Absolute motion's point of reference

[JesseM]

Actually Energy was defined and used here on earth, which is clearly a non-inertial frame

A small region of curved spacetime (like the region of a lab on Earth where physics experiments are typically done) is pretty much indistinguishable from flat spacetime--are you familiar with the http://www.einstein-online.info/spotlights/equivalence_principle to represent the G-force felt due to acceleration). For example, the path of light rays would be slightly curved in such a frame, and the coordinate speed of light would be slightly different from c, but the effect would be very tiny, so it's not too surprising that observers on Earth didn't notice these small corrections and just came up with the simpler equations that would apply in an inertial frame.

 

[yoelhalb]

No, each frame would predict the same thing about the readings on any physical device to measure G-force (i.e. any accelerometer). It just happens to be true that the reading of G-force does not correspond to the coordinate acceleration at a point on the object's worldline where the object is not instantaneously at rest in that inertial frame. However, the laws of physics are still the same in each frame because the way G-force relates to coordinate acceleration as a function of velocity is still the same in each frame. In every frame, if you know the coordinate velocity v and the coordinate acceleration dv/dt at some point on the object's worldline, the formula for calculating the measured G-force a at that point would be:

$$a = \frac{1}{(1 - v^2/c^2)^{3/2}} dv/dt$$

I don't understand that correctly, (maybe you would like to supply me with a source for that, if so then thanks in advance).
From where does the velocity v coming if this is coordinate system to use?
What I understand from your words that you need the to use the velocity of the uniform motion.
And if that is then he cannot claim resting and we must say that his time is the one that is getting slower.

 

[JesseM]

I don't understand that correctly, (maybe you would like to supply me with a source for that, if so then thanks in advance).

Sure, check out this textbook for example.

From where does the velocity v coming if this is coordinate system to use?
What I understand from your words that you need the to use the velocity of the uniform motion.

No, the velocity is just the instantaneous velocity which is the first derivative of the function x(t) that gives position as a function of time (coordinate acceleration is the second derivative of x(t), or the first derivative of velocity as a function of time v(t), which is why I wrote the coordinate acceleration as dv/dt). I asked you a few times before if you were familiar with the idea of first and second derivatives from calculus, can you please answer this question? Understanding of basic calculus is pretty essential for all of modern physics from Newton onwards (if you don't understand the basics I would say you don't really understand what the words 'velocity' and 'acceleration' even mean in physics), so if you're not familiar with this stuff that's really where you need to start.

 

[yoelhalb]

No, please be careful with your wording. Their coordinate speed is far greater than c, but light's coordinate speed is even greater than that. So they are still not going faster than light.

Can you explain me this?
Light has been mesuared to be about c here on earth, clearly less then billions of light years per day.

 

[JesseM]

Can you explain me this?
Light has been mesuared to be about c here on earth, clearly less then billions of light years per day.

Again I recommend you read up on the http://www.einstein-online.info/spotlights/equivalence_principle (especially the third paragraph). As always, though, you're going to have trouble understanding any discussion of velocity if you don't understand the basic idea that instantaneous velocity at any given time t in a particular coordinate system is the first derivative of the position as a function of time x(t) in that coordinate system, i.e. v(t) = dx/dt. If you'd like to learn about derivatives I'm sure people here can recommend some good sources, but if you just keep ignoring the issue it's going to start seeming like you are not so much interested in learning about velocity and acceleration in relativity as just in finding reasons to criticize it.

 

[Dale]

Can you explain me this?
Light has been mesuared to be about c here on earth, clearly less then billions of light years per day.

Sure, I will use units of light-years for distance and years for time so that c=1. I will use capital letters to indicate an inertial frame and lower case letters to indicate Earth's non-inertial frame which rotates once per (sidereal) day, and the Z=z axis is aligned with celestial north.

The transformation between the inertial and non-inertial frames is given by:
x=X cos(ωT) - Y sin(ωT)
y=X sin(ωT) + Y cos(ωT)
z=Z
t=T

A star at rest wrt Earth and located 1 light year away on the X axis would have the coordinates:
R=(X,Y,Z)=(1,0,0)
r=(x,y,z)=(cos(2300t), sin(2300t), 0)
so at t=0 this gives a coordinate speed of
|dr/dt| = |(0, 2300, 0)| = 2300 > 1

A ray of light leaving that star at T=0 in the Y direction would have the coordinates:
R=(X,Y,Z)=(1,t,0)
r=(x,y,z)=(cos(2300t) - t sin(2300t), t cos(2300t) + sin(2300t), 0)
so at t=0 this gives a coordinate speed of
|dr/dt| = |(0, 2301, 0)| = 2301 > 2300

If you go out further than 1 light year the effect becomes greater. The coordinate speed of the stars becomes much greater than c, but the coordinate speed of light is even greater than that.

 

[yoelhalb]

If you'd like to learn about derivatives I'm sure people here can recommend some good sources.

Thanks, but I already know it.

 

[yoelhalb]

Anyway, if C is "accelerating to the left" in the inertial frame of the ocean where A is moving at constant velocity to the left and B has constant velocity to the right, then he will be closer to A than B, but will remain between them (to the right of A, to the left of B) until he finally catches up to A. Just suppose that in the ocean frame, the horizontal axis is labeled with an x-coordinate, with -x being to the left and +x to the right. Then x(t) for A could be x(t)=-100*t (so for example at t=2 hours, A will be at x=-200 miles, where x=0 being the position where ABC started at t=0 hours) while x(t) for B could be x(t)=100*t. In this case if C is accelerating at 1 km/hour per hour, then C could have x(t)=-0.5*t2, which means it has v(t)=-1*t (so for example at t=1 hour, C is at position x=-0.5 miles with v=-1 mph, then at t=2 hours C is at position x=-2 miles with v=-2 mph, at t=3 hours C is at position x=-4.5 miles with v=-3 mph, until finally at t=200 hours both A and C meet at position x=-20,000 miles).

As C accelerates toward A (as long as A stays at the same motion) the difference in their relative speed will drop by one mile an hour, each hour. After 3 hours, A will be 600 mph away from B, as B is traveling 200 mph in respect to A. C will be 294 miles away from A, moving 99 mph away from A in the first hour, then 98 Mph in the second hour, 97 mph in the third hour.

A is still at rest, however C is slowing down as it is moving away from A.

After 100 hours, B is 2000 miles away, still going at a rate of 200 mph (A's 100 mph and B's 100 mph...to B, A is doing the same). C however is at rest in regard to A (C has reached 100 mph), having slowed down from 99 mph down to 1 mph in regard to A.

As each hour increases, C gains 1 mph in speed as it approaches A, until it eventually overtakes A. It can overtake A if it simply travels at 101 mph (or 1 mph in regard to A), but in this case, C will overtake A much sooner as it is accelerating toward A now.

While the boat never "literally" turns around (it is always facing the same direction) and might seem silly to assume it is going backward so fast and leave a wake BEHIND it...most questions of this nature actually start in featureless space where only a,b, and c are present.

In your scenario, it is easier to assume D is the at rest rate, where D is the Earth everyone is moving across. This is no more valid than anyone else's reference, but it has features one can refer to and all three can measure against.

D by the way is moving in F, the Milky Way (Skipping the solar system), which is traveling at about a million miles an hour toward Q, which is the Great Attractor...so, using D as a rest reference, and ignoring F and Q (and everything in between) makes things much simpler.

According to what you write it follows that C (from A's point of view) started with its direction to the right (since they are all together in the beginning and then C moves to the right side of A and A is the frame of reference), and then he changed directions and met A, that essentially means that he changed direction without rotating.
With this you are actually destroying the answer on the twin paradox.

Actually although this can really be, there are some instances that such a claim is invalid, I have no clue if a spaceship can be claimed to be backing up, but it is against physics and common sense to claim that a buggy can pull the horse, (and yes there might be something like that in space), actually special relativity in its answer on the twin paradox claims this to be true for any motion.
So there are situations that the direction is clear for all, and A's claim makes no sense and you would never believed it if some one would tell you such a story in real life, and I don't see why we have to believe it just to support an hypothesis that can never be tested.

 

[Dale]

C (from A's point of view) started with its direction to the right (since they are all together in the beginning and then C moves to the right side of A and A is the frame of reference), and then he changed directions and met A, that essentially means that he changed direction without rotating.
With this you are actually destroying the answer on the twin paradox.

Things change direction without rotating all the time. Throw a pencil straight up into the air and note that as it rises then falls it changes direction without rotating.

Your objection is irrelevant.

 

[yoelhalb]

Things change direction without rotating all the time. Throw a pencil straight up into the air and note that as it rises then falls it changes direction without rotating.

Your objection is irrelevant.

Then according to you two twins moving away and them moving back and meeting, and according to what you say they can change direction without rotation, so who will be younger?
Anyway a pencil can change directions, but a horse and buggy it is against common sense and physics to claim motion in 2 directions (is the principle of relativity a religion?).

 

[Dale]

Then according to you two twins moving away and them moving back and meeting, and according to what you say they can change direction without rotation, so who will be younger?

The one which underwent non-zero proper acceleration.

Anyway a pencil can change directions, but a horse and buggy it is against common sense and physics to claim motion in 2 directions

Show me a detailed derivation where the horse and buggy doesn't make sense. You are imagining a problem with relativity that does not exist.

 

[yoelhalb]

The one which underwent non-zero proper acceleration.

Why should anyone of them?

Show me a detailed derivation where the horse and buggy doesn't make sense. You are imagining a problem with relativity that does not exist.

Imagine A,B,C are togheter, C is a horse and buggy.
Now A and B move apart in a uniform motion, and C starts accelration till it meets A.
Since a horse and buggy can go only in one direction it must be that A is the one that moves.

 

[Dale]

Why should anyone of them?

Because otherwise they will not be able to reunite.

Imagine A,B,C are togheter, C is a horse and buggy.
Now A and B move apart in a uniform motion, and C starts accelration till it meets A.
Since a horse and buggy can go only in one direction it must be that A is the one that moves.

This is not even approximately a derivation. Please do an actual derivation using explicit expressions for the various worldlines, transformations, and derived quantities of interest. Which derived quantity do you think is wrong? E.g. the tension between the horse and buggy should always be positive, do you think you get a negative tension in some frame, if so then derive the tension in that frame.

 

[yoelhalb]

Again A,B,C are together at one point.
C is a horse and a buggy.
Now A and B start moving apart, each one seeing the other one moving 100 m/s, A moves to the left and B to the right (from each others perspective), like this A<----------->B.
In the same second C also started an acceleration of 1 m/s² to the left.
Since it is clear the direction of the horse and buggy is clear it follows that every one must agree that C is moving left only.
Now if A is the point of reference then C should never be to his right, but just to his left, and will never meet him again as long C is not rotating, since according to A's perspective A is at the point of reference and where the motion started.
But if B's claim that he his the point of reference then C should be next to B, (in the first second he will be 1 m apart, the second 3 m, etc), until after a long time he will meet A.
Now we have clear proof who is moving.

 

[Dale]

You really don't seem to understand what a derivation is.

So I will ask you again: what are the worldlines of A, B, and C? And which derived quantity is concerning you?

You have asserted several times that a horse and buggy cannot go backwards, but have not said why you believe that and have not shown that whatever is troubling you is actually predicted by relativity. If you cannot even form a coherent argument how do you expect to have a rational discussion.

 

[JesseM]

According to what you write it follows that C (from A's point of view) started with its direction to the right (since they are all together in the beginning and then C moves to the right side of A and A is the frame of reference), and then he changed directions and met A, that essentially means that he changed direction without rotating.
With this you are actually destroying the answer on the twin paradox.

Actually although this can really be, there are some instances that such a claim is invalid, I have no clue if a spaceship can be claimed to be backing up, but it is against physics and common sense to claim that a buggy can pull the horse, (and yes there might be something like that in space), actually special relativity in its answer on the twin paradox claims this to be true for any motion.
So there are situations that the direction is clear for all, and A's claim makes no sense and you would never believed it if some one would tell you such a story in real life, and I don't see why we have to believe it just to support an hypothesis that can never be tested.

Your argument has nothing specifically to do with relativity at all! In basic Newtonian physics, suppose that in the frame of the ground a car is accelerating down the road so its speed relative to the road is increasing. Then if I am an inertial observer moving at constant velocity down the same road, if the accelerating car's velocity relative to the road goes from below mine to above mine, then in my frame the car's direction will change without it turning around. It's not hard to see why this must true--when the accelerating car's speed relative to the road is lower than mine, if I am in front the distance between me and the accelerating car is increasing, so in my frame (where I am at rest) the accelerating car must be moving away from me; but then when the accelerating car's speed exceeds mine while I am still in front, the distance between me and the accelerating car is now decreasing, so in my frame the accelerating car must now be moving towards me.

For a more mathematical demonstration, suppose the accelerating car's position as a function of time in the ground frame is given by x(t) = (1.5 meters/second^2)*t^2, and my own position as a function of time in the ground frame is given by x(t) = (27 meters/second)*t, so I have a constant speed of 27 m/s and we both start at position x=0 meters at time t=0. You said you were familiar with derivatives, can you calculate the instantaneous velocity as a function of time (i.e. first derivative of x(t) with respect to t) for the accelerating car, and therefore figure out the time t at which the accelerating car's velocity exceeds that of my car?

Then in Newtonian physics, the coordinates of events in my own rest frame x',t' are related to the coordinates x,t in the ground frame by the following simple transformation:

x' = x - (27 m/s)*t
t' = t

And the reverse transformation:

x = x' + (27 m/s)*t'
t = t'

So if the accelerating car has x=(1.5 m/s^2)*t^2 in the ground frame, we can substitute x=x' + (27 m/s)*t' and t=t' to conclude x' + (27 m/s)*t' = (1.5 m/s^2)*t'^2, which means x'(t') in my frame is x'(t') = (1.5 m/s^2)*t'^2 - (27 m/s)*t'. Again, can you take the first derivative of this to find the velocity as a function of time of the accelerating car in my own rest frame?

 

[yoelhalb]

Your argument has nothing specifically to do with relativity at all! In basic Newtonian physics, suppose that in the frame of the ground a car is accelerating down the road so its speed relative to the road is increasing. Then if I am an inertial observer moving at constant velocity down the same road, if the accelerating car's velocity relative to the road goes from below mine to above mine, then in my frame the car's direction will change without it turning around. It's not hard to see why this must true--when the accelerating car's speed relative to the road is lower than mine, if I am in front the distance between me and the accelerating car is increasing, so in my frame (where I am at rest) the accelerating car must be moving away from me; but then when the accelerating car's speed exceeds mine while I am still in front, the distance between me and the accelerating car is now decreasing, so in my frame the accelerating car must now be moving towards me.

It has with the principle of relativity.
Before special relativity there was claimed to be absolute motion (such as the ether) so you would never claim that the car changed directions.
Such a claim was only introduced by Einstein, and it can never be proved, and as I show it is against common sense.

 

[Dale]

as I show it is against common sense.

You certainly have not shown any such thing. You have merely asserted it with no proof, derivation, nor even an explanation about why you might think such an absurd thing.

 

[yoelhalb]

You certainly have not shown any such thing. You have merely asserted it with no proof, derivation, nor even an explanation about why you might think such an absurd thing.

Then please explain it to me.
I will tell you the story and you will explain me just what is going on.
Imagine 3 objects are at together A,B,C.
Then A and B are moving away with a uniform motion A<----------->B, at 100 m/s.
C also starts to accelerate to the left, C is a horse and buggy, accelerating from C's view 1 m/s² from the initial point.
So in the first second after the beginning of the motion, (C has moved 1 m from the initial point) will C be 1 m to the left of B or 1 m to the left of A?, and how will A and B interpret this.

 

[JesseM]

It has with the principle of relativity.
Before special relativity there was claimed to be absolute motion (such as the ether) so you would never claim that the car changed directions.

Even before relativity, there'd be nothing stopping you from having a road moving inertially relative to the ether frame (after all the Earth is not the center of the universe, so we wouldn't expect the surface of the Earth to remain at rest relative to the ether), and a car moving relative to the road at just the right velocity so it was at rest in the ether frame. In this case, if you have a second accelerating car initially at rest relative to the road, but then accelerating in a constant way so that its velocity relative to the road eventually exceeded the inertial car's velocity, then naturally the accelerating car will turn around without rotating in the inertial car's frame--and here we have set things up so the inertial car's rest frame is the ether frame, so the accelerating car turns around without rotating in the ether frame too.

 

[Mentz114]

Then A and B are moving away with a uniform motion A<----------->B, at 100 m/s.

I presume you mean moving apart. In which case they must have experienced some acceleration before the state of motion you describe.

This period is crucial in working out if C will overtake A. But if the system was completely specified there would be no doubt about the positions of A,B and C, and all observers will agree.

 

[yoelhalb]

Even before relativity, there'd be nothing stopping you from having a road moving inertially relative to the ether frame (after all the Earth is not the center of the universe, so we wouldn't expect the surface of the Earth to remain at rest relative to the ether), and a car moving relative to the road at just the right velocity so it was at rest in the ether frame. In this case, if you have a second accelerating car initially at rest relative to the road, but then accelerating in a constant way so that its velocity relative to the road eventually exceeded the inertial car's velocity, then naturally the accelerating car will turn around without rotating in the inertial car's frame--and here we have set things up so the inertial car's rest frame is the ether frame, so the accelerating car turns around without rotating in the ether frame too.

My question is different then your example, since in my example both started at the same place, and my question is not because he passes him but because A must claim him backing up, (also a car can back up without rotating and is not the same as a horse and buggy), here is what your example might look like on Earth before relativity.
Imagine two horse and buggies are initially at the same spot, then suddenly one horse and buggy accelerates backward and then suddenly he passes the other horse and buggy, all without rotation.
(However nobody would make such a claim before relativity, and you would never believe such a story).

This is analogous to what I am speaking, A,B,C are initially together, then A and B move away with linear motion and A is to the left, if A is the point of reference then how can C (who is accelerating to the left) be to the right side of A.
Thus, clearly showing that although A moves with a linear motion B is the point of reference.

You might claim that C will never be to A's right, but this is not true, consider two ships moving away with a uniform motion do you think the water between them will be emptied out?.
So C might stay to his right, and thus proving that a is not the frame of reference, (Actually this is what I started the whole thread that there must be some global reference and not that every body can claim to be his own point of reference).

 

[ghwellsjr]

Then please explain it to me.
I will tell you the story and you will explain me just what is going on.
Imagine 3 objects are at together A,B,C.
Then A and B are moving away with a uniform motion A<----------->B, at 100 m/s.
C also starts to accelerate to the left, C is a horse and buggy, accelerating from C's view 1 m/s² from the initial point.
So in the first second after the beginning of the motion, (C has moved 1 m from the initial point) will C be 1 m to the left of B or 1 m to the left of A?, and how will A and B interpret this.

You claimed in post #67 that you already know about derivatives but now it is clear that you do not. If C is accelerating at 1 m/s^2, then after 1 second, C will have moved 1/2 m from the initial point, not 1 m. You seem to be getting the position confused with the speed which is 1 m/s after 1 second.

And your example has nothing to do with relativity. We can't even figure out what your issue is. You give us a multiple choice question where all the answers are incorrect. Or maybe I should say, the only way one of your answers could be correct is if we interpret the question in a way that I'm sure you didn't mean.

The way I think you mean is: After one second, A has moved to the left 50 m, B has moved to the right 50 m, and C has moved to the left 1 m. But then why are you asking us if C is 1 m to the left of A or 1 m to the left a B?

So you must have meant that A is moving to the left and is 100 m from the starting point and B is stationary so then the correct answer would be: C is 1 m to the left of B.

But you could have meant that B is moving to the right and is 100 m from the starting point and A is stationary so then the correct answer would be: C is 1 m to the left of A.

(Keep in mind, I am using your incorrect understanding of the actual position of C after 1 second.)

Can you see why your example is so confusing? You have been presenting this example since post #17, and you still haven't presented it in an unambiguous way that would allow us to respond in any meaningful way.

 

[ghwellsjr]

My question is different then your example, since in my example both started at the same place, and my question is not because he passes him but because A must claim him backing up, (also a car can back up without rotating and is not the same as a horse and buggy), here is what your example might look like on Earth before relativity.
Imagine two horse and buggies are initially at the same spot, then suddenly one horse and buggy accelerates backward and then suddenly he passes the other horse and buggy, all without rotation.
(However nobody would make such a claim before relativity, and you would never believe such a story).

This is analogous to what I am speaking, A,B,C are initially together, then A and B move away with linear motion and A is to the left, if A is the point of reference then how can C (who is accelerating to the left) be to the right side of A.
Thus, clearly showing that although A moves with a linear motion B is the point of reference.

You might claim that C will never be to A's right, but this is not true, consider two ships moving away with a uniform motion do you think the water between them will be emptied out?.
So C might stay to his right, and thus proving that a is not the frame of reference, (Actually this is what I started the whole thread that there must be some global reference and not that every body can claim to be his own point of reference).

You seem to think that special relativity is saying that every person, horse, buggy, ship, car, etc. can all claim to be a different point of reference all at the same time. But it does not say that. It says you can pick anyone to be the point of reference and analyze what everyone else is doing from that reference frame. Then, if you want, you can pick another one to be the point of reference and analyze everything from that reference frame and there are ways to convert the answers you get from one reference frame into another reference frame. The number you get for speed, positions and times can be all different in each reference frame but they will be consistent with each other with regard to the order of events. You can even pick a frame of reference for which there is no object.

So let's do it for your example. If we decide that the initial starting point where A, B, & C are stationary is the frame of reference, then we could say that A moves to the left at 50 m/s, B moves to the right at 50 m/s and C accelerates to the left at 1 m/s^2. In this case, after 1 second, C would be to the right of A by 49 m and to the left of B by 51 m. (Again, I'm using your incorrect understanding of position due to acceleration.)

Or we could decide to use the frame of A's motion as the reference frame. Then after 1 second, C would be to the right of A by 49 m and B would be to the right of C by 51 m.

Or we could decide to use the frame of B's motion as the reference frame. Then after 1 second, C would be to the left of B by 51 m and A would be to the left of C by 49 m.

Do you see that these all give the same answers, even though we assume different reference frames?

(These speeds are good enough because they are so slow, it would be a little more complicated if the speeds approached the speed of light.)

 

[Dale]

Then please explain it to me.

Sorry, I cannot read your mind and I cannot make sense out of nonsense. Post 83 by ghwellsjr outlines the same problems that I am having with your scenario. It is not up to me to try to guess your intentions and do both sides of the argument. If you have a point to make then it is up to you to convey it clearly and unambiguously.

To make your point you need to do the following:
1) explicitly state the worldlines of A, B, and C in some inertial frame (as I have requested 3 times now).
2) explain what condition prevents a horse and buggy from going backwards.
3) show that that condition arises in your example.

If you cannot do 1) and 3) then you should still at least be able to do 2). You have provided no explanation for why a horse and buggy cannot go backwards other than asserting "common sense". So, explain, what prevents a horse and buggy from going backwards, do you imagine that the horse explodes, if so what causes the explosion, if not then what else could prevent it from going backwards?

 

[JesseM]

My question is different then your example, since in my example both started at the same place, and my question is not because he passes him but because A must claim him backing up, (also a car can back up without rotating and is not the same as a horse and buggy), here is what your example might look like on Earth before relativity.

In my example the accelerating car's wheels were still rolling forwards when it was initially going backwards in the rest frame of the inertial car at rest in the ether, it wasn't "backing up" in the traditional sense of making its wheels go backwards. That's because in this frame the road was itself moving backwards (think of a treadmill), so even though the accelerating car was going forwards relative to the road, it was still going backwards in the frame of the inertial car until its speed relative to the road matched that of the inertial car, at which they were both at rest relative to the ether, and after that the accelerating car's continued acceleration would cause it to start moving forward relative to the ether.

Nothing about this example would change if you imagined that we replaced the two cars with two horse-and-buggies, and imagined that both started at the same position on the road. It would still be true that if the road was moving backwards at speed v relative to the ether, and the inertial horse-and-buggy was moving forward at speed v relative to the road, then the inertial horse-and-buggy would be at rest relative to the ether. And if the accelerating horse-and-buggy started out at rest relative to the road, then it would start out moving backwards relative to the ether. If it later accelerated until it was moving at a speed greater than v relative to the road, then it would be moving forward relative to the ether. Do you disagree?

 

[yoelhalb]

You seem to think that special relativity is saying that every person, horse, buggy, ship, car, etc. can all claim to be a different point of reference all at the same time. But it does not say that. It says you can pick anyone to be the point of reference and analyze what everyone else is doing from that reference frame. Then, if you want, you can pick another one to be the point of reference and analyze everything from that reference frame and there are ways to convert the answers you get from one reference frame into another reference frame. The number you get for speed, positions and times can be all different in each reference frame but they will be consistent with each other with regard to the order of events. You can even pick a frame of reference for which there is no object.

So let's do it for your example. If we decide that the initial starting point where A, B, & C are stationary is the frame of reference, then we could say that A moves to the left at 50 m/s, B moves to the right at 50 m/s and C accelerates to the left at 1 m/s^2. In this case, after 1 second, C would be to the right of A by 49 m and to the left of B by 51 m. (Again, I'm using your incorrect understanding of position due to acceleration.)

Or we could decide to use the frame of A's motion as the reference frame. Then after 1 second, C would be to the right of A by 49 m and B would be to the right of C by 51 m.

Or we could decide to use the frame of B's motion as the reference frame. Then after 1 second, C would be to the left of B by 51 m and A would be to the left of C by 49 m.

Do you see that these all give the same answers, even though we assume different reference frames?

(These speeds are good enough because they are so slow, it would be a little more complicated if the speeds approached the speed of light.)

The question here is simple (it is a logical and not a mathematical question), since C was initially together with A, and since C is an horse and buggy heading to the left, then if A is the point of reference then C should not never arrive to his right.
To illustrate this in real life, consider the typical relativity example.
You are in a train and there is a train next to it, then both trains start to move apart, so you claim that the other train moves away from you, but the people on the other train claim that you are moving.
Now let's change the example and instead of another train this time a horse and buggy is next to your train, and again your train and the horse and buggy move apart, so you think that the horse and buggy has moved.
But then you look out and you see that while your train and the horse and buggy still move apart, the horse is still facing your train, that means in other words that the buggy is pulling the horse away from you.
Can this be?.
So you are actually the one who moves.

 

[yoelhalb]

In my example the accelerating car's wheels were still rolling forwards when it was initially going backwards in the rest frame of the inertial car at rest in the ether, it wasn't "backing up" in the traditional sense of making its wheels go backwards. That's because in this frame the road was itself moving backwards (think of a treadmill), so even though the accelerating car was going forwards relative to the road, it was still going backwards in the frame of the inertial car until its speed relative to the road matched that of the inertial car, at which they were both at rest relative to the ether, and after that the accelerating car's continued acceleration would cause it to start moving forward relative to the ether.

Nothing about this example would change if you imagined that we replaced the two cars with two horse-and-buggies, and imagined that both started at the same position on the road. It would still be true that if the road was moving backwards at speed v relative to the ether, and the inertial horse-and-buggy was moving forward at speed v relative to the road, then the inertial horse-and-buggy would be at rest relative to the ether. And if the accelerating horse-and-buggy started out at rest relative to the road, then it would start out moving backwards relative to the ether. If it later accelerated until it was moving at a speed greater than v relative to the road, then it would be moving forward relative to the ether. Do you disagree?

So you say now that objects are not being moved apart by a force internal to the object, but rather by an external force such as the road, water or wind (for ships), and the object itself might actually be moving in the opposite direction.
(This is similar to what the Greek's thought about the stars and planets rotating every day around the world, that the universe carries them around the world, even though the planets have their own motion).

So now let's imagine this with a simple example, A and B are initially together, then A and B are being moved apart by an external force A<------------>B
this can be true even if A and B are both horse and buggies facing the opposite direction of the motion, (e.g. A faces the right, and B the left).
The reason is because of an external force, that's what you explained.
Now imagine the external force (road, water, wind, or universe) changes its direction and instead of moving apart the objects it reunites them, (without any acceleration or rotation, actually in our example we don't rotation since the horse are anyway facing the direction of unity).
Now WHO of them is younger?

 

[ghwellsjr]

The question here is simple (it is a logical and not a mathematical question), since C was initially together with A, and since C is an horse and buggy heading to the left, then if A is the point of reference then C should not never arrive to his right.
To illustrate this in real life, consider the typical relativity example.
You are in a train and there is a train next to it, then both trains start to move apart, so you claim that the other train moves away from you, but the people on the other train claim that you are moving.
Now let's change the example and instead of another train this time a horse and buggy is next to your train, and again your train and the horse and buggy move apart, so you think that the horse and buggy has moved.
But then you look out and you see that while your train and the horse and buggy still move apart, the horse is still facing your train, that means in other words that the buggy is pulling the horse away from you.
Can this be?.
So you are actually the one who moves.

The problem with your examples is that you don't say enough about what is going on. When you say that A and C move apart without specifying which one (or both) is accelerating and then want to draw some conclusions based on which way a horse and buggy are facing, it shows that you don't understand some basic principles of physics which have nothing to do with relativity.

I want you to consider another example: You get in a stopped train at the railroad station. You sit in a seat. The shades are pulled down so you can't see out the windows. You consider this to be your reference frame. After a while, you feel a new force pushing you backwards into your seat. Now you know that you are accelerating. That means you are starting to move forward. As long as you continue to feel the force pushing you back into your seat, you know you are gaining speed. After a while, the force pushing you into the back of your seat diminishes until it is gone. Now you know that you have stopped accelerating and you are traveling at a constant speed. But you also know that as soon as you first felt the force, you were no longer stationary in your initial reference frame. You have been and continue to be moving in your initial reference frame. You don't need to look at anything outside your train to know that you are now moving with respect to your initial condition. Do you understand and agree with all of this?

 

[yoelhalb]

Sorry, I cannot read your mind and I cannot make sense out of nonsense. Post 83 by ghwellsjr outlines the same problems that I am having with your scenario. It is not up to me to try to guess your intentions and do both sides of the argument. If you have a point to make then it is up to you to convey it clearly and unambiguously.

To make your point you need to do the following:
1) explicitly state the worldlines of A, B, and C in some inertial frame (as I have requested 3 times now).
2) explain what condition prevents a horse and buggy from going backwards.
3) show that that condition arises in your example.

If you cannot do 1) and 3) then you should still at least be able to do 2). You have provided no explanation for why a horse and buggy cannot go backwards other than asserting "common sense". So, explain, what prevents a horse and buggy from going backwards, do you imagine that the horse explodes, if so what causes the explosion, if not then what else could prevent it from going backwards?

My question is that it is impossible to happen ny internal forces, yet it is possible to happen by external forces (even while it is itself accelerating on the opposite direction).
Imagine the horse and buggy are not on Earth but traveling in water, then the water can surely take them backwards.
But if this is true, then the external force can also take them back without any acceleration or rotation, now when they will meet together who will be younger?.