Why do things in free fall accelerate?

[PeterDonis]

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]

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.

Spoiler: Maxima 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?

 

[DrGreg]

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]

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]

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]

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.

 

[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]

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.