# General relativity and tidal forces

1. Sep 18, 2012

### altergnostic

Tides on earth are described with newton's theory of gravitation. Relativistic effects on tides theoretically become measurable on very strong gravitational fields, possibly becoming twice as strong as tides predicted by newtonian gravity: http://adsabs.harvard.edu/abs/1983ApJ...264..620N

Tides are presumably outcomes of gravitational forces. Einstein ditched forces and the concept of inertia in GR (http://archive.org/stream/TheBornEinsteinLetters/Born-TheBornEinsteinLetters_djvu.txt). So how is GR used to calculate tidal forces? If different parts of the body travel different geodesics, this would cause the body to tear apart over time. How can tides be described with the geometry of space-time?

2. Sep 18, 2012

### Staff: Mentor

If the forces holding the body together are strong enough, they will accelerate the different parts of the body off their geodesics and onto non-geodesic worldlines that stay close enough that the body doesn't tear apart. The center of mass of the object follows a geodesic and the other parts of the body experience fictitious forces that tend to pull the body apart and are resisted by whatever forces hold the body together.

These fictitious forces are tides.

3. Sep 18, 2012

### pervect

Staff Emeritus
Tidal forces, when suitably defined, can be identified as being components of the Riemann curvature tensor.

Under most circumstances, taking the tidal force as one would measure it via Newtonian means ( a couple of accelerometers separated by a rigid rod) is an excellent approximation to (one of the) geometric definitions, which is related to the apparent relative acceleration of nearby geodesics which are initially parallel.

In fact, you write earlier (this is a very good insight)

The point is that when you measure the forces needed to hold a rigid body together, to keep it rigid, you are indirectly measuring "how fast" the geodesics would expand (accelerate away from each other) if said restoring forces did not exist.

MTW's textbook "Gravitation", and a number of other textbooks, take this approach, though MTW is perhaps the textbook which invites the reader to take it most seriously.

The full Riemann curvature can, given a local description of time (a frame of reference, for instance, more formally a timelike congruence of worldlines) be decomposed by the Bel Decomposition http://en.wikipedia.org/w/index.php?title=Bel_decomposition&oldid=512613685 into three parts. One part, called the electrogravitic tensor, describes static gravity, and givenby the above "tidal forces", so the Newtonian tidal tensor can be pretty much directly be linked to the electrogravitic part of the Riemann tensor.

Another part, called the magnetogravitic tensor, described frame-dragging effects (which affect moving bodies, but don't directly affect static bodies). A third part, the topogravitic tensor, describes spatial curvature.

The Bel decomposition, unfortunately, is usually given short shrift in textbooks, so it may be hard to find a formal treatment I was introduced to it rather informally here on PF, for instance.

4. Sep 19, 2012

### altergnostic

All right, so I think I understand the basis of how GR creates tides, even though I have a minor issue with fictitious forces counteracted by real forces in this particular case, but that's a subject for another thread, I don't want to get into this here.
My problem with this analysis is that different parts of the body are not only trying to travel different geodesics, but also, in order to stay rigid, the outer parts of the rotating orbiter must have a faster tangential component than the inner parts. This is a bigger problem when you consider a body in tidal lock, such as the moon. Different geodesics would tend to rip the body apart radially, but different velocities would tend to shear te body in the line of orbit. In all gravitational theories, the tangential component of the velocity can't be caused by the gravitational field and is a constant. Newton called it the body's innate velocity.

Think of it this way: a body is at a constant linear velocity, so all parts of the body travel at the same linear speed. Then it is captured by a gravitational field. It starts to orbit that planet and the linear velocity is not changed, but the body is accelerated into a curved motion (or continues to have a constant velocity in curved spacetime). The tangential component must still be the same for all parts of the body since gravity imparts no (fictitious) forces tangentially. This would tend to cause shearing, especially in a body in tidal lock. If it doesn't start to shear, presumably the tangential velocities are different on different parts of the body, with the far side faster then the near side (this would be true for all bodies, not only the ones in tidal lock), but how can that be? What forces act on the body to change the tangential (innate) velocities if gravity has no way to do so neither with Newton nor with GR? Are tides capable of changing tangential velocities in either GR or Newton's theory?

5. Sep 19, 2012

### pervect

Staff Emeritus
While "centrifugal" forces, i.e. forces due to rotation, do contribute to the strain on a rigid bar, they do not contribute to the Riemann curvature tensor, which is ultimately based on how fast geodesics separate (or converge). The "force-on-a-bar" idea is very useful, but it can only be used if/when the bar isn't rotating.

So in your orbiter example, either you'd need to imagine that your spacecraft was not rotating (in which case in 1/2 an orbit the outer side would be the inner side, assuming no frame dragging effects), or if your space-craft is tide-locked, you'd have to manually subtract the forces due to its absolute rotation (once per orbit, again assuming no frame dragging) from the measured strain on the bar to get the tensor components.

The Electrogravitic component of the Riemann tensor must be traceless. The centrifugal forces on a rotating sphere are not traceless, this is one way you can tell if a system is rotating.

Accurate measurements of the gravity tensor, typically using rather exotic means such as superconductors and SQUID's for the detectors, are an expensive, but semi-routine, part of modern prospecting. Some interesting references are http://www.physics.umd.edu, http://www.bellgeo.com/tech/technology_theory_of_FTG.html [Broken], and http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA496707/GRE/NASA_SGG.pdf [Broken]. These papers describe some of the modern techniques that are actually used to measure the gravity tensor. The last has some discussion of the physics as well, though it's oriented mostly towards Newtonian gravity.

The Wiki article is also mildly helpful, http://en.wikipedia.org/w/index.php?title=Gravity_gradiometry&oldid=508813691, giving a list of some of the basic systems that have been implemented.

Last edited by a moderator: May 6, 2017
6. Sep 19, 2012

### altergnostic

I'm not sure i follow. Forces due to rotation cause strain, but can only be used if the bar is not rotating??????

If the orbiter does not rotate on its axis, as in your first example, everything is fine, because there's no variation in tangential speeds. But if the body is in tidal lock, then how does the tangential velocities adjust to maintain the body in position? Or, if they don't adjust, how come there's no shearing?

7. Sep 19, 2012

### Staff: Mentor

Why do you say that? Sounds like an issue with geometry to me, like you think that different velocities mean different parts of the object are moving apart. What you're missing is that the object is rotating. Spin a pencil on your desk and you'll see that different parts travel at different speeds, but there is no shear.
A torque is produced if a body's tidal bulge is not aligned with the source of the bulge: http://en.wikipedia.org/wiki/Tidal_locking

8. Sep 19, 2012

### altergnostic

Sorry, I meant to say that equal tangential velocities would tend to cause shearing, not different velocities. I'm making a diagram to clarify.

9. Sep 19, 2012

### altergnostic

Ok, I'm having a hard time putting the problem into words so I drew a diagram that may help to clarify the whole thing:
http://www.pictureshoster.com/files/aix43ezq7zp99daihgzu.jpg
http://www.pictureshoster.com/files/aix43ezq7zp99daihgzu.jpg

If different parts of the body have different velocities, there's no shearing. But if all parts of the body have the same linear velocity, then a body in tidal lock should exhibit shearing.

If there's no shearing, it follows that different parts of the body have different tangential components. Suppose that a body is traveling in a straight line at a constant velocity. It passes near a second body and starts to orbit it. If nothing else happens, the body wouldn't start to rotate on it's own axis and in "1/2 an orbit the outer side would be the inner side" just like pervect said. For a body in tidal lock, something seems to affect linear velocities on the near and far side so that the far side orbits faster than the near side.

My question is what causes the changes in these tangential velocities. Or alternatively, why is there no shearing. Since gravity has no tangential component neither in Newton nor in GR, I'm lost.

At first I thought maybe it had something to do with the rigidity of the body and tidal forces. I thought that gravity would accelerate the near and far sides differently and force the body in tidal lock, accelerating the far side more than the near side, but this turned out to be a dead end.

I think the make up of the problem is clearer now with my diagram, and I believe my logic is consistent. A body in this situation must either shear or achieve different tangential velocities on the near and on the far sides. If this is the case, what is the cause of these changes?

Last edited: Sep 20, 2012
10. Sep 20, 2012

### D H

Staff Emeritus

You are looking in the wrong directions. First look at a spec of mass on the orbiting body at the point furthest from the central mass. The velocity of that spec of mass is a function of the radius of the orbiting body and the velocity of the orbiting body's center of mass. Now imagine what would happen if that orbiting body wasn't there; all you have is the spec of mass as an orbiting body. That free particle will follow a different path than would our spec of mass. In particular, it would move outward. The tidal force at this point is radially outward away from the central mass, not tangential. Similarly, the tidal force on the point closest to the central mass is radially inward, toward the central mass. In both cases, the tidal force is away from the center of mass of the orbiting body. There is also a lesser effect (about half as much) for a particle at the leading and trailing points on the orbiting body. A free particle at those leading and trailing points would follow different paths than will a particle fixed to the orbiting body, but now the tidal force is directed toward the center of mass of the orbiting body. The end result is that the tidal forces act to pull the object apart radially, squeeze it together tangentially.

Newtonian mechanics and general relativity agree on the above description so long as the central mass isn't particular massive or the distance to the central mass is sufficiently large. The reason for this agreement is that space-time is locally flat. "Locally" is a fairly large volume in these weak field circumstances, at least as far as physicists are concerned. (Mathematicians will disagree; local means infinitesimally small to them.) Make the central body massive enough or close enough and those weak field approximations become invalid. You'll start seeing effects that result from the curvature of space-time. Newtonian mechanics and relativity diverge at this point. Newtonian mechanics does not properly describe the extreme spaghettification that results from close proximity to extremely massive objects.

11. Sep 20, 2012

### pervect

Staff Emeritus
If you connect the "red x"s on the same point on a non-rotating body, you'll get an elliptical orbit more like the one I'll attach, the picture you drew isn't right.

If you go through the math (using just Newtonian theory), you'll get the results DW quoted.

In particular, if the radial Newtonian force is -GM/r^2, differentiating this with respect to r gives the well-known result for the radial tidal force 2GM/r^3. See also the wiki article, http://en.wikipedia.org/w/index.php?title=Tidal_tensor&oldid=332450104.

It takes more work to go through the math to get the compressive tidal forces, the results dW quotes are correct however.

The results for GR are formally similar to the Newtonian results in a local frame-field, if you replace the radial distance "r" with the radial coordinate "r" for the Schwarzschild metric. Local frame fields seem to confuse more readers than they should, the math to compute them is somewhat involved, but the end result is just the forces/fields/tensors that a local observer would measure with local clocks and local rulers.

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12. Sep 20, 2012

### altergnostic

I completely agree. But this does not address the problem. In the diagram, I hid the effects of tides, I want to focus on the tangential components of the velocities. I am aware that the specs would move apart, I actually said that explicitly in my first post.

What keeps the near and far side from moving apart, which was answered in the beginning of the post, are the forces holding the body together, so it is clear that tidal forces tend to elongate the body radially, and it is clear that neither tides nor gravity have mechanisms to act tangentially on the near and far sides to change these velocities.

Please be patient since my question has not been answered. Leave tidal forces out for a moment and consider this:

- A body Y moves in a straight line at a constant velocity V.
- All parts of the body have the same linear velocity V.
- This body starts to orbit a central mass M.
- M imparts a centripetal acceleration due to gravity on Y.
- Y feels no tangential forces from M, and no drag.
- Y achieves tidal lock: the far side has a faster tangential velocity than the inner side.

Now, if Y did not achieve tidal lock, and no forces other then gravity from M act on it, we may assume that there is no change in the tangential velocities and there would be no fixed near and far sides, the tides would travel, the body would appear to rotate as seen from M, the body would not be rotating in it's own axis locally (relative to it's own orbit) and everything is fine.

But if Y orbits in tidal lock, the far side orbits faster than the near side, and we must assume a change in the tangential velocities of each sides, namely, a positive acceleration on the far side and a negative one on the near side.

Gravity can't cause these tangential accelerations, so there's no way M causes it. The body Y also cannot impart forces on itself, so again, it cannot be the cause of these changes.

My question is: what causes the changes in the tangential velocities of the near and far sides?

13. Sep 20, 2012

### Staff: Mentor

The body Y cannot apply forces to itself, but one part of the body can apply forces to another part. Replace the body with a cloud of dust, such that each dust particle is connected to its neighbors with a spring. Now if two of the particles are on diverging geodesics, the spring between them will be stretched, applying a force to both particles that will accelerate them off their inertial path and onto other geodesics. Obviously this process can change the shape of the body/cloud; but it can also change the velocity of one part of the cloud relative to another, as in tidal locking.

14. Sep 20, 2012

### Staff: Mentor

Even though it acts only in a radial direction, gravity can produce tangential accelerations. Consider, for example, dropping a horizontally oriented bar towards the surface of the earth. Each end of the bar will experience a gravitational force pointing towards the center of the earth. These two vectors are not parallel; if I extend them far enough they will intersect at the center of the earth, so they are ever so slightly converging. Thus, both vectors have a small tangential component that is pushing the ends of the rod towards the middle and is resisted by the rigidity of the rod.

It's true that these forces are balanced so they cannot affect the center-of-mass movement of the rod - but that's true of tidal locking of a rotating body as well.

15. Sep 20, 2012

### altergnostic

That's a good point, but isn't that one of the reasons the body should resist changes on tangential velocities on the near and far sides? For example, if I try to push only the far side, these internal bonds (rigidity) would tend to transfer forces to the near side (and everywhere else). But the force is only transfered in this case, isn't it? That means that there must be an external force acting on the tangent in this case, and this force would be resisted by rigidity.

You mention tangential accelerations caused by gravity in a rigid rod falling horizontally towards the central mass. But in this case, the body is not in orbit, and if it was (even if slowly falling), the forces would be on the trailing and leading sides, wouldn't they? And this would only help the radial bulging anyway. These tangential components are not capable of altering tangential velocities on the near and far sides. A good analogy would be to hold the same rod vertically wrt earth and throw it straight towards the horizon (ignore drag). Would the far side of the rod start to speed up relative to the near side? Of course not. But that would be a rod in tidal lock.

Since there's no tangential forces in gravity, I don't see where these changes in tangential velocities come from. The stretching you mentioned with the springs, caused by gravity, is only radial, it would cause radial stretching (tidal bulges). Exactly how can this radial force change the tangential velocities of the near and far sides?

Last edited: Sep 20, 2012
16. Sep 20, 2012

### Mentz114

It might be constructive to look at some actual figures. For a test particle in the Schwarzschild vacuum the proper acceleration felt by an observer in a circular path u is

$$\frac{du}{d\tau} = \frac{m\,{\cos\left( \theta\right) }^{2}}{{r}^{2}-2\,m\,r}\ dr -\frac{m\,\cos\left( \theta\right) \,\sin\left( \theta\right) }{r-3\,m}\ d\theta$$
If the particle is entirely in the equatorial plane (EP, θ = π/2) both components are zero, and the orbit is a geodesic. Considering a small sphere in orbit, with its centre following the geodesic we can make some deductions about the forces required to keep the whole body together.

The acceleration in the θ-direction ( the one pointing up and down from the equatorial plane) will change sign between the two halves so it always acts towards the EP causing compressive stress. In the radial direction, leaving the EP in either direction will cause an increase in the acceleration in the +ve r-direction. I don't know what deformation this causes. Both components get smaller as the radius increases so it looks like there is also some stretching in the r and θ-directions.

The acceleration above is calculated from the Schwarzschild circular orbit geodesic in the coordinate basis with 4-velocity
$$u=\frac{\sqrt{r}\,\sqrt{m\,{\sin\left( \theta\right) }^{2}+r-3\,m}}{\sqrt{r-3\,m}\,\sqrt{r-2\,m}}\ \partial_t -\frac{\sqrt{m}}{r\,\sqrt{r-3\,m}}\ \partial_\phi$$
This is only a geodesic if sin(θ) = 1. So any circular path off the EP must apply forces to offset the accelerations engendered.

17. Sep 20, 2012

### zonde

Let's say that this body in tidal lock does not have perfect spherical symmetry. Then as it rotates it's gravity affects large mass M differently. Now let's say that body M is not entirely rigid so that there are slightly different responses to small body in different rotation angle positions.
Another possibility. Body in orbit itself is not entirely rigid and then radial stretch elongates the body in different directions as it rotates. If this response to stretch in different directions is different then shouldn't it be possible to dump angular momentum using that radial force?

18. Sep 21, 2012

### Staff: Mentor

When you're holding the rod vertically, it's exactly lined up with the radial gravitational force; both ends of the rod and the center of the earth lie in the same line, and the gravitational force acts along that line. But when you throw the rod it moves off that vertical line. Now the forces acting on the two ends are no longer exactly parallel (extend the vectors and they meet at the center of the earth, therefore are very slightly converging). Furthermore, the forces have very slightly different magnitudes, because the distance to the center of the earth is different for the two ends.

So we have forces of different magnitude acting in different directions on the two ends of the rod. Why shouldn't one end speed up relative to the other?

19. Sep 21, 2012

### altergnostic

Mentz114, Thanks for this. Correct me if i'm wrong, i take this to mean that anything above the equatorial plane suffers an accel towards the equatorial plane. That would be the flattening at the poles of tidal theory. As you also pointed out, there should also be stretching in the equatorial plane (i assume only on the near and far sides) which would be the tidal bulges. Still, i see no component that would cause different tangential forces on the near and far sides that could adjust velocities to cause a body to achieve tidal lock, is this correct?

20. Sep 21, 2012

### altergnostic

I have considered this, and i believe that would tend to cause shearing. If the tangential velocities on the near and far sides remain unchenged, these forces that you mention would work against this stretching, but not overcome it. If a body has almost no rigidity, it could not hold itself together and it would shear (near side ahead of the far side), stretch radially and tear apart.

Last edited: Sep 21, 2012