1
$\begingroup$

Let's say that, from my point of view, another observer is accelerating. Now, from his point of view, he is standing still: all he feels is an overall fictitious force of gravity, which is just a consequence of the the spacetime curvature at that point (caused by stress-energy tensor at the same point).

Now, if we accept this geometric view, It seems like acceleration cannot really exist in spacetime: every object keeps following a geodesic. When some external observer thinks that an object is accelerating, he should say instead that it is following a "curved geodesic".

Is this correct?

$\endgroup$
2
  • $\begingroup$ The word "acceleration" is overloaded: it has different meanings that conflict with each other in some conditions. Re-wording the question without using the word "acceleration" could be a valuable exercise. Hints: An observer following a geodesic feels weightless. An observer that is not following a geodesic does not feel weightless. Spacetime curvature affects which wordlines qualify as geodesics, but it doesn't affect the relationship between geodesic motion and weightlessness (ignoring finite-size effects). $\endgroup$ Aug 23, 2019 at 13:47
  • $\begingroup$ Would be also improved if you define what you mean by "curved geodesic" $\endgroup$
    – ohneVal
    Aug 26, 2019 at 8:57

6 Answers 6

2
$\begingroup$

The following answer refers to proper acceleration, as measured for example by an accelerometer at a given point, it coincides with the "g-force felt by" if you wish. Also I am ignoring effects due to a non-zero size of a body, that is I am speaking about point-like particles.

About this particular sentence:

It seems like acceleration cannot really exist in spacetime: every object keeps following a geodesic.

Sadly it is not. Geodesics are exactly the type of paths of free-falling observers, namely non accelerating observers. If you look at the geodesic equation $$\nabla_{\dot{\gamma}(\tau)} \dot{\gamma}(\tau) = 0$$ it is exactly demanding that the velocity of the path remains constant. This is however not saying that all particles follow geodesics. In principle a particle can follow any time-like curve in space-time, only the free case corresponds to geodesics.

To connect with measured acceleration you must define your observer, that means a tetrad frame at each point of the oberservers path. The observer's acceleration itself is then $$ a(\tau) = \nabla_{\dot{\gamma}(\tau)} \dot{\gamma}(\tau) = \nabla_{u(\tau)}u(\tau),$$ where $u(\tau)$ is called the 4-velocity and it is exactly the deviation from following a geodesic. (quantities above are all 4-vectors). If this observer wants to measure something in his frame, all it has to do is project said quantity into his world-line, so for the acceleration $a_p$, of some other word-line corresponding to some particle he would measure a 4-acceleration of $$u\cdot a_p = g(u,a_p) = g_{\mu\nu}\,u^\mu(\tau) a_p^\nu(\tau)$$

Notice that here the metric $g$ is assumed to be known. This is the object that has to solve Einstein field equations (homogeneous or not and any back-reaction on the geometry is ignored). So the metric, includes any effects from any energy-momentum tensor you might want to consider and tells according to your constraints what "free-falling" means. Any deviation from that is perceived as acceleration.

Another possible case is perhaps the relative acceleration appearing when you have a parametrized family of geodesics. Here one computes the rate of change of the projection of the deviation vector of the family of geodesics on to a given geodesic, explicitly, for a family of geodesics $X^\mu(\tau,s)$ where $\tau$ is the proper time and $s$ parametrizes the different geodesics: $$A^\mu = \frac{D}{d\tau} (T^\beta\nabla_\beta X^\mu ) $$

$\endgroup$
4
  • $\begingroup$ Thank you for the answer. I'm sorry I'm still confused: any acceleration is induced by a net force at that point. This force should be taken into account (in terms of potential energy perhaps) in the stress-energy tensor. And the stress-energy tensor bends the spacetime. So I see a clear connection between any acceleration and spacetime curvature. Where am I going wrong? (Conceptually) $\endgroup$ Aug 23, 2019 at 9:44
  • 3
    $\begingroup$ This answer would be clearer if it addressed the different meanings of the word "acceleration". When geodesics converge there is a perfectly correct usage which says the relative acceleration of particles on those geodesics is non-zero. It is ok to define acceleration such that free-fall equates to "zero acceleration", but it is also ok to define acceleration such that a dropped ball accelerates towards the ground. I think the original question is to do with confusion between the various possible definitions. $\endgroup$ Aug 23, 2019 at 9:55
  • $\begingroup$ @Andrew so is it correct to say this: if I (inertial observer, in free fall) see another observer accelerating, then from his point of view he feels that he is standing still but he feels a gravity force. Hence, both of us will deduce that he is not moving on a geodesic (and so is not an inertial observere): I can say this because I see him accelerating, while he can say this beacause he feels gravity. Correct? $\endgroup$ Aug 23, 2019 at 13:48
  • $\begingroup$ @FedericoToso This is neither correct nor incorrect but rather unclear. You need to define what your observation is more carefully. $\endgroup$ Sep 1, 2019 at 14:20
2
$\begingroup$

I think the confusion here stems from different types of acceleration: coordinate acceleration and proper acceleration. Coordinate acceleration is essentially the second derivative of position relative to the frame of an observer (so objects that you think are following curved paths have a coordinate acceleration), while proper acceleration is the acceleration that would be measured by an accelerometer.

For example if you are standing on the ground, your coordinate acceleration is zero relative to the Earth's surface. But an accelerometer would measure the acceleration due to gravity, so your proper acceleration is $9.81$ ms$^{-2}$ downwards.

Say there is are two spaceships floating in empty space. Both their coordinate accelerations and proper accelerations are zero. Now say one spaceship (A) switches on its rocket engines. Someone in the other spaceship (B) would measure a non-zero coordinate acceleration, since the velocity of A is increasing relative to B. Inside A, objects would also be pushed downwards as soon as the rocket engines are swithced on, and an accelerometer would measure an equal proper acceleration $a$.

According to Einstein's equivalence principle, being inside the spaceship A while it is accelerating at $9.81$ ms$^{-2}$ is indistinguishable from being inside the spaceship while it is sitting still on the surface of the Earth. The proper acceleration of both 'frames' is the same.

Let's say that, from my point of view, another observer is accelerating. Now, from his point of view, he is standing still

If I understand your meaning correctly, he is (1) standing still from his point of view if he is not experiencing any proper acceleration. In the language of general relativity, this means he will follow a (timelike) geodesic through spacetime. If the spacetime around him is curved, the geodesic he follows could be curved, and his coordinate acceleration will be nonzero.

Alternatively, you might mean that (2) you are in spaceship B from above and he is in spaceship A (he is accelerating away from you). In this case, he feels a proper acceleration. If he realises that this is caused by him switching on his rocket engines, he will know he is accelerating. However, if he somehow convinces himself he is actually on the surface of the Earth, he might think that he is perfectly still. However if he was on the surface of the Earth, then he would still not be following a geodesic in spacetime because he is not in freefall - geodesics around the Earth fall inwards, following the path of a falling object with no air friction.

If the second paragraph (2) is what you were asking about, then I am still a little confused what you mean by "from his point of view, he is standing still" - I suppose everyone is always perfectly still relative to themselves, but that's a bit of a physically useless fact.

all he feels is an overall fictitious force of gravity

(1): He does not feel gravity, even though it bends his path (as measured by you), because gravity is not a force in general relativity. In fact by Einstein's equivalence principle, if he has no proper acceleration then he will no different from floating freely in empty space.

(2): He feels the acceleration that he is experiencing thanks to his engines (his engines are constantly changing his speed, which means he is deviating from a geodesic path). If he thinks he is actually on the surface of the Earth, he will attribute this acceleration to the fact that he is deviating from a geodesic path (which falls inwards towards the Earth's centre) by being still on the surface. Proper acceleration and deviation from a geodesic are the same thing (they both describe the same physical effect).

Now, if we accept this geometric view, It seems like acceleration cannot really exist in spacetime: every object keeps following a geodesic.

Although most inanimate objects in space have no proper acceleration, it is still possible to accelerate in space, for example by firing rocket engines or using a light sail. Also, the strong magnetic fields around some planets (as well as neutron stars) cause forces on charged particle in the vicinity.

So acceleration can definitely still exist.

When some external observer thinks that an object is accelerating, he should say instead that it is following a "curved geodesic"

Below I'll use 'accelerating' to mean something that has nonzero proper acceleration.

This is incorrect. A geodesic is the equivalent of a 'straight line' in curved space, and is defined as the path of a non-accelerating observer. Geodesics are 'straight' if the spacetime is flat, and can be 'curved' if the spacetime is curved, and the geodesic happens to not follow a 'straight line' in the coordinates that you choose to use.

If an observer is not accelerating, he will always follow a geodesic path (which could be curved) through spacetime. If he is accelerating, he will not follow a geodesic, but he may or may not follow a curved path.

Also see the wikipedia page on proper acceleration

$\endgroup$
2
  • $\begingroup$ "If I understand your meaning correctly, he is standing still from his point of view if he is not experiencing any proper acceleration" - I think he means an observer in your spaceship A. That observer can consider himself standing still inside a gravitational field, as if he is standing still on the surface of a planet, that is, with proper acceleration. $\endgroup$
    – fishinear
    Apr 21, 2021 at 12:42
  • $\begingroup$ @fishinear I have edited my answer to cover that scenario as well $\endgroup$ Apr 21, 2021 at 14:55
0
$\begingroup$

1)
In general relativity you extend the concept of a straight line as a path that parallel transports its own tangent vector. If we define the tangent vector to a path $x^\mu (\lambda)$ as $dx^\mu / d\lambda$, the condition it is parallel transported is
$(D / d\lambda) (dx^\mu / d\lambda) = (dx^\nu / d\lambda) \nabla_\nu (dx^\mu / d\lambda) = 0$
where:
$D / d\lambda$ directional covariant derivative
$\nabla_\mu$ covariant derivative
$\lambda$ affine parameter

If $x^\mu (\lambda)$ is a timelike path, we can define $\lambda$ as the proper time $\tau$ and the tangent vector as the four-velocity $U^\mu = dx^\mu / d\tau$. Therefore the directional covariant derivative is the four-acceleration $A^\mu = (D / d\tau) U^\mu$.

As per definition of geodesic the four-accelation is zero. A free-falling object (geodesic) does not measure any acceleration.

2)
Instead an object at rest in a curved manifold, i.e. deviating from a geodesic, experiences an acceleration. For instance, an object on the surface of the earth measures the deviation from a geodesic as the Newtonian force of gravity.

3)
Maybe the question refers to the deviation between two geodesics. In a curved manifold two nearby geodesics accelerate from each other and the acceleration $A^\mu$ between them is
$A^\mu = (D^2 / d\lambda^2) S^\mu = R^\mu_{\nu \rho \sigma} T^\nu T^\rho S^\sigma$
where:
$S^\mu$ deviation vector
$T^\mu$ tangent vector
$R^\mu_{\nu \rho \sigma}$ Riemann tensor

Note: in a curved manifold you may have both geodesic and non-geodesic trajectories. In the latter you have for instance an object at rest or a spacecraft boosted by its engines.

$\endgroup$
0
$\begingroup$

The main issue here is a very unclear statement of the question. I will quote it and reply.

Let's say that, from my point of view, another observer is accelerating.

ok; this means that you have defined some measure of distance and time, and by this measure adopted by you, the location of the other observer has a non-zero second-rate of change.

Now, from his point of view, he is standing still:

Yes he can always adopt that way of considering his own situation if he likes.

all he feels is an overall fictitious force of gravity, which is just a consequence of the the spacetime curvature at that point (caused by stress-energy tensor at the same point).

This is totally confused. I think you have in mind an observer subject to no force other than gravity (e.g. no electromagnetic force). In this case he does not feel a force of gravity: he does not feel any force at all. The only way I can make sense of your statement is as a muddled way of saying that the acceleration which you say the other observer has might be attributed by you to a force of gravity on him.

Now, if we accept this geometric view, It seems like acceleration cannot really exist in spacetime: every object keeps following a geodesic. When some external observer thinks that an object is accelerating, he should say instead that it is following a "curved geodesic".

The word "acceleration", like the word "velocity" can only be defined as a relative concept. Someone can have non-zero acceleration with respect to one thing and zero acceleration with respect to another. I think what you are trying to say is that a bunch of things close to one another, all in free fall, will have zero acceleration relative to each other. (Here the term "free fall" means "no force is acting except gravity"). This is the sense in which there is "no acceleration" under gravity; it means nearby geodesics do not rapidly curve away or towards one another. Therefore the different parts of a small body all fall in unison without trying to approach or recede from one another; it follows that no internal forces are needed to keep them at the same separations from one another. But the distance from one object to another some distance away can have a second derivative: think of two rocks dropped from a tall building, for example. The distance between the rocks is constant, but the distance from the rocks to the ground changes with non-zero acceleration.

To understand general relativity, I advise that you keep referring back to everyday observations such as this one in order to keep track of what the geometric statements mean.

$\endgroup$
1
  • $\begingroup$ I think you misunderstand the question. With "from his point of view ... he feels an overall fictitious force of gravity " - he means from the point of view of an observer that is in a rocket that is accelerating. That observer will definitely feel a fictitious force of gravity: his feet are pressed to the floor of the rocket, and when he lets go of a ball, it will fall to the floor. According to the equivalent principle, he cannot locally distinguish that force due to his own acceleration from a force of gravity. $\endgroup$
    – fishinear
    Apr 21, 2021 at 12:37
0
$\begingroup$

Now, from his point of view, he is standing still: all he feels is an overall fictitious force of gravity, which is just a consequence of the the spacetime curvature at that point (caused by stress-energy tensor at the same point).

A body is following a geodesic trajectory exactly if that gravitational fictitious force is not present. We are not following geodesics in our everyday life on the surface of Earth for example.

There is a fictitious force of gravity in an accelerated frame, as a rocket keeping an artificial $g$ by letting the engines working (while there is fuel). So, it is not following a geodesic.

By the way, there is no space-time curvature in that rocket case. The Riemann tensor is zero for that frame.

Even when there is space-time curvature (the space around Earth where satelites orbit for example), the stress energy tensor is zero. It is not zero inside the planet.

$\endgroup$
0
$\begingroup$

In GR, we need to define the notion of acceleration in order to say that a particle is unaccelerated. This gives the dynamics. Geometrically speaking, this means the particle follows a geodesic.

Note that we can always define the velocity of a particle moving on some curve, but to define the acceleration we require a connection. The connection used in GR is the Levi-Civita connection.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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