Implications of the statement "Acceleration is not relative"by GregAshmore Tags: implications, statement 

#127
Feb1613, 04:33 PM

P: 2,892

Addressing the OP more specifically, maybe he is finding difficult to understand why if there is no two notions of velocity coexisting (an absolute and a relative velocity) in relativity, one does have both absolute and relative acceleration. While one justifies easily the first case because in relativity we have a spacetime rather than separate space and time and therefore one doesn't have any absolute space wrt wich define an absolute velocity, this doesn't work in the case of acceleration and ultimately it seems one has to settle with "because this is just the way it is".
This isn't something that is very often clarified. 



#128
Feb1613, 05:08 PM

P: 1,657

You can define a 4velocity [itex]V^\mu[/itex] by [itex]V^\mu = \dfrac{d}{d \tau} x^\mu[/itex] and you can similarly define a 4acceleration [itex]A^\mu[/itex]. Neither is more absolute than the other. However, you can always choose coordinates so that the spatial components of [itex]V^\mu[/itex] are all zero, but you can't always do that for the spatial components of [itex]A^\mu[/itex] 



#129
Feb1613, 06:13 PM

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At this point I guess I should wait for the OP to confirm if this gets any close to his line of thought. 



#130
Feb1713, 06:02 AM

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"To be sure, the accelerated coordinate systems cannot be called upon as real causes for the field, an opinion that a jocular critic saw fit to attribute to me on one occasion. But all the stars that are in the universe, can be conceived as taking part in bringing forth the gravitational field; because during the accelerated phases of the coordinate system K' they are accelerated relative to the latter and thereby can induce a gravitational field, similar to how electric charges in accelerated motion can induce an electric field." 



#131
Feb1713, 06:25 AM

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#132
Feb1713, 06:40 AM

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You certainly were not explicit at all about what you believe he means by the term "gravitational field" even when directly asked for clarification, and you seem to disagree with Einstein's use of the term although you quote him. If you wish to clarify what specifically you believe Einsetin refers to by the term "gravitational field" then we can continue the discussion. I don't understand your reluctance to clarify your position. Surely by now you realize how easily misunderstandings can arise in online forums. A request for clarification should always be taken seriously and complied with willingly. 



#133
Feb1713, 07:41 AM

P: 1,657

The first 3 aspects of gravity don't involve coordinates at all, and only mention physical forces, not fictitious forces. The 4th aspect to me is what the meaning of "fictitious forces" are, which is that the coordinate flows are not geodesics. The terminology "coordinate flow" is made up by me; I'm not sure what the technical term is, or if there is one. But in a similar way that specifying an initial point in spacetime, together with an initial velocity, determines a path through spacetimethe geodesic, specifying an initial point in spacetime, together with an initial velocity, determines a different path through spacetime, the coordinate flow, which is the solution to the component equation: [itex]\dfrac{dV^\mu}{d\lambda} = 0[/itex] (where [itex]\lambda[/itex] is the affine parameter for the path). This would be a geodesic, if it were flat spacetime and inertial cartesian coordinates were used. 



#134
Feb1713, 08:03 AM

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#135
Feb1713, 08:46 AM

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I just wanted to add a few words for GregAshmore concerning the distinction between coordinate and proper acceleration in the form of a thought experiment, in the hopes it might help make it more obvious.
I take it you understand the relative nature of velocity, but when two inertial objects have a constant velocity with respect to each other it cannot be said which is 'really' moving and which is at rest. It depends on which coordinates you choose. If one hits the gas to speed up, not only does the occupant feel the force, but no matter which coordinates you choose every observer will agree that its speed is changing. Hence anybody with a good understanding of relativity can calculate the force felt by the occupants. This is the consequence of proper acceleration. However, depending on the coordinate choice of the observer, not everybody will agree on how much or in what direction it is accelerating with respect to their coordinates. The acceleration is absolute, but how much acceleration occurs in relation to a particular coordinate choice is relative. Some observers will even say they are slowing down, which also requires acceleration. This is referred to as coordinate acceleration. To illustrate imagine sitting in a seat and tossing a rock straight up and catching it when it falls. That rock traces out a straight up and down line in your coordinates. Now imagine your seat is the back seat of a car moving at a constant 100 kph. The guy on the side of the road sees you toss the rock down the road and and the car carries you down with it to catch it. As far as the laws of physics are concerned the straight up and down path and the curved path up over the road are just as real. There are no 'real' paths in that sense, and the classical aether failed because it essentially sought to establish which path was real. It is effectively like an American arguing with the Chinese over which way is really up. The laws of physics allows everybody to agree on what is accelerating, though not necessarily by how much or what direction, but does not allow everybody to agree on what is moving at some constant velocity verses at rest. Gravity turns this relationship on its head. Note the force felt when accelerating. When on the surface of a gravitational mass, like Earth, the principle of equivalence tells us this weight we feel is the same force we feel when accelerating. Now if you jump off a roof then while accelerating toward the ground you feel no force, effectively weightless. Hence, in your frame of reference you are at rest, i.e., not accelerating, but the Earth is accelerating toward you. Yet everybody in the Universe can agree that your speed is changing, even if not by how much. In this case proper gravity is absolute, while how much gravity and coordinate paths are relative. So in this case gravity is absolute, but the coordinate acceleration due to gravity is relative. Now reconsider the twin paradox. If two observers experience the same amount of acceleration X time, then neither one will age any faster than the other. If two spaceship pass each other at a constant velocity, such that t=0 is mutually defined at the point of closest approach, then each will observe the others clock going slower than their own. To resolve this you accelerate your ship to catch up to the other, which requires applying an absolute force, i.e., proper acceleration. It doesn't matter whether you accelerate a little and take longer to catch up, or a lot to catch up faster, the effect is the same. You are the one that experienced a proper acceleration, hence your clock will be the one that appears to have slowed when you catch up. If you both apply the same proper acceleration to catch up to each other then no time dilation will be apparent under the definition given by t=0. Time is as fluid as the path of the rock in the back seat of the car, but like the rocks coordinate path it must transform from one coordinate choice to another by a well defined set of rules. Others have done an excellent job of articulating these quantitative rules here. 



#136
Feb1713, 09:17 AM

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#137
Feb1713, 09:29 AM

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#138
Feb1713, 11:15 AM

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#139
Feb1713, 12:27 PM

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I would agree that gravity is absolute for a different reason: tidal 'forces', physically; curvature geometrically. Tidal forces are detectable in a small region. 



#140
Feb1713, 01:57 PM

P: 221

I decided that I ought to do the calculations that I should have done Friday night, before reading any of the posts or papers referenced since then. Here's my best shot at the Twin Paradox. Later this evening, or maybe tomorrow night, I'll see how what I did compares with your suggestions.
I can't say that I did this without any outside influences since Friday night. I did see that the image that George referenced looks like a spacetime diagram. That may have triggered some thoughts about coordinates. I do all the calculations for the spacetime diagram but did not include an image of it. You all know what it looks like. Solution of the Twin Paradox  to the degree possible knowing only the Lorentz transform and the usage of the spacetime diagram. Given: G1. Earth and rocket are both at rest at same position. Earth clock and rocket clock are synchronised. G2. At time 0.0, rocket fires a pulse. G3. Earth and rocket separate at relative velocity 0.8c. G4. At distance 10 units from Earth, as measured in Earth frame, rocket fires a pulse. G5. Earth and rocket approach at relative velocity 0.8c. G6. Upon reaching Earth, rocket fires a pulse, coming to rest on Earth. G7. Gravitational effects of mass are to be ignored. Questions: Q1. Both the Earth and the rocket claim to be at rest throughout the episode. Can both make good on this claim? Q2. What are the clock readings on Earth and rocket at G6? Q3. Are the clock readings calculated for Q2 unambiguously unique? Solution: The questions are with regard to kinematics only: positions and times. With one exception noted later, the dynamics of the episode need not be considered. It will also be assumed that acceleration does not affect clocks. This assumption, together with G7, allows special relativity to be used in the attempt at a solution. It will also be assumed that the acceleration from rest to velocity V is instantaneous. Any effect assumption this might have on the calculated clock readings will be ignored for the purposes of this exercise. The Earth is at the origin of its own coordinate system. Likewise, the rocket is at the origin of its own coordinate system. By G1, the origins are coincident at the start of the episode. For convenience, the X axes of the two systems are colinear, and the relative velocity is along that axis, with positive V in the positive X direction. Four spacetime events will be considered. Event A: Corresponds to G2. Time and position are zero in both the Earth frame and the rocket frame. Velocity of the rocket frame is V. For convenience, the axes are set so that positive V is along coordinates ; positive V is made to be along the for convenience . For the moment, the Earth frame will be shown with orthogonal the velocity will be shown "to the right" on the spacetime diagram, Event B: Corresponds to G4, on the worldline of the Earth. Event C: Corresponds to G4, on the worldline of the rocket. Event D: Corresponds to G6. The worldlines of the Earth and rocket meet here, and become colinear. Notation: T represents time. X represents position. Events and frames are represented by lower case letters; event followed by frame. e represents Earth frame. r represents rocket frame. Example: Tbe represents the time at event B in the Earth frame. Times are given as the distance that light travels in one unit of time. T = ct. With this unit of time, and with velocity given as constant factor of c (v = Vc), the Lorentz transforms have the form: X' = g(X  VT) T' = g(T  VX) where g = 1 / Sqrt(1  V^2) With V = 0.8, g = 1.667 Calculate time in Earth frame at Events B and C. By the statement of G4, events B and C are simultaneous in the Earth frame. Measurements of distance in a frame are by definition taken at a single instant in the frame. Time in the Earth frame at Events B and C is distance from earth to rocket (as measured in Earth frame) divided by relative velocity. Tbe = Tce = 10 / V = 10 / 0.8 = 12.5. Calculate coordinates at Event B. Xbe = 0.0 (Earth is inertial; Xae = 0.0; position in an inertial frame does not change with time.) Tbe = 12.5 (As calculated above.) Xbr = 1.667 * (0.0  (0.8 * 12.5)) = 16.67 Tbr = 1.667 * (12.5  (0.8 * 0.0)) = 20.83 Calculate coordinates at Event C. Xce = 10.0 (By G4) Tce = 12.5 (As calculated above.) Xcr = 1.667 * (10.0  (0.8 * 12.5)) = 0.0 Tcr = 1.667 * (12.5  (0.8 * 10.0)) = 7.5 Note that calculation of Xcr confirms what is already known. Xcr must be zero because Xar = 0.0 and the rocket is inertial to this point. Earth or rocket must change frames at velocity reversal. The reversal of velocity in G4 must be represented by a change of frame in the spacetime diagram. Without a change of frame, the worldlines of Earth and rocket can never meet. Either the Earth or the rocket, or both, must change frames. The Earth cannot change frames: No unbalanced force acts on it; it is inertial. The rocket must change frames: It is acted on by an unbalanced force; it is not inertial. Setting up the new rocket frame. The rocket will be in its new frame during the approach in G5. The rocket approach frame will be represented by the addition of the lower case 'p' to the notation. The rocket must be assigned position and time coordinates in its approach frame. At the start of an exercise, coordinate values may assigned at will, due to the linearity of the Lorentz transform. In this case, the approach frame comes into play at an event in an ongoing episode, at Event C. To maintain correspondence with the physical reality, and taking into account the assumption of instantaneous acceleration, the coordinates of the rocket in the approach frame at Event C must match the coordinates of the rocket in the original separation frame at Event C. Transformation from the rocket approach frame to the Earth frame. The Lorentz transformation equations were derived with the origins of the two frames coincident. Therefore, Event C must be treated as a local origin for the purposes of transformation from frame to frame. Event coordinates relative to the local origin are transformed from frame to frame, as shown in the following equations. In these equations, replace the underscore with the symbol for the event to be transformed. To transform from the Earth frame to the rocket approach frame: X_rp = g((X_e  Xce)  V(T_e  Tce)) + Xcrp T_rp = g((T_e  Tce)  V(X_e  Xce)) + Tcrp To transform from the rocket approach frame to the Earth frame: X_e = g((X_rp  Xcrp) + V(T_rp  Tcrp)) + Xce T_e = g((T_rp  Tcrp) + V(X_rp  Xcrp)) + Tce As discussed above, Xce = 10.0 Tce = 12.5 Xcrp = 0.0 Tcrp = 7.5 Calculate the coordinates of Event B in the rocket approach frame. Xbrp = g((Xbe  Xce)  V(Tbe  Tce)) + Xcrp Xbrp = 1.667((0.0  10.0)  (0.8)(12.5  12.5)) + 0 Xbrp = 16.67 (Same as Xbr) Tbrp = g((Tbe  Tce)  V(Xbe  Xce)) + Tcrp Tbrp = 1.667((12.5  12.5)  (0.8)(0.0  10.0)) + 7.5 Tbrp = 1.667(0  (0.8)(10.0)) + 7.5 Tbrp = 1.667(0  8.0) + 7.5 Tbrp = 1.667(8.0) + 7.5 Tbrp = 5.83 (Compare 20.83 for Tbr) Calculate the coordinates of Event D. Time for approach is same as time for separation. (Relative velocity is the same.) Xde = 0.0 Tde = Tbe + Tbe = 25.0 Xdrp = g((Xde  Xce)  V(Tde  Tce)) + Xcrp Xdrp = 1.667((0.0  10.0)  (0.8)(25.0  12.5)) + 0.0 Xdrp = 1.667(10.0  (10.0)) + 0.0 Xdrp = 0.0 (Confirms inertial behavior of rocket from Event C to Event D: Xcrp = 0.0) Tdrp = g((Tde  Tce)  V(Xde  Xce)) + Tcrp Tdrp = 1.667((25.0  12.5)  (0.8)(0.0  10.0)) + 7.5 Tdrp = 1.667(12.5  (0.8)(10.0)) + 7.5 Tdrp = 1.667(12.5  8.0) + 7.5 Tdrp = 7.5 + 7.5 Tdrp = 15.0 (Confirms approach time equals separation time in rocket frame.) Answers to questions: Q1. Both the Earth and the rocket claim to be at rest throughout the episode. Can both make good on this claim? Yes, kinematically. Earth and rocket X coordinates are 0.0 throughout. Whether this makes physical sense dynamically is unknown, given limited knowledge noted above. Q2. What are the clock readings on Earth and rocket at G6? The Earth clock reads 25.0. The rocket clock reads 15.0. Q3. Are the clock readings calculated for Q2 unambiguously unique? Yes. There is only one way to construct the spacetime diagram, due to the unique noninertial behavior of the rocket. Visually, the Earth and rocket experiences are symmetric. Each sees the other move away and return. Nevertheless, the rocket is unambiguously younger than the Earth at reunion. 



#141
Feb1713, 03:37 PM

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#142
Feb1713, 04:50 PM

P: 863

The break in the symmetry under both special and general relativity occurs whenever an observer enters a noninertial state. Sitting motionless on the surface of the Earth is a noninertial state, which is why you feel weight. By the principle of equivalence you feel this same gforce when you accelerate under special relativity, which breaks the symmetry GregAshmore is wanting to absolutely maintain, leading to his difficulties. You cannot break this symmetry in either special or general relativity and pretend it is not broken. Once this symmetry, which applies only to inertial observers, is broken it is no longer "matter of free opinion" as to whether it is broken or not. GregAshmore is assuming the symmetry wasn't broken when his spaceship was accelerated. Whether this symmetry is broken by a rocket engine (SR) or a gravitational field (GR) makes no difference, though both break it in a different or inverse manner, breaking this symmetry requires one or the other. 



#143
Feb1813, 02:32 AM

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#144
Feb1813, 03:04 AM

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[addendum: kinematically the situation looks symmetrical (which is what I supposed you meant); however next George correctly highlights that of course there are visual differences that can be observed. That is pertinent for understanding the physics. This difference in observations has also been elaborated by Langevin in the article that I linked earlier.] 


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