# Gravity probe-B

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"?

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We intend to announce the final experimental results of GP-B through a NASA press/media event towards the end of 2007. At that time it is also our intention to have submitted a number of papers on the GP-B results for publication in peer-reviewed scientific and technical journals.

http://einstein.stanford.edu/

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

Not long to wait.

Garth

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

Not long to wait.

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?

Garth
Gold Member
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

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Garth
Gold Member
It's the geodetic effect I am interested in!

Garth

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

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

But what does "the same direction in space" mean? For Newton the answer was easy. Space and time were absolutes. A perfect gyroscope set spinning and pointed at a star would stay aligned forever. Not so for Einstein. Space-time is warped -- and may even be set in motion by moving matter. A gyroscope orbiting the Earth finds two distinct space-time processes -- frame-dragging and the geodetic effect -- gradually changing its direction of spin.

According to Einstein the Earth warps space-time. A second, much larger change in spin direction, the geodetic effect, follows from the gyroscope's motion through this space-time curvature. The phenomenon was foreshadowed in 1916 by W. de Sitter who predicted a minute relativistic correction to the complicated motions of the Earth-Moon system around the Sun -- an effect finally detected in 1988 through an elaborate combination of lunar ranging and radio interferometry data. For a gyroscope the predicted effect is a rotation in the orbit-plane of 6,600 milliarc-seconds per year -- quite a large angle by relativistic standards.

Gravity Probe B will measure the change to 1 part in 10,000 or better, the most precise qualitative check yet of any effect predicted by general relativity.

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.

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

Garth
Gold Member
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

pervect
Staff Emeritus
Yes MeJennifer - though there has been some confusion.

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.

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.
ere appears to be some coordina
Geodetic Precession on the other hand is a GR effect.

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.

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

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.

Garth
Gold Member
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.
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.
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.
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.
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.
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' .
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.
Indeed, they had a launch window of 1 second each day to put it into that orbit!

Garth

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pervect
Staff Emeritus
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.

(See MTW page 1119 equation 40.34)

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.

So observer B suffers 3X observer A's total precession.

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

Garth
Gold Member
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.

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

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.

Garth
Gold Member
As I said that 3-acceleration expression suffices in the PPN approximation.
From MTW page 118
Equations 40.33 describe in complete generality at the post-Newtonian level of approximation the precession of the gyroscope spin S relative to the comoving orthonormal frame that is rotationally tied to the distant stars.
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

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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,

The very concept of acceleration, therfore, implies "rotation" of veloctiy 4-vector. How then is the requirement of "no rotation" to be interpreted? Demand that the tetrad change from instant to instant by preciesly that amount implied by the rat of change of u, and by no additional arbitrary rotation. In orther words accept the inevitable pseudorotation in the timelike plane defined by the velocity 4-vector and the acceleration byht (2) rule out any ordinary rotation of the three space vectors.

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".

Garth
Gold Member
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:
A spinning object, free of all torque, but undergoing acceleration, changes its direction as this direction is recorded in an inertial frame of reference.

///////////////////////////

How is it possible for "no rotation" to appear as "rotation"? The answer is this: one pure boost, followed by another pure boost in another direction, does not have as net result a third pure boost; instead, the net result is a boost plus a rotation.

Garth

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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).

Garth
Gold Member
Well, let's get specific then.
Certainly! An illustrative example to think about.
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.
I claim that under that acceleration an on-board torque free gyroscope will precess relative to the fixed stars.
What is the axis of this rotation, and given an acceleration 'a', how many radians/second will it be?
My understanding of this case is, if we let the spaceship have a relative 3-velocity initially also in the direction of Sirius parallel to the 3-acceleration vector:

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

The gyroscope's intrinsic angular momentum 4-vector is always orthogonal to the 4-velocity.

It will precess in 3-space orthogonally to the plane defined by the gyro's axis and the ship-Sirius direction 3-vector by an amount (while speeds are non-relativistic):

$$\frac{a \sin(\theta)}{c} radians/sec$$

where $\theta$ is the angle between the direction of the gyro's spin axis and the ship-Sirius direction 3-vector and a is the magnitude of the acceleration.

Garth

pervect
Staff Emeritus
Certainly! An illustrative example to think about. I claim that under that acceleration an on-board torque free gyroscope will precess relative to the fixed stars.
My understanding of this case is, if we let the spaceship have a relative 3-velocity initially also in the direction of Sirius parallel to the 3-acceleration vector:

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

The gyroscope's intrinsic angular momentum 4-vector is always orthogonal to the 4-velocity.

It will precess in 3-space orthogonally to the plane defined by the gyro's axis and the ship-Sirius direction 3-vector by an amount (while speeds are non-relativistic):

$$\frac{a \sin(\theta)}{c} radians/sec$$

where $\theta$ is the angle between the direction of the gyro's spin axis and the ship-Sirius direction 3-vector and a is the magnitude of the acceleration.

Garth

Let's introduce a coordinate system, with Sirius lying in the 'x' direction.

y
^
^
^
-------> x (Sirius)

and define vectors using their components (t,x,y,z), in that order

Your claim, as I understand it, is that a gyroscope with a spin axis in the 'y' direction precesses.

Now for a spin axis in the 'y' direction, in a comoving frame we should have

$$S = \left( 0,0, L, 0 \right)$$

Let's use your formula (a^u) S to find the initial rate of change of S in a comoving frame.

Then
$$u = \left( 1,0,0,0 \right)$$
$$a = \left( 0, a, 0, 0 \right)$$

see MTW 6.4 pg 166 for the components of a, which we can see is orthogonal to u as it should be. Then

$$a \wedge u = \left( \begin{array} {cccc} 0&-a&0&0\\ a&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{array} \right)$$

We can see that $$\left( a \wedge u \right) \cdot S = 0$$

Why can we see this? Because (a ^ u) only has components (tx and xt). This is because a and u only have t and x components, hence a^u can only have t and x components. Because the wedge product is anti-symmetric, the tt and xx components are 0, only the tx and xt components exist, and they must be negatives of each other.

But the vector that we multiply by has only a y component. The only way to get a nonzero result is from y components in the wedge product. But all such components are zero.

Note that we also expect (0,0,L,0) to boost in the x direction to (0,0,L,0), i.e. we don't expect a boost in the x direction to change the y component of angular momentum given that it transforms as a 4-vector. Thus we don't really expect acceleration in the x direction (which is a series of small boosts) to change the y component of momentum. Which is what we just calculated in more detail.

i.e. there is no change in S, a gyroscope with its spin axis initially in the 'y' direction does not precess.

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Garth
Gold Member
Yes, I agree.

This is an interesting problem that I do not have time to do justice to.

My original statement was referring to a precession relative to the distant stars. (As we had been talking about GP-B.)

The tetrad of basis vectors itself undergoes a rotation in space-time relative to an inertial frame of reference.

The effect I am thinking of may, after a substantial velocity change has built up, be related to the relativistic aberration and the relativistic searchlight effect. The gyro would appear to precess relative to the distant stars, which themselves would precess in the frame of reference of the spaceship .

Or am I mistaken?

Garth

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I am waiting for the results!! (14th April)

Garth
Gold Member
I am waiting for the results!! (14th April)
Yes indeed! See the Alternative theories being tested by Gravity Probe B thread.

Perfect Returning to my previous musings about a constantly accelerating gyroscope observed from an inertial frame of reference.

Does not the its orthogonal tetrad of basis vectors not rotate but look warped from the inertial frame?

The spacecraft would experience this as the relativistic aberration of light from fixed stars by which the gyroscope's precession is being tracked.

In GP-B if the experiment was not in orbit but fell straight towards a non-rotaitng Earth would not the gyroscope still precess relative to the distant quasar?

If so, my question is, "Would it still precess if the experiment had only just begun to fall so it as under 'Newtonian' gravitational acceleration yet momentarily at rest w.r.t. the Earth?"

Garth

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