# I Sorry but I'm still going nuts over the twin paradox

#### PeroK

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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

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.
This is nonsense.

#### Dale

Mentor
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:
$$\tau_A = \int_0^T \frac{1}{\sqrt{1- \frac{v_A^2}{c^2}}} dt, \ \ \tau_B = \int_0^T \frac{1}{\sqrt{1- \frac{v_B^2}{c^2}}} dt$$
Shouldn't this be the inverse of the integrand?

• PeroK

#### PeroK

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Shouldn't this be the inverse of the integrand?
Thanks. Fixed.

• Dale

#### Ibix

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)
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?

They've spent the same coordinate time inertial and the same coordinate time non-inertial, at least in one frame. They've spent the same proper time accelerating under the same acceleration. The only thing that's different is the proper time spent moving inertially. So your method seems to say only that the one with less proper time is the one with less proper time.

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• #### Dale

Mentor
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
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.

Note, the reference that you provide is unpublished and therefore is not suitable for PF. Furthermore, in the end claims with experimental support, like Bailey's, always outweigh purely theoretical claims, like Gamboa's.

The time dilation of a clock, when measured in an inertial frame, depends only on the speed of the clock in that frame, not on the direction of its velocity, nor its acceleration, nor any higher derivatives.

• PeroK

#### Ibix

Note, the reference that you provide is both unpublished and therefore is not suitable for PF.
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.

#### stevendaryl

Staff Emeritus
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.
That's not true. Let's stick to the blank piece of paper for a while. There is an intrinsic notion of the distance between two points on a piece of paper. This notion of distance determines which paths are straight and which ones are curved. The same is true in Special Relativity. There is an intrinsic notion of something like distance between spacetime points. This is the spacetime metric. The metric determines which paths through spacetime are inertial and which ones are accelerated. This is geometric, prior to any notion of forces.

Now, once you put physics in, it is true in Special Relativity that accelerated paths require an expenditure of energy, while inertial paths do not. However, it isn't the forces that determine which paths are straight, but the metric that determines which paths require force to maintain them.

Think about a perfectly smooth spherical planet, with no friction. If you shove an object, and then let the object continue in a force-free manner, the object will follow "great circle" routes. A great circle route on a sphere is the shortest distance between two points on the sphere, if you're traveling along the surface of the sphere.

It's the geometry of spacetime that determines what are the inertial paths.

• Dale

#### stevendaryl

Staff Emeritus
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?
It's hard to make sense of your question. Yes, there is a physical distinction between an accelerated path and an inertial path, in the same way that there is a physical distinction between a straight line and a curved line on a piece of paper. However, you don't need to know the history of the line on the paper to know whether it's curved. You don't need to know what kind of forces produced the line. Whether it's curved or not is a geometric fact about the line.

Geometry is physical.

#### stevendaryl

Staff Emeritus
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?
It's the analogous question for two curves on a piece of paper. You have two points on the paper, $A$ and $B$. You draw two different curves connecting $A$ and $B$. Which curve is longer? The length of a curve is an intrinsic fact about the curve. You look at each curve independently to know how long it is. If you're only told how the distance between the two curves change, then you can't figure out from that how long each is.

• PeroK

#### Ibix

@SamRoss - I think the point being made here is that there is no difference between the statements that "an object was accelerated" and "an object has a curved worldline". They are the same statement.
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.
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.

The proper time in terms of the coordinates still depends only on the velocity (and the metric).

#### SamRoss

Gold Member
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.
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. When I looked at the FAQ and saw a flat red line and a bent green line, I thought "Okay, from the point of view of the red guy, he stays still while the green guy leaves and then comes back. 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?" Now, for someone who's more experienced with relativity, you might say that obviously there's some extra information, some dynamics, that were not explicitly mentioned in the build-up to the problem, but that's not obvious to everyone. Someone like me is left wondering, "Why can't I flip this diagram?" I strongly recommend some discussion be added to the FAQ concerning why it is not possible to flip the diagram as I did in my original post, to make it clear that something is different between the two observers. Furthermore, the distinction should be explained in terms of something more concrete than just the geometry itself. 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.

• weirdoguy

#### PeroK

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Re acceleration
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.
Please provide a reference where the paradox is presented like this. It's not the way it's ever been presented in my experience. It's usually shown as one twin blasting away from Earth in a rocket.

• weirdoguy

#### Orodruin

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Acceleration is everything -- and it's the only thing.
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.
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.
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.

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?"
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.

#### Ibix

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.
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

Gold Member
It's hard to make sense of your question.
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.
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... ...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... But he doesn't hold on to his ruler, does he? When he turns around, the ruler keeps going without him.

Here are three Minkowski diagrams showing the same scenario.
Ibix, I should have paid more attention to your post a while back. At first, when I saw this picture... ... it didn't seem to make sense. I didn't understand why the green line would ever leave the t' axis because I was not thinking of the moving reference frame as having an existence independent of the green guy. Reading through it again today, I realized I must be mistaken about the meaning of the diagrams. Now that I imagine him letting go of his ruler, it all makes sense! Please tell me I'm getting this right. The people don't hold on to their rulers forever, do they?

#### stevendaryl

Staff Emeritus
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...
Something to consider is the analogous question for Euclidean geometry. Suppose the two paths, the green path and the red path, instead of being paths in spacetime, are paths in space. Suppose that they are two different ways to get from one location, on the lower left, to another location, on the lower right. If two cars met in the lower left, and one car took the green path, and one car took the red path, then when they arrive at the lower right, they could compare distances as measured on the cars' odometers. The two odometer readings will be different. Is that a physical effect? 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?

#### SamRoss

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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?
Of course the odometers are simply measuring the distance along each path. I think my mistake was in taking too literally statements like "Alice's reference frame" and "Bob's reference frame". I thought Bob's reference frame was a frame that moved with him. Bob would always be at x=0 no matter what actions he took. 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. This misunderstanding prevented me from viewing world lines as fixed. I thought the world line would change shape when looked at from the point of view of a different observer.

• jbriggs444

#### jbriggs444

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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.
Nicely and accurately put!

Inertial reference frames are... inertial. They keep moving.

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.

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• SamRoss

#### Ibix

Simply stringing together a series of instantaneous tangent inertial frames does not always meet the requirements.
Indeed - see post #22 for one of the problems.

#### SamRoss

Gold Member
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.
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?

#### Orodruin

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Interesting. I would have thought you could string together an infinite number of inertial frames to get an accelerating one.
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

#### Dale

Mentor
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?
Here is one reference on the topic that I found recently:

Figure 1 identifies the main mathematical problem associated with naively stringing together reference frames.

• PeroK

#### stevendaryl

Staff Emeritus
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?
Here's the problem: Suppose that you are at rest in one frame up until time $t=0$, then you accelerate quickly so that you are now at rest in a second frame. The first frame uses coordinates $(x,t)$ and the second frame uses coordinates $(x',t')$. How can you piece together these two inertial frames so that you have a noninertial frame? Since "frame" is a little ambiguous, let me switch the problem to that of coordinate systems. How can you piece together two inertial coordinate systems, $(x,t)$ and $(x',t')$ to come up with a noninertial coordinate system?

The obvious thing would be to use the first coordinate system for events prior to $t=0$, and use the second coordinate system for events after $t=0$. Here's the problem with that. In the drawing below, I've drawn a representation of a region of spacetime. I've used the black lines to indicate the t-axis and the x-axis. I've used blue lines to indicate the t'-axis and the x'-axis. If you're wondering why the t' and x' axes seem tilted, that's just the Lorentz transformations. The t' axis is the line satisfying the equation $x' = 0$, which in terms of the original coordinates $(x,t)$ is the equation $\gamma (x - v t) = 0$ or $t = x/v$. So the t'-axis is a line with slope $1/v$ when plotted in terms of $x,t$. The x' axis is the line satisfying $t'=0$, which in terms of $(x,t)$ is the line $\gamma (t - \frac{vx}{c^2}) = 0$ or $t = \frac{vx}{c^2}$. So the x' axis is a line with slope $\frac{v}{c^2}$.

So our obvious way of combining coordinate systems is to use $(x,t)$ for events prior to the change of frames, and to use $(x',t')$ for events after the change. However, if you look at the chart, I've divided spacetime into numbered regions. Our rule works fine for regions 1, 3,4, 5, 7, and 8. However, there are two regions that are problematic: Region 6 is doubly-covered. According to coordinate system $(x,t)$, it covers event prior to the change of frames, and therefore should be described using the coordinates $(x,t)$. But according to the coordinate system $(x',t')$, it covers events after the change of frames and so should be described using the coordinates $(x',t')$. Both coordinate systems claim this region.

We have the opposite problem for region 2. It is not covered at all. According to the coordinate system $(x,t)$, it covers events that take place after the frame change, and so should be described using coordinates $(x',t')$. According to coordinate system $(x,t)$, region 2 takes place before the frame change, so should be described using coordinates $(x,t)$. Neither coordinate system claims this region.

There is no obvious way to come up with a coordinate system for an accelerated observer that
1. Covers every point in spacetime.
2. Doesn't double-cover any points in spacetime.
3. Has the observer at rest at all times. • SiennaTheGr8 and PeroK

#### PeroK

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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?
To add to the above, this issue came up in a different guise recently, in a paradox about length contraction:

#### Ibix

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.

"Sorry but I'm still going nuts over the twin paradox"

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