# Tidal Effects in the Schwarzschild Metric

1. Sep 22, 2010

### Passionflower

Assume we have a non rigid ball.

When this ball is radially free falling in a Schwarzschild metric the height increases while the width decreases due to tidal effects. How do we calculate the ruler width and height in terms of R and m?

When we have two of those balls radially free falling but one is falling at escape velocity (free fall from infinity) and the other is not, are the widths and heights identical or if not how do they relate to each other in terms of the difference in coordinate speeds?

If one of those balls is in a circular orbit with the same side always facing the 'center' of gravitation, does the width and height stay the same and thus can we say this ball is in an inertial frame, or is there precession? And what if this ball is instead in an elliptical orbit?

How in all those cases how does the radar distance relate to the ruler distance?

Let's first assume:

$$d\theta = 0, \, d\phi = 0$$
$$r_s = 2m$$

I know the radar distance of the length in coordinate time is:

$$\Delta t = 2 \left(r_2 - r_1 + r_s \ln \left[ \frac{r_s + r_2}{r_1 - r_s}\right]\right)$$

The conversion factor from coordinate time to proper time is:

$$\sqrt{1-r_s/r}$$

So clearly the radar distance depends on the location of the observer on the ball.

Now is the ruler distance this?

$$\Delta h = \sqrt{r_1 (r_1 - r_0)} - \sqrt{r_2(r_2-r_0)} + r_0 \ln { {\sqrt{r_1} + \sqrt{\sqrt{r_1} - r_0} \over {\sqrt{r_2} + \sqrt{\sqrt{r_2} - {r_0}}}$$

Is that right?

So now how does this relate, if at all to the tidal effect? Suppose we have a ball of height 1 with r1 and r2 near infinity, then at a given R how much is h of the ball?

But that is just the height, I do not know how to get the widths.

Now do those formulas work regardless of the coordinate speed of the ball or do they only apply when the ball travels at escape velocity?

Please no approximations I like to get the exact numbers

Last edited: Sep 22, 2010
2. Sep 22, 2010

### yuiop

In the other thread I think I have explained that the orbiting ball will not remain spherical, if it not rigid, due to tidal forces. I think it is also true that if the ball is not perfectly spherical and rigid, then it will tend to lock into a position with one the same side always facing the centre of gravitation, as is the case with our Moon where we never see the "dark side" from the Earth. This locking effect will probably be greater than any Thomas precession in most practical cases.

A sphere of unconnected coffee granules in orbit with negligible mutual gravitation will tend to smear out into a long string along the orbital path if each particle has sufficient orbital velocity to keep it at constant orbital radius. If the velocities of the individual particles is that required to keep the the particles on the same radial line from the centre of gravitation, then they will tend to stretch out into a line parallel to the radial line. This latter case is the one most similar to the case where the particles are elastically connected and there will be stress/strain pulling the elastically connected particles apart and preventing any precession.

Last edited: Sep 22, 2010
3. Sep 22, 2010

### starthaus

The Earth's gravity pulls on the closest tidal bulge incurred by the "Moon", trying to keep it aligned with Earth. As the "Moon" turns, under the Earth's gravity, this creates tidal friction, slowing the "Moon's" rotation down until its rotation period matches its orbital period exactly, a state called tidal synchronization. In this state, the "Moon's" tidal bulge is always aligned with Earth, which means that the "Moon" always keeps one face toward Earth. The "Moon" rotates nevertheless. There is no Thomas precession, there is no connection to ant Thomas precession for this particular case since the axis of the "Moon" does not precess at all.

4. Sep 22, 2010

### yuiop

I hope you put warning signs on the text books you own, that contain approximate formulas without any indication that they are approximations. Which book is the worst culprit for this?

5. Sep 22, 2010

### Passionflower

Good point, this bit me when I calculated the radar distance for a Born rigid object in a linear accelerating frame, I thought I dealt with exact formulas as it was not mentioned it was an approximation.

6. Sep 23, 2010

### sweet springs

Hi.
Let ball be free and still. The source of gravity is #1 also still, #2 running toward the ball, #3 running outward and #4 running transverse.　　　It seems that deformation of ball in co-moving local inertial system which is stick to the center of the free ball, are different in cases #1,#2, #3 and #4 as Rienard- Wiecherd potential in EM.
Regards.

7. Sep 23, 2010

### Mentz114

Passionflower,

I've worked out the tidal effects in the Schwarzschild spacetime for the locally Minkowski observer in the coordinate frame, and also for the Gullstrand-Painleve coordinates and I get exactly the same answer. This surprised me. I may have made a mistake. Do you want to see the calculations ?

Last edited: Sep 23, 2010
8. Sep 23, 2010

### Passionflower

Well anything that can help this topic move forward is good!

For clarity's sake it is probably best if we first stick to Schwarzschild coordinates, once everything is ironed out we can entertain other charts. At least that is my suggestion to avoid possible confusion.

9. Sep 23, 2010

### starthaus

The results should be coordinate - choice independent. The fact that you got the same results is an indication that you are doing things correctly. You should have only worried if you got different results.

10. Sep 23, 2010

### Mentz114

I did wonder**. Doesn't that answer your question (which I've emphasized) below ?

The tidal components are radial=2m/r3, phi, theta = -m/r3, but I don't know if this represents a velocity field.

This is hard. You need to calculate the kinematic decomposition of the congruence of circular orbits. I've done it for the most general case where the orbits need not be geodesics and for sure there is rotation and shear, but no expansion ( I think the last is common to all vacuum solutions). The expressions I've got for the acceleration are very long and unpleasant but there is a component in the -r direction of order ~1/r2 as expected.

The components of the shear tensor are equally intractable and I'm having a problem saying anything about them right now. Ditto the rotation, except it's in the right direction.

[Later]
Well, I have an interesting result for the tidal tensor as seen by a Hagihara observer, who is in orbit in the Schwarzschild spatetime. The components are,

$$T_{rr}=\frac{2\,m\,r-3\,{m}^{2}}{{r}^{4}-3\,m\,{r}^{3}}$$

$$T_{\theta\theta}=-\frac{m}{{r}^{3}-3\,m\,{r}^{2}}$$

$$T_{\phi\phi}=-\frac{m}{{r}^{3}}$$
which still add up to zero, but the tidal effects are not symmetrical anymore. It seems that tidal effects are frame dependent.

Confirmed here
http://en.wikipedia.org/wiki/Frame_...agihara_observers_in_the_Schwarzschild_vacuum

Last edited: Sep 23, 2010
11. Oct 7, 2010

### yuiop

The tidal acceleration dg between two particles is given by various references as:

$$dg = c^2 \frac{r_s}{r^3} \, dr$$

... where $r_s$ is the Schwarzschild radius, r is the instantaneous location of the falling object and dr is the differential Schwarzschild coordinate displacement of the falling object. Now for an object of proper length dL, which is reasonably short compared to the Swarzschild radius, Pervect and myself seem to be in agreement that dr = dL (1-rs/r). This suggests for a short fairly rigid falling object the differential tidal acceleration acting on it is:

$$dg = c^2 \frac{r_s}{r^3}(1-r_s/r) \, dL$$

An immediately obvious consequence of the above equation, is that at the event horizon the tidal acceleration on the falling object goes to zero and below the event horizon the stretching force becomes a compressive force. This may be due to Ricci curvature which tends to reduce volumes in strongly curved spacetime. Anyone any ideas on that?

It also suggests that a short rocket, free falling through the event horizon can detect when it is passing the event horizon by noting when the tidal strain on the rocket goes to zero.

The radius of maximum tidal stretching can found by differentiating the above equation and setting it to zero to find the turning point and then solving for r and the result is (4/3)rs. This means that the equation for the maximum tidal acceleration acting on a short body of proper length dL can be written as:

$$dg_{max} =\frac{3^3}{4^4} \frac{c^2}{r_s^2} \, dL$$

Now it seems the effective length contraction of the falling object acts to reduce the tidal stretching or spaghettification of the object as it falls, but the stretching is still significant enough to pose a formidable barrier to arriving at the event horizon in one piece for all but the largest black holes. For example a solar mass black hole has a Schwarzschild radius of about 3km. The speed of light is about 300,000 km/s and the height of an average person is about 1.8m or about 0.0018 km. Using the max equation:

$$dg_{max} =\frac{3^3}{4^4} \frac{c^2}{r_s^2} \, dL = \frac{27}{256}*\frac{300000^2}{9}*0.0018 \approx 2,000,000 km/s^2$$

which would almost certainly be fatal.

An astronaut falling towards a 40,000 solar mass black hole would experience a maximum tidal acceleration of about 10 m/s2 or 1g which should be no threat to survival.

Last edited: Oct 8, 2010
12. Oct 12, 2010

### Passionflower

Could you expand on this a little, I have trouble understanding this as you seem to be multiplying a distance with a coordinate.

Let's take a simple example, rs=1 and the rod is between r=10 and r=11 (I just want to understand what you propose so the length does not matter but if you feel it is too long just use 10 and 10.000001).

For the static user we use:

$$\sqrt {{\it ro}\, \left( {\it ro}-1 \right) }-\sqrt {{\it ri}\, \left( {\it ri}-1 \right) }+\ln \left( {\frac {\sqrt {{\it ro}}+ \sqrt {{\it ro}-1}}{\sqrt {{\it ri}}+\sqrt {{\it ri}-1}}} \right)$$

So what do you propose to use for a free falling rod?

This (1-1/r0)-(1-1/ri) ?

As I wrote before, from the book I use, the proper distance for a free falling rod at escape velocity is simply dr because the proper distance integrand * 1/gamma simply cancels out everything.

The Schwarzschild solution is a vacuum solution so it is Ricci flat. Only the interior Schwarzschild solution (e.g. the solution for the inside of a massive sphere) has Ricci curvature.

Last edited: Oct 13, 2010
13. Oct 12, 2010

### pervect

Staff Emeritus
The tidal stretching force on a short, free-falling rod should be -2M/r^3 (Mtw pg 822) -in geometric units. This force is independent of the radial component of the velocity of the rod, as mentioned in MTW.

I'm not sure what yuiop means by
unless he means what I wrote above.

If we look for the r coordinate difference of the two ends of the rod, at "the same time" *as judged in the local frame of the rod (assuming the rod is short enough to have one), for a short rod with length L we should get

$$\Delta_r = L$$

If we look for the r-coordinate difference "at the same time" *as judged by a static observer, we should get

$$\Delta_r / (1-2M/r) = L$$

** I think - it's a bit early, but as I recall that's what we worked out previously.

It's definitely untrue that the tidal force should go to zero at the horizon. MTW's result for the tidal stretching force comes from the geodesic deviation equation, and conceptually would be thought of as being measured in the local frame of the falling rod, so we'd want to use the defintion of simultaneity of the rod.