Ok. I'm enjoying this and hope to learn something here. I've attached two .jpg image files, LINEAR.jpg and CIRCULAR.jpg. (don't know how to post the image directly). Examine LINEAR.jpg (I hope the images show up!) The relatavistic approach velocity to the observer at A is Vr = Vx cos(theta) yes? (the object at B has no Vy component) This is what you would use to calculate "gamma" between the two objects correct? This, I suppose, would become a differential equation since the rate of approach is a varying function of time (lethe?). Now, examine CIRCULAR.jpg At no instant is there a relativistic velocity difference between the observer and the object (theta is always 90 degrees (1.57rad)) as the orbiting objects coordinate system rotates with it (as does the earthbound observers coordinate system. Is this an Inertial frame set?). If this is true, then the only difference in tick rates for clocks on earth and the orbiting object would be accounted for by gravitational time dilation per GR (the orbiting object is much further out of the earth's gravity well). Right? Please dissect and destroy this simplistic representation at will. Thanks. Linear Please Register or Log in to view the hidden image! Circular Please Register or Log in to view the hidden image! Ok. got it!
NOTE: This only applies for an object in geostationary orbit or one that is expending energy to maintain a fixed position over an earthbound reference. Also note that the orbiting objects coordinate system only maintains its orientation with respect to the earth if it rotates at the exact same rate as the earth.
superluminal, Please, search Internet for the definition of the notion "linear velocity" and "angular velocity". You have no clue what you are talking about. You are using wrong terms, wrong equation (like Vr = Vx cos(theta)), and make a wrong conclusions. 1. Linear velosity is exactly the velocity of body on orbit, or orbital velosity and is defined by general rule V = |Lim [r(t+Δt) - r(t)]/Δt| at Δt rushing to zero. V can be expressed as the following: V = dL/dt where L=L(t) is the length of trajectory passed to moment t (i.e. L is the passed part of circumference, at circular orbit). It alwasys is derected along current tangent to the orbit. For a circular orbit the linear velosity has no Vr component, at all - it has only Vφ-component (only angular component, or two current components Vx and Vy, where Vx=Vcos φ and Vy=Vsin φ; V has no radial component, at all, at circular orbit!) 2. The angular velocity is w = dφ/dt. It means that V = wR, where R is radius of the circular orbit. Because of geometrical property L= φR, V is equal to wR. 3. In SRT gamma-factor always contains the velocity of body, V. In case of orbital motion - the linear velocity V, not w! And all that is elementary Math ... Namely this elementary Math shows that no one satellite can have zero velocity in regard to points of serface of Earth! Assuming something other is ... stupidity!
Whoa Yuriy! Not so harsh! I agree with you! After a minutes thought I see the flaw in the reasoning. We are all so used to seeing these examples in terms on linear motion, that an orbital example my not be immediately intuitively obvious. I think you all are having a problem explaining yourselves. The V in the calculation of gamma is the velocity I measure with respect to a moving object. Period. As the observer, my motion is irrelevant. Therefore, when I measure the velocity of a spacecraft on geostationary orbit to be about 11,066 km/h (which I always will) that will be the V in gamma. OK?
"Whoa Yuriy! Not so harsh! I agree with you! After a minutes thought I see the flaw in the reasoning. We are all so used to seeing these examples in terms on linear motion, that an orbital example my not be immediately intuitively obvious. I think you all are having a problem explaining yourselves. The V in the calculation of gamma is the velocity I measure with respect to a moving object. Period. As the observer, my motion is irrelevant. Therefore, when I measure the velocity of a spacecraft on geostationary orbit to be about 11,066 km/h (which I always will) that will be the V in gamma. OK?" Not "OK", but indeed "Whoa!" Where from you have got that I am harsh? I am not. I am using different tools to attract your attention to the main thoughts you should cogitate on. That is it. But even that does not achive its goal very often. Just as in this case. Indeed, I wrote main principles of speaking about velocity and relative velocity here (see my post from 06:47 AM). Did you follow it? You do not, and it creates impossibility to understand what you just said in your last post... 1. "The V in the calculation of gamma is the velocity I measure with respect to a moving object" I guess you wanted to say "The V in the calculation of gamma is the velocity I measure for a moving object" 2. "As the observer, my motion is irrelevant" Contrary, as an observer, your motion is very relevant: if you are sitting on the surface of Earth you are measuring the velocity of satellite in respect to surface of Earth, i.e. the relative velocity we all are talking about. If you are sitting on the axis of Earth somehow in its center, you are measuring velocity of satellite in respect to this point. Those two measured velocities are very, very different. So not me, but you "are having a problem explaining yourselves". 3. But you have said "I measure the velocity of a spacecraft on geostationary orbit to be about 11,066 km/h" Because there exist only the one indeed geostationary orbit, and its radius is R = Re (gT²/4π²Re)^1/3 .......................... (**) (see my post from 15-01-05 11:45 AM) the linear velocity for it is V = 2πR/T = 2π(gRe²/4π²T)^1/3 = =2π(10*40.96*10^12/4*9.86*24*3600)^1/3 or approximately V = 3.1km/s relatively to the axis of Earth! As you know, the linear speed of points the surface of Earth due to Earth day-night rotation is Ve = 465.4m/s Therefore the relative velocity of geostationary satellite in respect to points of Earth surface is aproximately 2.66km/s All application of these facts to the discussion on this thread you can do by yourself...
Yuriy, you are indeed correct in each instance. My wording is poor sometimes. I mean to say the velocity I measure "for" a moving object, and that as an observer, I consider myself to be "at rest" and my measurement of the geostationary satellite will indeed reflect the fact that I am moving at approx 465.4m/s on the surface and will yield the approx. 2.66km/s as you state. Apologies for my imprecision.
by Yuriy: V = 3.1km/s relatively to the axis of Earth! ============================================================= This is where we have our misunderstanding. The Earth-Centered Inertial frame of reference I used, the frame of reference is an INERTIAL frame. It does not rotate. It is the frame in which all clocks in the GPS system are synchronized with each other, both Earth-surface clocks and satellite clocks. The ECI frame rotates within the ICRF frame when the ICRF frame is used. The Earth's surface THEN becomes a NON-INERTIAL Earth-Centered Earth-fixed (ECEF) frame where effects such as the Sagnac effect differentiate it from the ECI frame. The two frames are not Lorentz Invariant, as only one is an inertial frame. Here's a thought experiment for you to try, Yuriy. You seem to believe the geostationary satellite is in an inertial frame of reference always. In an inertial frame of reference, the physics from the observer located on a geostationary satellite would have to be Lorentz Invariant with the physics of an observer located on the Earth's surface. Let an observer located on the Earth's surface aim a laser at the geostationary satellite and fire it. The laser will MISS the satellite because the Earth observer aimed at the time delayed image of the satellite. There is a .1 second travel time for light each way, causing the laser beam to arrive .2 seconds BEHIND the satellite. Let an observer on the geostationary satellite aim a laser at a specific point on the Earth's surface. The laser beam will hit its mark because the Earth did not move from its observed position during the .2 second interval. In reality, the Earth would move in its orbit around the sun, but that can be discounted because the Earth/satellite system moves together in that orbit. So, in my way of thinking, the non-rotating ECI frame would be inertial with no relative motion between geo satellite and Earth surface, and the rotating ECI/ECEF frame would be non-inertial because the viewpoints of the satellite observer and the Earth observer would be different, not Lorentz Invariant. Does SR still apply in the non-inertial rotating frame? I know you will claim I am all mixed up, and I may be, but that is the way I preceive things, hehe. After all, I am not a trained physicist.
Superluminal, There is nothing you should apologize for. Simply try to follow the scientific logic and facts in your researches… 2inquisitive, 1. All numbers I calculated in my post were obtained in the inertial reference frame, axis Z of which is coinciding with axis of Earth day-night rotation and origin of which coincides with center of Earth. Sitting there, I see Earth and geostationary satellite rotating uniformly with period T = 24 hrs. Only using this system I can apply the Lorentz transformation, relativistic Doppler effect formulas and all other achievements of SRT. 2. I do not care, who and why has introduced the reference frames you are mentioning in your response: I do not connect my posts with GPS or any other concrete systems. Because all what I am concerning is the right application of SRT results. 3. One semantic notice: never say “reference frame is Lorentz-invariant”. This is absurd. Variables, events, equations can be Lorentz-invariant (each – in its special and well-defined sense), not RF. 4. There can not be in Nature any inertial RF rigidly tied with any point of Earth surface, no matter what a funny name you will chose for it. 5. “Does SR still apply in the non-inertial rotating frame?” Only in the following sense: let us consider inertial RF which has currently the same velocity (as a vector!) as our non-inertial one has instantly. Such inertial RF and all SRT conclusions in there are applicable to the instant situation in the considered non-inertial RF. 6. My advice: reconsider all your fantasies about aiming and shooting marks…
Excellant. Somebody wishes to actually seek an answer. Since I asked the question I'll not attempt to give a bonafide answer but based on the adjustment made for GPS (which isn't geosynchronous) but uses the local preferred rest frame that the velocity adjustment made for gPS is due to its velocity of orbit. As I calculated above using Yuriy's method of Relative Velocity" (V3 in the example, one gets a different value. Such that My thoughts are that the geosynchronous orbit will also show time dilation meaning tht relative velocity is out and gamma only is in. BTW. Thanks for your scientific post.
2Inquisitive, You are doing just fine. Don't fret about Yuriy's intense desire to claim GPS proves SRT. It proves the opposite. The issue is really quite simple. View the earths surface and the geosynchronous orbit as points on the radii of a rotating disk. Both have velocity. The surface velocity (V1) is obviously lower than the geo-orbit velocity (V2). The point is it will always be the case that the geo-orbit has the higher velocity and the SRT principle of swaping observers views at rest is a dead horse. Further you will find that taking each clock in reference to its absolute velocity of orbit/rotation with the reference being the earths axis, the V1, V2 gamma adjustments have different differential numbers numbers than if you claim V3 = (V2-V1) (Relative velocity) and apply SRT with V3 to compute gamma. You get the wrong gamma. Relative velocity (SRT) is out. Gamma is in.
by Dr. Neil Ashby: Suppose for a moment there were no gravitational fields. Then one could picture an underlying non-rotating reference frame, a local inertial frame, unattached to the spin of the earth, but with its origin at the center of the earth. In this non-rotating frame, a fictitious set of standard clocks is introduced, available anywhere, all of them being synchronized by the Einstein synchronization procedure, and running at agreed upon rates such that synchronization is maintained. These clocks read the coordinate time t. Next, one introduces the rotating earth with a set of standard clocks distributed around upon it, possibly roving around. One applies to each of the standard clocks a set of corrections based on the known positions and motions of the clocks, given by Eq. (28). This generates a ``coordinate clock time'' in the earth-fixed, rotating system. This time is such that at each instant the coordinate clock agrees with a fictitious atomic clock at rest in the local inertial frame, whose position coincides with the earth-based standard clock at that instant. Thus, coordinate time is equivalent to time that would be measured by standard clocks at rest in the local inertial frame [7]. When the gravitational field due to the earth is considered, the picture is only a little more complicated. There still exists a coordinate time that can be found by computing a correction for gravitational redshift, given by the first correction term in Eq. (28). ================================================================ OK, Yuriy, here is where I am confused. You speak of a rotating inertial frame in which you do your SR calculations. Quote by Yuriy: "1. All numbers I calculated in my post were obtained in the inertial reference frame, axis Z of which is coinciding with axis of Earth day-night rotation and origin of which coincides with center of Earth. Sitting there, I see Earth and geostationary satellite rotating uniformly with period T = 24 hrs. Only using this system I can apply the Lorentz transformation, relativistic Doppler effect formulas and all other achievements of SRT." End quote by Yuriy. Everything I have read, and I do admit I am a layman and may misunderstand something, everything I have read says a ROTATING frame of reference is a NON-INERTIAL frame of reference. I realize one can do SR calculations within this rotating frame by breaking it into small segments, but a rotating frame is a non-inertial frame from what I have read. In the first part of the quote by Dr. Ashby above, he is speaking of the Earth-Centered Inertial frame I have been referring to. It is a non-rotating frame attached to the center of the Earth. The second frame he speaks of is the non-inertial rotating frame, the Earth-Centered Earth-Fixed frame. Dr. Ashby is one of the designers of the GPS system and is still a consultant with the USNO that operates the system. Can you see where my confusion comes from? A link to the page I copied from: http://relativity.livingreviews.org/Articles/lrr-2003-1/node4.html
Now I understand where from comes your confusion. It is amazing how easy misunderstanding can appear between two talking people … from nothing! I said exactly what I meant, you heard exactly what I meant and concluded …exactly opposite to what I meant! Who is to blame on? My father always told me: “Remember son, dialog always means that there are two people who creates it”. I guess, we both are “guilty”... 1. Let me repeat what I said: “All numbers I calculated in my post were obtained in the inertial reference frame, axis Z of which is coinciding with axis of Earth day-night rotation and origin of which coincides with center of Earth. Sitting there, I see Earth and geostationary satellite rotating uniformly with period T = 24 hrs. Only using this system I can apply the Lorentz transformation, relativistic Doppler effect formulas and all other achievements of SRT." Let me divide it on set of separated simple sentences. All numbers I calculated in my post were obtained in the inertial reference frame. Axis Z of which is coinciding with axis of Earth day-night rotation Origin of which coincides with center of Earth. Sitting there, I see Earth and geostationary satellite rotating uniformly with period T = 24 hrs. Only using this system I can apply the Lorentz transformation, relativistic Doppler effect formulas and all other achievements of SRT." Where from you have got that this RF, I am speaking about, is "a rotating inertial frame"? (Your phrase: “You speak of a rotating inertial frame”) How I could see Earth and satellites rotated with period T = 24 hrs … if I am rotating too? I was speaking exactly about RF, which Ashby calls as ECI. 2. I saw the page you link on. As you can see, the basic initial equation (24) is … exactly the Einstein’s expression of the Minkowski interval for a rotating Earth! Therefore, entire GPS-program is totally based …upon SRT+GRT without any exceptions and/or changes! So, whole information, spread here by MacM about contradictions between GPS and SRT, is a pure BS… 3. BTW, Ashby did one small mistake: at transit from (27) to (28), he has put a wrong sign at term with v^2/c^2…. P.S. What the hell is "a rotating inertial frame"? Something you got from MacM?
Interesting how General Relativity is specifically mentioned. Special Relativity is not. Yet there is mention of "Motion" adjustment. I have pointed out how that motion adjustment is made. It is not SRT based. "Yuriy". You cannot steal Lorentz concept and claim it a victory for SRT, no matter how much you might like to. Lorentz and SRT both use the same mathematics when computing "Motion" affects. So the use of the "Lorentz" formulas does not equate to being SRT. It is Gamma only. The difference if you really do not understand is in the claims made by each view point. SRT: 1 - There is no absolute motion only relative motion. 2 - There is no special rest frame. 3 - Observer views are symmetrical and may be reversed. That is either one may equally consider themselves as at rest. 4 - All calculations are between observers in motion. I have already shown that if V1 is surface velocity, V2 orbit velocity and V3 = (V2-V1) that: A - GPS computes Gamma1 for V1 and Gamma2 for V2 where the observers use a local special rest frame. The time dilation affects of motion being Gamma3 = (Gamma2 - Gamma1.) B - Computing Gamma using V3 = (V2 - V1) and Gamma = 1/ (1 - V3<sup>2</sup>/c<sup>2</sup>)<sup>1/2</sup>; yields a wrong time dilation result. C - The use of the rest frame concept is not part of SRT. It causes (Rightfully) the orbit clock to always be the clock in motion with higher velocity and can never be reversed to claim it is at rest. THIS IS THE MOST CRITICAL COMPONENT OF THIS ISSUE. THE VIEWS ARE NOT REVERSABLE AND THEY SHOULD NOT BE REVERSABLE. THAT IS THE FLAW OF SRT THINKING. Even relative velocity is based on differential of two absolute velocities, even though one cannot directly tell which has the higher velocity. Testing however, can tell the difference and can be compensated for. That is the GPS system. It tests which actually has the higher velocity, makes the correction and forgets all about Einstien's kooky idea about relative velocity and either being at rest. D - Inspite of your yelling and unwarranted claims of authority on this issue. You are wrong. SRT is not used in GPS. SRT fails the test of reality. Gamma is used in GPS but is used based on frames of reference which are not allowed in SRT. So kiss off with the attacks. Suck it in. SRT fails. Lorentz wins.
by Yuriy: 1. Let me repeat what I said: “All numbers I calculated in my post were obtained in the inertial reference frame, axis Z of which is coinciding with axis of Earth day-night rotation and origin of which coincides with center of Earth. Sitting there, I see Earth and geostationary satellite rotating uniformly with period T = 24 hrs. also: Where from you have got that this RF, I am speaking about, is "a rotating inertial frame"? (Your phrase: “You speak of a rotating inertial frame”) How I could see Earth and satellites rotated with period T = 24 hrs … if I am rotating too? I was speaking exactly about RF, which Ashby calls as ECI. ============================================================== From your statements, I concluded (perhaps incorrectly) you were saying the WHOLE FRAME was rotating, just the geostationary were not rotating in reference to the Earth's surface, and that whole frame was inertial. In the ECI frame, the WHOLE FRAME is NOT considered to be rotating, just the GPS satellites DO have motion relative to the surface of the Earth. The surface of the Earth is mathematically represented by WGS-84, which is a part of the Earth-Centered Earth-Fixed frame. The ECI/ECEF frames THEN rotate within the International Celestial Reference frame, a non-rotating frame of reference gleaned by plotting the location of over 500 radio sources in the universe. Even though DR. Ashby states SR time dilation is taken into account, it does not seem to be reflected in the pre-launch adjustments to the satellite clocks to me. by Dr. Ashby: The negative sign in this result means that the standard clock in orbit is beating too fast, primarily because its frequency is gravitationally blueshifted. In order for the satellite clock to appear to an observer on the geoid to beat at the chosen frequency of 10.23 MHz, the satellite clocks are adjusted lower in frequency so that the proper frequency is: [-44.647 m/s] X 10.23 MHz = 10.229 999 99543 MHz This adjustment is accomplished on the ground before the clock is placed in orbit." From my understanding, General Relativity predicts the satellite clocks would beat fast by 44.65m/s per day due to gravitational effects. Special Relativity predicts the satellite clocks would beat slow by 7.2 microseconds per day due to relative velocity between the satellite clocks and the Earth-based clocks. Most universities claim the pre-set on the clocks before launch is 38 microseconds to account for SR and GR. That is not true, the pre-set is for -44.647 microseconds per day. A link: http://relativity.livingreviews.org/Articles/lrr-2003-1/node5.html
2inquisitive, You know, even now I can understand did you get what RF I am talking about, or do not? Please, forget, for a moment, any clocks adjustments, speak only about definition of the reference frames that you will appeal. We should for the beginning come to agreement about RF, and then we will easily recognize what corrections and adjustment should be done to compensate different effects: of SRT, of GRT, etc. My reference frame is well defined and, I hope, now you understand what it is. Am I right? Please, confirm it. After that, define all other RF you will use, no matter what for you will use them; just define them...
Talk is useless; post something about it yourself so we can all learn something, smile at each other, and be happy.
How do you know, MacM? Well, we know that yours is suck. Please Register or Log in to view the hidden image!
