# Lagrangian Point in General Relativity

• B
D.S.Beyer
TL;DR Summary
If gravity is a pseudo force, what is going on at the Lagrangian points in GR terms?
Is there a relationship between the Lagrangian ‘hill diagram’ and the spacetime curvature embedment graphs?

The Lagrangian map shows effective potential, which deals with centrifugal force. As centrifugal force is a fictitious force (and gravity is as well), I would assume the underlying phenomena to be an aspect of curved spacetime.

Would a Lagrangian point, without any mass in it, have a slight spacetime embedment due to the planet / sun system?

My gut says no, but it would be cool if it did. Thoughts?

Side note : Can anyone point me to an embedment graph of a 2-body system? Ideally something real like the sun / Earth system, or Earth / moon system. I'd like to see curvature as it relates to actual astronomical distances, not just a cartoon of a rubber sheet with some balls.

Staff Emeritus
I think you'd want to consider at least two massive bodies to have an interesting Lagrangian problem.

If you have a massive body orbiting another massive body, I don't believe there is an analytical solution for the metric.
We can certainly do an approximate analysis of the orbit in this case, though - it's well known that the body inspirals, due to the existence of gravitational radiation. It's also a very tiny effect.

So, basically, the mathematical perfection of the problem as a stable points is spoiled, but practically I'd expect there to be very little difference in the behavior in solar system three body problems, where Newtonian formula are an excellent approximation. I'm not too sure what what might happen in the more interesting strong field cases. I haven't seen or done any analysis.

D.S.Beyer
I think you'd want to consider at least two massive bodies to have an interesting Lagrangian problem.

If you have a massive body orbiting another massive body, I don't believe there is an analytical solution for the metric.
We can certainly do an approximate analysis of the orbit in this case, though - it's well known that the body inspirals, due to the existence of gravitational radiation. It's also a very tiny effect.

So, basically, the mathematical perfection of the problem as a stable points is spoiled, but practically I'd expect there to be very little difference in the behavior in solar system three body problems, where Newtonian formula are an excellent approximation. I'm not too sure what what might happen in the more interesting strong field cases. I haven't seen or done any analysis.

Okay, maybe let's simplify this a little.
Can we rationalize the existence of an L1 'like' saddle point in spacetime curvature between two massive bodies?
Similar to a Roche Lobe, but in spacetime?

Mentor
Can we rationalize the existence of an L1 'like' saddle point in spacetime curvature between two massive bodies?

The "saddle point" is not in spacetime curvature. It is in the Newtonian potential. That's not the same thing.

Mentor
Summary:: If gravity is a pseudo force, what is going on at the Lagrangian points in GR terms?
If you use the same coordinates in both cases then there is not any important difference.

In the Newtonian approach you have the fictitious centrifugal and Coriolis forces and the real gravitational force. The biggest difference in GR is that the gravitational force is also fictitious.

Of course there is a minor difference in that the exact value of all of these fictitious forces are slightly different than the Newtonian version.

Last edited:
D.S.Beyer
The "saddle point" is not in spacetime curvature. It is in the Newtonian potential. That's not the same thing.

I understand, (or think I understand) that the models of Roche Lobes, Lagrange Points, and Hill Spheres are of Newtonian potential, and that the embedment diagrams of spacetime (rubber sheet models) are representing the lengths of paths through spacetime, 2D slices of curved 3D space.

What I enjoy about the Newtonian potential models is how they show the interaction of multiple bodies, which creates the interesting topologies of saddles between objects. I'm not sure I have ever seen a embedment diagram that consists of more than a single, spherical, non rotating body.

Maybe the first question is : Can an embedment diagram be made with more than one massive object?
Follow up : Would the topological features be similar to Newtonian potential models?

Mentor
he embedment diagrams of spacetime (rubber sheet models) are representing the lengths of paths through spacetime

No, they're not. They're representing space, not spacetime; and space only in a particular system of coordinates. They are not good tools to use if you want to understand spacetime.

D.S.Beyer
No, they're not. They're representing space, not spacetime; and space only in a particular system of coordinates. They are not good tools to use if you want to understand spacetime.

Sorry. I misspoke. You are correct, it's 'space' not 'spacetime'.

(as an aside : Do you think that answer (not mine) that I linked to on stackexchange does a good job of explaining it? I often refer people to it as a way to begin grappling with the rubber sheet analogy. I would love to know if you think it's a sound place to start.)

Mentor
Do you think that answer (not mine) that I linked to on stackexchange does a good job of explaining it?

It does a reasonable job of describing the "rubber sheet" as a visualization tool. It does, IMO, a terrible job of explaining the serious limitations of that visualization tool.

I would love to know if you think it's a sound place to start.)

I don't think the rubber sheet analogy is a good place to start at all. The "shape of space" in those particular coordinates is not a good thing to be focusing on.

Dale
If gravity is a pseudo force, what is going on at the Lagrangian points in GR terms?
Since you emphase gravity being a pseudo force in GR: Note that even in Newtonian physics, the computation of the Lagrangian Points is based on the pseudo Centrifugal force, and its potential. The potential of which the LPs are saddles and maxima, is the sum of Newtonian Gravity and Centrifugal force (pseudo force in the rotating common rest frame of the two masses)

Sorry. I misspoke. You are correct, it's 'space' not 'spacetime'.
It's not your fault, but a very common misunderstanding based on the misleading rubber sheet analogy. I explained this here:

- Rolling balls on a rubber sheet can be used as a qualitative analogy for the gravity well (gravitational potential). That's why it gives the correct qualitative result. But that has nothing to do with explaining General Relativity and curved space-time, because it applies equally to Newtonian Gravity.

- The indented rubber sheet can be used as a qualitative visualization of the space (not space-time) distortion in General Relativity (Flamm's paraboloid). But that has has nothing to do with explaining how masses attract each other in General Relativity, which requires including the time dimension (space-time distortion). Flamm's paraboloid represents a distortion of spatial distances between coordinates, and could just as well be shown with the funnel upwards, so the rolling balls would give a wrong result. Therefore rolling some balls on the curved surface representing Flamm's paraboloid makes no sense.

- The local intrinsic curvature of space-time you are asking about is primarily related to tidal-effects, or gravity gradient.

I often refer people to it as a way to begin grappling with the rubber sheet analogy.
You should not refer anyone to the rubber sheet analogy, as an explanation for gravitational attraction in GR. This is a better analogy:

Can anyone point me to an embedment graph of a 2-body system? I
Not one that explains the gravitational attraction for orbiting bodies. Since you need the time dimension, you can only show just one space dimension for curved space time (2D), because your non-curved embedding space is 3D.

You can do this for a radial fall:
https://www.physicsforums.com/threa...lly-at-rest-begin-to-fall.995946/post-6416452

Last edited:
Dale
Gold Member
2022 Award
Also in GR gravitation is not a "fictitious force" (I prefer the notion of "inertial force", but that's semantics). The distinction between "fictitious forces" and a gravitational field is that the former you can completely compensate by a change of the reference frame to a local inertial frame, which is not possible for the latter, where you always have tidal forces also in the local inertial frame.

Abhishek11235 and alantheastronomer
Staff Emeritus
Also in GR gravitation is not a "fictitious force" (I prefer the notion of "inertial force", but that's semantics). The distinction between "fictitious forces" and a gravitational field is that the former you can completely compensate by a change of the reference frame to a local inertial frame, which is not possible for the latter, where you always have tidal forces also in the local inertial frame.

How do you regard the effect, usually called a force, that one feels in an accelerating elevator? The effect that you'd measure by standing on a scale and taking it's reading while you are in the accelerating elevator?

It's quite common in the popular literature, I think, to call this effect a force at the lay-level. The next observation to make is to point out that it's not a real force. We seem to be arguing about what we want to call this effect now. But hopefully we can get everyone to agree that this effect is not an actual force, at least, even if we don't quite agree on exactly what we want to call it.

Eventually, we might also agree about how this effect, which happens in flat space-time, is related to gravity, which happens in curved space-time, too. After a lot more discussion :).

Going back to the problem of the Roche lobe, I'm not sure how one would set up the problem in coordinate-free terms to do a full GR analysis. Probably what one would do is pick some coordinates first, then make some further approximations and not use full GR. The PPN formalism comes to mind, it would both define some coordinates, and make some useful approximations. See for instance https://en.wikipedia.org/w/index.php?title=Parameterized_post-Newtonian_formalism&oldid=976721689 as a general guide to what PPN formalism is about.

The end result would be a whole lot of work, and I would expect that it wouldn't generate a lot of insight, probably not even any experimentally observable effects under most circumstances.

vanhees71
Mentor
Also in GR gravitation is not a "fictitious force" (I prefer the notion of "inertial force", but that's semantics).
I tend to use the word “gravitation” to include the entire set of all phenomena modeled by GR, including tidal effects, time dilation, deflection of light, frame dragging, etc.
So I agree that “gravitation” is not an inertial force.

I use the term “gravity” specifically to refer to the part of gravitation that shows up as an inertial force. That arises from the Christoffel symbols, just like any other inertial force, and has all of the other characteristics of an inertial force. This is also the quantity most associated with the word “gravity” in Newtonian physics. So saying gravity is an inertial force is both valid and common.

How do you regard the effect, usually called a force, that one feels in an accelerating elevator? The effect that you'd measure by standing on a scale and taking it's reading while you are in the accelerating elevator?
I would call that the normal force. Inertial forces, including gravity, are not measurable.

vanhees71
D.S.Beyer
Thanks everyone for jumping into this, and getting into the weeds about visualizing spacetime.
Maybe we can approach this from another direction, with a more concrete example question.

If we put satellites at each of the Lagrangian points (of the sun/earth system), what adjustments must they make to their internal clocks to remain synchronous with clocks on earth?

There are already a few satellites at L1 (SOHO, ACE, WIND) and L2 (WMAP, Plank, Herschel). So this problem has, ostensibly, been solved. I am particularly interested in the temporal adjustments needed in the L4 and L5 spots.

This doesn’t exactly solve the visualization problem, but could offer some additional understanding to what is going on with the fabric of spacetime at those locations.

(Mods : Let me know if this should be a different thread.)

If we put satellites at each of the Lagrangian points (of the sun/earth system), what adjustments must they make to their internal clocks to remain synchronous with clocks on earth?
This is a good question. In the rotating common rest frame (of Sun, Earth and Satellite) there is no kinetic time dilation. So any difference in clock rates is due to gravitational time dilation, or the effective potential difference (gravitational + centrifugal potential).

There are already a few satellites at L1 (SOHO, ACE, WIND) and L2 (WMAP, Plank, Herschel). So this problem has, ostensibly, been solved. I am particularly interested in the temporal adjustments needed in the L4 and L5 spots.
I don't know the exact engineering solution used.

This doesn’t exactly solve the visualization problem, but could offer some additional understanding to what is going on with the fabric of spacetime at those locations.
The effective potential should have stationary points there, just like in the Newtonian treatment. But visualizing L4 and L5 is tricky, because you need 2 spatial dimensions.

For L1, L2, L3, it is simpler, because they are in line with both massive bodies, so you need only 1 spatial dimension (in the rotating frame).

Here is the space-propertime diagram and a geodesic freefall worldline for a single massive body in an non rotating frame:

The red path is the geodesic world-line of a free falling object, that oscillates through a tunnel through a spherical mass. Note that the geodesic always deviates towards the "more stretched" proper time, or towards greater gravitational time dilation. Gravitational time dilation has an extreme point at the center of the mass (gradient is zero), so there is no gravity there (but the maximal gravitational time dilation).

Here is a rough sketch of such a space-propertime diagram for two massive body in the rotating common rest frame (it's supposed to be a surface of revolution around the shown space axis, like above):

Note that since geodesics deviate towards the "fatter parts" (greater time dilation / lower potential), the shown L points are unstable in the radial direction. This is also true based on the Newtonian potential (below). You actually need the Coriolis force to explain why any of them are semi-stable:

https://www.math.arizona.edu/~gabitov/teaching/141/math_485/Final_Report/Lagrange_Final_Report.pdf

Last edited:
D.S.Beyer
D.S.Beyer
Here is a rough sketch of such a space-propertime diagram for two massive body in the rotating common rest frame (it's supposed to be a surface of revolution around the shown space axis, like above):

View attachment 272668

Note that since geodesics deviate towards the "fatter parts" (greater time dilation / lower potential), the shown L points are unstable in the radial direction. This is also true based on the Newtonian potential (below). You actually need the Coriolis force to explain why any of them are semi-stable:

This is, quite possibly, the most instructive visual I've seen online in many years.
It wonderfully grounds these 'space proper-time diagrams', which are easily one of the most abstract visuals that come up in discussions like this.

Let me see if I'm reading this correctly.
Disregarding the time dilation from centrifugal potential for a minute... the L points have the fastest clocks, then the earth, then the sun. And, based on the Newtonian potential maps, I would guess that the L4 and L5 points would have even faster clocks than anything else in the system.

This is, quite possibly, the most instructive visual I've seen online in many years. It wonderfully grounds these 'space proper-time diagrams', which are easily one of the most abstract visuals that come up in discussions like this.
Thanks. A intuitive explanation of these diagrams is in the later chapters of this book:

Disregarding the time dilation from centrifugal potential for a minute... the L points have the fastest clocks, then the earth, then the sun. And, based on the Newtonian potential maps, I would guess that the L4 and L5 points would have even faster clocks than anything else in the system.
This is true for all clocks that are at rest relative to the Sun-Earth and the L-points (rotating with them).

Not sure why you want to disregard the centrifugal potential. The Newtonian potential map shown above also includes the centrifugal potential. That why it falls off towards negative infinity, when the distance from the barycenter towards infinity. The pure mass-attraction potential approaches a constant value at infinite distance.

Also keep in mind that both diagrams show a rotating frame. For the space-propertime diagram above the propertime "streching" shows the clock rate of clocks at rest in the rotating frame, along the Sun-Earth-line. In the non-rotating frame, clocks along this rotating line would approach light-speed, if far enough from the rotation center, and thus have infinite kinetic time dilation. In the rotating frame, where these clocks are at rest, this is accounted by the centrifugal potential. That's why the propertime blows up to infinity at both ends (proper time rate goes to zero, potential goes to negative infinity just like in the Newtonian picture)

Last edited:
Mentor
I have calculated the time dilation factors in a Sun-centered inertial frame, and here are the results I get.

For the Earth and Sun, I am using the time dilation factor at the center of the object, assuming it to be a sphere of uniform density. Of course this is not really accurate, but it's a reasonable approximation for the time dilation factor. For a spherical object of radius ##R## and total mass ##M##, the time dilation factor at the center, according to GR, solely due to the object's own gravity, is:

$$\frac{3}{2} \sqrt{1 - \frac{2 G M}{c^2 R}} - \frac{1}{2}$$

where ##G## is Newton's gravitational constant and ##c## is the speed of light.

We then combine this (what "combine" means will be specified further below) with the time dilation factor due to the other object's gravity and the time dilation factor due to velocity (if any).

The time dilation factor due to the Earth's gravity, for objects other than the Earth itself, is

$$\sqrt{1 - \frac{2 G M_E}{c^2 r}}$$

where ##M_E## is the mass of the Earth and ##r## is the object's distance from the Earth's center. A similar equation applies for the time dilation factor due to the sun's gravity, with the sun's mass ##M_S## in place of the Earth's mass and the object's distance from the center of the Sun used in place of the distance from the center of the Earth.

The time dilation factor due to a velocity ##v## in the given inertial frame is

$$\sqrt{1 - \frac{v^2}{c^2}}$$

When multiple time dilation factors apply to the same object, we combine them by multiplying them together.

For the distances involved, we make use of the formulas from this Wikipedia article:

https://en.wikipedia.org/wiki/Lagrange_point

If we use ##R_H## for the radius of the Earth's Hill sphere, as given in the article, then for L1 and L2, we have ##r_E = R_H## and ##r_S = D_E \pm R_H## for the distances from the Earth and Sun, where ##D_E## is the Earth-Sun distance in meters (I used ##D_E = 1.49 \times 10^{11}##). For L3, we have ##r_E = 2 * D_E + R_3## and ##r_S = D_E + R_3##, where ##R_3## is the distance ##r## given in the L3 section of the Wikipedia article. For L4 and L5, we have ##r_E = r_S = D_E##.

We also note that, since all of the Lagrange point objects have the same orbital period as the Earth about the Sun, their velocities are all given by ##v = \omega r_S##, where ##r_S## is the distance from the Sun and ##\omega = 2 \pi / Y##, where ##Y## is the length of the Earth's year in seconds (I used ##Y = 3.1 \times 10^7##).

Putting all the above together, I come up with the following time dilation factors:

$$V_S = 1 - 3.2002517723617174 \times 10^{-6}$$

$$V_E = 1 - 1.6082657650073884 \times 10^{-8}$$

$$V_1 = 1 - 1.5038067791017795 \times 10^{-8}$$

$$V_2 = 1 - 1.5041770162760315 \times 10^{-8}$$

$$V_3 = 1 - 1.5035442113564557 \times 10^{-8}$$

$$V_4 = V_5 = 1 - 1.5035456768508482 \times 10^{-8}$$

If we just take these raw numbers, we have for the clock rates Sun < Earth < L2 < L1 < L4,L5 < L3. However, these results are really only valid to two or three significant figures, so the Lagrange point values are really indistinguishable at this accuracy and we can really only say Sun < Earth < Lagrange points.

D.S.Beyer
Gold Member
2022 Award
How do you regard the effect, usually called a force, that one feels in an accelerating elevator? The effect that you'd measure by standing on a scale and taking it's reading while you are in the accelerating elevator?
You mean that the scale shows a larger/smaller weight when accelerating upwards (downwards)? Seen from my restframe it's an inertial force or equivalently part of the gravitational force (in GR it's the same, i.e., it cannot be distinguished whether you take it as inertial or gravitational foce as far as local physics is concerned, that's the GR version of the equivalence principle).

My point of view is the following: I think it's not very important, how you name these "forces" or rather "interactions". The only important thing to remember is that in each spacetime point there's a locally inertial frame of reference, which is realized by a Fermi-Walker transported (i.e., non-rotating) tetrad of a free-falling point-like observer. Whether or not you have "purely inertial" or "real gravitaty" is determined by the curvature tensor, i.e., if it is 0 there's no gravitational field present, and this is a frame-independent and thus physical definition.

Gold Member
2022 Award
I tend to use the word “gravitation” to include the entire set of all phenomena modeled by GR, including tidal effects, time dilation, deflection of light, frame dragging, etc.
So I agree that “gravitation” is not an inertial force.

I use the term “gravity” specifically to refer to the part of gravitation that shows up as an inertial force. That arises from the Christoffel symbols, just like any other inertial force, and has all of the other characteristics of an inertial force. This is also the quantity most associated with the word “gravity” in Newtonian physics. So saying gravity is an inertial force is both valid and common.

I would call that the normal force. Inertial forces, including gravity, are not measurable.
I think that's a good terminology, but one always has to define it, because as this discussion shows, it's not so common to be as accurate even in university textbooks (let alone in the original research literature or, even worse, the popular-science literature).

Dale
Mentor
You mean that the scale shows a larger/smaller weight when accelerating upwards (downwards)? Seen from my restframe it's an inertial force or equivalently part of the gravitational force (in GR it's the same, i.e., it cannot be distinguished whether you take it as inertial or gravitational foce as far as local physics is concerned, that's the GR version of the equivalence principle).
Be careful. The reading on the scale is purely dependent on the real force (the normal force). The scale cannot detect inertial forces.

PeterDonis
Gold Member
2022 Award
As I said, it's semantics, and I guess we are in danger to get into endless (somehow useless) debates, but what do mean by "normal force" here?

Take a scale at rest on Earth. If I stand on it the gravitational force due to the presence of the Earth acts on me and there's an equal and opposite force (of electromagnetic nature on the fundamental level) which compensates this gravitational force. The reaction of the spring in the scale to this interaction is that it is shortened somwhat and that length difference is shown by the scale.

If now the accelerator is accelerated upwards this is due to some external force and this has to be compensated additionally by the scale's spring and that's why it reads "more weight". From my point of view in the frame accelerated relative to the Earth's rest frame it's an additional inertial force, which within GR however is indistinguishable from a gravitational force.

Mentor
the gravitational force due to the presence of the Earth

There is no such force in relativity. The only force acting on you according to relativity is the force of the Earth's substance pushing up on you. That's the force that the scale is indicating. And if you get in a rocket with a thrust of more than 1 g, so it can propel you upward, the force of the rocket's engine is what the scale indicates. There is never any "gravitational force" at all in relativity.

Mentor
As I said, it's semantics, and I guess we are in danger to get into endless (somehow useless) debates, but what do mean by "normal force" here?
It is not semantics. This is experimentally testable. Only real forces are measurable, inertial forces are not measurable.

The normal force is the contact force between the scale and the object being weighed.

Take a scale at rest on Earth. If I stand on it the gravitational force due to the presence of the Earth acts on me and there's an equal and opposite force (of electromagnetic nature on the fundamental level) which compensates this gravitational force. The reaction of the spring in the scale to this interaction is that it is shortened somwhat and that length difference is shown by the scale.
The only thing that the scale measures is the real contact force between your feet and the scale. It does not measure the inertial force. Consider the same measurement in the frame of a nearby inertial (free falling) observer. The inertial force is changed to zero, the real force is unchanged, and the scale reading is unchanged.

If now the accelerator is accelerated upwards this is due to some external force and this has to be compensated additionally by the scale's spring and that's why it reads "more weight". From my point of view in the frame accelerated relative to the Earth's rest frame it's an additional inertial force, which within GR however is indistinguishable from a gravitational force.
Same as above. A nearby inertial observer will have no inertial force but the same real force and the same scale reading. Therefore the scale does not detect the inertial force, it detects the real force only.

We have discussed this before, do you not recall?

If now the accelerator is accelerated upwards this is due to some external force and this has to be compensated additionally by the scale's spring and that's why it reads "more weight". From my point of view in the frame accelerated relative to the Earth's rest frame it's an additional inertial force, which within GR however is indistinguishable from a gravitational force.
The additional contact force by the scale, and the additional inertial force have opposite directions and act differently on the body (foot soles vs. whole body volume). They are not the same thing.

Staff Emeritus
Let me see if I'm reading this correctly.
Disregarding the time dilation from centrifugal potential for a minute... the L points have the fastest clocks, then the earth, then the sun. And, based on the Newtonian potential maps, I would guess that the L4 and L5 points would have even faster clocks than anything else in the system.

There's an interesting graphic in a Nasa domain webpage on the Lagrange points, https://map.gsfc.nasa.gov/ContentMedia/lagrange.pdf that graphs the generalized potential with a contour diagram, though it includes the centrifugal potential.

The graph doesn't / can't include the velocity dependent terms in the generalized potential, however.

It's a PDF file, so I can't easily just past the graphic here, it's figure 2. The graphic shows that L4 and L5 are peaks in the effective potential, a little bit "higher" than L3. L1 and L2 are troughs, which are actually saddle points, in the potential.

The reference also has the analytic form of the generalized potential.

If it weren't for the velocity dependent terms in the generalized potential, L4 and L5 would be unstable, like a marble on the top of a hill.

The approach to determine stability, though is more involved than looking at the graphic. From said website

Nasa said:
Usually it is enough to look at the shape of the effective potential and see if the equilibrium points occur at hills, valleys, or saddles. However, this simple criterion fails when we have a velocity dependent potential. Instead, we must perform a linear stability analysis about each Lagrange point. This entails linearising the equations of motion about each equilibrium solution and solving for small departues from equilibrium.

When you add in the velocity-dependent terms, due to the coriolis force, L4 and L5 become stable for high enough mass ratios between the primary and secondary.

The link between time dilation and the potential to Newtonian order is very direct. The square of the time dilation , modulo possible factors of G and c depending on one's unit conventions, is that square of the time dilation is equal to ##|g_{00}|##, as ##d\tau^2 = g_{00} \, dt^2## for a body at rest. For a single central mass, ##|g_{00}| = 1 - 2U = 1 + 2 \Phi##, where U > 0 and ##\Phi < 0##, i.e. U and ##\Phi## are the same quantity with different signs.

If you include the usual factors of G and c, for a central body, ##U(r) = \frac{GM}{c^2 r}##. To Newtonian order, you can simply add the potentials from multiple bodies together.

Finding the approximate effects of GR would involve using a post-Newtonian approximation. But generally, I'd expect these corrections would be negligible.

Keeping the factors of c and G, we can write U(r) in a non-rotating frame with a single central mass as ##\frac{GM}{rc^2}##. And to Newtonian order, we can just add the potentials together from the various bodies.

Staff Emeritus
Oh, something i wanted to add. Knowing the time dilation to Newtonian order, we can use the principle of maximal aging - or it's big brother, the principle of extremal aging- to determine the equations of motion to Newtonian order.

E.F Taylor's book "Exploring Black Holes" goes into this approach in much more detail. A second edition is available on the Author's website for free , nowadays. http://www.eftaylor.com/exploringblackholes/

Much of the introductory work on said principle is in the first chapter.

PeterDonis
There's an interesting graphic in a Nasa domain webpage on the Lagrange points, https://map.gsfc.nasa.gov/ContentMedia/lagrange.pdf that graphs the generalized potential with a contour diagram, though it includes the centrifugal potential.

The graph doesn't / can't include the velocity dependent terms in the generalized potential, however.

It's a PDF file, so I can't easily just past the graphic here, it's figure 2. The graphic shows that L4 and L5 are peaks in the effective potential, a little bit "higher" than L3. L1 and L2 are troughs, which are actually saddle points, in the potential.

The reference also has the analytic form of the generalized potential.

If it weren't for the velocity dependent terms in the generalized potential, L4 and L5 would be unstable, like a marble on the top of a hill.
And for the Sun-Earth the saddle points L1-3 are actually unstable, despite the velocity dependent terms.

In short, Lagrange points are complicated enough in Newtonian physics. Expecting intuitive visualizations of them based on General Relativity is maybe asking too much.

Last edited:
Gold Member
2022 Award
There is no such force in relativity. The only force acting on you according to relativity is the force of the Earth's substance pushing up on you. That's the force that the scale is indicating. And if you get in a rocket with a thrust of more than 1 g, so it can propel you upward, the force of the rocket's engine is what the scale indicates. There is never any "gravitational force" at all in relativity.
That's one possible point of view. For me the gravitational interaction is an interaction as any other. From your point of view all phenomena we call "gravitation" were just inertial forces in non-inertial reference frames in Minkowski space. That's definitely not the case according to GR in the geometrical interpretation, where the presence of any energy-momentum-stress distribution leads to a spacetime with curvature. Gravitation is only equivalent to inertial forces in a local sense!

In the geometrical interpretation of the gravitational field, the gravitation of the Earth leads to a non-flat spacetime and the gravitational interaction between a test body and the Earth can only locally be compensated by changing to a free-falling reference frame. Of course, the "gravity I feel" is in fact the reaction force of the Earth on me, which is of electromagnetic nature.

As I said before, all this is just semantics and doesn't lead to much deeper understanding of GR. I think, however, it's important to stress that gravitation is not simply inertial forces only but that this is a local concept as is the equivalence principle (in its various weak and strong forms).

Gold Member
2022 Award
The additional contact force by the scale, and the additional inertial force have opposite directions and act differently on the body (foot soles vs. whole body volume). They are not the same thing.
I've not said that they are the same thing.

Gold Member
2022 Award
It is not semantics. This is experimentally testable. Only real forces are measurable, inertial forces are not measurable.

The normal force is the contact force between the scale and the object being weighed.

The only thing that the scale measures is the real contact force between your feet and the scale. It does not measure the inertial force. Consider the same measurement in the frame of a nearby inertial (free falling) observer. The inertial force is changed to zero, the real force is unchanged, and the scale reading is unchanged.

Same as above. A nearby inertial observer will have no inertial force but the same real force and the same scale reading. Therefore the scale does not detect the inertial force, it detects the real force only.

We have discussed this before, do you not recall?
Yes, you can locally transform the gravitational force away due to the equivalence principle. The only point I want to make is that one has to stress the word "locally" and that you cannot globally transform away a "true" gravitational field.

Mentor
As I said before, all this is just semantics
Again, this is not just semantics. The equivalence principle has clear experimental consequences. A scale measures the real contact force, not inertial forces and not (local) gravity.

The only point I want to make is that one has to stress the word “locally" and that you cannot globally transform away a "true" gravitational field.
Agreed. So make that point. The “only semantics” point is incorrect.

vanhees71
Gold Member
2022 Award
But then you should admit that it's wrong to say gravitational forces (I prefer the term gravitational interaction though since forces are something I want to restrict to the use in Newtonian physics only) are purely fictitious. This is indeed what's only local as is the equivalence principle. A lot of unnecessary discussion has occurred in the literature only because the local meaning of the equivalence principle hasn't been considered, e.g., the question whether a free falling charged body radiates or whether it doesn't.

In this sense it's right that it's not only semantics, because the claim that gravitational interactions are only like fictitious forces is inaccurate at best if not plain wrong.

Mentor
But then you should admit that it's wrong to say gravitational forces (I prefer the term gravitational interaction though since forces are something I want to restrict to the use in Newtonian physics only) are purely fictitious. This is indeed what's only local as is the equivalence principle.
Fully agreed, I have no objection to admitting that the inertial force designation is only local.

Do you similarly admit that the local designation of gravity as an inertial force is not purely semantic but has clear experimental consequences as described above?

vanhees71