DrRocket Posted February 18, 2011 Share Posted February 18, 2011 (edited) There have been some questions surrounding special relativity, particularly "time dilation" that might benefit from a global and somewhat abstract perspective on spacetime. This is a bit long, and I thought might therefore be appropriate in a new thread. Yes, keeping various reference frames straight can be confusing, and generates tons of confusion plus the odd "paradox". But there is a way around that mess, and you can thank Minkowski for it. After Einstein built on the work of Lorentz and Poincare to produce special relativity, Minkowski re-cast special relativity in terms of the geometry of a 4-dimensional space with a metric of signature +,-,-,- (or equivalently -,+,+,+). Einstein took that idea and from it constructed general relativity. So, what follows is essentially "special relativity from a general relativity perspective". For the moment, forget about "time". Also forget about "space". You live in a 4-dimensional world called spacetime. It is an affine space, but for clarity pick an arbitrary point and call it the origin. Spacetime is now a 4-dimensional real vector space. Spacetime comes equipped with a non-degenerate inner product, <.,.> We need some theorems that I will state as "Facts". Fact: Given a non-degenerate inner product on 4-space there are vectors [math]x_1, x_2, x_3, x_4[/math] such that [math]<x_i,x_j> = 0 , i \ne j [/math] and [math] <x_i, x_j> = \pm 1[/math]. These vectors form a basis for the vector space and are called an orthonormal basis. A vector [math]x[/math] for which [math]<x,x>=0[/math] is called a null vector. A vector [math]x[/math] for which [math]<x,x> > 0[/math]is called timelike, and a vector vector [math]x[/math] for which [math]<x,x> < 0[/math]is vcalled spacelike. Fact: The number of vectors in an orthonormal basis for which [math] <x_i, x_i> = 1[/math] is the same for any orthonormal basis. If an orthonormal basis has just one such vector the signature of the inner product is said to be +,-,-,-. Minkowski spacetime is 4-space with a metric of signature signature +,-,-,- Special relativity is really just the study of the geometry of Minkowski spacetime. A linear transformation, [math]\Lambda[/math] that leaves the Minkowski inner product invariant, ie. [math] < \Lambda x, \Lambda x>= <x,x>[/math] is called a Lorentz transformation. If in addition, given an orthonormal basis [math] e_1,e_2,e_3,e_4[/math] the sign of [math]<x,e_j>[/math] and [math]< \Lambda x, e_j>[/math] are the same for every null vector [math]x[/math] and each[math] e_j[/math] then [math]\Lambda[/math] is called orthochronous (preserves the direction of time). Hence the term "Lorentz transformation" will mean "orthochronous Lorentz transformation". These are the Lorentz transformations of special relativity. Given an observer and a second one in uniform motion relative to the first one, the relative velocities uniquely determine a Lorentz transformation that give an orthonormal basis for the second observer in terms of an orthonormal basis for the first observer – spacetime is transformed to spacetime but the space/time distinction is lost in the process Consider a smooth timelike curve in spacetime, i.e. a map [math] \phi : [0,1] \rightarrow \mathbb R^4[/math]. Then given the Minkowski metric one can consider the arc length, [math] \int_0^1 |\frac {d \phi (\tau}{d \tau}| d \tau[/math] where [math]|x| = \sqrt {<x,x>}[/math]. A timelike curve is called a world line, and the arc length is called the proper time of the world line. Fact: Given a standard clock which follows a world line in spacetime the time recorded on the clock is the proper time of the world line. The only time measured by ANY clock is the proper time of the world line of the clock. So, just what is "time" ? Consider some observer sitting in his own reference frame at the origin. His wrist watch keeps track of proper time along a world line with coordinates [math](\tau, 0,0,0)[/math]. It then follows immediately that [math]\tau[/math] is proper time. It is also the usual time coordinate. Consider some clock in the same reference at a different fixed "spatial" location [math](\tau, x,y,z)[/math] The proper time for that clock is also [math]\tau], which is still the coordinate time for that reference frame. So, proper time and coordinate time, referred to any single reference frame are the same thing What does a different observer see ? He sees the same proper time for any world line, because the Lorentz transformation preserves the Minkowski inner product. But be has a different orthonormal basis so his mix of space and time is different. This is the source of "time dilation" and "length contraction". Confusion arises from the natural human tendancy to think in terms of space and time as distinct. They are not distinct. They are not invariant. They are coordinate-dependent. Edited February 24, 2011 by DrRocket Link to comment Share on other sites More sharing options...

michel123456 Posted February 19, 2011 Share Posted February 19, 2011 What is your question? Confusion arises from the natural human tendancy to think in terms of space and time as distinct. That is certain. Now that you mention it, I guess in the "twin paradox", we always forget to mention that the traveler twin is not only younger but contracted too... (I can't believe I referred to the twins again, sorry for that). Link to comment Share on other sites More sharing options...

DrRocket Posted February 19, 2011 Author Share Posted February 19, 2011 That is certain. Now that you mention it, I guess in the "twin paradox", we always forget to mention that the traveler twin is not only younger but contracted too... (I can't believe I referred to the twins again, sorry for that). The contraction only applies between two reference frames in relative motion. But the traveling twin really will be younger even after the trip is over and the twins are re-united. The reason for the younger twin is quite simple using general relativity with its Lorentz/Minkowski metric. I'll show you how: For simplicity let's start and stop the trip at the South Pole, which along with the Earth is in orbit, free fall. Twin A stays there. Twin B travels to some star and returns. Twin A, in free fall, has a world line that is a spacetime geodesic. Twin B, accelerates and decelerates several times and therefore has a world line that is not a geodesic. Both world lines start at the South pole at the same spacetime point and end there later at another common spacetime point. Now use the metric to calculate the arc length of the two world lines. The geodesic of twin A will be of maximum length, while the non-geodesic of twin B will be shorter. ( On a lorentzian manifold, unlike a purely Riemannian manifold with a positive-definite metric, geodesics maximize rather than minimize length). That "length" is proper time. Ergo twin B isyounger. Link to comment Share on other sites More sharing options...

michel123456 Posted February 20, 2011 Share Posted February 20, 2011 The contraction only applies between two reference frames in relative motion. But the traveling twin really will be younger even after the trip is over and the twins are re-united. (...)The geodesic of twin A will be of maximum length, while the non-geodesic of twin B will be shorter. ( On a lorentzian manifold, unlike a purely Riemannian manifold with a positive-definite metric, geodesics maximize rather than minimize length). That "length" is proper time. Ergo twin B isyounger. Isn'it a wonderful property for an object standing at rest? Doesn't that mean that "at rest" has unique properties compared to all other observers? I may need your point in some other thread. What is the reason that make you conclude that in this thought experiment time "behaves" differently than space and has separated effects? If "contraction only applies between two reference frames in relative motion", in which I agree, why is it different for time? Or, to put the same question otherwise, if time dilation produces a gap in the age of the twins, why length contraction produces no effect at all? Isn't there some contradiction with your very logical statement (in which I agree) that : "Confusion arises from the natural human tendancy to think in terms of space and time as distinct. They are not distinct. They are not invariant. They are coordinate-dependent." Link to comment Share on other sites More sharing options...

swansont Posted February 20, 2011 Share Posted February 20, 2011 What is the reason that make you conclude that in this thought experiment time "behaves" differently than space and has separated effects? If "contraction only applies between two reference frames in relative motion", in which I agree, why is it different for time? Or, to put the same question otherwise, if time dilation produces a gap in the age of the twins, why length contraction produces no effect at all? It does. Time dilation is actually a change in frequency; the net shift is the frequency change integrated over the entire path. But at the conclusion of the trip, the clocks are running at the same frequency. The analogous measurement for length will be an odometer, which counts the number of unit lengths the twin travels. The traveling twin will measure a shorter distance traveled for the trip, even though a meter stick will have the same length as the earth twin's meter stick once he has returned. When each twin takes his own value for distance/time, they will agree on the speed of the travel. 1 Link to comment Share on other sites More sharing options...

michel123456 Posted February 20, 2011 Share Posted February 20, 2011 So time has accumulated the gap. But the gap in spatial dimension has disappeared, it remains only as a measurement (the velocity). Do we have to conclude that time & space have a different behaviour? And that " the natural human tendancy to think in terms of space and time as distinct" is correct. Link to comment Share on other sites More sharing options...

swansont Posted February 20, 2011 Share Posted February 20, 2011 So time has accumulated the gap. But the gap in spatial dimension has disappeared, it remains only as a measurement (the velocity). Do we have to conclude that time & space have a different behaviour? And that " the natural human tendancy to think in terms of space and time as distinct" is correct. The time comparison has a gap. So does the distance comparison. Link to comment Share on other sites More sharing options...

michel123456 Posted February 20, 2011 Share Posted February 20, 2011 The time comparison has a gap. So does the distance comparison. During the travel, did the twin contract? Link to comment Share on other sites More sharing options...

swansont Posted February 20, 2011 Share Posted February 20, 2011 During the travel, did the twin contract? According to an observer in another frame, yes. Link to comment Share on other sites More sharing options...

michel123456 Posted February 20, 2011 Share Posted February 20, 2011 During the part of the travel when he contracted, did time dilated? Link to comment Share on other sites More sharing options...

swansont Posted February 20, 2011 Share Posted February 20, 2011 During the part of the travel when he contracted, did time dilated? According to an observer in another frame, yes. Link to comment Share on other sites More sharing options...

lemur Posted February 20, 2011 Share Posted February 20, 2011 Time dilation is actually a change in frequency The frequency of what? Electron oscillations in the atoms composing the objects? If so, wouldn't this change the relationship between the frequency of the particles and their motion relative to each other? I.e. the particles themselves would be "existing faster" but their faster existence would be relative to the same relations between energy and momentum that characterized them when their existence/frequency was slower, right? Or am I still misunderstanding something fundamental about this concept? Link to comment Share on other sites More sharing options...

michel123456 Posted February 20, 2011 Share Posted February 20, 2011 During the return phase of the travel, did the twin expand? (I mean return to its original size) Link to comment Share on other sites More sharing options...

swansont Posted February 20, 2011 Share Posted February 20, 2011 The frequency of what? Electron oscillations in the atoms composing the objects? If so, wouldn't this change the relationship between the frequency of the particles and their motion relative to each other? I.e. the particles themselves would be "existing faster" but their faster existence would be relative to the same relations between energy and momentum that characterized them when their existence/frequency was slower, right? Or am I still misunderstanding something fundamental about this concept? The frequency (or rate) of time, which includes the oscillator in your clock. Remember that this happens to observers in other frames, not in the frame of the clock. Any property you look at has to be transformed into the other frame. During the return phase of the travel, did the twin expand? (I mean return to its original size) Contraction is only speed dependent. It's still there during the return trip. It goes away when you return to the same frame. Link to comment Share on other sites More sharing options...

DrRocket Posted February 20, 2011 Author Share Posted February 20, 2011 When each twin takes his own value for distance/time, they will agree on the speed of the travel. Yes, since in special relativity the factors for time dilation and length contraction are the inverse of each other. In fact 4-velocity, in units where c=1, is always 1. The reason is pretty simple. Take a curve in spacetime. Parameterize it by arc length. Then, as one knows from basic calculus, the speed along the curve is 1. But in Minkowski space arc length is just proper time, so the "parameter speed" is also physical speed. Link to comment Share on other sites More sharing options...

michel123456 Posted February 20, 2011 Share Posted February 20, 2011 (...) Contraction is only speed dependent. It's still there during the return trip. It goes away when you return to the same frame. Contraction goes away. But time dilation don't? Link to comment Share on other sites More sharing options...

swansont Posted February 20, 2011 Share Posted February 20, 2011 Contraction goes away. But time dilation don't? No, as I explained before, the contraction for distance and dilation of the frequency do go away. The accumulated measure of distance is the analogous measurement to time, which persists. The contraction of the twin is not the analogue of the dilation of the clock. Link to comment Share on other sites More sharing options...

lemur Posted February 20, 2011 Share Posted February 20, 2011 The frequency (or rate) of time, which includes the oscillator in your clock. Remember that this happens to observers in other frames, not in the frame of the clock. Any property you look at has to be transformed into the other frame. So dilation only refers to the differential between two observers with regards to how each appears to the other. So once two clocks diverge, relative to one another they never become synchronized again naturally unless they reverse the differential that caused them to diverge in the first place? Link to comment Share on other sites More sharing options...

DrRocket Posted February 20, 2011 Author Share Posted February 20, 2011 So dilation only refers to the differential between two observers with regards to how each appears to the other. So once two clocks diverge, relative to one another they never become synchronized again naturally unless they reverse the differential that caused them to diverge in the first place? Right. Two accurate clocks, co-located, will measure the same time intervals, but if they start out with a difference in "displayed" time of one hour they will always be one hour apart. That is why you can find banks of accurate clocks that show " London time", "New York time", "Los Angeles time" and "Hong Kong time". Dilation refers to two reference frames in relative motion to each other. When there is no relative motion, there is no dilation. It is awkward, but possible to handle acceleration in special relativity. You must first identify an inertial reference frame. Then the time effect with respect to an accelerating object is determined by the instantaneous speed relative to the inertial frame. Then you have to integrate to get the net effect. But when that is done, the difference in time intervals between a "stationary" clock and the accelerating clock are persistent, even though when re-united the two clocks continue to measure future time intervals identically -- just like London time vs Hong Kong time. Link to comment Share on other sites More sharing options...

lemur Posted February 20, 2011 Share Posted February 20, 2011 Right. Two accurate clocks, co-located, will measure the same time intervals, but if they start out with a difference in "displayed" time of one hour they will always be one hour apart. That is why you can find banks of accurate clocks that show " London time", "New York time", "Los Angeles time" and "Hong Kong time". Dilation refers to two reference frames in relative motion to each other. When there is no relative motion, there is no dilation. It is awkward, but possible to handle acceleration in special relativity. You must first identify an inertial reference frame. Then the time effect with respect to an accelerating object is determined by the instantaneous speed relative to the inertial frame. Then you have to integrate to get the net effect. But when that is done, the difference in time intervals between a "stationary" clock and the accelerating clock are persistent, even though when re-united the two clocks continue to measure future time intervals identically -- just like London time vs Hong Kong time. So there's some speed/route to travel between Hong Kong and London where you don't have to reset your watch? How long would that route take for the traveller and how long would their friends in London have to wait for them to arrive? That's sounds like a good question for an exam (which I would fail). Maybe you could make it harder by making them actually plan the route through spacetime and calculate the energy needed for a vehicle of a given mass. Good thing for me it's easier to come up with math problems than to solve them:) Link to comment Share on other sites More sharing options...

michel123456 Posted February 21, 2011 Share Posted February 21, 2011 (edited) (...) But when that is done, the difference in time intervals between a "stationary" clock and the accelerating clock are persistent, even though when re-united the two clocks continue to measure future time intervals identically -- just like London time vs Hong Kong time. Why persistent? In order to re-unite the two clocks, don't you measure a negative acceleration (deceleration) exactly the same as the original acceleration? Why the one counts (when the 2 clocks get away from each other) and the other don't (when they get together). There is something terribly asymmetric in this situation. Edited February 21, 2011 by michel123456 Link to comment Share on other sites More sharing options...

DrRocket Posted February 21, 2011 Author Share Posted February 21, 2011 (edited) Why persistent? In order to re-unite the two clocks, don't you measure a negative acceleration (deceleration) exactly the same as the original acceleration? Why the one counts (when the 2 clocks get away from each other) and the other don't (when they get together). There is something terribly asymmetric in this situation. No, negative acceleration is still acceleration. You don't get time dilation for acceleration and time "contraction" for negative acceleration. Both represent deviations from a free fall (geodesic) path in spacetime, and non-geodesics have smaller proper time than geodesics, between the same points in spacetime. You seem to be trying to understand relativity based on some notion of "common sense" and intuition from everyday Newtonian experience. That is not going to work. There is no substitute for studying the theory using a good text. There are several that might be suitable, depending on your background in mathematics. If you can handle Misner, Thorne, and Wheeler at any level, that is one very good choice. Edited February 21, 2011 by DrRocket Link to comment Share on other sites More sharing options...

michel123456 Posted February 21, 2011 Share Posted February 21, 2011 (edited) No, negative acceleration is still acceleration. You don't get time dilation for acceleration and time "contraction" for negative acceleration. Both represent deviations from a free fall (geodesic) path in spacetime, and non-geodesics have smaller proper time than geodesics, between the same points in spacetime. That's what I say, it is asymmetric. And I am not comfortable with the idea. --------------------------------- edited. You don't get time dilation for acceleration and time "contraction" for negative acceleration. And what about length contraction? Edited February 21, 2011 by michel123456 Link to comment Share on other sites More sharing options...

DrRocket Posted February 21, 2011 Author Share Posted February 21, 2011 (edited) That's what I say, it is asymmetric. And I am not comfortable with the idea. Your comfort is not part of the theory. And what about length contraction? Length contraction is special relativistic effect that depends on speed, not velocity (once things are in standard configuration with axes aligned). There is no "length expansion"; the length of an object is the maximum in its rest frame. My earlier comment is reinforced by these questions. You need to study a good book on relativity. Trying to "get comfortable" using Newtonian intuition. It will not work. Edited February 22, 2011 by DrRocket 1 Link to comment Share on other sites More sharing options...

michel123456 Posted February 22, 2011 Share Posted February 22, 2011 Then, how do you account for this comment: Contraction is only speed dependent. It's still there during the return trip. It goes away when you return to the same frame. (emphasis mine) How does it "go away"? I have understood that when the 2 clocks come back together, we all agree that their frequency will be the same again. How is it possible to obtain such a result accounting only on time dilation? There seems to be some inconsistency. Link to comment Share on other sites More sharing options...

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