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Dale
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Nope, additional information is required. The accelerometer readings would be sufficient.SamRoss said:Can you tell me which observer experienced a greater proper time?
Nope, additional information is required. The accelerometer readings would be sufficient.SamRoss said:Can you tell me which observer experienced a greater proper time?
Dale said:Nope, additional information is required. The accelerometer readings would be sufficient.
SamRoss said:Let's say there are two observers, A and B. They start out at the same point in space-time. Next, their distance from each other increases at a constant rate. After that, their distance from each other decreases at a constant rate until they are back at the same position. Can you tell me which observer experienced a greater proper time?
No, because you have insufficiently defined the geometry. It is exactly the same in regular geometry: You draw two curves. First the distance between them increases, then it decreases. Which curve is longer? You do not know because your specification is incomplete. You need to specify which curve has a kink in it because that kink is part of the geometry.SamRoss said:Let's say there are two observers, A and B. They start out at the same point in space-time. Next, their distance from each other increases at a constant rate. After that, their distance from each other decreases at a constant rate until they are back at the same position. Can you tell me which observer experienced a greater proper time?
Federation 2005 said:Acceleration is everything -- and it's the only thing. That comes straight out of the action principle for the law of inertia. For the action principle, you use the negative of proper time as the action. So "least action" -- which is what gives you inertial motion -- means "greatest proper time". So, as a consequence, that means that for any two trajectories that cross paths at the start and end of an interval, the one which has more inertial motion and has spent more tine in 0G will have less action -- meaning more proper time; while the one which has less inertiality in its motion (and more acceleration) will have more action (and less proper time)
It should actually be possible to take the function a = A(s), that maps one's proper time s to one's acceleration a in one's rest frame acceleration and derive, from it, the trajectory as a function of coordinate position and time. That, is, one should be able to prove a theorem like this:
Theorem: Given the initial position r(0) = 0, and initial velocity v(0) = V, with proper time s set to 0 at t = 0, and given the acceleration (in one's instantaneous rest frame) a = A(s) as a function of proper time, then there is a unique trajectory r(t) for which each of these are true. Assume that A is continuous.
Let T > 0 be the end time, S the proper time and (without any real loss of generality) assume r(T) = 0. Then T - S can be expressed entirely as a functional of A such that (i) T - S > 0 if A is non-zero for any proper time between 0 and S (inclusive); (ii) T - S = 0 if A is 0 between proper times 0 and S.
Proof:
Exercise and your next published paper. The details are very hairy; lots of coupled differential equations that need to have explicit solutions by quadrature found, before you can get an actual (integral) expression for T - S in terms of the function A.
For instantaneous changes in velocity (which are technically illegal since they are unphysical) you have to use delta functions for A and that WILL produce a non-zero contribution (as you can see by treating the delta function as a limit of smooth functions and examining the limiting value of T - S as you allow the smooth functions to approach the delta function).
Reference:
The twin paradox: the role of acceleration
J. Gamboa, F. Mendex, M. B. Paranjape, Benoit Sirois
https://arxiv.org/abs/1807.02148
This sets the story straight and gets rid of all the folklore myths that are STILL being perpetuated even by professional and mainstream physicists on this issue; and the faulty analyses usually seen for this problem; that contribute to the misunderstanding of the issue.
Shouldn't this be the inverse of the integrand?PeroK said:Take any inertial reference frame (IRF). Let ##v_A(t)## and ##v_B(t)## be the velocities of A and B in that IRF, with ##t## the coordinate time in that IRF. In general, the proper time experienced by A and B during the period ##t = 0## to ##t = T## is:
$$\tau_A = \int_0^T \frac{1}{\sqrt{1- \frac{v_A^2}{c^2}}} dt, \ \ \tau_B = \int_0^T \frac{1}{\sqrt{1- \frac{v_B^2}{c^2}}} dt$$
Thanks. Fixed.Dale said:Shouldn't this be the inverse of the integrand?
Twin A accelerates with some proper acceleration A for some proper time T, waits one (proper) year, accelerates at A in the opposite direction for proper time 2T, waits another year, accelerates in the first direction at A for proper time T, then waits. Twin B does the exact same thing, except waiting two years of proper time between acceleration phases. Which one will be younger when B returns?Federation 2005 said:So, as a consequence, that means that for any two trajectories that cross paths at the start and end of an interval, the one which has more inertial motion and has spent more tine in 0G will have less action -- meaning more proper time; while the one which has less inertiality in its motion (and more acceleration) will have more action (and less proper time)
The claim is clearly incorrect as it violates the clock hypothesis. The clock hypothesis has been experimentally tested and validated for accelerations up to 10^18 g. See: Bailey et al., “Measurements of relativistic time dilation for positive and negative muons in a circular orbit,” Nature 268 (July 28, 1977) pg 301.Federation 2005 said:Acceleration is everything -- and it's the only thing. ...
Reference:
The twin paradox: the role of acceleration
J. Gamboa, F. Mendex, M. B. Paranjape, Benoit Sirois
https://arxiv.org/abs/1807.02148
As far as I can tell on a quick read, the paper simply does detailed calculations of the stay-at-home's proper time between the end of "during the outbound inertial leg" according to the outbound inertial frame and the beginning of "during the inbound inertial leg" according to the inbound inertial frame, using Rindler coordinates. I'm not sure I've seen the calculations done explicitly, but it's hardly revolutionary. That there is proper time unaccounted for if one just uses two inertial frames is textbook stuff and has certainly appeared in posts here. It's just that calculating it explicitly (rather than taking the total proper time for the stay-at-home and subtracting the time accounted for "during" the inertial phases) requires a lot of maths.Dale said:Note, the reference that you provide is both unpublished and therefore is not suitable for PF.
SamRoss said:As I said later in that quote - I don't remember my exact wording - something else must be specified in order for that freedom to disappear. Your own reply implies this. The freedom will go away only if I specify the lengths of the lines I want you to draw. If I don't tell you how long I want the lines to be then you can do whatever you want. In the twin paradox, unless you say which twin was under the influence of the force then you won't be able to justify picking one specific twin as the one with the shorter world line.
SamRoss said:Are we doing pure math or physics? How can we specify the lengths of the lines without looking at what's actually going on in the situation? How can we justify making one line longer or shorter than the other unless there is some physical distinction between the two observers in the real world?
SamRoss said:Let's say there are two observers, A and B. They start out at the same point in space-time. Next, their distance from each other increases at a constant rate. After that, their distance from each other decreases at a constant rate until they are back at the same position. Can you tell me which observer experienced a greater proper time?
Assuming you have the initial position and velocity in the coordinate frame, then yes. You just integrate. If you don't have the initial velocity (or at least some equivalent boundary condition) then no.Federation 2005 said:It should actually be possible to take the function a = A(s), that maps one's proper time s to one's acceleration a in one's rest frame acceleration and derive, from it, the trajectory as a function of coordinate position and time.
Orodruin said:No, because you have insufficiently defined the geometry... Your information needs to be sufficient to identify both the spacetime geometry (in SR assumed to be Minkowski space) and the world-lines as geometrical objects in that spacetime.
SamRoss said:Exactly. My whole point is that too often when the twin paradox is described, the description seems to be incomplete, just as I have described it. The distance between two observers increases and then decreases - that's all you get - yet one of them turns out to be older.
No, it is not, as has been explained repeatedly in this thread already. Geometry is the only thing. Specifying the geometry and curves in the spacetime is part of the specification.Federation 2005 said:Acceleration is everything -- and it's the only thing.
Sorry, but this is absurd. The typical specification of the twin paradox clearly states which twin turns around, which is the very physical statement you are ignoring that makes the situation well defined. You left this out in your description.SamRoss said:My whole point is that too often when the twin paradox is described, the description seems to be incomplete, just as I have described it. The distance between two observers increases and then decreases - that's all you get - yet one of them turns out to be older.
This again completely ignores that one path in spacetime has a bend and the other does not. This is like asking why you cannot make a curve with a bend in it straight by rotating the paper you drew it on. Again, everything is in the geometry.SamRoss said:Why can't I flip that? Why can't I say that from the point of view of the green guy, he stays still and the red guy leaves and comes back?"
A spacetime diagram is just a displacement-time graph (we take them a bit more literally in relativity is all). It's a plot of ##x(t)## for an object. The slope of the line is ##dx/dt##, which is the velocity. The slope changing is therefore the same as the velocity changing - which is acceleration.SamRoss said:If it is possible to "see" the dynamical distinction somehow encoded into the geometry, as you clearly do, then that should be added as well.
stevendaryl said:It's hard to make sense of your question.
PeroK said:Please provide a reference where the paradox is presented like this.
Orodruin said:The typical specification of the twin paradox clearly states which twin turns around, which is the very physical statement you are ignoring that makes the situation well defined. You left this out in your description.
Ibix said:Here are three Minkowski diagrams showing the same scenario.
SamRoss said:I think I just had an "aha!" moment. Tell me if I'm on the right track. In descriptions of different observers' reference frames, people often say things like, "Pretend Alice is holding a ruler in her hand that she uses to measure distances relative to her and Bob is holding a ruler to measure distances relative to him." Well, I didn't realize that they ever let go of their rulers! I thought that even when changing direction, the green guy would keep his ruler in his hand. That's why, to me, the diagram shown in the FAQ...
View attachment 245184
...describes a completely symmetric situation, with the red guy arbitrarily chosen as the one at rest. If the green guy continues to hold on to his ruler, then from his point of view the situation would look like what I put in my original post...
stevendaryl said:Did the odometers work differently along the green and red paths? Or do the odometers just accurately reflect the difference in lengths of the two paths?
Nicely and accurately put!SamRoss said:Now I'm realizing that the reference frame might have been named after Bob because he happened to be moving along with it at first, but it will not follow him if he decides to change direction.
Indeed - see post #22 for one of the problems.jbriggs444 said:Simply stringing together a series of instantaneous tangent inertial frames does not always meet the requirements.
jbriggs444 said:There is such a thing as an accelerating reference frame but it usually takes more than the trajectory of a single object to define one. There are rules involving continuity, not double-mapping events and such that must be followed. Simply stringing together a series of instantaneous tangent inertial frames does not always meet the requirements.
Measurements of time and space referenced against an accelerating frame are even weirder than those referenced against inertial frames.
This is the issue. There is no unique way of doing this and ”string together” is not a very well defined concept. I would suggest readibg up on curvilinear coordinates in general for Euclidean spaces before trying to look at non-inertial reference frames in relativity. One major issue is the relativity of simultaneity, which makes it non-trivial to define what ”string together” actually means.SamRoss said:Interesting. I would have thought you could string together an infinite number of inertial frames to get an accelerating one.
Here is one reference on the topic that I found recently:SamRoss said:Could you direct me to where I can read about measuring time and space in an accelerating reference frame and how this is different from stringing together inertial frames?
SamRoss said:Interesting. I would have thought you could string together an infinite number of inertial frames to get an accelerating one. Could you direct me to where I can read about measuring time and space in an accelerating reference frame and how this is different from stringing together inertial frames?
SamRoss said:Interesting. I would have thought you could string together an infinite number of inertial frames to get an accelerating one. Could you direct me to where I can read about measuring time and space in an accelerating reference frame and how this is different from stringing together inertial frames?
See post #22. It's just stringing two inertial frames together, but immediately you see that parts of spacetime (where the party was) get lost from the picture. Other parts get double-counted, in fact.SamRoss said:Could you direct me to where I can read about measuring time and space in an accelerating reference frame and how this is different from stringing together inertial frames?
SiennaTheGr8 said:It's maybe worth emphasizing that constructing whole coordinate systems for an accelerating traveler is a different exercise from tracking the accelerating traveler's aging (elapsed proper time). For the latter, I believe it's always okay to divide the trip into infinitely many instantaneous inertial frames (for which the traveler is momentarily at rest) and integrate over the traveler's corresponding proper-time infinitesimals.
Is there any construction that can give an unambiguous answer when the two travelers are not colocated? To remove the ambiguity we need an additional arbitrary choice of a simultaneity convention, do we not?stevendaryl said:Yes, that is true. What that construction does not do, however, is to give an unambiguous answer to the question: How old is this traveler compared with that traveler? (Unless they are in the same place and time.)
Nugatory said:Is there any construction that can give an unambiguous answer when the two travelers are not colocated?