# Gravity probe-B

1. Mar 20, 2007

### magnetar

Gravity probe-B!

The gravity probe-b mission had finished collection date for a long time .Why not announce any result ??

Do you think that the results will consistent with "general relativity"?

Last edited: Mar 20, 2007
2. Mar 20, 2007

### neutrino

http://einstein.stanford.edu/

3. Mar 20, 2007

### Garth

But the first results are promised in April, on Saturday the 14th at the APS conference in Jacksonville Florida.

Not long to wait.

Garth

4. Mar 24, 2007

### rbj

do you think we might have heard a hint or a leak if the results were similar to those of the Michaelson-Morley experiment? if the results are not within noise of what GR says they should be, that would be hard to contain, no?

5. Mar 24, 2007

### Garth

They have kept a 'blind' element in the results by not including the proper motion of the guide star against a distant quasar until the last minute.

They have kept their cards very close to their chest!

However we already know the proper motion of the guide star to 0.001"/yr from The stars of Pegasus from the Bright Star Catalogue, 5th Revised Ed. (Preliminary Version) (Hoffleit+, 1991, Yale University Observatory) as distributed by the Astronomical Data Center at NASA Goddard Space Flight Center.

IM Pegasi

RA J2000 : 22h 53m 2.3s
DEC J2000 : +16° 50' 28"
Proper motion in RA : -0.018 arcsec/y
Proper motion in DEC : -0.024 arcsec/y

mag : 5.64
MK spectral class : K1-2II-III

The Proper motion in RA will affect the E-W precession and the
Proper motion in DEC will affect the N-S precession.

The experiment measures the precession of the gyros to four decimal places of arcsec/yr, so the final convolution of the tracking a precession data will only increase the accuracy of the result in that fourth decimal place.

On the one hand, as you say, it would be hard to contain a non-GR result for so long, but on the other hand the fact that they are taking so long over processing the data might be an indication that they have found an anomaly and are checking and double checking their results before publication.

We wait and see.

Garth

Last edited: Mar 24, 2007
6. Mar 24, 2007

### magnetar

Last edited by a moderator: Apr 22, 2017
7. Mar 24, 2007

### Garth

It's the geodetic effect I am interested in!

Garth

8. Mar 24, 2007

### magnetar

This phenomenon, known as frame-dragging, was first predicted in 1918 as a natural consequence of Einstein's general theory of relativity, which describes the effects of gravity on space and time. But it had been unproved by experiments or observation until recently, when Italian researchers suggested the effect might be present near spinning neutron stars. The MIT team then applied a similar idea to several black holes in our galaxy-------

excerpt from http://web.mit.edu/newsoffice/1997/blackholes.html

Last edited by a moderator: Apr 22, 2017
9. Mar 24, 2007

### pervect

Staff Emeritus
From http://einstein.stanford.edu/content/story_of_gpb/gpbsty3.html [Broken]

The geodetic precession is closely related to Thomas precession (if that helps any).

It's of interest to Garth because his dark-horse (though peer-reviewed) theory of gravity, SCC http://arxiv.org/abs/gr-qc/0212111 makes an experimental prediction different from GR that will be tested by gravity probe B.

You can probably find a lot of threads on SCC here at PF if you look - being published in a reputable journal, it's "fair game" for discussion on PF.

Last edited by a moderator: May 2, 2017
10. Mar 24, 2007

### MeJennifer

This statement intrigues me.

While Thomas precession is an effect due to the hyperbolic nature of space and is a compensation for the non commutative Einstein addition formula (see for instance: "Ungar A A - Beyond the Einstein addition law and its gyroscopic Thomas precession (Kluwer,2001)(ISBN 0792369092)"), I thought the geodetic precession had something to do with curvature?

No?

11. Mar 25, 2007

### Garth

Yes MeJennifer - though there has been some confusion.

In four-space, SR's Minkowski space-time, an object moving relative to an observer has a four-vector inclined to the four-vector of that observer. Therefore, if a force accelerates a gyroscope, its four-vector gradually leans over relative to its original direction and its spin axis will consequentially precess. This is Thomas precession. It is a SR effect.

Geodetic Precession on the other hand is a GR effect.

Consider the 'curved funnel' model of the space component of curvature around a gravitating mass. At any one particular distance around the central mass the space curvature can be modelled by a 'slice' of a cone. That cone can be constructed by cutting a thin cake slice out of a circle and gluing the edges together.

The space curvature component of geodetic precession, described by the Robertson parameter $\gamma$, is the angle of the slice cut out of the full circle.

The other component of the geodetic precession is a time dilation effect.

Another way of describing the GP-B geodetic effect is to say 'it is the amount the gyros 'lean over' into the slope of the curvature of space-time' as they orbit the Earth. If that helps.

In the GP-B experiment SCC predicts the same frame-dragging E-W precession but only 2/3 the N-S precession as GR. We shall know soon!

Garth

12. Mar 25, 2007

### pervect

Staff Emeritus
Some of the confusion here appears to be mine, which I apologize for, but I don't want to pass up the opportunity to learn something and hopefully become less confused.

Suppose we have observer A, in flat space-time, in a powered circular orbit. The centripetal acceleration is supplied by his rocket. Because the direction of his acceleration is constantly changing, he will experience Thomas precession.

Suppose we have observer B, in curved space-time, in an unpowered circular orbit. From a Newtonian viewpoint, the centripetal acceleration is supplied by gravity. From a GR viewpoint, he is simply following a geodesic.

I was under the (possibly erroneous) impression that observers A and B experienced the same amount of precession in the same direction. I really ought to do a calculation to check this, but it will have to wait.

This appears to be the standard explanation of geodetic precession (and standard is good).

I ran across http://prola.aps.org/abstract/PRD/v42/i4/p1118_1 though, which seems to blur the distinction being made here, and to suggest that some of the distinction comes about because of the particular coordinate system being assumed.

Experimentally the distinction is important because the two effects occur at right angles for GP-B. They occur at right angles only for a satellite in polar orbit. In fact the polar orbit was chosen to make this happen.

13. Mar 25, 2007

### Garth

Note: observer A would experience Thomas precession even if she were accelerating in a straight line - because her 4-vector would be rotating in space-time.
In that situation the Thomas precession of observer A is equal and opposite in direction to the 'time dilation' component of the Geodetic precession of observer B.

On top of that GR would predict a 'space curvature ("missing slice") component' precession for observer B that was twice his 'time dilation component' precession. (See MTW page 1119 equation 40.34)

So observer B suffers 3X observer A's total precession.
Although I do not have access to the paper itself I can understand what they are doing. From an orbiting coordinate system the Earth is spinning even if it is non-rotating wrt the fixed stars. It could be confusing because the two precessions in their case are equal yet have fundamentally different origins. This becomes obvious if the satellite's orbit is not in the plane of the Earth's rotation, as indeed it is not with GP-B.

Geodetic precession will be experienced by a gyroscope orbiting the equator of a non-rotating Earth, but IMHO it is confusing to call this 'frame dragging' .
Indeed, they had a launch window of 1 second each day to put it into that orbit!

Garth

Last edited: Mar 25, 2007
14. Mar 25, 2007

### pervect

Staff Emeritus

Surely v x a would be zero if v were in the same direction as a? Implying no precession for an observer accelerating in a straight line (where a is always in the same direction as v) as MTW's 40.33b reduces to -(1/2) v x a for this case.

OK, I can see that now. Thanks for the correction.

15. Mar 26, 2007

### Garth

a is actually the 4-acceleration, in equation 40.34 the components are evaluated in the PPN approximation where the 3-acceleration suffices. See MTW exercise 6.9 page 175.

Garth

16. Mar 26, 2007

### pervect

Staff Emeritus
40.33 (pg 1118) has already been re-written in a 3d vector form - it's not a 4-vector equation (and the cross product only makes sense in 3 dimensions).

One could regard the Lorentz boost as a kind of rotation, but there isn't any spatial rotation in the normal 3-d sense in the case of constant acceleration in the same direction that I can see. And I've looked at exercise 6.9 as you suggest.

I.e. if we accelerate in the x direction with a constant acceleration, a gyroscope with its spin axis in any of the x, y, or z directions will not precess in any spatial direction (assuming the force accelerating the gyroscope is applied at the center of mass).

The most that will happen is that time dilation might affect the apparent angular momentum carried by the gyroscopes, but this is not a 3-d rotational effect.

17. Mar 26, 2007

### Garth

As I said that 3-acceleration expression suffices in the PPN approximation.
From MTW page 118
Where we are confusing ourselves is over two sorts of acceleration.

In the case of a gyroscope fixed to the Earth's surface the significant Thomas precession is due to the gyro's acceleration when carried around by the Earth's rotation.

However, my point in #13 was referring to a linear acceleration that was the subject of MTW Exercise 6.9.

Here a constant force acting through the centre of mass produces a acceleration constant in direction however, the 4-velocity of an linearly accelerating body actually rotates in an inertial space-time frame.

Now, the four-angular momentum S of a rotating body has the property of always being orthogonal to its 4-velocity (eq. 6.21), therefore S has to precess under acceleration $\mathbf{a}$ to keep orthogonal to that 4-velocity $\mathbf{u}$:

$$\frac{d\mathbf{S}}{d\tau} = (\mathbf{u} \wedge \mathbf{a})\cdot d\mathbf{S}$$

Garth

Last edited: Mar 26, 2007
18. Mar 26, 2007

### pervect

Staff Emeritus
I would agree that there will be a "pseudorotation" in the timelike plane formed by a and u, as described in MTW,

However, I don't believe that it is been demonstrated (and I don't believe that it is true) that an observer in hyperbolic motion, Fermi-Walker transporting his basis vectors (or equivalently, using a set of gyroscopes he carries with him to orient his basis tetrads) will observe any apparent rotation of the distant "fixed stars".

19. Mar 26, 2007

### Garth

I disagree.

Looking more closely at the case of the Thomas precession of a gyroscope attached to the Earth's surface, in MTW eq. 40.34.

The acceleration concerned is the upwards acceleration $\bigtriangledown U$ of a freely falling frame so that it becomes at rest with the Earth's surface. The velocity is that of the gyroscope being carried around by the Earth's rotation.

Again, has this Thomas precession applied to a spinning electron moving in a circular orbit around an atomic nucleus not been observed? Does it not account for a factor of two in the effective energy of coupling of spin and orbital angular momentum of a atomic bound electron?

MTW page 1145 \$41.4:
Garth

Last edited: Mar 26, 2007
20. Mar 26, 2007

### pervect

Staff Emeritus
Well, let's get specific then.

Suppose one accelerates at a constant rate towards a "guide star", say Sirius, in a flat Minkowski space-time, while using an array of gyroscopes to transport one's basis vectors (the mathematical description of this operation would be Fermi-Walker transporting these basis vectors).

I say that there should be no rotation of the "fixed stars" (assumed, for the sake of this example, to all be stationary relative to each other in some inertial frame).

You appear to be claiming that there will be. What is the axis of this rotation, and given an acceleration 'a', how many radians/second will it be?

For starters, can we agree that the guide star will always be "ahead" of the accelerating spaceship, i.e. that the Fermi-Walker transported basis vector intitally pointing towards the guide star will always point towards the guide star? (I suppose I should add for completeness that I'm assuming the guide star is far enough away that one doesn't reach it).