# Geodesics and acceleration

1. Jul 11, 2013

### DiracPool

I have a related question which may broaden the image for conceptual clarity. Imagine an object (me) moving through "flat" outer space, far from any gravitational bodies. We can say that the geodesic I am travelling along is essentially straight or flat, as is its worldline, correct? So me moving at a constant velocity simply moves through space (time) in a straight line.

Now we introduce a gravitational body into the picture, say a moon-sized object. Ok, so now the spacetime geodesic is no longer flat, it is curved in the vicinity of the moon. I am cruising along in my flat spacetime straight geodesic when all of a sudden I start to feel this curved space develop in front of me. My geodesic begins to become more and more curved as I slingshot past the moon in a curved path.

Ok, that's all fine and good, but my question is why do I not maintain a constant velocity while I am following this curved geodesic? Why is there acceleration involved? Where does that come from? Is there some sort of physical scenario where one would follow a curved geodesic where they maintained constant velocity? I'm guessing no, but why?

Similarly, does the rate of change of the geodesic, or simply the degree of curvature, determine the value of the acceleration? For instance, say instead of the moon I go cruising by an Earth-sized object. The curvature of the spacetime geodesic/worldline is more pronounced or "steep" there, if you will, therefore there is more gravity and a greater acceleration impelled upon a passing object. My question is why does a greater curvature compel a greater acceleration? Why isn't just the path of the moving body affected by the curvature while it continues to travel at a constant velocity.

Edit: Now I'm going for "dumb question" broke so forgive me: It doesn't have anything to do with centripetal acceleration caused by the curved path, does it?

Last edited: Jul 11, 2013
2. Jul 11, 2013

### A.T.

You don't "feel" anything in free fall, unless there are extreme tidal forces.
Geodesics don't "curve", they are locally straight. It's only globally that the seem to change direction.
You do maintain a constant velocity, in a local inertial frame.
It's only coordinate acceleration (see my previous post). It's a result of the choice of an non-inertial reference frame.

3. Jul 11, 2013

### soothsayer

4. Jul 11, 2013

### DiracPool

I don't see how these two threads are related at all, except perhaps superficially. The thread you posted deals strictly with the relationship between mass and curved space, not why curved space-time causes acceleration.

5. Jul 11, 2013

### soothsayer

If you read past the first page or so, you'll see that a lot of the OP's underlying questions were about how gravity could cause acceleration if it wasn't a "force". We spent a lot of time trying to explain it. If I were to respond to the OP in this thread, I would really just be regurgitating the things I wrote in the thread I just linked.

6. Jul 11, 2013

### DiracPool

Hmmm, perhaps I'm mistaken, it looks as though that thread has evolved significantly since my participation in it. Maybe I can learn something from it

7. Jul 11, 2013

### pervect

Staff Emeritus
Let's go into some detail as to how you might "feel" the curved space-time develop. If you're by yourself, and you're reasonably small, you probably won't feel a thing, as others have pointed out.

But suppose you have a friend, following a parallel course.

Both of you are cruising along following geodesics. Then you hit a curved patch of space-time (or, as we more usually say, pass close to a massive body).

You'll notice that your friends course starts to diverge from yours. Rather than maintaining a constant distance, he'll start to accelerate away from you. If you have several friends, you'll notice that the rate of acceleration is proportional to the distance.

This relative acceleration, between you and your friend (or friends) is called geodesic deviation, and the magnitude of it can be expressed by the geodesic deviation equation

$\frac{D^2 x^a}{D t^2}= R^a{}_{bcd} u^b x^c u^d$

where x is the separation vector between you and your neighbor, and u^b and u^d are unit time like vectors for you and your neighbor.

R is a rank 4 tensor, and it turns out to be the famous Riemann tensor, that describes space-time curvature.

You can look in wiki and other sources for more details under this name.

But the basic idea is just that your measured acceleration (geodesic deviation) is proportioal to one particular component of space-time curvature multiplied by distance between you and your friend. And the distance goes up quadratically with time (or bi-linearly with time, when you consider time as being the product of your time and your friends time).

8. Jul 12, 2013

### pervect

Staff Emeritus
I should add a bit to my previous post. Geodesic deviation is how GR explains the acceleration between you and your friend in a nearby rocket. It's entirely based on curvature. You can imagine two great circles on a sphere (the Earth, for instance, if you ignore it's bulge) and, because great circles on a sphere are geodesics, apply a 2-space + 1-time version of the geodesic deviation equation to calculate said relative acceleration.

In a Newtonian context, though, you'd explain the relative acceleration between you and your friend by tidal forces.

There is a very close relation between the two - in fact you can usually think of the Riemann as being a "bigger" version of the Newtonian tidal force tensor. It is a good general approach, that you'll find in textbooks such as MTW's "Gravitation", there's a few specific situations where you can get into trouble though.