B Why do things in free fall accelerate?

jbriggs444

Homework Helper
In stronger spacetime curvature you need more force to stay at the same altitude
This is the second time I've seen this picture presented in this thread. I believe it to be an incorrect over-simplification.

Spacetime curvature is essentially equivalent to tidal gravity -- the rate at which gravitational acceleration changes with position. If you want to compare an acceleration here against a state of rest over there, what matters is, roughly speaking, the integral of curvature over a path.

The local acceleration of gravity is independent of the local spacetime curvature.

My mistaken impression has been corrected.

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Nugatory

Mentor
If you want to compare an acceleration here against a state of rest over there, what matters is, roughly speaking, the integral of curvature over a path.
But here we are comparing the deviation between the geodesic worldline of an object in freefall and the non-geodesic worldline of an object hovering with constant Schwarzschild $r$ coordinate at the point where the two worldlines intersect. That's a measure of the force required to hold the hovering object on its worldline, and it is indeed greater in a more strongly curved spacetime.

Ibix

The local acceleration of gravity is independent of the local spacetime curvature.
The proper acceleration of an observer hovering at constant Schwarzschild $r$ (with $G=c=1$) is $$a=\frac{M}{r^2\sqrt{1-2M/r}}$$The Kretchmann curvature invariant, which is what I was thinking of as "the strength of the curvature", is $K=R_{ijkl}R^{ijkl}=48M^2/r^6$. You can eliminate $r$ between those two equations and see that the acceleration needed to hover behaves as $K^{1/3}$ for small $K$ and as $K^{1/4}$ for large $K$. So the proper acceleration needed to hover increases in more strongly curved spacetime.

Intuitively, something like this must be true. Else, why would you feel heavier closer to a gravitating mass?

FactChecker

Gold Member
2018 Award
GR-curved space-time implies that the geodesics are not the same as they would be without GR. Without GR, we would think that an object held stationary at a constant altitude was following a geodesic, $A_{nonGR}$. But we know that the true GR-geodesic, $B_{GR}$, curves away from that. The increasing separation between $A_{nonGR}$ and $B_{GR}$ appears as "acceleration" to one who does not take GR into account. So an object that is left to follow the true (GR) geodesic, $B_{GR}$, will appear to be accelerating. That fact requires no calculations at all. The details of how much "acceleration" there is and in which direction requires calculations.

Dale

Mentor
The local acceleration of gravity is independent of the local spacetime curvature.

My mistaken impression has been corrected.
I don’t know. I am more or less onboard with your “mistaken” impression. At each point on the earth the local spacetime is approximately flat and yet you are accelerating upwards. So locally I don’t think that curvature explains the acceleration. The acceleration is (IMO) explained by real forces. The floor pushes up on your feet, the ground pushes up on the floor, the bedrock pushes up on the ground ... Every part of the surface of the earth is accelerating upward due to unbalanced real forces acting on it.

What curvature explains is why every point on that surface can accelerate away from each other without the distance changing.

Z3kr0m

I was yesterday on presentation of Kip Thorne about Gravitional waves, and he said that the acceleration is due to the changes of time flow (or gradient or something), so I started to not understand this phenomenon again. Can somebody explain me this please?

PeroK

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I was yesterday on presentation of Kip Thorne about Gravitional waves, and he said that the acceleration is due to the changes of time flow (or gradient or something), so I started to not understand this phenomenon again. Can somebody explain me this please?
You might have to ask Kip what he meant.

A.T.

... he said that the acceleration is due to the changes of time flow (or gradient or something),...
He probably meant that the gradient of the gravitational time dilation is related to the strength of the gravitational field.

See the images here:

This might also help:

A.T.

So the proper acceleration needed to hover increases in more strongly curved spacetime.
Is this true in general, for example inside a spherical mass? Just because "stronger attractive gravity" coincides with "stronger tidal gravity" for the exterior Schwarzschild solution, doesn't mean it's a general relation or even causation.

PeterDonis

Mentor
locally I don’t think that curvature explains the acceleration. The acceleration is (IMO) explained by real forces.
The acceleration itself is explained by real forces, yes; curvature doesn't give you any proper acceleration.

But the magnitude of the proper acceleration that is required to follow a stationary worldline does depend on the curvature; for example, @Ibix gave a relationship between the proper acceleration of a stationary worldline in Schwarzschild spacetime and the Kretzschmann curvature invariant.

PeterDonis

Mentor
So the proper acceleration needed to hover increases in more strongly curved spacetime.
As @A.T. has indicated, this statement is too strong; it is true for the particular curved spacetime you give as an example (with the caveat that the concept of "hovering" only makes sense outside the horizon), but not for a general curved spacetime. The obvious counterexample, as @A.T. noted, is the interior of a spherically symmetric mass; the proper acceleration goes to zero at the center, but the curvature is not zero there, and is in fact (I believe) larger there than anywhere else.

pervect

Staff Emeritus
You can get it directly from the metric. If you are at rest relative to the Earth, then you can calculate your "proper" acceleration, which requires a force to sustain.
The formula may not make a lot of sense without a fair amount of background knowledge of tensors, but I'll try to give a summary. It may not make any sense at all unless one knows what a tensor is. Informally, a rank 0 tensor is just a number, a scalar, that all observers agree on, a rank 1 tensor is a vector, and a rank 2 tensor is rather like a matrix. There are some conditions on how tensors transform that I won't get into.

Tensors have components, which are numbers. Components are influenced by the coordinates chosen, the tensor itself is regarded as an entity that represents a physical phenomenon, independent of any choice of coordinates. This is slightly oversimplifed, one actually needs to choose the coordinates and additionally a set of basis vectors, but I will assume that one is using what is called a "coordinate basis", in which case knowing the coordinates also specifies the basis vectors.

A rank 0 tensor has 1 component, a rank 1 tensor has 4 components, and a rank 2 tensor has 16 components.

General relativity and special relativity have significantly different paradigms. In General relativity, the acceleration of a body in free fall is always zero. In Newtonian mechanics, the acceleration of a freely falling body is nonzero, and is due to "gravity", which is regarded as a force.

The trajectory of a body can be specified by knowing the path that the body takes through space-time. One can specify the path by writing $x^i(\tau)$, where $x^i$ are the coordinates of the body, and $\tau$ is proper time.

$x^i$ represents the position of the body, both in space and in time.

It's important to know here what proper time, $\tau$ is. That's the sort of time that a clock (a wristwatch) on the body measures, as opposed to the value of the time coordinate, which in GR has the status of a label that one attaches to an event that is more or less arbitrary. Proper time is a rank 0 tensor, because it's a number that everyone agrees on, regardless of coordinates. The time coordinate of a body is not a tensor, because it depends on the coordinate choices, and tensors are geometric entities that are independent of coordinate choices.

Then the four-velocity $u^i$ is a vector, whose components are given by $u^i = \partial x^i / \partial \tau$, the partial derivative of the position $x^i$ with respect to proper time.

The acceleration 4-vector can be calculated from $u^i$ by another sort of derivative, the covariant derivative. This would be, in the tensor notation of General relativity

$$a^b = u^a \nabla_a u^b$$

The symbol $\nabla_a$ represents taking the covariant derivative. The covariant derivative of the 4-velocity $u^a$ is a rank 2 tensor, with 16 components. $u^a$ has four components, it's covariant derivative has 16. Contracting (another tensor operation) of this rank 2 tensor with the 4-velocity gives one back a 4-vector.

This is very terse, I haven't really explained what the covariant derivative about, but it's rather like a gradient operation, as you might guess from the symobl.

So, the result of this formula gives the acceleration 4-vector, $a^b$. The magnitude of this four-vector gives you the number that represents the magnitude of the proper acceleration can be computed from the 4-acceleratio and the metric tensor

$$A^2 = g_{ab} a^a a^b$$

here A is the magnitude of the proper acceleration, so a^2 is the squared magnitude of the acceleration, while $a^i$ is the accleration 4-vector we calculated previously.

Because of the use of tensors, A is a rank 0 tensor, which means that it's defined in a manner that's independent of coordinate choices. It may take some knowledge of tensors to appreciate fully how that is possible.

Ibix

As @A.T. has indicated, this statement is too strong; it is true for the particular curved spacetime you give as an example (with the caveat that the concept of "hovering" only makes sense outside the horizon), but not for a general curved spacetime.
Agreed - in a non-static spacetime, "hovering" doesn't even make sense, I think.
The obvious counterexample, as @A.T. noted, is the interior of a spherically symmetric mass; the proper acceleration goes to zero at the center, but the curvature is not zero there, and is in fact (I believe) larger there than anywhere else.
You and @A.T. are correct as usual. I considered the case of a constant density sphere of radius $R$. The Kretschmann scalar is discontinuous at the surface, jumping from $12R_s^2/R^6$ to $15R_s^2/R^6$ presumably because the stress-energy tensor is discontinuous here, then rises smoothly to the centre of the sphere. The general expression for $K(r)$ inside the sphere is messy and not particularly interesting - Maxima code is in the spoiler tags if you want to see.
Code:
load(ctensor);
/* Lazy way of defining constant density interior Schwarzschild metric - */
/* set up for exterior Schwarzschild and edit. Note that this is the     */
/* metric for a sphere of radius rg and Schwarzschild radius Rs.         */
ct_coordsys(exteriorschwarzschild);
lg[1,1]:-(sqrt(1-Rs*r^2/rg^3)-3*sqrt(1-Rs/rg))^2/4;
lg[2,2]:(1-Rs*r^2/rg^3)^(-1);

/* Derive the Kretschmann scalar */
cmetric(false);
christof(false);
riemann(false);
lriemann(false);
uriemann(false);
rinvariant();

/* What is the curvature invariant at the surface? */
ksurface:ratsimp(substitute(rg,r,kinvariant));

/* What is the curvature invariant at the centre? */
substitute(0,r,ratsimp(kinvariant/ksurface));
substitute(alpha*rg,Rs,%);
kcentre:ratsimp(%);

/* What is the curvature through the interior? Define */
/* alpha=Rs/rg and rho=r/rg for convenience.          */
substitute(alpha*rg,Rs,ratsimp(kinvariant/ksurface));
substitute(rho*rg,r,%);
krho:ratsimp(%);
plot2d([substitute(0.1,alpha,krho),
substitute(0.02,alpha,krho),
substitute(0.01,alpha,krho)],
[rho,0,1],
[legend,"R=10Rs",
"R=50Rs",
"R=100Rs"],
[xlabel,"r/R"],
[ylabel,"K(r)/K(R)"]);
That said, I find myself thinking that the proper acceleration of a hovering observer in a static spacetime ought to have some relationship to the local curvature. I think we can define "hovering" in a coordinate-free way as a worldline that is the integral curve of the timelike Killing vector. And the proper acceleration is some kind of measure of how hard you have to work to follow that path, which feels to me like it ought to be related to how non-flat spacetime is where the hovering observer is.

But writing a generic static spherically symmetric metric, $g_{\mu\nu}=\mathrm{diag}(g_{tt}(r),g_{rr}(r),r^2,r^2\sin^2\theta)$, leads to a simple expression for the modulus of the proper acceleration$$a=\frac{dg_{tt}/dr}{2\sqrt{g_{rr}}g_{tt}}$$but a much more complicated one for the Kretchmann scalar which includes $g_{rr}$, $g_{tt}$, and derivatives of both. So apparently my intuition is wrong - perhaps because the acceleration seems to depend on the spatial variation of the metric coefficients related to the timelike direction while curvature takes into account the full Riemann tensor?

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DrGreg

Gold Member
But even in flat spacetime Rindler observers "hover" (from their point of view) above a Rindler horizon, with ever-increasing proper acceleration (diverging to $\infty$) the closer they are to the horizon. They are each following the flow of a Killing field but the spacetime curvature is zero.

PeterDonis

Mentor
writing a generic static spherically symmetric metric, $g_{\mu\nu}=\mathrm{diag}(g_{tt}(r),g_{rr}(r),r^2,r^2\sin^2\theta)$, leads to a simple expression for the modulus of the proper acceleration
$$a=\frac{dg_{rr}/dr}{2\sqrt{g_{rr}}g_{tt}}$$
Shouldn't it be $d g_{tt} / dr$ in the numerator?

$$a = \sqrt{g_{rr}} a^r = \sqrt{g_{rr}} \Gamma^r{}_{tt} u^t u^t = \sqrt{g_{rr}} \frac{1}{2} g^{rr} \frac{d g_{tt}}{d r} \frac{1}{g_{tt}} = \frac{1}{2 \sqrt{g_{rr}} g_{tt}} \frac{d g_{tt}}{d r}$$

Ibix

Shouldn't it be $d g_{tt} / dr$ in the numerator?
Yes. Thanks - typo now corrected above.

PeterDonis

Mentor
typo now corrected above
I think you corrected the wrong thing--you put $\sqrt{g_{tt}} g_{tt}$ in the denominator, instead of $d g_{tt} / dr$ in the numerator.

Ibix

I think you corrected the wrong thing--you put $\sqrt{g_{tt}} g_{tt}$ in the denominator, instead of $d g_{tt} / dr$ in the numerator.
I'd just spotted that. Re-corrected.

Karl Coryat

Back to the original discussion: Surely, one reason why people have trouble with GR basics is confusion over the word "accelerate." Most people think of acceleration as a perceptible change in velocity, so it's counterintuitive to imagine that a person sitting in a chair on Earth is accelerating upward. My question: Is it correct to say that ordinary velocity change is a special case of acceleration (i.e., that acceleration doesn't necessarily mean a change in meters-per-second), or is it more correct to say that a person sitting in a chair is experiencing a velocity change — like a person on a merry-go-round — only that the seated person's acceleration is a change in their velocity through spacetime? If that's the case, are there units for expressing one's velocity through spacetime at any given instant?

PeroK

Homework Helper
Gold Member
2018 Award
Back to the original discussion: Surely, one reason why people have trouble with GR basics is confusion over the word "accelerate." Most people think of acceleration as a perceptible change in velocity, so it's counterintuitive to imagine that a person sitting in a chair on Earth is accelerating upward. My question: Is it correct to say that ordinary velocity change is a special case of acceleration (i.e., that acceleration doesn't necessarily mean a change in meters-per-second), or is it more correct to say that a person sitting in a chair is experiencing a velocity change — like a person on a merry-go-round — only that the seated person's acceleration is a change in their velocity through spacetime? If that's the case, are there units for expressing one's velocity through spacetime at any given instant?
There is an analogy with circular motion where the acceleration is toward the center of the circle but there is no change in the distance to the center.

Also, velocity is frame dependent. If you accelerate for a time in flat spacetime then the net result is not a change in absolute velocity, but a change in your inertial reference frame.

In the case of sitting in a chair, therefore, the upward acceleration does not imply an absolute motion in that direction. But it does imply a continuous change of local inertial reference frame.

Ibix

Is it correct to say that ordinary velocity change is a special case of acceleration (i.e., that acceleration doesn't necessarily mean a change in meters-per-second), or is it more correct to say that a person sitting in a chair is experiencing a velocity change
The problem is defining acceleration as a perceptible change in velocity without saying velocity with respect to what. For a free-falling observer (who feels no force) the person in the chair is clearly accelerating upwards (and, in fact, feels an upwards force from the chair).

Newton regards the person in the chair as "at rest" (or moving inertially, more precisely). Einstein regards the free-faller as moving inertially. So the person in the chair is genuinely accelerating - they can feel it, and inertial observers can see it.

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A.T.

Surely, one reason why people have trouble with GR basics is confusion over the word "accelerate."
The only around this is to always be explicit, if one means "coordinate acceleration" (change in velocity) or "proper acceleration" (measured by accelerometer).

FactChecker

Gold Member
2018 Award
Back to the original discussion: Surely, one reason why people have trouble with GR basics is confusion over the word "accelerate." Most people think of acceleration as a perceptible change in velocity, so it's counterintuitive to imagine that a person sitting in a chair on Earth is accelerating upward.
If a person wants to understand GR basics, he may just have to learn something about geodesics in spacetime and understand that acceleration is a deviation from a geodesic in spacetime.

"Why do things in free fall accelerate?"

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