviously curved - if I draw a circle and measure the ratio of its circumference to its radius I always get the flat space value of $2\pi$. Time isn't obviously curved either because the curvature is small (though atomic clocks can measure it) but once you multiply by $3 \times 10^8$ the effect is big enough to make objects move in parabolae.

One day I will write a canonical Q/A explaining why objects accelerate towards the centre of the Earth. I have actually started this a couple of times, but finding a way of describing the physics that is universally accessible has proved challenging so far.

• I upvoted Valter's answer, but this one completes the answer with notations I could follow without (much) reference material, so I upvoted it as well. – Edouard Jul 20 '19 at 23:30