Einstein’s special relativity is sometimes popularized with statements like: “moving clocks run slower than stationary clocks and moving rods are length contracted relative to stationary rods”. The problem is that special relativity also states that there can be no absolute motion; so how can one define “moving” and “being stationary”?
The usual answer is that all motion is relative and you can take any inertial frame and declare it the “reference frame” against which all other motions can be measured. This however cannot mean that all other inertial frames, moving relative to this one, must now have slower clocks and contracted rods.
To illustrate this, consider two flashes happening at the same spot, one after the other, say with a ten seconds interval as timed in the reference frame. Two identical vehicles happen to pass in opposite directions, just as the first flash occurs. Assume that the vehicles maintain identical (but opposite) speeds and the occupants measure the distance traveled and the time it took before the second flash was observed (seen). Because light travels at the same speed in all directions in every inertial frame, the observers in the vehicles must get the exact same results.
Now the dilemma: The two vehicles were moving relative to each other and special relativity predicts that their clocks and rods must behave differently due to their relative speed. However, if the vehicles would stop and the occupants compare results, they will find that, within experimental error, they recorded the same distances and the same times.
At ordinary road speeds, this is probably an impractical experiment – the errors will be larger than the effect being looked for. Put the same experiment in space, with ultra fast spacecraft and ultra sensitive equipment, and the results must be identical.
For the scientists out there: how do you explain this apparent paradox in special relativity?
SL: Your Aerospace Watchdog
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Scruffy wrote “I thought I had made it quite clear that SR predicts…etc” but no, you have only made it clear what you believe will happen. Opinions vary.
As I indicated in my previous, those ( like John Bell ) who prefer the Lorentz – Poincare reality and disdain Einstein’s version with its unrealistic reciprocity etc. tend to expect the string to break, but those who closely follow the detail of Einstein’s SR with its geometrical kinematic structure expect no breakage. ( In Bell’s case he felt the observer dependancy could violate causality in EPR type experiments, but all his colleagues disagreed with him.)
As for Wikipedia, apart from its main reputation for notorious unreliability, I would point out that there are references that support either side and some of those listed are irrelevant to the problem.
Also look at “Lorentz Contraction” in Wiki, where you will find a classic presentation of the geometric argument, where the various contracted lengths are different “simultaneity slices” through the “world tube” of the object – which would obviously predict no string breaking but rather that the inter-rocket distance shortens exactly as the string-ends.
If you do any survey of textbooks on or including SR, you will find that the rocket/string problem is hardly ever mentioned and that the rotating disc paradox never is. There is a good reason why these are avoided in favour of “easy” explanations of the twin paradox & pole and barn problem etc. It is because they lead into deep and dark waters at the foundations of relativity and have no widely accepted resolution.
Con, your post does not address the issue of simultaneity or the synchronization of clocks that are separated in space. Michelson, Morley, LIGO and light clocks use interferometers and/or two-way measurements, which are not under dispute.
The issue is that the one-way speed of light depends on synchronization of clocks, which are conventional, not absolute. Also, synchronization does not concern the rate of the clocks, but how they are set to read the same time (the definition of simultaneity in an inertial reference frame).
The question is, does length contraction also depend on clock synchronization. If so, then Lorentz contraction is a convention. Looking at solutions to the “ladder paradox”, it seems to be the case.
SL: Your Aerospace Watchdog
rockinggoose wrote:
I thought I made it quite clear that SR predicts that the string would break, but as many others, I’ve got some reservations! This is in line with your view, except for your original statement that SR cannot solve this. I suppose it depends on what you mean by “solve”, but SR makes a definite prediction.
SL: Your Aerospace Watchdog
Con Morton wrote about inertial frames and light clocks (i.e. the standard SR stuff), but does not address the issue at hand: synchronization of clocks and hence simultaneity are conventional and may make length contraction just a convention as well.
Actually I thought the clocks were well synchronized in that the each recorded the same time duration when in the same reference frame. The first paragraph really just sets the table and is an reiteration of time dilation and mass increase that is proven almost daily in particle accelerators around the world. The crux is in the second part when one of the traveling clocks is rotated so that its axis is parallel to the direction of motion. In this case the question becomes. Does the clock read the same time in the same frame independent of the pointing direction of the clock? The fist cut without the Lorentz contraction results in the clock not agreeing with itself which is contrary to logic and numerous tests, Michelson, Morely and LIGO for example. The second cut invoking contraction results in the clock not knowing which direction it is pointing and agreeing with itself. In this case the moving clock in question is synchronized with itself being the same clock and the observers clock is only required to measure the same duration between ticks as only ratios are being observed.
There were a few word drawings with the origonal post that did not come through.
Incidentally the string breaks and if the ships are long enough it breaks befor the pilot of the trailing ship knows he is moving, providing the pilots are in the noses of the ships.
Thanks…..Con
I think your answer is somewhat self-contradictory ! If you are not clear as to whether the analyses represent reality or not then you cannot be sure the string will break, for such a breakage would definitely be physically real.
The problem is that the earlier views in this thread suggest that the Lorentz contraction is not a physical shrinkage but an artifact of measurement – which would accord with the “impossible” reciprocity whereby length A < length B and length B < length A.
Also with the multitude of different lengths for arbitrary observers and with Einstein's kinematic (ie. non-dynamic) derivation.
All of which suggest that the string would not break but that the whole assembly would behave like an irregularly shaped rod – and therefore the rockets would be measured as getting closer as the string gets shorter.
Unfortunately this conflicts with translational symmetry in that the rocket trajectories should be able to be translationally superimposed.
To claim that the string breaks is also to revert to Lorentian relativity that pre-dated Einstein, where the contraction is regarded as physically real as thermal expansion or contraction.
There is definitely a deep contradiction in this problem that (Wikipedia notwithstanding !) divides opinion about equally. Which is why I said the issue was still as yet unresolved.
[Another even worse paradox is the Ehrenfest problem of the rotating disc, on which conflicting papers are still being published on an almost monthly basis]
Con Morton wrote about inertial frames and light clocks (i.e. the standard SR stuff), but does not address the issue at hand: synchronization of clocks and hence simultaneity are conventional and may make length contraction just a convention as well.
SL: Your Aerospace Watchdog
Anonymous wrote:
“Surely the really awkward problem for Special Relativity is the Bell spaceship problem … SR cannot solve this question !”
I disagree with your statement, but I agree that the way scientists solve it may be suspect due to the clock synchronization issue.
If the two rockets are identical and have the exact same acceleration profiles, special relativity predicts that the string will break, without a doubt. Read the various Wikipedia analyses and references in there. To me the fuzzy thing is still the conventionality of simultaneity and my mind is nor clear as to whether the analyses represent reality or not.
SL: Your Aerospace Watchdog
Surely the really awkward problem for Special Relativity is the Bell spaceship problem (originally by Dewan & Beran 1959) where two identical rockets are connected nose to tail by a thin string and accelerate identically. Does the thread eventually break due to Lorentz contraction ? SR cannot solve this question !
Posted by Con
This goes back to your first post.
If I consider my frame my reference frame then any object that has a relative motion to me is in a different frame. Any object that is without relative motion to me is in my reference frame. The relative motion that I can measure is either towards me or away from me, inertial motion only as we are talking about SR.
Your moving vehicles are moving in relation to the light source, point A which is the reference frame. Let them be B and C. Also lets place identical light clocks at A and in B and C and positioned such that the light paths in each clock is at 90 degrees to the track of B and C. Note that the Acme Light Clock Co. specifies that the maximum deviation between any two clocks is less than one pico second per year when measured under their standard conditions at their laboratory location. Now suppose that B is moving from right to left at 0.86603c (Vab) with respect to A and C is moving from left to right at 0.86603c (Vac) with respect to A and that they have both passed A. I have chosen velocities so that Y (gamma) equals 2 for this example. Now an observer at A notes the clock in C ticks one time for every two ticks of the reference clock at A, looking at the clock in B also ticks one time for every two ticks of the reference clock at A. As far as A is concerned the clocks in B and C agree with each other and they are running at half the rate of the clock at A. Of course B and C do not agree. First the relative velocity of B with respect to C or C with respect to B must be determined. As these are relativistic velocities they do not simply add. Then Vr = (Vba+Vac)/( 1+ Vba*Vac) which equals 0.98797441c and Y equals 49 so B sees C’s clock ticking once for every 49 ticks of his and C sees B’s clock ticking once for every 49 ticks of his. However if B and C return to A’s reference frame, in the same manor then their clocks will agree tick for tick and show the same time, they will agree tick for tick for A’s clock but will only show one half the elapsed time from the original time they passed A.
Light clocks.
Transverse stationary Moving at 0.86603c
The stationary light clock counts seconds, one second each time the light pulse travels from one mirror to the other, the vertical distance between mirrors is 3*10^8 meters (L). The stationary observer sees the moving clock a bit differently. After one second the clock has moved horizontally 0.86603c and the light pulse has traveled a distance of c but has only traveled a vertical distance of H. As this forms a right triangle H is easily found as (1-0.86603^2)^0.5 = 0.5c or 0.5 seconds of the moving clocks time. Thusly the observers clock measures one second for every 0.5 seconds for the moving clock, Y = 2.
Rotating the clock so the length is parallel to the velocity vector not only shows the time dilation but the length contraction.
When stationary the results are the same as the vertical clock. When the clock is moving the results are similar to any race. The clock moves at 0.86603 c, a light pulse bounces off mirror 1 and travels towards mirror 2 at a velocity of c as seen by the stationary observer. The difference in velocity as seen by the observer is c – 0.86603c = 0.13397c. The time it take for the light pulse to reach mirror 2 is then L/0.13397 = 7.464357 seconds and the return time is L/(c+0.86603c) = 0.535897 for a total of 8.000 seconds as seen by the stationary observer. This is obviously wrong for several reasons. Reason says that the two clocks are moving at the same rate relative to the observer and should read the same as does SR. Experiment also says that they read the same, Michelson and Morley performed this experiment several time in the 1980s and much to their frustration the clocks always read the same. With L unchanged the moving clock runs to slow, if L is shortened so that L’ = L/Y = L/2 then the two clocks agree as the do in experiment. Lorentz determined the math required to reconcile this shortly after the Michelson/Morley experiments which formed the underpinnings of SR. This effect is also used in the ring laser gyros, three of which are used in thousands of inertial navigation systems used in military and commercial aircraft and missile systems world wide. This only leaves one more SR effect to comment on, the increase of mass with relative velocity. There are several ways to calculate and measure this effect. One is to use the Lorentz transform to determine Y and then multiply rest mass by Y, this is the easiest. Another is to determine the energy required to accelerate the rest mass to some velocity and then convert the energy to mass via e=mc^2, tedious when working at relativistic velocities but is results in the same answer as using Y. Note that at velocities approaching c the energy gain is immense. It is a good thing the pilots of the vehicles mentioned above were very good, if they had crashed head on at point A instead of passing for every kg of rest mass on one of the vehicles the resulting crash would release the equivalent of 49 kg of mass converted to energy. Interestingly this aspect of SR is a cornerstone of high energy physics and is tested and verified everytime a particle accelerator is used.
Thanks…..Con Morton
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Thanks, Burt.
I need to retune my instincts on clocks in gravitational fields. My terminology was indeed off, as I should have realized if I had used an electric field analog.
I also should have thought of the bottom of the well as the point at which the object would come to rest with a little friction included. Then that clearly is the center of the Earth.
So are you saying that the falling person both ages more slowly and has more fun doing it? :)
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Hi Fred, you wrote: “Inside a uniform density Earth, the field (GM/r2) increases linearly from zero.”
GM/r^2 is not the field, but the gradient of the gravitational potential (field). The Newtonian potential outside of a spherical, homogeneous body is just -GM/r. On the inside it is -0.5*GM/R * (1+r^2/R^2), where R is the radius of the body and r the coordinate from the center. It gets more negative as r goes to zero (a deeper ‘gravitational well’ than at the surface). The gradient on the outside is the familiar -GM/r^2 and on the inside it is -GM*r/R^3.
Clock rates (dtau/dt) are determined by the potential, not the gradient. It comes from GR, where static clocks in Schwarzschild coordinates run at a fraction sqrt[1-2GM/(rc^2)] (approx. 1-GM/(rc^2) in the low field limit) of clocks at ‘infinity’.
For David’s “twin falling through the Earth”, the low field approximation of the instantaneous clock rate is:
dtau/dt = 1-0.5GM/(Rc^2)*(1+r^2/R^2)-0.5*v^2/c^2, where dt refers to the clock at ‘infinity’.
The gravitational time dilation term has the same sign as the velocity time dilation term, indicating an error in your rationale.
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
I’m not sure it’s a blunder, Burt.
It’s not a gravitational well in the sense that you are experiencing an inverse square field going outward from the surface of Earth.
Inside a uniform density Earth, the field (GM/r2) increases linearly from zero. That’s because the mass outside radius r produces a net zero field, and M, the mass inside r, is directly proportional to r3.
[Note added in edit]: In other words, the surface of the Earth is at the bottom/meeting point of two gravitational wells, since the field decreases in both directions from there.
Fred Bortz — Science and technology books for young readers (www.fredbortz.com) and Science book reviews (www.scienceshelf.com)
Hi Fred, you wrote: “Burt, “feeling” the acceleration is not a very precise term. ”
On the contrary: if I strap an accelerometer to my butt, I’m sure myself and that piece of technology will agree to a very large extent.
The acceleration that is important here is not relative to any specific reference frame, but relative to a ‘free-fall’ spacetime geodesic (‘proper acceleration’). Without worrying about the effect of gravity on ageing, let’s modify David’s ‘experiment’ slightly.
If you sit on a tower that reaches all the way up to orbital altitude and I’m in a circular orbit at the same altitude, who will age slower? Me, of course, who is ‘weightless’ and in a spacetime geodesic.
If I stay in orbit and you stay on the tower (given the proper life support), I’m sure you will live longer than me, purely due to ‘weightless illness’ or whatever. Nevertheless, you feel the acceleration and I don’t.
Huh? Fred, I just noticed that you blundered seriously in this paragraph:
“This time, the main contribution to age difference is that the traveler spends most of the trip in lower gravity (proportional to the distance from the center, assuming a uniform density planet). That means his/her clocks and biological processes run faster, so s/he ages more than the surface bound person“.
Clocks below Earth’s surface always tick slower than clocks on the surface. At the center of Earth, gravitational acceleration is zero, but the ‘depth of the gravitational well’ is greatest, meaning the slowest lapse of time.
Don’t worry, we all blunder now and then…
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
The question of twin paradox is seen constantly on physics, relativity and science bolgs and books. Common explanations is that travelling twin experiences acceleration while stay home twin does not. Other explanation is similiar to explanation of accleration seen in rotating bodies that one is rotating in relation to the whole universe while other is at rest. The problem of twin paradox should be considered similiar to Mach’s principle. Both lack proper explanation.
The reason is basically movement against the background of space is not accepted by physicists because of ether problem. If we thought that one object or twin is moving through space and other is not then the problem is partially solved. Actually looking at Lorentz equation acceleration does not come into picture at all. Time dilation is calculated on velocity alone. Moreover acceleration deceleration of travelling twin could be made instantaneous so that the travel time is based mostly at high constant velocity. Acceleration of particles in linear accelerators cannot be used to calculate their time dilation but velocity can. Acceleration story does not solve twin paradox.
According to physicists space can be curved warped made into black holes however we still do not accept travel simply against background fixed space. There another major reason for that…the constancy of speed of light in all reference frames. So to all of those giving these unsatisfactory explanations for twin paradox by drawing space time diagrams etc good luck you will not have a solution till you have answered questions like:
1 What is Time?
2. What causes Time?
3. Why Time slows in gravity and with motion?
5. Why time is not really a dimension?
6. How gravity works?
7. Why gravity is only attractive?
8. Acceleration and Twin paradox
9 The mechanism of length contraction?
10 What is inertia?
11. Why speed of light remains constant?
12. Why black holes cannot have singularity?
13. Is the universal accelerating or time is slowing down?
14. Arrow of time
Look for answers at:
http://www.timephysics.com
Hi Burt.
I now see that:
The space traveller in the g accelerated rocket with a global warming engine (patent pending) after 30 years will reach the edge of the universe, but he will do so some time later than the hardy research assistant I have just despatched on a light beam (method so secret that I am not even patenting it). The latter will feel himself to have made the trip in no time at all, and will have plenty of time to examine and reflect on the nature of the edge of the universe while he awaits the arrival of the g accelerated rocket. Their respective arrivals are not at the same event, only at the same place.
Sincerely,
Christopher
Hi, Burt.
Thank you for this kind and careful reply. Do you never sleep?
I need to reflect on these very puzzling things.
Sincerely,
Christopher
Hi Christopher, you asked:
“… reply about particles in centrifuges. Is it convenient for you to give me the references for this?”
My source is the book “Gravitation”, by Misner, Thorne and Wheeler, with the specific case discussed on p. 1055. There are unfortunately no authoritative web based articles (that I know of) freely available on these tests.
On your previous post, where you wrote:
“I think the answer is to describe things from the viewpoint of a particular observer, in terms of what is physical for him.”
Of course, for a traveler (be it particle or human) it’s proper time that counts. The continuously accelerated observer (at 1g) can circumnavigate the present observable universe about 300 times in 30 years of his time, not the time of us Earthlings! This sort of space travel is pure SciFi, because there will most probably never exist an engine that can do that.
Back to your latest post. Your ‘man in the white hat’ should rather be located in inter-galactic space. Then I have told you that for the sibling on Earth’s equator, the umpire will observe her to lose some 10 second in every billion of his seconds. The sibling in geostationary orbit will lose about the same amount.
On the other hand, the umpire will observe the fly-away sibling to lose an average of over half a second every second, depending on his instantaneous relative speed. It will be small initially and become more severe, like losing 0.8 second for every second of inter-galactic time after 5 years (at the maximum speed of ~0.98c). Overall, the fly-away sibling will age ~2.3 years for every 5 years of the umpire’s time (which is for all practical purposes Earth time, within 1 part in 10^8).
I hope this eases the headache! You will be able to do all these sums when you have studied your “Relativity 4 Engineers” ;-)
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)
Hi Burt.
Thank you for your kind and very unexpectedly quick reply about particles in centrifuges. Is it convenient for you to give me the references for this? I would be grateful.
It is not altogether easy to translate from experiments on particles in centrifuges to scenarios of adventures of astronauts. The problem is in keeping a clear head about who is observing what. The interpretations of the observations on particle decay in centrifuges do not compare the effects of speed and acceleration in quite the terms that are needed to make clear what is really affecting what. They are interpreted in terms of verbalistic but not really physical ideas. The authors of these experiments and reports are happy to confirm certain orthodoxies, but they are in fact just repeating the muddled and verbalistic thinking that gives rise to those orthodoxies, instead of giving scientific accounts of the physical observations.
You note that the laboratory workers recorded longer lives for the more rapidly moving particles. Longer lives as recorded between certain events in the laboratory. But these particles were moving at the time. I am asking about the case when the triplets start and finish sitting stationary together drinking tea. Yes I agree that the effect of gravity will be small in comparison with the hypothetically potential effects of speed. But that doesn’t yet deal with the problem that I have stated. So far your comments haven’t indicated that gravity might have any relevant effect at all, yet it obviously does affect the problem at least as to osteoporosis.
The problem is to compare apples with apples, and blueberries with blueberries. This needs a careful statement of what the umpire in the white hat will observe in terms of what is physical for him. I think you have not made that careful statement.
Your reply implies or says that acceleration will not matter while speed will have a big effect, but you haven’t shown that by a proper comparison such as is needed for a scientific answer. This is partly because you have not explicitly compared the experiences of the three triplets as the man in the white hat would see them. Yes, your reply correctly echoes the orthodox views about this, but they also fail to make the comparisons that scientific method demands. That is why so much is written in this blog, with so many different opinions, some of them rather rancorous at times, I note. They think that basic scientific method can be abandoned because they have a superior mathematical understanding. But it is not safe to abandon basic scientific method, no matter how clever one might be. I think you will agree with me about this.
Sincerely,
Christopher
Hi Christopher. You wrote:
“But I think for natural understanding we do need to have a clear view of the “effects” of speed and acceleration separately, and I think we need triplets for that.”
It has been shown in centrifuge experiment with decaying particles that the half-life of the particles are only influenced by the rotational speed and not by the centripetal acceleration. It was done by simply rotating the particles in two centrifuges with different radii, spun so that the particles experienced the same speed, but different accelerations. The different accelerations made no difference to the decay times. However, when spun so that the centripetal accelerations were the same and the speeds different, the particles with the higher speed decayed slower, as predicted by SR.
Back to your ‘triplets scenario’. Wherever you “park” the two stay-at-home siblings, the gravitational time dilation working on them will be very, very small. At the surface of the Sun, it is ~50 seconds per million seconds and on the surface of Earth it is ~10 seconds per billion seconds. This is how much slower our clocks tick compared to a clock somewhere in inter-galactic space and at rest relative to the solar system. For Earth, it makes very little difference being on the surface or up in geostationary orbit – it is the depth of the Sun’s gravitational well that determines the rate of the clocks, because it overwhelms Earth’s own gravity well depth by about 14 to 1.
So how much correction must on make for an Earth observer in 20 years? At 10 parts per billion, I get ~6 seconds in 20 years. Hardly worth the bother if you consider the velocity induced effect is more than 10 years in 20 years!
Regards,
Burt Jordaan (www.Relativity-4-Engineers.com)