Twin paradox without accelerative frame.

[stevendaryl]

Because describing the current the state of the world should not depend on what I did once upon a time, and even less so on what I will do in the future, just on my current state as an observer. And any other observer that is momentarily identical to me should agree with me.

The problem with this is that changing rest frames shifts your notion of which events are "now". You could try saying:

If an observer is at rest in frame A up to time T, and is at rest in frame B afterward, then you use A's coordinates for events that took place before time T and use B's coordinates for events that took place after time T.

But the problem is that there are some events that take place after time T, according to frame A, but before time T, according to frame B. These events are not described at all in the patched-together coordinate system.

Another problem is that there are some events that take place before time T, according to frame A, but after time T, according to frame B. These events would appear twice in the patched-together coordinate system.

So even though it sounds plausible that an accelerating observer need only be concerned with what he's doing now, rather than what he's doing in the future or the past, it's not clear how to make that consistent.

 

[Dale]

Because describing the current the state of the world should not depend on what I did once upon a time, and even less so on what I will do in the future, just on my current state as an observer. And any other observer that is momentarily identical to me should agree with me.

Why? I don't recall ever reading any of those points as being a requirement of a coordinate system. Do you have a reference justifying any of that?

 

[Dale]

You and DaleSpam are both free to continue using it of course, but try not to completely distract people interested in learning the subject from the much simpler and objective system of direct application of the Lorentz transforms for the momentarily comoving inertial reference frame of the observer.

I also prefer to use a momentarily comoving inertial frame (MCIF). However, a non-inertial observer is not always at rest in any of its MCIF's, and therefore none of the MCIF's are a candidate for being "the observer's frame".

I think that the utility of "the observer's frame" is rather limited, but if you insist on using such a concept then you have to actually look at the gory details of non-inertial frames.

 

[PAllen]

Because describing the current the state of the world should not depend on what I did once upon a time, and even less so on what I will do in the future, just on my current state as an observer. And any other observer that is momentarily identical to me should agree with me.

A separate point from those made so far: you assume momentarily identical means momentarily identical velocity. But that is not the only kinematic property of a body. There is also proper acceleration. You arbitrarily assume this should make no difference for a 'preferable' notion of simultaneity.

 

[WannabeNewton]

You arbitrarily assume this should make no difference for a 'preferable' notion of simultaneity.

This is a very important point!

When you synchronize two clocks all you're doing is defining what it means for mutual clock readings to be simultaneous. But, as PAllen noted earlier, the clock synchronization convention is operational. There has to be some kind of communication between the two clocks in order to synchronize them according to whatever convention has been adopted. As such you can't use a single event in your vicinity to determine if your clock is synchronized with another clock i.e. you can't use just this local event to define what it means for your clock reading and other clock readings to be simultaneous-simultaneous events are space-like separated so they can't causally influence one another. As such a determination of distant simultaneity by means of communication to and from distant clocks will involve past and future events in your vicinity. If you want to say "now we know that my clock is synchronized with your clock so whenever my clock says $t_0$ your clock will also say $t_0$ simultaneously, thus we have defined simultaneity of events" you will necessarily involve past and future events on your world line. This goes for both inertial observers and non-inertial observers. Hence the behavior of the world line at just a single event isn't enough to determine simultaneity meaning proper acceleration and rotation will both influence the determination, as PAllen noted. Whether you choose to use the MCIF's set of simultaneous events or the non-inertial observer's set of simultaneous events is an arbitrary choice-nothing in nature says one is right and the other is wrong. However both will necessarily involve past and future events on the world lines of the respective observers.

 

[PeterDonis]

So while choice of simultaneity may be optional, if you have an IRF and you want to make Newtonian mechanics work, your choice of simultaneity conventions isn't arbitrary, you must choose the required option.

This is a valid point, but AFAIK nobody has spent much time trying inertial frames with simultaneity conventions other than the standard one, the one you're referring to here. The issue is what to do with *non*-inertial frames.

 

[georgir]

A separate point from those made so far: you assume momentarily identical means momentarily identical velocity. But that is not the only kinematic property of a body. There is also proper acceleration. You arbitrarily assume this should make no difference for a 'preferable' notion of simultaneity.

Even then, radar measurements do not limit themselves to considering the observer's momentary acceleration for specifying his 'now', but the observer's acceleration profile over his whole timelime.

And right here is your problem: you think that "the current state of the world" is a well-defined concept. It isn't. The world does not have a unique "current state". You, at a particular event on your worldline, have a current state, but that's because you are a localized object. The different ways of assigning a "current state" to the rest of the world, at events spacelike separated from a particular event on your worldline, are arbitrary conventions; none of them make any difference to any physical predictions.

This, by the way, points out another flaw in your argument: you are assuming that a given observer is *forced* to use a certain simultaneity convention (which is what assigning a meaning to "the current state of the world" amounts to). That is a necessary premise for your further claim that two observers who are momentarily comoving must use the same convention, which is what your claim that "the current state of the world" must be the same for two momentarily comoving observers amounts to. This claim is false because the premise is false: any observer, in any state of motion, can choose whatever simultaneity convention he likes. Simultaneity conventions are just that: conventions, arbitrary choices that do not affect any physical predictions.

You are right that "current state" can be defined as any cauchy surface, and I never said the opposite. But a flat plane of simultanety (or space, as the case might be) is the most obvious and simple approach, and it should be explained to new people before you jump to the complex curved "now" of radar measurements. That's all I'm saying. I have never claimed that radar measurements are wrong or lead to different predictions or anything like that, I just think they are an unnecessary complication in most cases. Anyway, I'm out of this mess, I hope the OP managed to figure it out...

 

[pervect]

This is a valid point, but AFAIK nobody has spent much time trying inertial frames with simultaneity conventions other than the standard one, the one you're referring to here. The issue is what to do with *non*-inertial frames.

Ah, sorry - sometimes in these long threads, I skip over things more than is really good.

 

[ghwellsjr]

I have never claimed that radar measurements are wrong or lead to different predictions or anything like that, I just think they are an unnecessary complication in most cases.

But radar method will tell you that the velocity of the subject changed a while before C changed his, and then again a while after, and that's the wonky part that will never fly.

What does "never fly" mean if not that it's wrong?

 

[georgir]

What does "never fly" mean if not that it's wrong?

Ok, really, I did not mean that it is wrong, just that it does not make any more sense than the simpler case. But I did get carried away there, and expressed myself quite badly. Busted.

 

[PAllen]

One point to make about the so called 'problem' of radar coordinates that a future event on your world line affects what you consider as simultaneous now is as follow:

- All concept of what is going on elsewhere 'now' for me is extrapolation, inherently unobservable.

- The only simultaneity conclusions that can be based on observation are for space like separated events both in your past light cone. Only then do you know about the events you are considering simultaneous.

- Radar, as I think of it, simply makes the above explicit. Simultaneity is undefined until the events are both in your past light cone. The so called problem then becomes the triviality: your past motion (all of it) influences how you map the simultaneity of your past light cone.

Thus the 'problem' only arises talking about something unknowable based on extrapolation. Radar simply say you better extrapolate not only from the last you know about distant objects, but also extrapolate your world line until all of interest is in your past light cone.

 

[phyti]

I would say that if in frame A you have object B traveling with velocity $\vec v_{BA}$ and object C traveling with velocity $\vec v_{CA}$, then the separation velocity (in frame A) of object C from object B is just the difference $\vec v_{CA} - \vec v_{BA}$, which is a vector.

However, this separation velocity does not equal the the velocity of C in B's rest frame $\vec v_{CB}$ (which is what we normally mean by the relative velocity of C and B). For that, you need a vector version of the "relativistic velocity addition" formula, which I'm too lazy to look up at the moment.

Velocity is expressed as a signed fraction of c.
Given a common frame U, with observer A moving at +a, and B moving at +b, the expression for A measuring the velocity of B is

v=(b-a)/(1-ab).

For a=.4, b= .6, v=.38.

For a=.5, b=-.5, v=.-8.

 

[phyti]

Here is a modified version of the graphic in post 83. The actual times aren't noted for Red in order to emphasize the increasing then decreasing frequency of its clock as perceived by Green. The increase begins after t = -1. Green interprets the motion of Red as a combination of accelerating toward him and a g-field at t = 3. Leaving the original Red path (dashed line) allows a visual comparison of ticks via the black lines between them.

https://www.physicsforums.com/attachments/66056

 

[PAllen]

Velocity is expressed as a signed fraction of c.
Given a common frame U, with observer A moving at +a, and B moving at +b, the expression for A measuring the velocity of B is

v=(b-a)/(1-ab).

For a=.4, b= .6, v=.38.

For a=.5, b=-.5, v=.-8.

jtbell was specifically referring to the general, non-colinear case, to which your formula does not apply.

 

[Trojan666ru]

[/CENTER]

Can you see that the Proper Times on both stopwatches and the signals that go to and from them are identical in both frames even though the Coordinate Times and Coordinate distances are different? Both of these frames contain identical information. The boxes don't know that we have drawn two different diagrams with different coordinates. All they know is what they can see when, according to the Proper Times on their own stopwatches, the signals arrive at their own locations.

That's all same. It's like drawing something on a rubber sheet and then stretching it. Every markings will be in order but the diagram just seem stretched.

But if somehow i can look at the green box with a telescope in real experiment, will it be farther away from me?

Or suppose in the real experiment, we have real rubber sheet like spacetime diagram with coloumns and rows, too large enough to hold the entire setup. From this new setup, will the green box move away from its specific coloumns in rubber sheet or the rubber sheet along with the coloumns & box seems stretched away?

 

[PAllen]

That's all same. It's like drawing something on a rubber sheet and then stretching it. Every markings will be in order but the diagram just seem stretched.

But if somehow i can look at the green box with a telescope in real experiment, will it be farther away from me?

Or suppose in the real experiment, we have real rubber sheet like spacetime diagram with coloumns and rows, too large enough to hold the entire setup. From this new setup, will the green box move away from its specific coloumns in rubber sheet or the rubber sheet along with the coloumns & box seems stretched away?

Not quite sure what you're asking, but if you change motion dramatically, distant objects optically appear to change distance (either by parallax or image size; also by apparent brightness versus 'known' intrinsic brightness).

 

[ghwellsjr]

Can you see that the Proper Times on both stopwatches and the signals that go to and from them are identical in both frames even though the Coordinate Times and Coordinate distances are different? Both of these frames contain identical information. The boxes don't know that we have drawn two different diagrams with different coordinates. All they know is what they can see when, according to the Proper Times on their own stopwatches, the signals arrive at their own locations.

That's all same. It's like drawing something on a rubber sheet and then stretching it. Every markings will be in order but the diagram just seem stretched.

One diagram may look like a stretched version of another diagram but remember, we're talking about one coordinate of time and the other coordinate of distance. So a better analogy is of two very long rulers moving with respect to each other and synchronized clocks (or maybe calendars in this example) placed all along their lengths.

But if somehow i can look at the green box with a telescope in real experiment, will it being farther away from me?

No. I think you're be fooled by the analogy of the rubber sheet. When we are talking about the distance between two objects, they must be taken at the same Coordinate Time, not the same Proper Time. I think you are looking at the position of the red box when its Proper Time is 0 and comparing the distance to the green box when its Proper Time is also 0 and getting a distance of over 4.6 light-years which would be stretching its Proper Distance in their mutual rest frame (prior to accelerating) of 4 light-years. Rather you should be looking at the positions at the Coordinate Time of 2 years where the red box is at the Coordinate Distance of -2.75 light-years and the green box is at 0.75 light-years for a distance between them of 3.5 light-years which is less than 4 light-years.

Furthermore, if you track the distance between them as you move up the diagram, they only get closer together until they cross each other. But remember, distance is a coordinate effect and it will be different in different frames. It is not something that is directly observable to the red box looking at the green box through a telescope, how could it be if it is different in each frame? The red box doesn't know which frame we are using to determine the distance.

However, the red box can shine a laser beam on the red box and measure how long it takes for him to see it reflected back to him and analyze how far it was at the average time between sending and receiving the laser beam. Isn't this the technique you specified in your first post for the blue coordinator to confirm that the two boxes were equal distances away from him?

And the red box can continue to do the same technique throughout the entire scenario to ascertain the distance to the green box as a function of his own Proper Time and he will never see it farther away than 4 light-years.

Or suppose in the real experiment, we have real rubber sheet like spacetime diagram with coloumns and rows, too large enough to hold the entire setup. From this new setup, will the green box move away from its specific coloumns in rubber sheet or the rubber sheet along with the coloumns & box seems stretched away?

As I mentioned before, there's no rubber sheet but you could do it with two long rulers with synchronized clocks (or calendars) along their lengths. Then each observer can keep track of his own position relative to both of the rulers and their clocks as a function of his own Proper Time and then they can all plot their results on two separate diagrams and they will look just like the two diagrams in post #81.

 

[phyti]

jtbell was specifically referring to the general, non-colinear case, to which your formula does not apply.

I didn't imply the expression was used outside SR. In the context of the current example, with two observers converging/diverging relative to a common frame, it isn't necessary for a different/special 'velocity addition' form if you use the correct sign.

 

[PAllen]

I didn't imply the expression was used outside SR. In the context of the current example, with two observers converging/diverging relative to a common frame, it isn't necessary for a different/special 'velocity addition' form if you use the correct sign.

Even in SR, it only applies for col-linear motion. What jtbell was discussing was motion in two arbitrary directions (but pure SR). There are many ways to express the general result. I gave one answer after Jtbell's post.

A more practical answer than the one I gave (which was aimed apply in GR as well as arbitrary coordinates) is that the relative gamma factor between two velocities in SR is:

γ [relative] = γ(u)γ(v)(1 - u v cosθ)

where θ is the spatial angle between the velocities. Then, obviously, √(1-(1/γ^2)) gives the relative speed.

I leave it as an exercise that this gives the standard result when θ is 0 or 180 degrees.