JerryWise
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Reged: 12/26/03
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This has to be simple for those more informed.
If observer "A" traveling at .8 C does a flyby of the Earth and the Sun they will observe a shorter distance between the two objects than observer "B" on Earth. The calculations based on the distance observed by "A" would not support a stable orbit at the masses and distances observed during the flyby. Were the distance measurements observed in "A's" frame of reference applied to "B's" frame of reference the Earth would spiral into the Sun.
We know "A" has valid distance measurements within his frame of reference. We also know "B" has valid distance measurements in his frame of reference. One's frame supports stable orbital mathematics, the other does not. How is this one addressed?
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Jerry
LX200ACF 14", Tak FS 152 & TOA 150
AP-1200 & Mach1
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Jarad
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Loc: Atlanta, GA
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Both will see a stable orbit, but not the same shape. Let's assume the earth's orbit is a perfect circle. To the observe moving at 0.8 C, it appears to be squashed in their direction of motion into an ellipse, and would appear to move fastest when moving along the long sides closest to the sun (where earth's motion is perpendicular to their motion, so not compressed). So the earth would see a circular orbit with constant velocity, and the 0.8C visitor would see an elliptical orbit where the moves faster when it's closer to the sun, and slower when further away. Both fit a stable orbit, although not the same orbit.
Jarad
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LesB
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Who would be the non-inertial observer in this case or would that apply?
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"The genius of humanity is to establish an identity which lies at an ever-increasing distance from our organic nature." Ray Tallis
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Jarad
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I was assuming he meant both were in rest frames (no accelleration, SR, not GR).
I presume the situation will be more complex with accelleration, but the end result should still be that both obervers see a situation of a stable orbit (although the shape, speed, and masses of the objects will differ).
Jarad
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JerryWise
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Loc: Lexington, SC
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At this time, I was addressing Special Relativity. The mass is what I was curious about. And not from what the observer "A" sees but what he calculates math wise for the orbit. In the example, observer "A" will pass the two objects in basically a straight line (or great circle). He will observe passing Earth at X time. He will observe passing the Sun at Y time. Since the distance between the passings is less from his observations (made in his frame of reference at .8C) he will use that distance in his orbital calculations. If he calculates the orbits using the mass of the two bodies as they were when he left Earth to begin his flight, the Earth could not maintain the orbit it had when he left. His math will show an unsustainable orbit unless the mass is altered too in Special Relativity.
If we alter mass, do we not then present additional problems for general relativity?
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Jerry
LX200ACF 14", Tak FS 152 & TOA 150
AP-1200 & Mach1
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Pess
(Title)
Reged: 09/12/07
Posts: 1910
Loc: Toledo, Ohio
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The problem, as you present it, mixes apples and oranges and wonders why you got a lemon. (this fruit diet my wife has me on is making me delusional) 
In simple terms you can't use Newtonian physics to compute a relativistic effect.
For example, using simple geometry you can measure the length of any line by the formula:
L^2= X^2 + Y^2 + Z^2
This is simply three dimensional Cartesian coordinates and the Pythagorean theorem applied.
If you attempt to use this formula to compute orbital mechanics, you'll find it won't work if there is significant differential velocity while you take the measurement.
However, in Einsteins universe the length of a rod can not be computed this way since any relative motion between the measurer & measureee would not take into account relativistic changes.
Einstein tells us that we live in a 4 dimensional universe where we require length, width, height, AND duration to define something.
Therefore our 'simple' geometry to measure a length in his universe is:
L^2= X^2 + Y^2 + Z^2 - (ct)^2
where ct is the speed of light times the time coordinate (time of existence measured in seconds). So since 'c' is given in meters per second and 't' is given in seconds this variable reduces to a length measurement.
Plug this in and your planets and sun will still happily circle each other.
Pesse (Off for some orange slices.) Mist
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JerryWise
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Reged: 12/26/03
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Loc: Lexington, SC
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Thanks Pess. I knew it would be simple but simple in a way beyond me. So would this mean Newtonian physics will work for observer "B's" frame of reference on Earth but not "A's" on the spaceship?
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If you attempt to use this formula to compute orbital mechanics, you'll find it won't work if there is significant differential velocity while you take the measurement.
Just thought back over this too. Would not a differential velocity not indicate a preferred frame of reference to compute the differential from?
Man, this is good stuff. Sorry about the additional edits but it does bring up some points.
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Therefore our 'simple' geometry to measure a length in his universe is:
L^2= X^2 + Y^2 + Z^2 - (ct)^2
where ct is the speed of light times the time coordinate (time of existence measured in seconds). So since 'c' is given in meters per second and 't' is given in seconds this variable reduces to a length measurement.
What is t the "time of existence measured in seconds" the existence of? I miss that.
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Jerry
LX200ACF 14", Tak FS 152 & TOA 150
AP-1200 & Mach1
Edited by JerryWise (09/28/08 09:23 PM)
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Pess
(Title)
Reged: 09/12/07
Posts: 1910
Loc: Toledo, Ohio
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Quote:
Just thought back over this too. Would not a differential velocity not indicate a preferred frame of reference to compute the differential from?
The entire underlying foundation of Einsteins relativity sits on the postulate that there is no 'preferred' frame of reference.
You can be in a ship and zip past the solar system at .8c, or you can be sitting in a stopped ship while the solar system zips by you at .8c.
Both are exactly the same and equally valid.
Pesse (It's what puts the relative in relativity.) Mist
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JerryWise
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Reged: 12/26/03
Posts: 6880
Loc: Lexington, SC
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Quote:
The entire underlying foundation of Einsteins relativity sits on the postulate that there is no 'preferred' frame of reference.
You can be in a ship and zip past the solar system at .8c, or you can be sitting in a stopped ship while the solar system zips by you at .8c.
Both are exactly the same and equally valid.
Pesse (It's what puts the relative in relativity.) Mist
I understand the principal. Seems if Newtonian Physics worked to some extent in one frame of reference using simple calculations it should work in any frame without Relativistic considerations.
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Jerry
LX200ACF 14", Tak FS 152 & TOA 150
AP-1200 & Mach1
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LesB
Carpal Tunnel
Reged: 12/20/04
Posts: 1687
Loc: Z-Hills, FL
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I'll argue that the observer "A" was an inertial observer when he began his flight and the failure lies in the adjustment to relativistic speeds. Lorentz tranforms should come into play at this point and I am assuming uniform motion.
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"The genius of humanity is to establish an identity which lies at an ever-increasing distance from our organic nature." Ray Tallis
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JerryWise
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Reged: 12/26/03
Posts: 6880
Loc: Lexington, SC
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Quote:
the Lorentz transformation converts between two different observers' measurements of space and time, where one observer is in constant motion with respect to the other.
This looks good too.
I think where I'm getting a bit off base is going by the given "the laws of Physics are the same in any frame of reference". Observer A and B are in two different frames of reference. There is nothing relative between them. No "with respect to the other".
One observer is on the planet in orbit around the star. He makes an observation the star has a quantifiable mass, the planet has a quantifiable mass, and the distance between them is quantifiable as well as the rotational velocity. These four parameters should balance in any frame of reference for a successful orbit as observed.
If a second observer (A) passes this star and planet at .8C one of the parameters (distance between the star and planet) changes. Observer B's calculations have no bearing on observer A but calculations used by observer B should yield the same results for the same conditions to exist in both frames of reference. They both see a sun/planet system in orbit governed by the laws of physics. Simple Newtonian Physics should work in both frames of reference to the extent it works in either as the laws of physics are identical in both. Two objects orbiting their barycenter should function the same in any frame of reference. If they did not, would this not create the dreaded "special frame of reference"?
(This is what yall get for recommending "the Teaching Company" DVDs.)
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Jerry
LX200ACF 14", Tak FS 152 & TOA 150
AP-1200 & Mach1
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llanitedave
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Posts: 10483
Loc: Amargosa Valley, NV, USA
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What you're missing is that Newtonian physics never provides exact solutions. It's an approximation that is very close to correct at low velocities and masses, but becomes more and more inaccurate at higher velocities. At 0.8c, the Newtonian solution is way off.
The relativistic calculations will be correct at any speed.
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"S.O.E." (Sauron's Other Eye) 16" Royce conical mirror: A permanent work in progress.
10" Homebuilt dob, old Coulter mirror
Next Project: The "Eye of Sauron" Observatory!
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George Jones
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Reged: 09/27/08
Posts: 18
Loc: Saint John, NB, Canada
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There is a way to do this within special relativity, but the force must be handled in a relativistic way. If I remember correctly (I've forgotten the reference), this gives precessing "ellipses", with the amount of precession predicted different than that given by general relativity. The results are more accurate than Newton/Galileo, but less accurate than general relativity.
I have developed an interactive Java animation for the extremely (general) relativistic case of orbits about spherical black holes. Some folks here might find it fun to play with.
The animation takes as input from the user the initial r coordinate, and the initial speed and angle with respect to horizontal for a platform hovering (and experiencing an incredibly large "g-force") at that r value.
The animation sits in a frame in a bunch of hyperlinked html files. These files contain a number of short explanations and write-ups for some suggested virtual experiments that can be performed with the package.
Experiment 1: Falling Into a Black Hole
Experiment 2: Escape Velocity
Experiment 3: Investigation of Stable Circular Orbits
Experiment 4: Investigation of Unstable Circular Orbits
Experiment 5: Investigation of Spiral Orbits
Experiment 6: Gravity Bends Light Rays!
Experiment 7: Boundaries Between Orbits of Various Kinds
Experiment 8: Orbital Precession and Closed Orbits
I've had a lot of fun (and some frustrations) developing and playing with the package. I have attached a zipped file that contains all the necessary files. To use the animation, unzip the zipped file, making sure that all files are extracted into the same folder. (A new folder should probably created before doing this.) Click on the file blackHoleOrbits.html to view and interact with the animation.
The animation is not at a professional level, i.e., it has glitches and warts.
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Regards,
George
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Jarad
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Reged: 04/28/03
Posts: 3858
Loc: Atlanta, GA
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Quote:
One observer is on the planet in orbit around the star. He makes an observation the star has a quantifiable mass, the planet has a quantifiable mass, and the distance between them is quantifiable as well as the rotational velocity. These four parameters should balance in any frame of reference for a successful orbit as observed.
If a second observer (A) passes this star and planet at .8C one of the parameters (distance between the star and planet) changes.
Actually, they all change. Time is dilated, so velocities change. Mass measurements are also affected by relativity, so the apparent masses change. And distances change as you have already noted. But the orbital equations will still balance.
Quote:
Observer B's calculations have no bearing on observer A but calculations used by observer B should yield the same results for the same conditions to exist in both frames of reference. They both see a sun/planet system in orbit governed by the laws of physics.
And they do.
Jarad
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JerryWise
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Reged: 12/26/03
Posts: 6880
Loc: Lexington, SC
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Quote:
What you're missing is that Newtonian physics never provides exact solutions. It's an approximation that is very close to correct at low velocities and masses, but becomes more and more inaccurate at higher velocities. At 0.8c, the Newtonian solution is way off.
I don't think I'm missing this. Certainly at .8C Newtonian physics would fail miserably. The observer traveling at .8C is doing so relative to an observer B. Or, Observer B is moving backward at .8C. Both are in a steady state. Physics will be the same for both observers, even imperfect physics. Neither holds a special observational platform. They are observing the simple orbiting of a planet and star and the physics working at these masses and velocities should work exactly the same for both observers. If Newtonian Physics will give a close approximation of orbits at the extremely slow speeds (relative to C) of the planets orbital velocities it should do so in either frame of reference. Is that not correct? If not, one of the observers is in a special state and special states cannot exist. The only difference between the two observers is the method of measurement between the planet and star. One used his passing of the planet and star as his yardstick. The other, traditional methods. Time space dilation means nothing to either observer. It only means something when comparing differences between the two observer's observations.
So Newtonian physics is imperfect at light speeds but it should be equally imperfect in both frames of reference. If the base line distance is different, something else is different. Taking a simple base line measurement at .8C imputes no special modification to physics at any frame of reference. It only has meaning comparing frames of reference.
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Actually, they all change. Time is dilated, so velocities change. Mass measurements are also affected by relativity, so the apparent masses change. And distances change as you have already noted. But the orbital equations will still balance.
Ok, that would fit. Thank you Jarad. It is the Mass measurements that would balance out the imperfect but adequate Newtonian Physics at the slow (relative to C) orbital velocities of the planet/star. I wasn't aware observed masses changed as the observer took a baseline measurement at .8C. It does make an interesting little thought game.
There had to be another change in the parameters. Thinking about the transiting measurement approaching C triggered the above question. As the speed increased and the distance diminished it would move the Earth's orbit into the Sun and the barycenter to the Sun's core. We know that can't be because the physics are the same in any observational platform. And besides, it wouldn't be a good thing.
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Jerry
LX200ACF 14", Tak FS 152 & TOA 150
AP-1200 & Mach1
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Qkslvr
Pooh-Bah
Reged: 06/23/06
Posts: 1054
Loc: NE Ohio, US
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Aren't some of the distant galaxies moving at high speed in relationship to the suns orbit around the milky way?
What about the orbits in high G near the SM black hole at galactic center?
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Mike
N8/CG-5/40D
Coming sometime/Maybe FrankenRebel
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George Jones
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Reged: 09/27/08
Posts: 18
Loc: Saint John, NB, Canada
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There is more to special relativity than time dilation, Lorentz contraction, and relativistic mass. These are special cases of more general concepts.
The bound orbits for Newtonian gravity are found by using Newtonian gravity in Newton's second law F = m a. To do this relativistically, a (special) relativistic generalization of Newtonian force and Newton's second law must be found. This can be done.
Now, assume a planet's orbit is viewed from two frames. The values of the various physical quantities depend on the frame from which they're viewed, but the laws of physics are same in all frames.
Time, positions, velocities, and accelerations are frame-dependent, but so are expression for force. Special relativity gives a prescription (called a Lorentz transformation) for transforming values of these quantities in frame to values in another frame.
Let F and F' be the forces as viewed in the two frames, and a and a' be the acceleration in the two frames. Solve F = m a in one frame to find the orbit of the planet in one frame. Solve F' = m a' to find the orbit in the other frame. This what's meant by the laws of physics are the same in all (inertial) frames.
The two orbits found in this way will be related by a Lorentz transformation of positions. Physics is consistent.
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Regards,
George
Edited by George Jones (09/29/08 10:16 AM)
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Pess
(Title)
Reged: 09/12/07
Posts: 1910
Loc: Toledo, Ohio
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Well, I thought I made it pretty clear but apparently not.
Let's try another example.
Suppose directly in front of you two people hold the end of a rope with a ball tied on each end. One person is to your right and the other, your left.
I ask you to measure the rope without moving from your chair.
You use whatever method for measuring from a distance you feel comfortable with and come up with 10 feet as an answer.
A----------B
You
10' observed length
..or Rope Length^2 = (10)^2 + (0)^2
rope length thus equals 10 feet
Now I ask one of the people holding the rope to step away from you about 2 feet so the rope is now oblique to you.
I again ask you to measure the length of rope without moving from your chair.
You NOW report the rope 'appears' to be about 9.8 feet long but since you observed that the rope is still taunt between person A and person B you think physics has been violated since you did not observe the people approach one another...yet you are perplexed because the rope is now shorter!
the answer is now obvious. You did not take into account the third dimension and once you do the length comes out the same.
..or Rope Length^2 = (9.8)^2 + (0)^2 + (2)^2
rope length still equals 10 feet.
Your problem above was measuring a 3 dimensional length by two dimensional math. When you incorporate the third dimension things become accurate.
Same is true with your measurements in a 4D universe. You must incorporate the fourth dimension in your math or you are using Newtonian 3-dimensional math which is wrong.
There are other ways to approach the answer to this question but this is the most straightforward and easiest to understand.
Pesse (Time to go fishing.) Mist
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JerryWise
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Reged: 12/26/03
Posts: 6880
Loc: Lexington, SC
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I liked the Geese thing better.
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Jerry
LX200ACF 14", Tak FS 152 & TOA 150
AP-1200 & Mach1
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JerryWise
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Reged: 12/26/03
Posts: 6880
Loc: Lexington, SC
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Quote:
Well, I thought I made it pretty clear but apparently not.
............
Sorry Pess. My fault but I just can't seem to get my feeble mind around your analogies and fit them to the stated circumstances. It would be very easy to pre-suppose a knowledge level way over my head in these things. So I apologize for not picking up on your examples.
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Jerry
LX200ACF 14", Tak FS 152 & TOA 150
AP-1200 & Mach1
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