Prediction for Gravity Probe B

[grav]

Grav,

The attitude of probe "body" or housing or whatever one calls it is precisely controlled and known so that can indeed determine the true precession of the gyros inside. It used telescopes to keep a precise stellar fix and keep itself aligned as much as possible, and I assume measure any deviation that did occur.

-Richard
I was thinking this morning that the only thing a controlled housing could do was to keep the gyro aligned in some way. This might actually be important. If the gyro and housing were allowed to turn as they pleased, then they will turn together, and no precession will be measured (relative to the housing itself). If it were aligned so that the same side always faces the Earth, then a positive precession on one side of the Earth will be cancelled by the negative precession on the other, and no precession will be measured. However, if it were aligned so that the gyro is always vertical with the Earth's equatorial plane, it will have a positive precession on one side of the earth and a positive precession on the other as well, since it would be turning the same way on both sides in respect to the Earth (clockwise or counter-clockwise). Looking at publius' statement above, it seems that a stellar fix would indeed produce this, whereas the same side would always be facing the same fixed stars, and the gyro would be kept vertical in respect to the Earth's equatorial plane. So now we're in business. :dance:

Okay. So first we need to find the precession caused by the Earth. GR would predict (3GM/c2)/d for the amount of the orbit per orbit. My relation says (v/c)*(R/d). It is probably more precisely [(GM/c2)*(Requa-Rpole)]1/2, but they come out the same for uniform fluid bodies through the production of an equatorial bulge. The extra distance travelled per orbit due to the precession is (v/c)(R/d)(2pid)=(v/c)(2piR), regardless of the distance of orbit. For the sun, this extra distance is (v/c)(2piR)=(2piR/T)(2pi/R)/c=(2piR)2/Tc=29088.33622 meters, where R=6.96*108 meters and T=2.193*106 seconds. This is the extra distance Mercury travels per orbit due to the precession caused by the sun. It is also the extra distance Venus travels, and Earth, and Mars, and so on. For Earth, it is found to be 61.819774 meters, where R=6.36*106 meters at the poles and T=86164 seconds. This is much smaller than that of the sun, but still measurable. In fact, it should be very noticable for close range orbits since they would orbit very fast, so that the number of orbits would add up quickly and so would the extra distance over time.

So gravity Probe B will orbit at 640 km from Earth. This would give it an orbit of about 7000 km from its center and a velocity of vB=(GME/d)1/2=7546 m/s, where ME=5.9736*1024 kg. The time per obit is then PB=(2pid)/vB=5828.56 sec. The precession per orbit for a satellite orbitting at the equator at this distance would be pr=(vE/c)(RE/d)[(360 degrees per orbit)*(3600 arc-seconds per degree)]=1.8216 arc-seconds per orbit. Not very much, right? But let's check this over time, say, a year. It would be pr=(vE/c)(R/d)(360 degrees per orbit*3600 arc-seconds per degree)][(3.1536*107 sec per year)/P]=(1.8216 arc-seconds per orbit)[(3.1536*107 sec per year)/P]=9856 arc-seconds per year=2.7378 degrees per year. This should be quite noticable. Are any satellites that orbit the equator noticably this far ahead in their orbit each year, or behind if orbitting in the opposite direction of the Earth? Are any corrections made in this amount in a year's time?

The precession of Gravity Probe B would be greater on its closest side and less on its other, and so is proportional to (vE/c)[RE/(d-Rgyro)]-(vE/c)[RE/(d+Rgyro)]=(vE/c)(RE)(2Rgyro)/(d2-Rgyro2)1/2, which is approximately (vE/c)(RE*2Rgyro)/d2. This would add an extra spin in this amount to the gyro, but integrated over its circumference, and so is smaller by an amount of 1/2, or 1/3, or 2/pi, or something of that nature (I haven't done the calculation yet). Let's just call this fractional ratio for integration z for now. Since the probe will be travelling over the poles, the integration is the same for this circumference, so now we have a fraction of z2. So the total extra spin added to Gravity Probe B would be z2[(vE/c)(RE*2Rgyro)/d2](360*3600)(3.1536*107/P)=z2(9856 arc-seconds per year)(2R[gyro[/sub]/d)=z2Rgyro*(2.816*10-3 arc-seconds per year per meter). So if, for instance, z=1/2 and Rgyro=1 meter, then the gyro would turn an extra (1/2)2(1 meter)(2.816*10-3)=7.04*10-4 arc-seconds in a year's time (or less by this amount, depending on which way it is rotating).

This last part needs a lot of work, so don't take it too literally yet, please, but I thought I might mention the possibility. Regardless of whether my formula for the precession is correct or if that for GR stands, this effect on rotation has one very profound effect. It causes bodies to rotate differently in the presence of gravity. If atoms are oriented randomly, some electrons turning one way might radiate faster while those oriented in the other radiate slower. This would be related to the factor of z as well. This might not cancel out over all of the atoms, but if the atoms radiate over time, then the differential between the minimum and maximum times becomes 1/(P-textra)-1/(P+textra)=2textra/(P2-textra2)1/2, which comes to approximately 2t/P. I'm not sure if this is the correct differential I should be using here, but one gets the idea. Anyway, from the formulas in the previous paragraph, one can see that this varies with the inverse of the square of the distance from a gravitating object, the same as gravity itself. So atoms, then, might precess more quickly on average in proportion to the gravity present. In other words, they age faster. This goes along with the GR prediction (and apparent verification) of time dilation in a gravitational field (I'm still working on the details).

[hhEb09'1]

Looking at publius' statement above, it seems that a stellar fix would indeed produce this, whereas the same side would always be facing the sun, and the gyro would be kept vertical in respect to the solar system's plane of rotation, and to the Earth's as well. So now we're in business. :dance: Same side of the housing would be facing the sun?

And publius's statement mentioned stellar fix, which I'm pretty sure means to the "fixed stars". Also, the solar system plane of rotation and the Earth's plane of rotation are different by over twenty degrees, you can't be vertical to both of them at the same time.

[grav]

Same side of the housing would be facing the sun?

And publius's statement mentioned stellar fix, which I'm pretty sure means to the "fixed stars". Also, the solar system plane of rotation and the Earth's plane of rotation are different by over twenty degrees, you can't be vertical to both of them at the same time.
Well, basically they both mean the same thing in this case, but not quite, you're right, since a fix on the sun would be angled slightly differently above or below the center of the Earth. The further away the star the telescope is centered on, the better, I suppose. In any case, it would work as long as the probe did not turn with its revolution around the Earth (as with tidal lock), but maintained the same orientation to the fixed stars, like you said. As far as the orientation itself, I guess the calculations would be simpler and the effect the greatest if it is vertical to the Earth's plane of rotation and not the solar system's after all. I was thinking about them as being close to the same, for some reason.

[hhEb09'1]

My main question there was, do you--in your interpretation--have the housing always facing the sun, or always facing the stars?

[publius]

Well, basically they both mean the same thing in this case, but not quite, you're right, since a fix on the sun would be angled slightly differently above or below the center of the Earth. The further away the star the telescope is centered on, the better, I suppose. In any case, it would work as long as the probe did not turn with its revolution around the Earth (as with tidal lock), but maintained the same orientation to the fixed stars, like you said. As far as the orientation itself, I guess the calculations would be simpler and the effect the greatest if it is vertical to the Earth's plane of rotation and not the solar system's after all. I was thinking about them as being close to the same, for some reason.

Grav,

Bingo. That's the whole point of the "stellar fix" -- to make sure you are not rotating as you orbit. That is, your little X-Y-Z axes remain fixed.

NASA calls that orbital mode "stellar inertial". Many times, you do want to be "tidally locked" with the earth and keep the same side facing, like when you have instruments, cameras, antennas, etc. There's a word for that I which I forget.

NASA has the "logs" of some of the Apollo flights posted, with transcripts of the whole mission. You'll see references to this stuff, where they have to "get their bearings" from various sources, and fire the reaction control rockets to get the spacecraft in the desired orbital mode. And they used stellar fixes to calibrate. In the parking orbit, one of the astronauts had to use a little telescope and get a fix on a number of stars. You'll see the transcipts of this.

-Richard

[grav]

My main question there was, do you--in your interpretation--have the housing always facing the sun, or always facing the stars?
Oh, yes. That's important, isn't it? :) In my examples, even though it may seem I said otherwise (and I guess I did :shifty: ), the calculations, if correct, are actually based on an orientation of the gyro with the fixed stars and positioned vertically with the Earth's equatorial plane. Sorry about that, and thank you. I have edited the original post to reflect that. Surprisingly, however, a vertical orientation with the ecliptic, although off by about a whopping 23 degrees, would only vary by about 8% in the calculations.

[grav]

Okay. So we have a precession on the rotation of the gyro of z2*Rgyro*(2.816*10-3 arc-seconds per year per meter). I should have stated this as z*z'*Rgyro*(2.816*10-3 arc-seconds per year per meter), since it turns out that z for the integration around the gyro and z' for the integration around the orbit are indeed different. z' for the orbit is simple. It is the summation of the cosine of the angle from zero to ninety degrees divided by the number of integrations (the number of points integrated, to get the average). In other words, at the south pole, the precession is zero. As it passes the equator, the precession is at a maximum, or one. At the north pole, it is again zero, and a maximum at the equator again. The value of the precession from minimum to maximum, zero to one, is simply the cosine of the angle from the center of the Earth to the probe to that of the equatorial plane. The integration comes to 2/pi. We now have a precession on the gyro of z*(2/pi)*Rgyro*(2.816*10-3 arc-seconds per year per meter). The integration for the value of z is more complex, but I will work on it next.

[publius]

Grav,

More homework: :)

Compare your prediction of the gyro precession with GR. There are actually two precessional effects on the gyro's spin axis, the "geodetic" effect and the gravitomagnetic effect(Lense-Thirring, or LT for short). IIRC, the geodetic precession is actually greater than the gravitomagnetic precession.

Remember this is a different effect than the GR precession of the periapsis of an orbit around a spherically symmetric mass. And remember that GR formula for that, 3R/d as you write it, the 'd' is not actually the average radius, but the "semilatus rectum" of the ellipse (look that up, a picture of what that is is worth a thousand words trying to define it).

For low eccentricity 'd' is very close to the semi-major axis, but for a highly elliptical orbit, it would be much smaller, making the precession much greater.

So that's your homework -- look up the actual LT formula, as well as the geodetic formula predictions and compare to your own. Don't get the
3R/d confused with LT and the geodectic effect.

-Richard

[grav]

Grav,

More homework: :)

Compare your prediction of the gyro precession with GR. There are actually two precessional effects on the gyro's spin axis, the "geodetic" effect and the gravitomagnetic effect(Lense-Thirring, or LT for short). IIRC, the geodetic precession is actually greater than the gravitomagnetic precession.

Remember this is a different effect than the GR precession of the periapsis of an orbit around a spherically symmetric mass. And remember that GR formula for that, 3R/d as you write it, the 'd' is not actually the average radius, but the "semilatus rectum" of the ellipse (look that up, a picture of what that is is worth a thousand words trying to define it).

For low eccentricity 'd' is very close to the semi-major axis, but for a highly elliptical orbit, it would be much smaller, making the precession much greater.

So that's your homework -- look up the actual LT formula, as well as the geodetic formula predictions and compare to your own. Don't get the
3R/d confused with LT and the geodectic effect.

-Richard
Well, I've completed some of my homework assignment. The dog ate the rest. No, really. ;) It appears that the Lense-Thirring effect has been verified to within 99% accuracy in accordance to GR using the satellites, LAGEOS l and LAGEOS ll. Where that leaves me, I do not know yet. At first I thought that was it. But then I thought, if the precession is actually caused by the equatorial bulge, then of course that would have already been figured in for the Earth. The equatorial bulge of the Earth is well known, and it would have been an obvious factor of the gravitational pull of Earth for a satellite orbitting at the equator. So I broke down and wrote a computer program that integrated the gravitational pull of a body along the equator at some distance from the center of a massive spinning body. The spin of a uniformly dense body would determine the extent of the equatorial bulge itself. The acceleration does become greater, but not by much. It turns out to be a'=a[1+(.6)(Requa-Rpole)R/d2] for d>>R. For a satellite orbitting at d=Requator, it is difficult to tell what what it would be exactly.

This formula is much too small for the precession, however. If it were (.636)(Requa-Rpole)/d=(2/pi)(Requa-Rpole)/d instead of (.6)(Requa-Rpole)R/d2, though, it would match the precession for that of the sun and its orbitting planets. But it is not. It is (.6)(pi/2)R/d smaller (unless I have made an error somewhere). The way I figure it, the extra acceleration becomes an extra velocity for the orbit, where a'd=v'2. So v'2=v2(1+z), where z is the fraction for precession, of course. We then get v'=v(1+z)1/2. Since z is small, v'=v(1+z/2). Finally, then, the precession caused purely by an equatorial bulge would be (v'-v)/v=z/2=(.3)(Requa-Rpole)R/d2.

But then I also thought about what you said before, publius, about how a0 might be -a0 in the direction of orbit. This would cancel out the total precession for a circular orbit, so that only highly elliptical ones become noticable. This makes sense, but in a round about kind of way. The precession that I am attempting to calculate applies only to the orbit itself. That is, the entire orbit will precess. We can tell this by measuring the degree of the precession of the orbit each time it passes closest to the orbitted body. Now what if the orbitting object is also precessing in its line of orbit, but in the opposite direction? This would make no difference to that of the precession of the orbit itself. It would still precess in the positive direction each time it passed the closest point of orbit. The precession of the orbitting body would simply give it a longer overall period. So now I must find some way to determine what the period should be other than through Newtonian means. In any case, if the orbit is elliptical, or even close to circular but not quite, we can always tell the precession of the orbit according to its point of closest approach. If it is too circular, however, the precession will look like an extra velocity of the orbitting object itself. Add that to a negative velocity for the precession in the line of orbit, and the total precession disappears.

It seems that a negative precession could only be caused by the Doppler effect, whereas a stronger gravity would be produced by the side spinning towards the orbitting body, pulling it backwards in that direction. Although it would also seem that a slightly greater positive velocity of the orbitting body itself due purely to the equatorial bulge might cause the entire orbit to precess as well. I am not sure. But in any case, I am now working on a program for integrating for the precession caused by a Doppler shift of gravity due to spin. I have found so far that this should indeed produce both types of precession, but I am not done working it all out yet. I'll keep you posted. :)

[publius]

Grav,

The difference in sign of acceleration I was talking about was with the gravitomagnetic force vis Nduriri's botched GEM equations. He was making the sign of the gravitomagnetic force on a moving mass be the same as the magnetic force on a moving charge. In fact, it has to be opposite, and this can be done by defining B_g to point in the opposite direction (ie "left hand rule"). So for a prograde orbit, it would point anti-radial, and radial for a retrograde orbit.

I wasn't saying anything about the direction of the MONDian a0. :)

B_g is too darn small to make much of a difference in orbits (although some of those papers I've linked to are looking for the tiny effects) anyway.


-Richard

[grav]

Okay. I have integrated the acceleration on an orbitting body due to a Doppler shift across many points of the gravitating body. Since the formula for the Doppler shift is (v+vo)/(v-vs), the gravitating body actually doesn't need to spin, since the velocity of the object in orbit will cause a Doppler shift of its own (a relative speed), so we have (v+vo)/v=(c+vo)/c=1+vo/c. So all we need to do is find the angle of the vectors between the line of travel of the orbitting object to each point in the the gravitating body. If the orbitting object is currently at some distance (d) from its center along the x axis, and is orbitting along the x,z plane, so that its axis of revolution lies along the y axis, the cosine of this angle is q=z/(d2+R'2-2dx)2, if I calculate correctly, where R'=(x2+y2+z2)1/2, and is the distance to each point from the center of the gravitating body.

So the first thing I did was to find points in a grid pattern so that they are spread uniformly, and skip the ones that did not lie within the sphere of radius R for the gravitating body. I then used the summation of (q+vo/c)(GM/d'/d')(d/vo2) for all of the points, where d'=[(d-x)2+y2+z2]1/2, and is the distance from the orbitting object to each point. The result is [(GM/c2)/d]1/2. However, I also noticed that if I integrate over [q+(vo/c)2](GM/d'/d')(d/vo2) instead, I get (GM/c2)/d as a result, one third of what Einstein predicts. If the second one is indeed correct, though, I wouldn't know how (v/c) becomes (v/c)2 in order to do so, or where the factor of three comes from.

[publius]

Grav,

Okay your "doppler shift" thing gets the precession ~ 1/sqrt(d), which is larger that ~1/d, and the difference is (v/c) vs (v/c)^2 in the integration. Well, relativistic effects (ie differences from classical behavior) start at order (v/c)^2, not (v/c). You will see this in all the relativistic formula, from SR on to GR.

And I'll remind you again, that the 'd' in the GR formula is actually the semilatus rectum of the orbital ellipse-- look that up, well if you draw a line perpendicular to the semi-major axis passing through a focus, the semi-latus rectum is the length where that intersects the curve of the ellipse. For a circle, that is just the radius, but for a long skinny ellipse, this would be very short compared to the semi-major axis.

And actually that itself (3R/d) is just a (close) approximation for orbits much greater than R where all the "crazy stuff" wouldn't be that strong.

Grav, if you want to see how this 3R/d thing comes about, I'll find something for you. Basically it comes from the corrections to the equations of motions. Do you have a mechanics text available?

If you do, or can get one, I want you (well, if you like :) I hope you would) to see how Netwonian gravity results in ellipses for the simple 2-body (or reduced 2-body where one mass is very small compared to the other so the big mass' motion can be ignored).

One does this in polar coordinates, and gets equations of motion for r(t) and theta(t). Now, enter the GR corrections for the equations of motions. This changes things slightly, and the 3R/d precession comes about from those changes.

To appreciate how it works, it's nice to know how Newtonian orbits work, so you'll see how the GR perturbations work.

-Richard

[publius]

Grav,

You know what you're thinking with your "doppler effect" thing, and I wonder if it would help to see the actual GR "acceleration" so you could compare with your doppler shifted g acceleration.

See those threads about the Schwarzschild metric (where I got a little hot under the collar! :o ) to see the derivation. Anyway, this is the GR *coordinate* acceleration we see from a far away reference frame, stationary with respect to the source. Now, this is for a radial free fall, no L, and with L, things would be different -- I would have to work a while, but what you get is velocity dependent component that is directed along the velocity vector, plus the radial component. This does not have that non-radial component in it:

g(r) = -GM/r^2 * [1 - 4R/r + 3(R/r)^2]; where R = 2GM/c^2 as always.

That is for a radial free-fall from infinity with zero initial velcocity -- heck we need a velocity dependent version that we can start with from scatch. Hold on..............

g(r) = -GM/r^2[ (1 - 2R/r +(v/c)^2) r - 4(v/c)^2 *( v * r) v ]

The bold r and v are *unit vectors*, radial and along the velocity, and the bold (v * r) is the dot product of these two. Note how all the extra stuff (save for 2R/r) comes in at (v/c)^2

That is the general GR coordinate equation of motion for a test particle. That sucker will give you the precession and all that.

ETA: This is the general equation of motion for the test particle in the Schwarzschild metric: spherically symmetric mass distribution, non-rotating. Rotation will give us the Kerr Metric, and B_g effects, but it is so small for the sun we don't have to worry for anything but ultra-precise stuff. The non-spherical moments will perturb things some more, but then the other planets perturb probably more. Anyway, that will give an idea of how GR changes the equations of motion.

-Richard

[grav]

g(r) = -GM/r^2[ (1 - 2R/r +(v/c)^2) r - 4(v/c)^2 *( v * r) v ]
It looks like I need to learn to read your equations better. I ran this equation according to the way I thought it might read and came up with exactly zero. I stopped between integrations to make sure it was calculating, and got a number other than zero, so it was only the end result. Is there any possible way you could expand the last part of this to show the meaning of the bold? The vector in the line of travel of the orbit is q (along the z plane), so I guess that would be the first r, right?

[publius]

Grav,

Maybe I need to learn how to write equations that can be read better! I'm sticking to pure text, and it's hard enough for me to even put bold in.

The bold v and r are unit vectors. The Newtonian -GM/r^2 multiplies the whole thing in brackets.

The first term inside (1 - 2R/r + (v/c)^2) mulitplies the Newtonian along the radial direction.

The second term is along the direction of v. The bold term is again a unit vector along the test mass's velocity vector. The bold ( v * r) is the dot product of these two unit vectors, the unit vector in the direction of the velocity dotted into the unit radial vector.

It may save some numerical trouble to write things in the "1 over r cubed" form, to save from needlessly calculating unit vectors when you already have the vectors themselves:

g(r) = -GM/r^3[ (1 - 2R/r +(v/c)^2) r - 4/(c)^2 *( v * r) v ]

Now, the bold v and r are the vectors themselves. Now to start a numerical solution of that, just specify the intitial v and r vectors and go to town. Note the second term is non-radial, directed along the velocity vector.


-Richard

[grav]

Maybe I need to learn how to write equations that can be read better! I'm sticking to pure text, and it's hard enough for me to even put bold in.

No. It's me. I don't know what dotting is, why I keep coming up with zero for the integration, if this is the integration you are showing or the end product, or even how you got that last equation from the first. I would like to run it through my program, but I'm not sure how it would translate into my language, which is strictly algebra and geometry. When things start getting dotted, I start getting confused. :eh:

[publius]

Grav,

Another thing about a numerical integration of that equation of motion. You want R/r to be large enough to get stong enough non-Newtonian effects, but not so large as to start causing the really strange behavior near the event horizon.

For example, the sucker is going to slow down greatly near the horizon (R/r ~ 1) in our time coordinates (which is the thing I was getting hot under the collar about. :) ) Keep r large enough at first to avoid that. I've seen various simulation programs and they all have numerical trouble near the horizon. Also "close in" you will get very chaotic looking "orbits".

If you keep r large enough, but not too large, you should get a nice precessing ellipse.

-Richard

[publius]

No. It's me. I don't know what dotting is, why I keep coming up with zero for the integration, if this is the integration you are showing or the end product, or even how you got that last equation from the first. I would like to run it through my program, but I'm not sure how it would translate into my language, which is strictly algebra and geometry. When things start getting dotted, I start getting confused. :eh:

Grav,

When you said integrating, I thought you were running a numerical solution to the differential equation of motion (ie a numerical gravity simulator). That is what my g(r) equation is, the vector acceleration as a function of r and v.

There is no integration of contributing mass elements. This the GR acceleration field for a spherical mass M. When M is centered at the origin, r is the position vector of the test mass. v is its velocity vector.

But I take you were NOT running a numerical integration of the motion? Well, that explains that. :)

-Richard

[grav]

Grav,

Another thing about a numerical integration of that equation of motion. You want R/r to be large enough to get stong enough non-Newtonian effects, but not so large as to start causing the really strange behavior near the event horizon.

For example, the sucker is going to slow down greatly near the horizon (R/r ~ 1) in our time coordinates (which is the thing I was getting hot under the collar about. ) Keep r large enough at first to avoid that. I've seen various simulation programs and they all have numerical trouble near the horizon. Also "close in" you will get very chaotic looking "orbits".

If you keep r large enough, but not too large, you should get a nice precessing ellipse.

-Richard
Oh, I guess I should have clarified. This is not a similation program for an actual orbit. It is an integration of the acceleration of gravity from many points within a gravitating body to some distance from it, along the line of motion of the orbit, or perpendicular to the gravitating body.

[grav]

Okay. I see what's happening here. Apparently, q on its own integrates to zero, so multiplying it times any constant integrates to zero as well. In your equation, publius, I changed the r's to q's, and the rest is constant, so I basically multiplied the whole thing times q and came up with zero. With mine, [q+(vo/c)](GM/d'2)(d/vo2), for d is large, d is approximately d', so (GM/d'2)(d/vo2) becomes 1, since GM=vo2d. That leaves just q+vo/c for the integration. This is really just the sum of two separate integrations, q and vo/c, so since the integration of q is zero, and vo/c is a constant, this becomes simply vo/c=[(GM/c2)/d]1/2. For (vo/c)2, then, it would be the square of that result, or (GM/c2)/d.

[grav]

Grav,

When you said integrating, I thought you were running a numerical solution to the differential equation of motion (ie a numerical gravity simulator). That is what my g(r) equation is, the vector acceleration as a function of r and v.

There is no integration of contributing mass elements. This the GR acceleration field for a spherical mass M. When M is centered at the origin, r is the position vector of the test mass. v is its velocity vector.

But I take you were NOT running a numerical integration of the motion? Well, that explains that. :)

-Richard
I just realized that I would probably have to run an orbit similator as well in order to find that amount of the precession which is given to the object in orbit and the amount that is given to the orbit itself. For all I know, they could cancel out completely for a circular orbit, which is what I am using here, but the difference might really only be noticable in highly elliptical orbits, as you mentioned earlier, publius.