3
$\begingroup$

In general relativity, it's the curvature of spacetime that gives the effect of gravity, due to objects following geodesics. What I learnt about geodesics is that they are like straight lines in Euclidean space. But the Wikipedia article on Geodesics in general relativity says that geodesics depend on objects velocity, which contradicts what I originally thought:

I originally thought that, since there's no gravitational force, an object with zero net force of could still experience effect gravity. Now (at least in Euclidean space) if you aren't experiencing force, hence no acceleration, your path (that you'll eventually gone through) shouldn't change as velocity varies, as velocity in this case should merely determine how fast you gone through a certain length of the path.

Wouldn't this be the same for non-Euclidean space? Like if a geodesic is a straight line, wouldn't the path that I'll be taking also be independent from velocity? Am I wrong with my statement above? Or is there something that I've ignored (maybe the curvature due to velocity...as kinetic energy?).

$\endgroup$
1
8
$\begingroup$

In flat space, your path through 4-D space-time depends on your velocity. If you have velocity of 0, your path through 4-D space-time will be purely in the time direction. If you have a velocity of c in the "x" direction, your path through 4-d space-time will look like x=ct. If we focus on just the "t" and "x" directions, the graph of this path looks different from the graph of the path x=ct/2 or x=0.

An example of how this affects curved space is the space around the Earth, our most familiar massive body. With 0 velocity, you will fall to the Earth radially (straight "down"). With enough velocity, your geodesic could become a circle (orbit!) around the Earth. If you wanted to look at the time dimension, your geodesic would look like a spiral, due to your velocity.

Because of the addition of the "time" direction, you can think of velocity as changing the direction in 4-D space. Even if your velocity is only along the "x" direction, changing the speed changes the relationship between the "x" direction and the "t" direction, hence changing your angle in 4-D space.

$\endgroup$
3
  • $\begingroup$ If I understand correctly (correct me if I don't), does this mean since what I refer as changes in velocity, when substitute it into relativity and requiring it to satisfy $ds^2=dt^2-dx^2$, variation in x direction also effects in t direction. So as the result, what I refer as "changing velocity (in magnitude)" actually corresponds to varying the velocity's angle. $\endgroup$ Aug 20 '21 at 7:38
  • 1
    $\begingroup$ Yes, changing the velocity magnitude results in a different direction in "t" and "x" space. From special relativity, when you move faster in the "x" direction you move slower in the "t" direction, from the perspective of an outside observer at rest. Consider time dilation and length contraction! To clarify, varying the magnitude varies the "angle" in 4-D space. $\endgroup$
    – Alwin
    Aug 20 '21 at 7:46
  • 2
    $\begingroup$ @Andrew.Wolphoe The idea that velocity is an angle is a core part of the whole spacetime concept. There's more on that topic here: physics.stackexchange.com/q/595578/123208 $\endgroup$
    – PM 2Ring
    Aug 20 '21 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.