Why can't Gravitational Accelerations vanish everywhere?

[beyondthemaths]

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?

 

[jedishrfu]

I imagine a simple pendulum bob would show that as it would point toward the center of the Earth hence in different directions in different places.

 

[WannabeNewton]

Take two particles separated by some small distance and drop them to the floor; if the height of the fall is large enough you will see them converge towards each other. This means there is tidal gravity i.e. the gravitational field is not uniform hence we cannot transform it away everywhere by going to a freely falling frame. We can only do so locally in a freely falling frame that exists for a small period of time in a small region of space.

 

[beyondthemaths]

I see, I still have one question then. In the question above, it is also mentioned that " This should then be interpreted as meaning that the sum of the inertial
and the gravitational acceleration could be made equal to zero everywhere. " Why sum? How was this interpreted? I mean interpretation is something that follows from what preceded, I couldn't relate this sentence I just quoted to what preceded.

 

[WannabeNewton]

https://www.physicsforums.com/threads/what-is-the-nature-of-gravity.756377/#post-4770244

 

[Emilie.Jung]

@WannabeNewton I am also interested in the question asked, which is why sum? I didn't understand how the link you provided settles this?

 

[PeterDonis]

In the question above, it is also mentioned that " This should then be interpreted as meaning that the sum of the inertial
and the gravitational acceleration could be made equal to zero everywhere. " Why sum?

The interpretation offered is not correct. "Inertial acceleration" and "gravitational acceleration" are not the same thing and you can't add them together or use one to offset the other. More precisely, you can add them mathematically, but the sum has no physical meaning.

What the OP calls "inertial acceleration" is acceleration you actually measure with an accelerometer. It is a direct observable. (The more usual name for it in GR is "proper acceleration", or just "acceleration", since it is the only kind of acceleration that is an invariant, independent of coordinates.)

What the OP calls "gravitational acceleration" is a coordinate-dependent phenomenon; you can make it vanish, at least locally, by choosing appropriate coordinates. (It is a special case of what, in GR, is called "coordinate acceleration".) So even if the "sum" of inertial and gravitational acceleration, calculated mathematically without regard for the physical meaning of the two, happens to be zero in one system of coordinates, you can make it change just by changing to another system of coordinates. But changing coordinates doesn't change any physics, so this "sum" can't have any physical meaning.

A better way of interpreting the fact noted in the OP would be to say that, in flat Minkowski spacetime, we can find a system of coordinates such that objects in free fall, i.e., with zero proper acceleration, have zero coordinate acceleration everywhere; i.e., you can make "gravitational acceleration" vanish everywhere, globally, by choosing appropriate coordinates. In a curved spacetime, we can't do this globally; we can only do it locally, in a small patch of spacetime. In any global system of coordinates in a curved spacetime, freely falling objects--objects with zero proper acceleration--will have nonzero coordinate acceleration in at least some regions.

 

[Emilie.Jung]

@PeterDonis thanks very much for your answer. I understand your argument very nicely if I considered e.g. when you said " you can make "gravitational acceleration" vanish everywhere, globally, by choosing appropriate coordinates." is correct. However, I can't tell you that I know what you mean by "appropriate" coordinates here. I would appreciate an example or something if this is not difficult to give.

 

[PeterDonis]

I can't tell you that I know what you mean by "appropriate" coordinates here.

"Appropriate" coordinates are just standard inertial coordinates in flat spacetime: $t$, $x$, $y$, $z$, with the metric $\eta_{\mu \nu} = - dt^2 + dx^2 + dy^2 + dz^2$.

 

[David Lewis]

PeterDonis wrote: " What the OP calls inertial acceleration is acceleration you actually measure with an accelerometer. "

David Lewis wrote: I suppose an accelerometer will also measure gravitational acceleration, and that it's unable to distinguish between inertial and gravitational acceleration.

 

[Ibix]

PeterDonis wrote: " What the OP calls "inertial acceleration" is acceleration you actually measure with an accelerometer. "

David Lewis wrote: I suppose an accelerometer will also measure gravitational acceleration, and that it's unable to distinguish between inertial and gravitational acceleration.

No - this is the point Peter is making. If you are in a space-station, or dropped out of a hovering rocket over an airless planet, you are in free-fall. Your accelerometer (a bathroom scale, essentially) will read zero. You will be able to detect tidal effects with sufficiently precise instrumentation, but no acceleration.

However, if you are sitting in a chair on Earth or in the rocket under power then your accelerometer will have a non-zero reading. This is because the Earth or the rocket floor is accelerating you out of your free-fall path.

 

[David Lewis]

In free fall (near the Earth's surface) your accelerometer will initially display the sum of inertial acceleration (9.81 m/s^2) and gravitational acceleration (-9.81 m/s^2).

 

[Ibix]

No. It will display your proper acceleration - zero.

What force is causing the "inertial acceleration" you are talking about?

 

[PeterDonis]

I suppose an accelerometer will also measure gravitational acceleration

You suppose incorrectly. You need to think more carefully about the definitions of the terms you are using.

In free fall (near the Earth's surface) your accelerometer will initially display the sum of inertial acceleration (9.81 m/s^2) and gravitational acceleration (-9.81 m/s^2).

Why do you think "inertial acceleration" will be +9.81 m/s^2? What definition of "inertial acceleration" are you using? On the definition I gave, "inertial acceleration" in free fall is always zero; that includes the case you are describing. You are correct that "gravitational acceleration" in your scenario is - 9.81 m/s^2, but as I've already pointed out, "gravitational acceleration" is just a coordinate effect which can be made to vanish, locally, by an appropriate choice of coordinates; it's not what an accelerometer measures.

 

[Emilie.Jung]

@PeterDonis , I ve read some place that when we are in Minkowski space we have$\frac{d^2}{d\tau^2}x(\tau)=0$ and then we we do general coordinate transformation, we get $\frac{d^2}{d\tau^2}x^{\mu}+\Gamma^{\mu}_{\nu\lambda}\frac{dx^{ni}}{d\tau}\frac{dx^{\lambda}}{d\tau}=0$, the author calls the second term "inertial acceleration" does this contradict with what you said?

 

[PeterDonis]

I ve read some place

What place?

 

[PeterDonis]

Book called lectures on GR by Papapetrou

Can you give a chapter/page reference?

 

[Emilie.Jung]

See this @PeterDonis https://books.google.com.lb/books?i...generally by the Christoffel symbols.&f=false

 

[PeterDonis]

See this

Ok, it looks like he is using the term "inertial acceleration" here to refer to what I am calling "coordinate acceleration", and what the OP appears to refer to as "gravitational acceleration". His terminology is confusing, however, since he uses the term "gravitational acceleration" to refer to the same thing. (The OP appears to be quoting from the same Papapetrou book, just a little further on from the passage you quote.) Basically what he seems to be saying is that this thing--the term involving $\Gamma^{\mu}_{\nu \lambda}$ in his equation 18.3--can be called either "inertial acceleration" or "gravitational acceleration", or a combination of both, depending on how we're looking at it.

This provides a good example of why ordinary language is the wrong tool to use when discussing physics. When you look at the math, it's obvious that the term involving $\Gamma$ is coordinate-dependent. That's true whether spacetime is flat or curved, and whether you think of the term as "inertial acceleration" or "gravitational acceleration" or whatever. This coordinate dependence is the reason why the usual term for this thing, if you have to use ordinary language to describe it, is "coordinate acceleration".

But coordinate-dependent things are the wrong things to focus on when trying to understand the actual physics. What Papapetrou appears to be doing in the quoted passages is to try to match up ordinary Newtonian intuitions about non-inertial frames and "gravitational fields" with the correct relativistic math as best he can. IMO it would be better just to say straight up that Newtonian intuitions are wrong and that trying to understand GR in terms of them is not going to work well.

Furthermore, when you look at the math, you can also see that the term involving $\Gamma$, in and of itself, tells you nothing about the actual physical observable in question: the reading on an accelerometer carried by the object, i.e., the proper acceleration (which is what I called "inertial acceleration" in earlier posts--but it's clear now that that's not how Papapetrou is using that term, so I'll avoid it from now on). The proper acceleration is given by the sum of both terms in equation 18.3 of Papapetrou, i.e., it is given by

$$
\frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu \lambda} \frac{dx^\nu}{ds} \frac{dx^\lambda}{ds}
$$

If this sum is zero, the object is in free fall, feeling zero acceleration. If this sum is nonzero, the object is feeling acceleration, the magnitude of which is given by the sum. And this is true regardless of the coordinates we choose.

 

[Emilie.Jung]

@PeterDonis I am confused. I have to say. Then what happens to your answer here:

A better way of interpreting the fact noted in the OP would be to say that, in flat Minkowski spacetime, we can find a system of coordinates such that objects in free fall, i.e., with zero proper acceleration, have zero coordinate acceleration everywhere; i.e., you can make "gravitational acceleration" vanish everywhere, globally, by choosing appropriate coordinates. In a curved spacetime, we can't do this globally; we can only do it locally, in a small patch of spacetime. In any global system of coordinates in a curved spacetime, freely falling objects--objects with zero proper acceleration--will have nonzero coordinate acceleration in at least some regions.

How can this answer (that answered my question about why sum and OP's question what experience does this tell us) be translated after knowing that the book we're using means different things?
Also, notice there that the author says gravitational=inertial acceleration after using Principle of Equivalence, so would this mean that what you said previously can still hold? Now that I am lost a little, I would appreciate if you could translate your previous answers and I will read them as if I am reading those for the first time. Thanks Pete.

 

[PeterDonis]

How can this answer (that answered my question about why sum and OP's question what experience does this tell us) be translated after knowing that the book we're using means different things?

Nothing about what I posted changes. To translate what I said into math, remember that I used "coordinate acceleration" to refer to the term involving $\Gamma$ in the math--i.e., the term involving $\Gamma$ in Papapetrou's equation 18.3. In flat Minkowski spacetime, we can find coordinates (standard inertial coordinates) in which $\Gamma$ vanishes everywhere; and if $\Gamma$ vanishes everywhere, then of course the term involving $\Gamma$ in Papapetrou's equation 18.3 must vanish everywhere.

Papapetrou is using the terms "inertial acceleration", "gravitational acceleration", and "sum of inertial and gravitational acceleration", all to refer to the term involving $\Gamma$. But he leaves out the most important fact, which is that that term is coordinate-dependent. My main point has been to emphasize that, and to emphasize that coordinate-dependent things are the wrong things to focus on if you want to understand the physics.

the author says gravitational=inertial acceleration

And I said that was confusing terminology which is best ignored. You need to look at the physics and the math that describes the physics, not at ordinary language terminology.

I would appreciate if you could translate your previous answers

I'm not sure at this point what I should be answering, since one of my key points has been that the terminology in which the question is phrased in the first place is bad terminology. What piece of physics--not terminology--do you want clarified? Remember that none of the terms "inertial acceleration", "gravitational acceleration", or "sum of inertial and gravitational acceleration", as Papapetrou uses them, describe any physics; they only describe a coordinate-dependent quantity (the term involving $\Gamma$). Can you phrase the question in a way that does not use any of those terms?

 

[Emilie.Jung]

I am afraid I can't but use the terminology. Maybe my question is simply, is it okay to call the term involving $\Gamma$ an inertial acceleration and still use your previous answers to understand things? If not, what is the exact reason that drove the author to say that this term involving $\Gamma$ is an inertial acceleration to start with? @PeterDonis

 

[PeterDonis]

I am afraid I can't but use the terminology.

I think you're giving up too easily. See further comments below.

is it okay to call the term involving $\Gamma$ an inertial acceleration and still use your previous answers to understand things?

Not really, since I used "inertial acceleration" to mean something else in earlier posts (I used it to mean proper acceleration).

what is the exact reason that drove the author to say that this term involving $\Gamma$ is an inertial acceleration to start with?

I couldn't say. As I said, I find his terminology confusing, and I can't speculate on why he chose it.

Let me try to help by restating at least some of what the OP appears to be getting at, without using any of the confusing terminology. Consider two groups of astronauts:

Group 1: The astronauts are all floating freely in empty space, with no significant gravity anywhere, and are all at rest relative to each other at some instant of time.

Group 2: The astronauts are all floating freely in space above the Earth, at different altitudes and along different radial lines from the center of the Earth, and are all at rest relative to each other at some instant of time.

Obviously the key difference between the two groups is that group 2 is in the presence of significant gravity while group 1 isn't. What difference does that make on how they move and what they experience? What does this difference tell us about our ability to find coordinates in which $\Gamma$ vanishes everywhere in the two cases? (Remember that all of the astronauts in both groups are in free fall, with zero proper acceleration, so Papapetrou's equation 18.3 applies to them exactly as it is written. I think this sort of comparison is what Papapetrou had in mind in the section of the book we've been referring to.)

 

[Jilang]

I am reminded of the underground spherical cave. So we drill down into a spherical body of uniform density a few kilometres and hollow out a nice spherical cave with a say a kilometre diameter. Hey presto - no tidal forces and a uniform field. (Not to be confused with a spherical cave at the centre of the body).

 

[pervect]

I am reminded of the underground spherical cave. So we drill down into a spherical body of uniform density a few kilometres and hollow out a nice spherical cave with a say a kilometre diameter. Hey presto - no tidal forces and a uniform field. (Not to be confused with a spherical cave at the centre of the body).

I don't understand why you think there would be no tidal forces. Using Newtonian gravity, there should be both net forces and tidal forces inside the spherical cave if the cave isn't at the center of the spherical body.

 

[Jilang]

It's quite easy to prove that the field is uniform, objects will fall in parallel trajectories. Consider first the field of the whole sphere then subtract the field of the missing part.

 

[PeterDonis]

It's quite easy to prove that the field is uniform

The field inside a spherical mass of uniform density is not uniform. It depends on the radial coordinate $r$, even though the density itself does not. This case is worked out explicitly in MTW.

 

[Ibix]

Jilang is referring to the field within a spherical hollow within a spherical body. In Newtonian gravity, the field within the hollow is (somewhat surprisingly, to me at least) uniform and parallel to the symmetry axis - so no tidal forces. As Jilang notes, this is easy enough to prove by calculating the field internal to a spherical object ($\propto-\vec{r}$) and subtracting off the field due to the mass removed in creating the hollow.

I've no idea if that's the case or not in a full GR treatment. My guess is "not", since my understanding is that you can't superpose the fields from two objects like that.

 

[Jilang]

I would be interested to learn whether the space would be flat or not in that scenario,

 

[David Lewis]

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.

 

[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.