Relative speed and time dilation

[dayalanand roy]

1. If A is static and B is moving away from A at velocity v, from Einsteinian point of view, is A also moving away from B at the same velocity? If yes, is the similarity of movements and velocities of both A and B valid in every context? If B is experiencing a time dilation from the view point of A, is A also experiencing a similar time dilation from the view point of B?

2. If the answer to the above points are yes, then If A and B are atomic clocks, A resting on the ground and B traveling away from B on board an airplane, both clocks should experience the same time dilation and when brought together again, both should again show the same time. But as far as I have understood from the literature about these experiments, only the clock B runs behind the A- only B shows time dilation.

3. In my view, the similarity of movements of A and B should have validity in limited contexts. If they are vehicles, they will not burn equal fuels; if they are men, they will not get equally tired; and if they are vehicles or men or clocks, they should not feel equal time dilation. (I don't know what is the reality.)

 

[russ_watters]

1. If A is static and B is moving away from A at velocity v, from Einsteinian point of view, is A also moving away from B at the same velocity? If yes, is the similarity of movements and velocities of both A and B valid in every context? If B is experiencing a time dilation from the view point of A, is A also experiencing a similar time dilation from the view point of B?

Yes.

2. If the answer to the above points are yes, then If A and B are atomic clocks, A resting on the ground and B traveling away from B on board an airplane, both clocks should experience the same time dilation and when brought together again, both should again show the same time. But as far as I have understood from the literature about these experiments, only the clock B runs behind the A- only B shows time dilation.

No. The first scenario is a snapshot in time and the second scenario is fully developed. They are different. And the thing that makes the movements different is what makes the dime dilation happen.

3. In my view, the similarity of movements of A and B should have validity in limited contexts. If they are vehicles, they will not burn equal fuels; if they are men, they will not get equally tired; and if they are vehicles or men or clocks, they should not feel equal time dilation. (I don't know what is the reality.)

Right: they aren't the same, they are different. And what makes them different makes them different.

 

[DaveC426913]

1. If A is static and B is moving away from A at velocity v, from Einsteinian point of view, is A also moving away from B at the same velocity? If yes, is the similarity of movements and velocities of both A and B valid in every context? If B is experiencing a time dilation from the view point of A, is A also experiencing a similar time dilation from the view point of B?

Yes.

2. If the answer to the above points are yes, then If A and B are atomic clocks, A resting on the ground and B traveling away from B on board an airplane, both clocks should experience the same time dilation

The devil is in the details:

and when brought together again, both should again show the same time.

No. One of them had to turn around and race back toward the other. That breaks the symmetry.

But as far as I have understood from the literature about these experiments, only the clock B runs behind the A- only B shows time dilation.

Because B underwent observable* acceleration. Twice. Once in each direction.
* literally - the passenger in the windowless cockpit of B can feel the acceleration in both cases, and knows he is not in an inertial FoR.

3. In my view, the similarity of movements of A and B should have validity in limited contexts. If they are vehicles, they will not burn equal fuels; if they are men, they will not get equally tired; and if they are vehicles or men or clocks, they should not feel equal time dilation. (I don't know what is the reality.)

If A and B have symmetrical experiences, then neither of them has remained stationary. Thus, when they come back together, their clocks will match.

[ EDIT ] Doh. Russ beat me by a minute.

 

[PeroK]

2. If the answer to the above points are yes, then If A and B are atomic clocks, A resting on the ground and B traveling away from B on board an airplane, both clocks should experience the same time dilation and when brought together again, both should again show the same time. But as far as I have understood from the literature about these experiments, only the clock B runs behind the A- only B shows time dilation.

In fact, in this case, it depends whether B travels east or west. The clock A is not at rest. It is on the surface of a spinning Earth. It's moving east at about 1,000km/h (if it is close to the equator) relative to a clock at the North Pole, say. Let's use this clock at the North Pole as a reference clock.

If the clock B flies round the Earth westward at 600km/h, then effectively it reduces its speed to 400km/h eastward (relative to the North Pole).

When the clocks A and B meet, they will both be behind the reference clock but A will be more behind than B.

Note: to be precise, for this experiment you would need to take some gravitational time dilation into account as well. But, as a thought experiment, we can ignore that,

 

[Ibix]

1. If A is static and B is moving away from A at velocity v, from Einsteinian point of view, is A also moving away from B at the same velocity?

From B's point of view, yes. Other frames will disagree.

If yes, is the similarity of movements and velocities of both A and B valid in every context?

I don't know what you mean by "similarity of velocity". I suspect you are aiming at symmetry between A's rest frame and B's rest frame.

If B is experiencing a time dilation from the view point of A, is A also experiencing a similar time dilation from the view point of B?

I'd say that if B's clocks are time dilated in A's frame and vice versa. No-one ever experiences time dilation - their own thought processes tick at the same rate as their clocks.

2. If the answer to the above points are yes, then If A and B are atomic clocks, A resting on the ground and B traveling away from B on board an airplane, both clocks should experience the same time dilation and when brought together again, both should again show the same time.

No. All of your previous paragraph assumes inertial motion in flat space, in which case they could never meet up again. If you want them to meet up you need at least one of the to accelerate, and your chain of reasoning falls apart.

But as far as I have understood from the literature about these experiments, only the clock B runs behind the A- only B shows time dilation.

Both parties will be able to infer time dilation in the other clock for at least some of the experiment. The phenomenon you are describing is better called differential aging.

3. In my view, the similarity of movements of A and B should have validity in limited contexts. If they are vehicles, they will not burn equal fuels; if they are men, they will not get equally tired; and if they are vehicles or men or clocks, they should not feel equal time dilation. (I don't know what is the reality.)

Here you are invoking a special frame, the local rest frame of the surface of the Earth, where one has to work to move and may relax at rest. However this is a consequence of the Earth being very large and very heavy and surrounded by atmosphere, so there's always friction that slows you back yo the Earth's frame. Relativity is not about that. It's about the underlying physics, not a special case with a very heavy object whose mass dominates everything around it.

By the way, you marked this thread as A level, meaning that you expect postgraduate level answers. In that case the answer is that in general $\int_{\gamma_1}\sqrt {ds^2}\neq\int_{\gamma_2}\sqrt {ds^2}$ where the gammas represent different paths. Is that what you wanted?

 

[Dale]

If yes, is the similarity of movements and velocities of both A and B valid in every context?

Hmm, this is not a correct way of stating the first postulate. The first postulate is that all inertial frames are equivalent. It doesn’t directly deal with movements of objects A and B.

But as far as I have understood from the literature about these experiments, only the clock B runs behind the A- only B shows time dilation.

And this result can be obtained by applying the laws of physics in any inertial frame.

In my view, the similarity of movements of A and B should have validity in limited contexts. If they are vehicles, they will not burn equal fuels; if they are men, they will not get equally tired.

And all of the measurable facts can be found using any inertial frame.

 

[dayalanand roy]

Yes.

No. The first scenario is a snapshot in time and the second scenario is fully developed. They are different. And the thing that makes the movements different is what makes the dime dilation happen.

Right: they aren't the same, they are different. And what makes them different makes them different.

Many thanks. But the answers to point 1 and 3 seem contradictory.

Yes.

No. The first scenario is a snapshot in time and the second scenario is fully developed. They are different. And the thing that makes the movements different is what makes the dime dilation happen.

Right: they aren't the same, they are different. And what makes them different makes them different.

Many thanks. But the answers to point 1 and 3 seem contradictory.

 

[PeroK]

Many thanks. But the answers to point 1 and 3 seem contradictory.Many thanks. But the answers to point 1 and 3 seem contradictory.

"Seem" contradictory or "are" contradictory. You need to be careful with looking at fuel consumption. A bird, for example, may find it easier to fly around than to hover. So, in that case, the bird that moves around (relative to the air) may use less energy than a bird that hovers (relative to the air).

Likewise, if there is a strong wind blowing, it may be easier to be blown along than to stand still.

Or, if you are watching two people on a moving walkway and one starts to walk "backwards" at the speed of the walkway, so that they are not moving relative to you, then who is really moving?

And, finally, as I pointed out above, we are all moving with the surface of the Earth in any case. If you are traveling west, you are actually using fuel to go more slowly (relative to the centre of the Earth of the North Pole), than someone "at rest" on the Earth's surface.

 

[dayalanand roy]

Yes.The devil is in the details:No. One of them had to turn around and race back toward the other. That breaks the symmetry.Because B underwent observable* acceleration. Twice. Once in each direction.
* literally - the passenger in the windowless cockpit of B can feel the acceleration in both cases, and knows he is not in an inertial FoR.If A and B have symmetrical experiences, then neither of them has remained stationary. Thus, when they come back together, their clocks will match.

[ EDIT ] Doh. Russ beat me by a minute.

Many thanks.
But i want to know that is time dilation caused by movement only, or by acceleration? And when B is accelerating, in either direction, for him A is accelerating too. Or suppose B is making a circular movement, constantly changing direction and reaches A. Then for B, A is also constantly changing direction. In should not matter, according to Einstein, that the passenger in the cockpit of B is feeling the acceleration and the one in A not. In a sense, both have symmetrical experience. Still, only clock B feels time dilation.

 

[PeroK]

Many thanks.
But i want to know that is time dilation caused by movement only, or by acceleration? And when B is accelerating, in either direction, for him A is accelerating too. Or suppose B is making a circular movement, constantly changing direction and reaches A. Then for B, A is also constantly changing direction. In should not matter, according to Einstein, that the passenger in the cockpit of B is feeling the acceleration and the one in A not. In a sense, both have symmetrical experience. Still, only clock B feels time dilation.

The concept of time dilation that you are talking about refers to inertial reference frames. If one or more of the clocks is accelerating, then you need to be more careful how you calculate the time dilation.

 

[dayalanand roy]

In fact, in this case, it depends whether B travels east or west. The clock A is not at rest. It is on the surface of a spinning Earth. It's moving east at about 1,000km/h (if it is close to the equator) relative to a clock at the North Pole, say. Let's use this clock at the North Pole as a reference clock.

If the clock B flies round the Earth westward at 600km/h, then effectively it reduces its speed to 400km/h eastward (relative to the North Pole).

When the clocks A and B meet, they will both be behind the reference clock but A will be more behind than B.

Note: to be precise, for this experiment you would need to take some gravitational time dilation into account as well. But, as a thought experiment, we can ignore that,

Thanks. What if B is moving eastward?

 

[PeroK]

Thanks. What if B is moving eastward?

Then it's traveling even faster (1600km/h) relative to our reference clock at the North Pole.

Let's look at the whole thing:

Clock A is at the North Pole (reference)
Clock B remains at a point on the equator
Clock C flies westward round the equator (relative to the Earth)
Clock D flies eastward round the equator relative to the Earth

In this case, all time dilation is, in fact, absolute; as only clock A at the North pole is not executing circular motion (which involves constant centripetal acceleration).

The clocks will keep time in the order A-C-B-D, with A being the fastest and D being the slowest. And, all will agree on this.

In this particular case, I used the essentially inertial reference frame at the North Pole to analyse the experiment.

This is not a case of symmetric time dilation.

 

[dayalanand roy]

From B's point of view, yes. Other frames will disagree.
I don't know what you mean by "similarity of velocity". I suspect you are aiming at symmetry between A's rest frame and B's rest frame.
I'd say that if B's clocks are time dilated in A's frame and vice versa. No-one ever experiences time dilation - their own thought processes tick at the same rate as their clocks.
No. All of your previous paragraph assumes inertial motion in flat space, in which case they could never meet up again. If you want them to meet up you need at least one of the to accelerate, and your chain of reasoning falls apart.
Both parties will be able to infer time dilation in the other clock for at least some of the experiment. The phenomenon you are describing is better called differential aging.
Here you are invoking a special frame, the local rest frame of the surface of the Earth, where one has to work to move and may relax at rest. However this is a consequence of the Earth being very large and very heavy and surrounded by atmosphere, so there's always friction that slows you back yo the Earth's frame. Relativity is not about that. It's about the underlying physics, not a special case with a very heavy object whose mass dominates everything around it.

By the way, you marked this thread as A level, meaning that you expect postgraduate level answers. In that case the answer is that in general $\int_{\gamma_1}\sqrt {ds^2}\neq\int_{\gamma_2}\sqrt {ds^2}$ where the gammas represent different paths. Is that what you wanted?

Thanks. I am sorry to mark it A level. It was a mistake. I am not a physicist.
I am excited to hear the term differential aging, because I haven't read it before in the said experiments, but I have been thinking in the same direction. Could you please quote some reference about that?
What i am not able to follow is if their movements are symmetrical, when B is accelerating, the A should also seem to accelerate from view point of B. So their acceleration should also be symmetrical. And both should feel same consequences. Or else we should distinguish between the movements.
Thanks again.

 

[PeroK]

Thanks. I am sorry to mark it A level. It was a mistake. I am not a physicist.
I am excited to hear the term differential aging, because I haven't read it before in the said experiments, but I have been thinking in the same direction. Could you please quote some reference about that?
What i am not able to follow is if their movements are symmetrical, when B is accelerating, the A should also seem to accelerate from view point of B. So their acceleration should also be symmetrical. And both should feel same consequences. Or else we should distinguish between the movements.
Thanks again.

Acceleration is absolute, so we can distinguish.

 

[dayalanand roy]

Hmm, this is not a correct way of stating the first postulate. The first postulate is that all inertial frames are equivalent. It doesn’t directly deal with movements of objects A and B.

And this result can be obtained by applying the laws of physics in any inertial frame.

And all of the measurable facts can be found using any inertial frame.

Thanks.

 

[Nugatory]

Thanks. I am sorry to mark it A level. It was a mistake.

Fixed.

 

[dayalanand roy]

The concept of time dilation that you are talking about refers to inertial reference frames. If one or more of the clocks is accelerating, then you need to be more careful how you calculate the time dilation.

Thanks

 

[dayalanand roy]

Then it's traveling even faster (1600km/h) relative to our reference clock at the North Pole.

Let's look at the whole thing:

Clock A is at the North Pole (reference)
Clock B remains at a point on the equator
Clock C flies westward round the equator (relative to the Earth)
Clock D flies eastward round the equator relative to the Earth

In this case, all time dilation is, in fact, absolute; as only clock A at the North pole is not executing circular motion (which involves constant centripetal acceleration).

The clocks will keep time in the order A-C-B-D, with A being the fastest and D being the slowest. And, all will agree on this.

In this particular case, I used the essentially inertial reference frame at the North Pole to analyse the experiment.

This is not a case of symmetric time dilation.

Thanks

 

[Ibix]

I am excited to hear the term differential aging, because I haven't read it before in the said experiments, but I have been thinking in the same direction. Could you please quote some reference about that?

It's quite straightforward. The amount of time you experience is just the interval along your path through spacetime (because $\Delta s^2=c^2\Delta t^2$ in your rest frame) and, just like the distance along your path through space, two paths starting in the same place and ending in the same place need not have the same "length". Any textbook on relativity will discuss this (Taylor and Wheeler's Spacetime Physics is a popular choice), although I'm not sure if they use the term "differential aging". Google can easily find the term in the literature, but I'm not certain how widely used it is.

What i am not able to follow is if their movements are symmetrical, when B is accelerating, the A should also seem to accelerate from view point of B. So their acceleration should also be symmetrical. And both should feel same consequences. Or else we should distinguish between the movements.

Their movements are not symmetrical. One clock will have to fire a rocket to turn round. The other doesn't fire a rocket.

 

[Sorcerer]

Thanks. I am sorry to mark it A level. It was a mistake. I am not a physicist.
I am excited to hear the term differential aging, because I haven't read it before in the said experiments, but I have been thinking in the same direction. Could you please quote some reference about that?
What i am not able to follow is if their movements are symmetrical, when B is accelerating, the A should also seem to accelerate from view point of B. So their acceleration should also be symmetrical. And both should feel same consequences. Or else we should distinguish between the movements.
Thanks again.

If I recall correctly, there was recently a discussion here that involved general relativity and Mach's principle where it was mentioned that that sort of reciprocal acceleration doesn't apply in general relativity.

I believe it was this thread:

Exceeding the speed of light (stars seem to exceed c in our Earth's rotating reference frame)

 

[timmdeeg]

What i am not able to follow is if their movements are symmetrical, when B is accelerating, the A should also seem to accelerate from view point of B. So their acceleration should also be symmetrical. And both should feel same consequences.

As was pointed out this situation is not symmetrical. B accelerates away, A remains at his location. B can see that the distance between him and A increases accelerated. But this doesn't necessarily require proper acceleration of A.
Instead you can make the situation symmetrical in case both, A and B accelerate away and come back.

 

[Janus]

Thanks. I am sorry to mark it A level. It was a mistake. I am not a physicist.
I am excited to hear the term differential aging, because I haven't read it before in the said experiments, but I have been thinking in the same direction. Could you please quote some reference about that?
What i am not able to follow is if their movements are symmetrical, when B is accelerating, the A should also seem to accelerate from view point of B. So their acceleration should also be symmetrical. And both should feel same consequences. Or else we should distinguish between the movements.
Thanks again.

As pointed out earlier there are ways to distinguish which of the two are accelerating. Imagine that both A and B have two clocks each which are separated by some fixed distance along a line parallel to their relative motion. As long as A or B remains inertial (not accelerating), an observer at rest with respect to either would note that the clocks he is at rest with respect to run at the same rate. If however, B were to accelerate, then an observer sharing that acceleration would note that B's two clocks would not tick at the same rate. The clock in the direction of the acceleration would tick faster than the other clock. This is what PeroK was alluding to when he said that you had to be careful when applying time dilation to accelerating frames.

 

[vanhees71]

Well, as I pointed out zillions of times, such questions are better answered by formulae rather than prose. The time measured by an ideal clock is the proper time along its world line:
$$\tau=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}.$$
If you have two observers/clocks, just calculate this proper time of each of them to see which one ages more than the other when meeting again.

 

[m4r35n357]

Quite, furthermore (for the OP), acceleration is not the key here; you can have differential aging even when neither observer is accelerating.

 

[DaveC426913]

Well, as I pointed out zillions of times, such questions are better answered by formulae rather than prose.

Problem is, those who understand such formulae are also those who don't struggle with such things as relativistic effects.

Those who are asking questions because they don't understand it are not likely to be satisfied, or perhaps even understand, the formulae.

 

[Sorcerer]

Problem is, those who understand such formulae are also those who don't struggle with such things as relativistic effects.

Those who are asking questions because they don't understand it are not likely to be satisfied, or perhaps even understand, the formulae.

In those cases I wonder if it may help to use a similar but less complex mathematical structure as an example. For example, maybe, a two dimensional vector instead of a massive tensor, or the fundamental theorem of calculus in one dimension rather than the divergence theorem.

 

[Mister T]

If B is experiencing a time dilation from the view point of A, is A also experiencing a similar time dilation from the view point of B?

There is more to this claim than you realize. If you arrange a situation where A and B reunite then your claim is false.

 

[vanhees71]

Problem is, those who understand such formulae are also those who don't struggle with such things as relativistic effects.

Those who are asking questions because they don't understand it are not likely to be satisfied, or perhaps even understand, the formulae.

My point is to explain these things in the most simple form, which is mathematics and not many words which are not accurate enough to provide a clear information on these rather abstract notions of modern physics. The good thing is that the math is not that difficult in the case of special relativity. It's mostly linear algebra and a bit differential geometry (admittedly for a pseudo-Euclidean rather than Euclidean affine space).

 

[Sorcerer]

My point is to explain these things in the most simple form, which is mathematics and not many words which are not accurate enough to provide a clear information on these rather abstract notions of modern physics. The good thing is that the math is not that difficult in the case of special relativity. It's mostly linear algebra and a bit differential geometry (admittedly for a pseudo-Euclidean rather than Euclidean affine space).

It seems like you can get away with explaining a lot of special relativity with just elementary algebra and elementary calculus, along with carefully constructed physical examples.

 

[vanhees71]

I try my best...

 

[DaveC426913]

What you typed:

It's mostly linear algebra and a bit differential geometry

What I read:
"It's mostly ☠☠✂⚙ and a bit of ✂☠☠☠"

 

[dayalanand roy]

It's quite straightforward. The amount of time you experience is just the interval along your path through spacetime (because $\Delta s^2=c^2\Delta t^2$ in your rest frame) and, just like the distance along your path through space, two paths starting in the same place and ending in the same place need not have the same "length". Any textbook on relativity will discuss this (Taylor and Wheeler's Spacetime Physics is a popular choice), although I'm not sure if they use the term "differential aging". Google can easily find the term in the literature, but I'm not certain how widely used it is.
Their movements are not symmetrical. One clock will have to fire a rocket to turn round. The other doesn't fire a rocket.

Thanks.
But it will have to fire a rocket to move away too. So its receeding movement too is not symmetrical.
I am trying to learn the very physical basis of time and relativity( though i am a student of biology).
Kindly see the replies above. It is confusing to me for why the two movements are symmetrical and why not symmetrical?

 

[Janus]

Thanks.
But it will have to fire a rocket to move away too. So its receeding movement too is not symmetrical.
I am trying to learn the very physical basis of time and relativity( though i am a student of biology).
Kindly see the replies above. It is confusing to me for why the two movements are symmetrical and why not symmetrical?

In most twin paradox scenarios we assume that the acceleration magnitude is extremely high for a very short duration (a near instantaneous change in velocity). This means we also assume that our accelerating twin moves almost not at all during the acceleration period. In other words, after our rocket has reached its top velocity, it is still essentially right next to our Stay at home twin. But when he does the acceleration during the turn around, he is far removed from the Stay at home twin. Going back to my example with our observers A and B, we can assume that they are walking in the same direction before B changes direction, and are essentially sharing the same space. At the moment immediately after B changes direction from that that A is walking, this doesn't change, so he doesn't see A as shifting in time. But once he gets some distance between himself and A, then any change in direction he makes has a effect on A's relative position from his perspective.
Grab a couple of paper plates, put them on the floor a few feet apart. Stand on one while facing the other. Without moving from your spot, turn 180 degrees. The plate that was in front of you goes from being in front to being behind you. The plate you are standing doesn't change its position relative to you at all.

This is similar to what is happening to clocks for someone undergoing an acceleration. Nothing special happens to clocks that are co-located with you, but clock removed from you in the direction of the acceleration run fast, while clocks in the opposite direction run slow.

Now in a real situation you can't change speeds instantaneously, but do so over some time while you travel some distance. In this case, as our rocket accelerates away from the stay at home clock, it would note that not only was the stay at home clock running slow due to to the increasing relative speed but also because of the increasing distance from the rocket in the direction opposite that of the acceleration.
The rocket also sees a compound effect in the "stay at home" clock during the turn around acceleration. But all this really does is complicate the calculations to find the final results, which is why we tend to assume conditions where we can ignore these additional complications.

 

[Ilythiiri]

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html
This one was good enough for me, when i got a similar question as OP(time dilation/lightspeed but i was searching in BH context). Simple and comprehensive - prose, examples and pictures that is (:

Quotes to prove relevance:
"...Terence sits at home on Earth. Stella flies off in a spaceship at nearly the speed of light, turns around after a while, thrusters blazing, and returns. (So Terence is the terrestrial sort; Stella sets her sights on the stars.)
When our heroes meet again, what do they find? Did time slow down for Stella, making her years younger than her home-bound brother? Or can Stella declare that the Earth did the travelling, so Terence is the younger? ..."

"... Why so many different analyses? Are the relativists just trying to bamboozle their opponents, like the defence attorney who just has to stir up doubt about the plaintiff's case without giving his own theory of events? Not at all; the physical theory should and does tell a single coherent story here.
Relativity pays the price of permissiveness. It says to us, "Pick whichever frame you like to describe your results. They're all equivalent." No wonder that one analysis ends up looking like three or four. ..."

 

[dayalanand roy]

In most twin paradox scenarios we assume that the acceleration magnitude is extremely high for a very short duration (a near instantaneous change in velocity). This means we also assume that our accelerating twin moves almost not at all during the acceleration period. In other words, after our rocket has reached its top velocity, it is still essentially right next to our Stay at home twin. But when he does the acceleration during the turn around, he is far removed from the Stay at home twin. Going back to my example with our observers A and B, we can assume that they are walking in the same direction before B changes direction, and are essentially sharing the same space. At the moment immediately after B changes direction from that that A is walking, this doesn't change, so he doesn't see A as shifting in time. But once he gets some distance between himself and A, then any change in direction he makes has a effect on A's relative position from his perspective.
Grab a couple of paper plates, put them on the floor a few feet apart. Stand on one while facing the other. Without moving from your spot, turn 180 degrees. The plate that was in front of you goes from being in front to being behind you. The plate you are standing doesn't change its position relative to you at all.

This is similar to what is happening to clocks for someone undergoing an acceleration. Nothing special happens to clocks that are co-located with you, but clock removed from you in the direction of the acceleration run fast, while clocks in the opposite direction run slow.

Now in a real situation you can't change speeds instantaneously, but do so over some time while you travel some distance. In this case, as our rocket accelerates away from the stay at home clock, it would note that not only was the stay at home clock running slow due to to the increasing relative speed but also because of the increasing distance from the rocket in the direction opposite that of the acceleration.
The rocket also sees a compound effect in the "stay at home" clock during the turn around acceleration. But all this really does is complicate the calculations to find the final results, which is why we tend to assume conditions where we can ignore these additional complications.

Thanks a lot sir.
It is really educating and i am grateful that you have taken this much pain to give such a long response.
Great!