Why can't Gravitational Accelerations vanish everywhere?

[Nugatory]

In ordinary language, it might be argued that an accelerometer will measure the sum of your change in velocity and gravitational acceleration and it can't distinguish between the two.

I don't understand that. Let's consider a few different cases, and for each one answer the question "What does the accelerometer read, and what is the 'change in velocity' and 'gravitational acceleration' that we are summing to get that reading?":
1) I am sitting in my chair typing this.
2) I am in freefall after jumping off a tall building and before I hit the ground.
3) I am weightless in Earth orbit.
4) I in a spaceship firing its engines in such a way as to produce an acceleration of 1g inside the ship.

 

[PeterDonis]

In Newtonian gravity, the field within the hollow is (somewhat surprisingly, to me at least) uniform and parallel to the symmetry axis

Can you give more details or a reference to how this is calculated?

 

[Ibix]

Can you give more details or a reference to how this is calculated?

Sure. I think it only works for a sphere of uniform density, which I see I didn't point out.

The Newtonian gravitational force on a mass $m$ due to some mass of uniform density $\rho$ with an arbitrary volume $V$ is\vec{F}=\int_V-\frac{G\rho m dV}{r^3}\vec{r}where $\vec{r}$ is the vector from the elementary volume $dV$ to the mass $m$, and $r$ is its magnitude. In the case that the volume is solid except for one hollow we can write\int_V - \frac{G\rho m dV}{r^3}\vec{r}=\int_{V_S} - \frac{G\rho m dV}{r^3}\vec{r} - \int_{V_H} - \frac{G\rho m dV}{r^3}\vec{r}where $V_S$ is a solid volume (i.e. $V$ with the hollow filled in) and $V_H$ is the solid volume that would exactly fill the hollow. We are interested in the case where the mass $m$ is inside the hollow - inside both $V_S$ and $V_H$. Assuming that $V_S$ is a sphere centered at the origin and $V_H$ is a sphere centered at $(x,y,z)=(0,0,z_0)$ then the shell theorem let's you write down the two integrals easily:\begin{eqnarray*}\vec{F}&=&\int_{V_S} - \frac{G\rho m dV}{r^3}\vec{r} - \int_{V_H} - \frac{G\rho m dV}{r^3}\vec{r}\\ &=& -\frac{4}{3}\pi G\rho m \left( \begin{array}{c}x\\y\\z\end{array} \right) + \frac{4}{3}\pi G\rho m \left( \begin{array}{c}x\\y\\z-z_0\end{array} \right)\end{eqnarray*}Obviously the position dependence cancels out so the force is uniform, directed parallel to the z axis, and proportional to $-z_0$. In the case $z_0=0$ it reduces to zero gravity inside a spherical shell, which is a classic result.

I've seen a reference for this calculation done for electrostatics rather than gravity - obviously the maths is equivalent. I can probably track it down if you like.

 

[PeterDonis]

I think it only works for a sphere of uniform density

Yes, I see, if $\rho$ is constant then the dependence of the force on position is linear in Newtonian gravity, so you can subtract out the force due to the hollow. But if $\rho$ is not constant, then the dependence of the force on position is no longer linear, so just subtracting out the force due to the hollow doesn't work.

Also, if we use the GR equations, the dependence of the force on position is not linear even in the uniform density case, so the conclusion of a uniform field inside the hollow no longer holds even for that case.

 

[David Lewis]

To Nugatory:
Sitting in chair = Accelerometer will read g (9.81 m/s^2 acceleration straight up).
Free fall = Readout will initially display zero, and then increase until it shows g
when you reach terminal velocity, assuming accelerometer is oriented vertically.
Weightless in orbit = Readout will display zero.
Accelerating spaceship = Readout will display the component of
the vector sum of gravitational acceleration and acceleration caused by the engine
that is parallel with the accelerometer's measurement axis.

 

[Nugatory]

To Nugatory:
Sitting in chair = Accelerometer will read g (9.81 m/s^2 acceleration straight up).
Free fall = Readout will initially display zero, and then increase until it shows g
when you reach terminal velocity, assuming accelerometer is oriented vertically.
Weightless in orbit = Readout will display zero.
Accelerating spaceship = Readout will display the component of
the vector sum of gravitational acceleration and acceleration caused by the engine
that is parallel with the accelerometer's measurement axis.

Yes, I understand that. I'm asking how each of these values is the sum of something that that you're calling "gravitational acceleration" and something that you're calling "change of velocity" - that is, in each of these cases what two values are you adding to get these results?

 

[PeterDonis]

Accelerating spaceship = Readout will display the component of
the vector sum of gravitational acceleration and acceleration caused by the engine
that is parallel with the accelerometer's measurement axis.

You didn't give a number for this case. What actual number will the accelerometer read? (That's in addition to Nugatory's question, which is also a good one.)

 

[David Lewis]

I'm adding acceleration vectors. Call them a and g. In the following 2 examples, the magnitudes of a and g are equal.

Example 1: If the spaceship is accelerating horizontally near the Earth's surface at a=9.81 m/s^2, the readout will display 9.81 m/s^2 when the accelerometer is horizontal (i.e. parallel to the flight path) or vertical, and (square root of 2 * 9.81 m/s^2) if the accelerometer is oriented 45 degrees above horizontal.

Example 2: If the spaceship blasts off vertically and accelerates at a rate of a=9.81 m/s^2, the readout will display (a+g = 2 * 9.81 m/s^2) if the accelerometer is oriented vertically.

 

[PeterDonis]

In the following 2 examples

Neither of these are Nugatory's examples. What are a and g for those?

 

[Emilie.Jung]

@PeterDonis Pete, is it true that freely falling body have zero proper acceleration? (Your version of proper acceleration?)

 

[Ibix]

But if $\rho$ is not constant, then the dependence of the force on position is no longer linear, so just subtracting out the force due to the hollow doesn't work.

To be pedantic about your wording, it works in the sense that it gives the correct (Newtonian) answer, but at least one of the integrals is not spherically symmetric so does not have a simple form from which position dependence will cancel.

Also, if we use the GR equations, the dependence of the force on position is not linear even in the uniform density case, so the conclusion of a uniform field inside the hollow no longer holds even for that case.

Also, I understood that simply superposing solutions doesn't quite work in GR. Is that right?

 

[Nugatory]

@PeterDonis Pete, is it true that freely falling body have zero proper acceleration? (Your version of proper acceleration?)

A freely falling body has zero proper acceleration, and what you're calling his "version" is the only generally accepted definition of the term.

 

[A.T.]

In ordinary language, it might be argued that an accelerometer will measure the sum of your change in velocity and gravitational acceleration and it can't distinguish between the two.

You can define all kind of quantities, and derive all kind of relationships with them, where they cancel each-other. But the simple fact is:

The reading of an accelerometer is fully determined by the sum of non-gravitational interaction forces acting on it and it's mass.

That's a frame independent result. Change in velocity relative to some frame of reference or the local strength of the gravitational field do not come into this.

 

[PeterDonis]

Pete, is it true that freely falling body have zero proper acceleration? (Your version of proper acceleration?)

There is only one "version" of proper acceleration. Yes, a freely falling body has zero proper acceleration; that's the definition of "free fall".

 

[PeterDonis]

it works in the sense that it gives the correct (Newtonian) answer

Yes, what I meant is that the correct Newtonian answer if $\rho$ is not constant is not a uniform field.

I understood that simply superposing solutions doesn't quite work in GR. Is that right?

Yes.

 

[pervect]

In attempt to describe the consequences of the Equivalence Principle, this is almost said:
When there are gravitational accelerations present, as for example in the
gravitational field of the earth, the space cannot be the flat Minkowski space. Indeed,
in the Minkowski space we can have
$$\Gamma^{\lambda}_{\mu\nu}=0$$
everywhere. This should then be interpreted as meaning that the sum of the inertial
and the gravitational acceleration could be made equal to zero everywhere. This does,
however, not correspond to our experience about gravitational accelerations: When
gravitational accelerations exist, it is not possible to make them vanish everywhere.
We can only make them vanish at one point, or approximately in a small region, by the
use of an appropriate coordinate system. Therefore, when a gravitational field is
present, the space will be necessarily a curved Riemannian space. The gravitational
field will then appear as the expression of the fact that we are in a curved riemannian
space and no longer in the flat Minkowski space.

For the sentence in bold, what experience that tells us so about gravitational accelerations (that we can not make them vanish everywhere except in a small region)? How do we know that?

Well, if we don't try to go "beyond" the maths, the answer is pretty clear. I'm not sure if it will be helpful to the OP or not, but I thought I'd mention it anyway.

Assume $\Gamma^{\lambda}{}_{\mu\nu}$ is zero everywhere. And apply Einstein's field equations.

There's a well known relationship (see for instance https://en.wikipedia.org/w/index.php?title=List_of_formulas_in_Riemannian_geometry&oldid=699598160) that gives the Riemann curvature tensor from the Christoffel symbols $\Gamma$.

$$
R^\ell{}_{ijk}=
\frac{\partial}{\partial x^j} \Gamma^\ell{}_{ik}-\frac{\partial}{\partial x^k}\Gamma^\ell{}_{ij}
+\Gamma^\ell{}_{js}\Gamma_{ik}^s-\Gamma^\ell{}_{ks}\Gamma^s{}_{ij}$$

from which it also follows (same Wiki reference)

$$R_{ij}=R^\ell{}_{i\ell j}=g^{\ell m}R_{i\ell jm}=g^{\ell m}R_{\ell imj}
=\frac{\partial\Gamma^\ell{}_{ij}}{\partial x^\ell} - \frac{\partial\Gamma^\ell{}_{i\ell}}{\partial x^j} + \Gamma^m{}_{ij} \Gamma^\ell{}_{\ell m} - \Gamma^m{}_{i\ell}\Gamma^\ell{}_{jm}.$$

Evaluating this, we find that if the Christoffel symbols are zero, the partial derivatives of them are also zero, and thus the Riemann and the Ricci tensors are zero. The Ricci tensor being zero implies that the Einstein tensor is zero. Einstein's field equations imply that when the Ricci tensor is zero, the stress-energy tensor is zero.

Therfore, given Einstein's field equations, we can conclude that if the Christoffel symbols are zero, the Riemann, Ricci, Einstein, and stress-energy tensors are zero. Basically, zero Christoffel symbols implies no gravity which from the field equations implies no matter ( a zero stress-energy tensor).

This isn't really terribly surprising.

If you're not familiar with the math, Baez's "The Meaning of Einstein Equation", http://math.ucr.edu/home/baez/einstein/, is probably the most helpful introduction that may give a bit of physical significance to said equations.

 

[Jilang]

I thought that the test of General Relativity was that is resulted in Newtonian gravity if the field was weak and the velocities were low?

 

[PeterDonis]

I thought that the test of General Relativity was that is resulted in Newtonian gravity if the field was weak and the velocities were low?

Yes. But the meaning of "weak" and "low" depends on how accurate your measurements are. We can detect departures from Newtonian gravity in the Earth's field (for example, Gravity Probe B).

 

[Jilang]

Thanks, I can see that now. Wouldn't relativistic effects also apply to bodies accelerated by a force other than gravity too?

 

[PeterDonis]

Wouldn't relativistic effects also apply to bodies accelerated by a force other than gravity too?

Yes.

(Also, in GR gravity isn't a force, and doesn't accelerate anything--more precisely, it doesn't cause any proper acceleration. Non-gravitational forces do cause nonzero proper acceleration, and in GR, the term "force" is reserved for those.)

 

[Jilang]

Apologies for the sloppy wording, but I am guessing you know what I meant. So it is the strength of the field that causes the departure from the uniform Newtonian picture in the spherical cave?

 

[PeterDonis]

So it is the strength of the field that causes the departure from the uniform Newtonian picture in the spherical cave?

It depends on which case you're talking about. If you're talking about the (highly unrealistic but consistent with the laws of physics) case of constant density, where the Newtonian answer is a uniform field in the cave, yes, if the overall field is strong enough for GR effects to be significant, then those effects cause the field inside the cave to be non-uniform, because the dependence of the field on position is no longer linear.

If you're talking about the (much more realistic) case of non-constant density, then the field inside the cave is not uniform even in the Newtonian case, because if the density is not constant, the dependence of the field on position is not linear even in the Newtonian case.

 

[David Lewis]

Nugatory's Example 4:
I am in a spaceship firing its engines in such a way as to produce an acceleration of 1g inside the ship:

If the spaceship is in deep space then g = 0, and a = 9.81 m/s^2.
If the spaceship is near the Earth's surface then g = 9.81 m/s^2, and a = 0

a is the rate of change in velocity (dv/dt)
g is acceleration due to gravity (weight/mass)

 

[PeterDonis]

acceleration due to gravity (weight/mass)

Equating your "g" to "weight/mass" is not correct. Weight is a direct observable (for example, the reading on a scale you're standing on). An astronaut standing on a scale inside the spaceship in Nugatory's example 4 reads his normal weight on the scale, regardless of whether the ship is in deep space or near the Earth's surface. So the astronaut's weight/mass is 1 g, again regardless of whether the ship is in deep space or near the Earth's surface.

Given that you have agreed that an accelerometer cannot distinguish your "a" from your "g", as the above example illustrates, I am confused as to why you keep insisting on separating them.

 

[pervect]

If we model a sphere-with-a-cavity as a constant density sphere of positive mass, containing another constand density sphere of matter of negative mass (that creates the hole because the negative mass cancels out the positive), using Newtonian gravity we can get a classical expression for the force. I'm rather skeptical that the force inside the cavity will be constant if the cavity is not at the center of the sphere, though.

 

[David Lewis]

PeterDonis wrote: "...the astronaut's weight/mass is 1 g ...regardless of whether the ship is in deep space or near the Earth's surface."

David Lewis wrote: Weight in my formula is the force exerted on a body by gravity, so (under that definition) an astronaut's weight in deep space would be zero.

 

[PeterDonis]

Weight in my formula is the force exerted on a body by gravity

I would point out that, with this definition, you are using Newtonian physics, not GR. In GR, gravity is not a force and an object moving purely under the influence of gravity has zero weight, because it has zero proper acceleration. (Newtonian physics works as an approximation to GR for weak fields and slow motion, but it's only an approximation.)

I would also point out that your use of the term "weight" is nonstandard; objects in free-fall orbit about the Earth are standardly referred to as "weightless", even though gravity is exerting a force on them (on the Newtonian view). By your definition, such an object would have a "weight" equal to mg. (Btw, you haven't answered how you would assign a to this case, Nugatory's example 3.)

Further, the sign you are assigning to "weight", by your definition, is wrong. The force of gravity pulls things down (on the Newtonian view); but you are defining g as pointing in the upward direction. Gravity doesn't push things up, so your definition seems inconsistent.

 

[David Lewis]

Nugatory's Example 3: I am weightless in Earth orbit.

Say your altitude is 100 km:
a = -9.53 m/s^2
g = +9.53 m/s^2
weight = your mass x acceleration due to gravity (mg)
accelerometer reading = zero

a = rate of change of velocity (dv/dt)
g = weight/mass

 

[PeterDonis]

weight = your mass x acceleration due to gravity

But, once again, the sign is wrong. Gravity pulls things down; it doesn't push them up.

In fact, in Newtonian terms, the "acceleration due to gravity" is what you are calling a, not g. In Newtonian terms, in a frame in which the Earth is at rest, which is the frame you appear to be using, there is nothing that corresponds to your g. So what cancels out a to make the accelerometer reading zero?

 

[David Lewis]

You're correct, the sign is wrong, but to simplify calculations I arbitrarily assigned to all downward pointing acceleration vectors (such as the acceleration of a freely falling object) a negative sign. That forces me to assign to vector g a positive sign.