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The difference between felt, and unfelt motion.

  1. Aug 6, 2005 #1
    I’m currently reading “The Elegant Universe”. I’ve just finished the sections on Special and General Relativity, and I want to make sure I understand as much as I can before going on to Quantum Mechanics.

    He uses wonderful analogies to describe most aspects of Relativity, and I feel that I have a good conceptual grasp on most of them. He states many times, however, that an object that can feel acceleration will know that it is the one in motion, and therefore will agree that time has passed more slowly than a stationary object. There are no good analogies in the book to help me understand why this is.

    For instance, in the example of George and Gracie floating through space and sycronizing their clocks as they pass each other. He says that if George was wearing a jet pack, and turned it on so that he could catch up with Gracie, he would feel that he was accelerating and would agree that his clock is ticking more slowly than Gracie’s... I don't get it.

    Another thing that seems paradoxical to me: What if, after synchronizing their clocks, they were both wearing jetpacks turned them on at the exact same time, just long enough to get them floating back toward each other. When they met back up, would they not each argue that they had the faster clock, and that less time had passed for the other?

    Any help would be appreciated.

    Last edited: Aug 6, 2005
  2. jcsd
  3. Aug 6, 2005 #2
    No, when George and Gracie remeet, if they have both experienced the same amount of acceleration, their clocks will still be synchronized. Think of it in terms of the reference frame that they will end up in: Say there is a third person, Gary. Initially, say that Gary is moving at half the speed of George from Gracie's point of view (and half the speed of Gracie from George's point of view). Gary will see both George and Gracie moving at the same speed, but in opposite directions. So let's look at it from Gary's perspective. Gary is sitting still out in space and sees George approaching him from one direction and Gracie approaching him from the other at the same speed. Just as George and Gracie pass Gary, they all synchronize their clocks, then continue on their way, but George and Gracie also agree that in one hour they will both turn their jet packs on and return to Gary. From Gary's point of view, George accelerates back to him, and so does Gracie, so both of their clocks will be slower than his. George and Gracie will agree with Gary that their clocks are slower because they accelerated and Gary didn't. But George and Gracie's clocks will still both be synchronized, although they will disagree with Gary's.
    Last edited: Aug 6, 2005
  4. Aug 6, 2005 #3


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    In special relativity, the laws of physics look the same in any inertial (non-accelerating) reference frame, and each inertial reference frame is equally valid. And in every inertial reference frame, clocks which are moving faster in that frame are ticking slower. But certain questions must have the same answer in every frame--for example, if two clocks pass each other in the same location, every frame will make the same prediction about what each clock reads at the moment they meet. So suppose George and Gracie are moving apart at constant velocity, then George turns on his jetpack and accelerates towards Gracie, then coasts until they meet. It's not true in all frames that George's clock is ticking slower after he accelerates--for example, there's a frame where George starts out moving very quickly, but when he turns on his jetpack he slows down and is actually at rest after he turns it off and Gracie is coming towards him. In this frame, when they are initially moving apart George is moving faster than Gracie, so his clock is ticking even slower than hers during this phase, but then it's ticking faster than Gracie's as they move together. But it will still work out that this frame predicts that George's clock will be behind Gracie's when they reunite, because of how much slower it was ticking when they were initially moving apart. No matter what frame you pick, if two clocks start out at the same point in spacetime and then meet again at another point in spacetime, the path between these points that leads to the maximum time being ticked by a clock travelling that path is a straight line through spacetime (ie moving through space at constant velocity). This is sort of analogous to the fact that if you pick two points on a flat paper and draw various paths between them, the path with the shortest length will always be a straight line.
    If they are moving symmetrically in some inertial frame, so that in that frame each one's speed at any given moment will be the same, then assuming their clocks were synchronized when they initially passed each other moving apart, then their clocks will read the same time when they reunite, because they were ticking at the same rate at every moment in this frame. In other frames the clocks were not always ticking at the same rate, but other frames will always make the same prediction about what their clocks read when they reunite.
    Last edited: Aug 6, 2005
  5. Aug 7, 2005 #4
    The example given by Greene of two clocks being synchronized as they pass each other is misleading - You can set them to read the same time at the instant of passing - but they are not necessarily calibrated in the sense that they will keep the same time thereafter - the only method proposed by Einstein is that initially both clocks are at rest in the same frame and are synchronized. If one clock is subsequently moved, the two clocks will no longer be in sync.

    So using Einstein's example in his 1905 paper, if two clocks A and B are at rest in the stationary frame, and they are initially separated by a distance L, then if A moves toward B, when they meet, A will have accumulated less time - likewise if instead B moves toward A, B will be found to have accumulated less time upon meeting A. In the example of two clocks meeting in space, you would not know whether B had moved or whether A had moved - or whether both had moved.
    Last edited: Aug 7, 2005
  6. Aug 7, 2005 #5
    All of these answers make sense to me, and have helped a lot. I still have one more thing to work out though:

    Lets start off by having George and Gracie both stationary in space. They both synchronize their clocks to read the same time. Then George turns on his jet pack to move away from Gracie. From George’s perspective, since light always travels at the same rate, Gracie’s clock will be slowing down despite the fact that he knows he is the one in motion. It will still appear to him that light has to travel farther for her, than for him.

    Keeping his eye on her clock he turns his jet pack on to stop his motion. Now their clocks should be ticking at the same rate again but from George’s perspective Gracie’s clock should be behind. So George turns back on his jet pack in order to float back to Gracie. Once again from George’s perspective, light will have to travel farther for Gracie than for himself. So, once again her clock should be ticking slower, and should still be behind his when they meet back up again.

    We’ve already established that they should agree when they meet back up that les time should have passed for George. This is easy to understand from Gracie’s perspective but, at some point George must have the opportunity to witness Gracie’s clock moving faster than his.
  7. Aug 7, 2005 #6


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    One thing to keep in mind is that how fast a moving clock ticks in a given inertial reference frame is not the same as how fast you would see the clock ticking if you were at rest in that frame. For example, if there's a clock moving towards me, then if the time-coordinate of one tick is two seconds later than the time-coordinate of the previous tick in my frame, this clock is ticking at twice the normal rate in my frame; but since the clock is moving towards me, the light from the later tick will have less of a distance to travel than the light from the earlier one, so if I look through my telescope the clock will appear to be ticking a bit faster than that--depending on the velocities I may even see it ticking faster than my own (actually this may be true no matter what the approach velocity, I'm not sure). This is known as the Doppler effect, you may have heard of it before.

    Anyway, George doesn't have a single inertial reference frame, so you can't really talk about how fast Gracie's clock is ticking at different times in his frame (although you can talk about how fast her clock is ticking in the two different frames where he is at rest during the outbound coasting phase vs. the inbound coasting phase--in each of these frames, Gracie's clock is ticking at a constant slow rate, but George's own clock was ticking even more slowly during the leg of the trip where he wasn't at rest in the frame you're looking at). But if you want to know what George will actually see throughout the trip using light signals, this problem is discussed in the Doppler shift section of this page on the twin paradox. The answer is that George will see Gracie's clock running slow as they move apart, but then because of the Doppler effect, he'll see her clock running fast as they move towards one another (the page has a spacetime diagram to help you visualize the reason for this here).
  8. Aug 7, 2005 #7
    Explanations based upon what each twin observes usually wind-up giving the same time dilation between moving frames - but for the wrong reasons - no amount of observation of the other guys clock can affect the reading on your clock - Einstein recogonized there is real physical effect associatiated with a clock that has been put in motion with respect to a frame in which it was brought to sync with another clock which remains in that frame. We verify this every day with high speed muons and hi speed pions and GPS satellites in orbit around the earth. Einstein was above all - intuitive - his interpretation of the physical effects to be attributed to the transforms in Part IV of his original paper was the breakthough than distinquished SR from all previous attempts to explain MMx - and it follows from a very simply substitution, namely x = vt in the transform. Einstein gave the time difference a physical reality - notice he does not say that clock A must travel a round trip to return to its original location at the time of synchronization - the moving clock will lose time right from the start - every experiment with actual particles and precision clocks confirms this result - how amazing since at the time (1905) there was no way to verify this surprising conjecture

    Jesse and I disagree on this (at least in part) - with all due respect to Jesse, I believe that the original source (Einstein's 1905 paper) is only subject to one interpretation - as you get more involved in SR you will find many different views - all concur that there is no paradox - but different authors arrive at this conclusion for different reasons.
    Last edited: Aug 7, 2005
  9. Aug 7, 2005 #8


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    What quote from the paper do you think supports this interpretation?
    Again, what quote do you think supports this idea? Are you claiming that Einstein would have said that if clock A and B are in sync in one location, then B starts moving at constant velocity away from A, that B is "objectively" ticking slower as it moves away? If there is a third observer C who has always been moving at the same velocity that B was moving after it accelerated, then in his frame B comes to rest after it accelerates so it should be ticking faster than A--would you say that this frame is "objectively" less valid than the frame of A?
    No experiment supports the idea that any inertial frame's perspective is any less valid than any other's.
  10. Aug 9, 2005 #9
    Jesse - we have been all through this before - after the clocks are brought to sync - and one clock is put in motion - there is no longer reciprocity - the symmetry is destroyed - a high speed pion doesn't have to turn around and return to its launch point to experience a longer lifetime - its lifetime is extended on a one way mission - every experiment supports the proposition that when one clock is moved - the two frames are no longer symmetrical - there is not a single experiment to show that a clock system attached to a high speed muon would measure time in the earth system as slowed -

    In your query re frame C - if B moves off from A, so that A remains stationary, B will click slower than A, not faster - when B comes to rest in C, B will run at the same rate as C - this is exactly the sitution with GPS - one clock A on the ground - the other B put in motion - if B is not preset to run faster than A to account for the approx 18,000 mph velocity boost, each time B passes overhead it will have lost time wrt to A. If C is another GPS satellite that has not been velocity corrected - then when B arrives at the same altitude and velocity, B will run at the same rate as C, and both B and C will run slower that A (ignoring the difference in gravitaional potential)
    Last edited: Aug 9, 2005
  11. Aug 9, 2005 #10
    One more point Jesse - Einstein refers to the observation of lengths as being reciprocal, he does not make the same statement about the rate of clocks, to wit:

    "A rigid body which, measured in a state of rest, has the form of a sphere, therefore has in a state of motion--viewed from the stationary system--the form of an ellipsoid of revolution with the axes.....It is clear that the same results hold good of bodies at rest in the stationary'' system, viewed from a system in uniform motion."

    Length contraction is apparent and reciprocal, but not time dilation:

    "Further, we imagine one of the clocks which are qualified to mark the time t when at rest relatively to the stationary system, and the time when at rest relatively to the moving system, to be located at the origin of the co-ordinates of k, and so adjusted that it marks the time. What is the rate of this clock, when viewed from the stationary system?"....Between the quantities x, t, and , which refer to the position of the clock, we have, evidently, x=vt .....whence it follows that the time marked by the clock (viewed in the stationary system) is slow.... From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B"

    I repeat - Einstein says that B will measure A's clock as having lost time when A arrives at B but does not say that A will measure B's clock as having lost time when A arrives at B.
  12. Aug 9, 2005 #11


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    But this is silly, because he STARTED with a postulate that no inertial frame of reference is unique. Thus, VIA SYMMETRY, it is automatically implied that the other situation is true using SR's postulate. He doesn't need to explicitly state this.

    And since when did Einstein wrote his SR's paper in English to allow you to play with the quotation game?

  13. Aug 9, 2005 #12


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    The symmetry is destroyed between A and B, but not between A and C, neither of whom has accelerated. Why do you think C's frame is not just as valid as A's? To make the situation even more symmetrical, we could imagine that C also has a probe D that was travelling alongside it until the moment it crossed paths with A and B, and that at the moment they all crossed paths (with all their clocks showing the same reading at that moment), B accelerated to be at rest relative to C, while D accelerated to be at rest relative to A. This situation would be perfectly symmetrical, so would you nevertheless say that B and C are "objectively" ticking slower than A and D? Or would you say that both B and D are ticking slow since they are the ones that accelerated, despite the frame that there is no single frame where both of them are ticking slow? Either way, you won't find any justification for this in Einstein's writings.
    What if C is another planet that has been travelling at that velocity since its creation? Is C's frame still somehow less valid than the earth's, in your opinion? If so, why? If not, why does it matter whether C is a planet or just a satellite or a muon, or even just an abstract "frame" that doesn't have any specific object at rest in it?
    Einstein is only talking about moving a single clock to the same position as another clock which is at rest in the frame the first clock was originally at rest in too, so he's still talking about comparing the readings on nearby clocks, not about which of two distant clocks is "really" running slow. If I have a clock A' at rest at position A and another clock B' at rest at position B that's synchronized with A' in their mutual rest frame (which is also the frame we're using to talk about positions A and B), and then I move A' from its original position to the position of B', then when we compare the readings of A' and B' once they are next to each other, A' will be behind B'. This is all Einstein is saying in the quote above, and of course I agree with it, but it says nothing about A' objectively running slower after it accelerates, all it says is that its time will be behind B' when they meet. From the persective of another frame where A' came to rest after it accelerated and B' was moving towards it at constant velocity, the reason A' is behind B' when they meet is that they were not in sync at the moment A' accelerated, in fact B' was significantly ahead of A' at that moment, so even though A' was ticking faster after it accelerated it still ends up being behind B' at the moment they meet. Since Einstein was only talking about comparing the readings on nearby clocks, not about deciding which of two moving clocks is "objectively" ticking faster, this does not in any way validate your weird idea that even though Einstein explicitly said all inertial frames were equal, he really meant that some inertial frames are more equal than others.

    Anyway, I'm starting off on a road trip today and won't be back until late August, but I'll reply to any messages when I get back, and maybe someone else will be willing to continue this debate with you in the meantime.
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