3 ships and a rather long piece of string - SRT

[Prince_James]

Andbna:

Let me ask you:
What is the difference between 2 ships connected by a string (or fiber optic cable), and 2 ships communicating via radio transmission?

Virtual photons are exchanged at the speed of light. Light is invariable. Thus the string would remain invariable in all reference frames when judged like this.

 

[Prince_James]

Quantum Heraclitus:

I believe that SRT would postulate that each ship would measure the string differently in relation to its speed. That is to say, that even matters such as distance and size (but not mass or the speed of light) would be variable at different speeds.

 

[Quantum Quack]

Let me ask you:
What is the difference between 2 ships connected by a string (or fiber optic cable), and 2 ships communicating via radio transmission?

The radio is faster, easier, and won't break as easily.
Otherwise, one can think of the string as a large number of radio transmitters, each staying in contact via exchange of virtual photons (holding the molecules together,) just as the radio transmissions exchange photons.

-Andrew

The reason why the cable is important rather than using light or other EM is that the cable is provable as mass.
The cable is a material connection between all ships regardless of velocity.

The issue is about non-simultaneity as required by SRT for relative v RF's
Non-simultaneity must state that all ships are in a different world time line and none are together in a single dynamic present moment. It states that all ships must be simultaneously in a different world time line or differnet "NOWs" [present]

The string or cable either travels towards the future or the past but is still connected to each ship so the signal sent must by virtual of this SRT fact be travelling likewise.
IN fact all three ships could actually share the same flight path and actually be superimposed with each other and yet they would not collide or even notice the other ship is in the same location.
This is because the location is now made up of non-simultaneous NOWS so therefore neither ships observers NOW is observable from the other ships.
If three ships were programed to collide at a given point between two inertial stars using different vectors and velocities would they collide according to SRT.
In fact that would make a great gendanken....hmmmmm

How would you get three ships to collide simultaneously at a given point between two inertial stars if each ship is given a different vector and relative Velocity to each other and the stars in question?
The collision event must be a simultaneous NOW for all three ships and the stars.

 

[Quantum Quack]

Quantum Heraclitus:

I believe that SRT would postulate that each ship would measure the string differently in relation to its speed. That is to say, that even matters such as distance and size (but not mass or the speed of light) would be variable at different speeds.

I think you are correct that this would be the SRT position.

 

[funkstar]

The domain of the Special Theory is Euclidean 4-space , i.e. "flat" in some sense. But, thinking of spacetime as a manifold, we have no reason to assume it is globally flat, in the same sense (this was Ben's earlier point). This seems to imply that the Special Theory is a local theory, as indeed it is

Ah, see, there's the rub - I was simply thinking of R^4, not manifolds in general. Then everything makes a lot more sense.

Thanks for the (snipped) explanations, and patience.

 

[BenTheMan]

Virtual photons are exchanged at the speed of light. Light is invariable. Thus the string would remain invariable in all reference frames when judged like this.

A better way to say this is that photons would take the shortest path between the ships. If they are constrained to travel along the wires, then they can no longer take the shortes path.

Quantum Quack---I think you are more or less completely confused about what you are saying.

 

[QuarkHead]

funkstar, you're welcome. In fact we can model the special theory on $$R^4$$ to show something rather wonderful. Let me first give a definition. Suppose I choose a coordinate system for some test particle A such that it has zero displacement on spatial axes. I will call this the state of being "at rest", though this is an abuse, since we can never be at rest w.r.t the time axis (sadly).

I will call a coordinate system an inertial frame if, for any material particle B in uniform motion relative to A, there is a coordinate transformation that will bring B to rest in it's own coordinate system. This transformation will, of course, be Lorentz.

Now for simplicity let's only consider displacements on the x spatial axis, and call our space $$R^2$$. Since this is a flat space, we can assume that $$R^2 = (t,x)$$ is our inertial frame. Assume that A is at rest in this frame.

Now consider a time-dependent displacement of A on the x-axis i.e. A has unifom velocity relative to $$R^2$$. It is easy to see that A', as we now call it, can be brought to rest by a rotation about the origin $$R^2 \to R^2'$$. Thus, Lorentz is realized as a coordinate rotation.

And for as many different velocities as I care to consider, there will be a corresponding $$R^2_i$$ given by the Lorentz rotation, where each of these different $$R^2_i$$'s is each isomorphic to each other and to $$R^2$$.

But this is just the definition of a manifold! So the set of Lorentz transformations is a manifold.

Moreover, since any 2 or more consecutive rotations is again a rotation, and since the rotation $$r$$ followed by the rotation $$-r$$ is equivalent to the zero rotation, the Lorentz set is a group.

So the set of Lorentz transformations is a manifold and also a group. This is the definition of a Lie group!!!

Incidentally you can use my definition of an inertial frame above to easily show that the so-called Twin Paradox is not a paradox, has nothing to do with acceleration, and most certainly nothing to do with General Relativity.

Gosh, look at the time. I gotta run...

 

[Prince_James]

Bentheman:

A better way to say this is that photons would take the shortest path between the ships. If they are constrained to travel along the wires, then they can no longer take the shortes path.

Correct, but it could only do so at the speed of light.

This actually makes an interesting thought experiment:

Would a string of fiber optic cable force all the ships into the same reference frame? As any signal from either ship would be visible to all the others and would have to be reported precisely the same due to the constant nature of C...

 

[Prince_James]

Quantum Heraclitus:

I think you are correct that this would be the SRT position.

To be quite honest, I have no idea what SRT would predict int erms of spatial dilation on the string from speed and such.

 

[andbna]

Virtual photons are exchanged at the speed of light. Light is invariable. Thus the string would remain invariable in all reference frames when judged like this.

(For QQ as well: )
Yes, the virtual photons are exchanged at light speed, but that means that the minimum time for any 'signal' (be it a movement of the end piece,an electron being added to the end of the wire, etc...) takes a minimum time of t=L/c to propogate from one end to another, where L is the length of our string.
Because of this, the ends will not 'know' about what's happening at the other instantaniously, which is the entire point.

This is in contrast to a standard rigid body, in which we assume that forces, velocities, etc... affect the entire body at the same time. Ultimatly, even if we used an iron rod instead of a string, we must remember that it too is a softbody, with it's atoms of iron being rigid body portions. While we may approximate it as a rigid body in most cases, we cannot use this approximation on a very long rod and expect useable resaults, anymore than we can approximate the energy requried to get an object travelling at 0.9c using Newtonian mechanics.

This means that if an acceleration is applied to one end of the string, it will not be felt by the other for L/c amount of time, so for sufficiently long string, there can easily be periods in which some of our 'radio transmitters' are accelerating while the others are still at rest, and as such, we will observe some of them to have slower tick rates than others.
Hence, we must recognize the string as a softbody composed of neumerous individual objects which do not necessarily share a referance frame.

The reason why the cable is important rather than using light or other EM is that the cable is provable as mass.
The cable is a material connection between all ships regardless of velocity.

My 2 radio transmitters have mass, and they represent a physical connection between the ships just as much as the string does (albiet, much much weaker, but a cable won't stand up to a rocket engine anyways.)
They also simplify calculations

The string or cable either travels towards the future or the past but is still connected to each ship so the signal sent must by virtual of this SRT fact be travelling likewise.

Right, some transmitters on the way are travelling at a different velocity than others. Not much different than bouncing a radio signal off a satelite orbiting earth (signal originated from a ground station) in order to communicate with a Mars lander.

How would you get three ships to collide simultaneously at a given point between two inertial stars if each ship is given a different vector and relative Velocity to each other and the stars in question?

If 2 (or more) events occur at the same time and position relative to some observer, they occur at the same time and position relative to any observer.

-Andrew

 

[Quantum Quack]

If 2 (or more) events occur at the same time and position relative to some observer, they occur at the same time and position relative to any observer.

Doesn't this mean that for the two or more ships to share the same event at the same time and location must also share the same world time line to do so?
btw thanks for taking the time to post.

 

[BenTheMan]

Would a string of fiber optic cable force all the ships into the same reference frame? As any signal from either ship would be visible to all the others and would have to be reported precisely the same due to the constant nature of C...

No I don't think so. Any signal along the wire must propogate at a speed less than or equal to c. You're just describing a ``perfect telephone''!

 

[Quantum Quack]

From what I understand the issue of a fibre optic cable joining 3 ships has posed some interesting thoughts.
The one that intrigues me the most at present is that if light speed is invariant it can only be deemed so over a given distance.
Now if we assume that the cables velocity is variable from one end to the other as some of the cable is in another reference frame and the rest is somewhere in between then how could we say that lights speed in invariant when the distances needed for that invariants can become infinitely divisible and microscopic.

As the cable takes up the velocity of one of the RF the cable must progressively speed up along it's length in a form of acceleration. I guess.

So maybe the answer lies in how an accelerating RF is viewed.
However where I think it gets complicated is that for light speed to be invariant it must have a distance to travel with in an inertial frame...am I mistaken?
And if the cable cannot be broken down into relative inertial frames as the velocity and vector even, may be an infinite gradient, then how can the light speed be deemed invariant?
sheesh ! hard to put into words...

 

[Quantum Quack]

Another possible scenario could be that Ship B is connected to Ship A by way of a winching cable rolled up inside Ship A
As ship B 's relative v is say .6c the cable exiting Ship A has a velocity of .6 c
So a message is sent to Ship B along this cable that has a v or .6 c to reach the ship also at v=.6 'c'
Say the cable is not undergoing any acceleration.

The light message is sent via this cable to ship B

how would this be handleed by SRT? I wonder?

I think we would still have a signal of light travelling into the future as the issue of non-simultaneity is intractible.
But it would be in that future immediately the signal acquired the cables velocity of .6c to maintain invariants.

 

[Prince_James]

Bentheman:

No I don't think so. Any signal along the wire must propogate at a speed less than or equal to c. You're just describing a ``perfect telephone''!

I do believe that all relations to C are held the same in all reference points, regardless of whether C is at all slowed down.

Don't quote me on that.

In fact, let's see if anyone knows whether that is the case fully.

 

[Prince_James]

Andbna:

(For QQ as well: )
Yes, the virtual photons are exchanged at light speed, but that means that the minimum time for any 'signal' (be it a movement of the end piece,an electron being added to the end of the wire, etc...) takes a minimum time of t=L/c to propogate from one end to another, where L is the length of our string.
Because of this, the ends will not 'know' about what's happening at the other instantaniously, which is the entire point.

This is in contrast to a standard rigid body, in which we assume that forces, velocities, etc... affect the entire body at the same time. Ultimatly, even if we used an iron rod instead of a string, we must remember that it too is a softbody, with it's atoms of iron being rigid body portions. While we may approximate it as a rigid body in most cases, we cannot use this approximation on a very long rod and expect useable resaults, anymore than we can approximate the energy requried to get an object travelling at 0.9c using Newtonian mechanics.

This means that if an acceleration is applied to one end of the string, it will not be felt by the other for L/c amount of time, so for sufficiently long string, there can easily be periods in which some of our 'radio transmitters' are accelerating while the others are still at rest, and as such, we will observe some of them to have slower tick rates than others.
Hence, we must recognize the string as a softbody composed of neumerous individual objects which do not necessarily share a referance frame.

You are correct that there are multiple reference frames to be addressed in a string as you've stated. But that being said, as we're dealing with time=l/c, with c being invariant in all reference frames, then it would seem every reference frame in the string would report light the same way, and of course, those people at the end of the strings.

Let's also assume that at any point the observers on the spacecraft can see the light travelling down the string/fiber optic cable (let's switch to the latter as it seems to make this a lot more interesting). If that is the case, and again no reporter is going to ever disagree on C, then it would seem everything would share a single isntance, violating the fact that they ought to be in a different reference frame.

 

[BenTheMan]

I do believe that all relations to C are held the same in all reference points, regardless of whether C is at all slowed down.

Well, c can never be ``slowed down'', as this violates the first postulate (c is always the same in all reference frames).

 

[Prince_James]

BenTheMan:

I should have said "the beam of light".

Scientists have been able to slow down light to a few miles an hour by trapping it in dense clouds of gas and such.

 

[andbna]

Doesn't this mean that for the two or more ships to share the same event at the same time and location must also share the same world time line to do so?
btw thanks for taking the time to post.

No, not exactly. It merely means that thier 3 worldlines must each intersect at one point.
When they do collide, if we assume the cockpits all join together (and ignor the debris jettisoned in the collision,) then what we have is 1 ship which can now be treated as one referance frame.

Now if we assume that the cables velocity is variable from one end to the other as some of the cable is in another reference frame and the rest is somewhere in between then how could we say that lights speed in invariant when the distances needed for that invariants can become infinitely divisible and microscopic.

Well, truly, it's no suprise that we need to calculate time dilation at the atomic level. I can cut the cable into atoms no? Each atom, on it's own would be it's own referance frame. When assembled as a cable, there isn't anything real that makes it's referance frame merge with that of the other frames of the cable.
We, as humans, merge them to simplify calculations, because it usualy produces resaults that are accurate eanough for our purposes.
This is where the concept of rigid, versus soft bodies comes into play. A perfect rigid body is 1 pure structure; an entire macro-referance frame. It does not exist in reality (even a steel cube is still composed of atoms, and it can be deformed with a strong eanough force,) but again, it simplifies calculations.
Soft bodies however can move relative to each other. Generaly we only deal with thier relative movements on a newtonian scale (too small to account for relativity,) again to simplify calculations.
Now, a softbody which is moving relativisticaly relative to itself (as the string between the rockets) defies our usual methodes of simplification. At this point, it's best to devide the string into sections which are moving approximatly the same speed and treat them as individual referance frames.

In conclusion: the real question is not "why can we devide the string into inumerous small referance frames?" but "how can we unite these inumerous small referance frames for the purpose of simplicity?"

Let's also assume that at any point the observers on the spacecraft can see the light travelling down the string/fiber optic cable (let's switch to the latter as it seems to make this a lot more interesting). If that is the case, and again no reporter is going to ever disagree on C, then it would seem everything would share a single isntance, violating the fact that they ought to be in a different reference frame.

Light travels at c always, but the clock rates are not constant. Hence, different observers will observe others to be suffering from some time dilation. Even ends of the fiber optic will, though it may not be relevant for calculating the time it takes for the signal to come out one end.

Anyway, my enitre point being: what you shove in one end of a cable doesn't come out the other for some time.


Excuse me if this post lost coherancy near the end, and sorry I can't address your other post QQ; it's getting late.
-Andrew

 

[Prince_James]

Andrew:

Light travels at c always, but the clock rates are not constant. Hence, different observers will observe others to be suffering from some time dilation. Even ends of the fiber optic will, though it may not be relevant for calculating the time it takes for the signal to come out one end.

No problemo on the coherence, I understood enough.

Now let's talk about the clocks in relation to the light beams.

Clearly SRT puts forth the idea that clocks will run at different rates at different speeds and time itself dilates from such.

However, none of the observers, no matter how slow or fast their clocks are going, will ever disagree on how fast the light is going. An observer stationary compared to one at 1/3rd C and 1/2 C will both accurately talk about the light beam. Accordingly, it is an absolute, and therefore...it would seem all could pinpoint the exact same time, despite being in different refrence points.