Twins Paradox - but with a different spin

In summary, the classical Twin Paradox is a thought experiment involving two twins, A and B, who are standing on each other's shoulders and facing in opposite directions. They are both tethered to a pole with synchronized clocks and jetpacks. Both twins then accelerate in opposite directions around a 1ly circumference path and maintain a constant velocity of .5c for 100 years. From the perspective of the pole, both twins will experience time dilation and age at the same rate. However, from the perspective of each other, they will see each other's clocks as running slower due to their relative motion. This scenario is not equivalent to the more traditional Twin Paradox because both twins are constantly accelerating, making their reference frames non-inert
  • #1
qualiaoftao
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While attempting to wrap my head around the classical Twin Paradox, I asked myself the following. So far I have not been able to reach a well reasoned answer... which is no big surprise. :)

Twins (A and B) are oriented in space such that A could be said to be standing on B but facing in the opposite direction. Both twins have a jetpack and clock. They are both tethered to a pole which is positioned 1ly/2∏ away. On this pole is mounted a third clock.

All three clock are syncronized and the twins ignite their jetpacks and begin to accelerate in the opposite direction around the 1ly circumference path. They both finally reach and maintain a constant velocity of .5c in the first revolution and continue as this velocity for 100 years.

Can one say the twins have equivalent accelerated reference frames? Maybe equivalent but opposite? I am unsure as to what constitutes a reference frame.

From the perspective of the pole, the twin's clocks should be slower. But what of the twins with respect to each other? Shouldn't they both see the other's clock as slower and the other twin age at a slower rate?
 
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  • #2
If I understand what you are saying correctly, then they should both age equally because the situation is symmetrical from the pole's rest frame.
 
  • #3
The two revolving observers are not inertial, and neither can meaningfully uses the Lorentz transform (except over a tiny region for a moment of time).

If they all got back together, at the pole (for example; I assume the pole is inertial), the two revolving clocks would be slower by the same amount compared to the pole clock.

As to perceptions, the pole clock would see the revolving clocks going slow at all times. The revolving clocks would see the pole clock going fast at all times. (Because of non-inertial motion, there is no symmetry in perception - each of these does not see the other slow).

As for how the revolving clocks see each other, this would vary moment to moment, and also depend on what you mean by 'see'. You could mean 'directly see', in which case Doppler effect would would be a major contributor. You could also mean, factoring out Doppler, in which case you would still interpret periodic variation each other's clocks, but the function would be different (from the 'raw' observation, not factoring out Doppler).
 
  • #4
The pole and initial rest state of the twins is the only valid inertial frame (of those you mentioned) from which to analyze your scenario in Special Relativity. You are correct that in this frame, both twins will experience the same time dilation and will age the same and their clocks will always have the same time on them. They will be about 86.6 years old after 100 years according to the pole's clock. They will always see the pole clock running faster than their own. They will actually see each others clock as running slower than their own as they separate and then running faster as they approach but at the same time each time they get together.

You cannot consider each twin as being in a valid frame according to SR because they are constantly accelerating which means they are not inertial. In the more normal Twin Paradox scenario, one twin travels away at a constant speed and during this time they each can consider the other ones clock to be time dilated but when the traveling twin stops and turns around, he becomes non-inertial and so is no longer at rest in the same frame.
 
  • #5
I would agree that if I were observing from the perspective of the pole, both twins should age the same to me. But what of A's perspective of B? It is my understanding that since there is no absolute positioning, it is equally fair for A to say he is not moving and both the pole and B are moving relative to him. This is the original Twin Pardox. The explanation for the paradox states that since one twin is actually undergoing acceleration (an accelerated reference frame) that the reference frames for the two twins is are not equivalent. Thus it is the accelerated twin which ages more slowly.

However in my example, both twins are moving identically, except for direction which is opposite. I think each twin would observe the other as aging more slowly since they are moving relative to one another... but this can not be.
 
  • #6
qualiaoftao said:
But what of A's perspective of B? It is my understanding that since there is no absolute positioning, it is equally fair for A to say he is not moving and both the pole and B are moving relative to him.

No, this is not valid because A's reference frame is not inertial. He has no right to claim he is at rest because he is constantly accelerating (circular motion, even at constant speed, involves acceleration which can be measured).
 
  • #7
elfmotat said:
No, this is not valid because A's reference frame is not inertial. He has no right to claim he is at rest because he is constantly accelerating (circular motion, even at constant speed, involves acceleration which can be measured).

Yes, I agree and it is why I did not put it in my original post. But am I wrong to say that A could claim an inertial frame so long as everything else (pole and B) are accelerated equally in the opposite direction of A's acceleration? Something a kin to free fall?

For example, if one plots the path of a projectile that undergoes acceleration due to gravity, one can also hold the projectile as motionless and move the paper in the opposite direction and successfully plot the exact trajectory? Think moving a pen along the path against stationary paper and then move the paper in the opposite direction against a stationary pen?
 
  • #8
qualiaoftao said:
Yes, I agree and it is why I did not put it in my original post. But am I wrong to say that A could claim an inertial frame so long as everything else (pole and B) are accelerated equally in the opposite direction of A's acceleration? Something a kin to free fall?
No. If A carries an accelerometer it will read something other than 0. Therefore A is a non-inertial observer. A is definitely not in free fall.
 
  • #9
DaleSpam said:
No. If A carries an accelerometer it will read something other than 0. Therefore A is a non-inertial observer. A is definitely not in free fall.

I am sorry, I failed to express my thoughts sufficently. I was not attempting to say that A was in a free fall.

Yes, I agree A is being accelerated and it can be measured. And due to this, A is a non-inertial observer. That being said... can A's frame of reference to transformed to an inertial frame by adjusting the acceleration of all other relevant objects in the frame in the opposite direction.

ie: When you plot the path of a projectile under the influence of gravity, you can move the pen over stationary paper or move the paper under a stationary pen... you obtain the same plot.
 
  • #10
No. If you could find an inertial frame where A was at rest then A wouldn't be able to measure any acceleration.
 
  • #11
ghwellsjr said:
...
You cannot consider each twin as being in a valid frame according to SR because they are constantly accelerating which means they are not inertial. In the more normal Twin Paradox scenario, one twin travels away at a constant speed and during this time they each can consider the other ones clock to be time dilated but when the traveling twin stops and turns around, he becomes non-inertial and so is no longer at rest in the same frame.

Ouch. Ghwellsjr, you don’t really mean that do you? You can’t use the Lorentz Transformation, but that’s not the only transformation in SR.
 
  • #12
qualiaoftao said:
A is a non-inertial observer. That being said... can A's frame of reference to transformed to an inertial frame
No. By definition "A's frame" is a frame where A is at rest. And an intertial frame is one where any object at rest in the frame is inertial (has 0 reading on an accelerometer). Therefore, since A is at rest in A's frame and since A is non-inertial A's frame is non-inertial.
 
  • #13
Ok... let me see if I got this...
If both A and B traveled in the same direction, they would both age the same with respect to the pole and both twin's clocks would have the same time.

In this example, they are moving in opposite direction. They would perceive each other's clocks as moving slower for half of the trip, as they moved away from each other. As their paths start to converge, they would perceive the others clock as moving faster and as their paths converged, their clocks would once again be in sync. So the net difference would be 0 with respect to one another.

The fact that they were accelerated negates any time differential that would have occurred due to traveling at different velocities relative to each other.

My egg is scrambled. :)
 
  • #14
It is actually a little more complicated than that. The fact that they are accelerated means that you cannot use the standard time dilation formulas or Lorentz transform formulas to determine what happens in A and B's respective frames. Those formulas only apply to inertial frames. This is why it is important to always know what situations a given physics formula covers, no formula covers everything.

In non-inertial frames all sorts of weird things can happen. In general, time dilation becomes a function of position as well as speed. There are ways to calculate what happens in non-inertial frames. The most systematic way is to use tensors and differential geometry, but it requires a little bit of notation and math concepts that you may never have encountered before.
 
  • #15
qualiaoftao said:
Ok... let me see if I got this...
If both A and B traveled in the same direction, they would both age the same with respect to the pole and both twin's clocks would have the same time.

In this example, they are moving in opposite direction. They would perceive each other's clocks as moving slower for half of the trip, as they moved away from each other. As their paths start to converge, they would perceive the others clock as moving faster and as their paths converged, their clocks would once again be in sync. So the net difference would be 0 with respect to one another.

The fact that they were accelerated negates any time differential that would have occurred due to traveling at different velocities relative to each other.

My egg is scrambled. :)

Please see my post #3, and tell me what you don't understand.
 
  • #16
qualiaoftao said:
Ok... let me see if I got this...
If both A and B traveled in the same direction, they would both age the same with respect to the pole and both twin's clocks would have the same time.

In this example, they are moving in opposite direction. They would perceive each other's clocks as moving slower for half of the trip, as they moved away from each other. As their paths start to converge, they would perceive the others clock as moving faster and as their paths converged, their clocks would once again be in sync. So the net difference would be 0 with respect to one another.
You have to make a distinction between perception (what each observer sees of the other clocks) and the times on the clocks as defined by a reference frame. Perception does not depend on any reference frame nor any theory such as Special Relativity or Lorentz Ether Theory or having any theory at all. So what you said about perception is correct, it's what I said in post #4, and to be precise, you should have said at the end, each time they came back together, their clocks would once again appear to be in sync because according to your chosen reference frame, that of the stationary pole, they are always in sync, they always have the same time on them. But this statement is dependent on a theory, in this case SR, and it is dependent on our chosen frame of reference because another frame of reference could have the twin's clocks out of sync during the major part of the cycle where they are not together.
qualiaoftao said:
The fact that they were accelerated negates any time differential that would have occurred due to traveling at different velocities relative to each other.

My egg is scrambled. :)
It's not just the fact that they were accelerated, it's the particular way that they were accelerated that results in them both being at the same speed (which is what time dilation depends on) according to our chosen reference frame and the fact that they are continually being brought back together that negates any time differential. If they were accelerated identically but in different directions such that they never came back to the same location, then in our chosen reference frame they would have clocks that remained in sync but in other reference frames this would not be the case.
 
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  • #17
MikeLizzi said:
ghwellsjr said:
...
You cannot consider each twin as being in a valid frame according to SR because they are constantly accelerating which means they are not inertial. In the more normal Twin Paradox scenario, one twin travels away at a constant speed and during this time they each can consider the other ones clock to be time dilated but when the traveling twin stops and turns around, he becomes non-inertial and so is no longer at rest in the same frame.
Ouch. Ghwellsjr, you don’t really mean that do you? You can’t use the Lorentz Transformation, but that’s not the only transformation in SR.
I keep hearing this but nobody ever comes through with such a transformation. Why don't you be the first to show the OP one of the infinite ways there are of doing this. And then we'll let the OP explain it to me.
 
  • #18
PAllen said:
The two revolving observers are not inertial, and neither can meaningfully uses the Lorentz transform (except over a tiny region for a moment of time).
I do not fully understand the Lorentz transform but think I get the main jest... glad it is not applicable. :)

If they all got back together, at the pole (for example; I assume the pole is inertial), the two revolving clocks would be slower by the same amount compared to the pole clock.
I thought I understood this but I may not.. see below.

As to perceptions, the pole clock would see the revolving clocks going slow at all times. The revolving clocks would see the pole clock going fast at all times. (Because of non-inertial motion, there is no symmetry in perception - each of these does not see the other slow).
Here is where I have trouble. Why do the non-inertial observers see the inertial clock as running faster? If they were not accelerating and thus inertial, the pole's clock would appear to be moving slower, correct? I do not understand how acceleration would change this.

As for how the revolving clocks see each other, this would vary moment to moment, and also depend on what you mean by 'see'. You could mean 'directly see', in which case Doppler effect would would be a major contributor. You could also mean, factoring out Doppler, in which case you would still interpret periodic variation each other's clocks, but the function would be different (from the 'raw' observation, not factoring out Doppler).
What I mean by "see" or "perceive" would be when the non-inertial twins path of travel converged, they would directly observe the other's clock and physical condition (aging). Here I am assuming they would detect a difference but it appears by your comments and the comments of other, my assumption is incorrect.
 
  • #19
ghwellsjr said:
I keep hearing this but nobody ever comes through with such a transformation. Why don't you be the first to show the OP one of the infinite ways there are of doing this. And then we'll let the OP explain it to me.

Yep. I get frustrated with folks who do that too. I put a summary on my website (because I'm terrible with latex).

http://www.relativitysimulation.com/Documents%20/AccelerationTransformation%20in%20SR.html

The hard part is going to be when you ask me to solve a problem with it.
 
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  • #20
qualiaoftao said:
Here is where I have trouble. Why do the non-inertial observers see the inertial clock as running faster? If they were not accelerating and thus inertial, the pole's clock would appear to be moving slower, correct? I do not understand how acceleration would change this.
Acceleration is fundamentally different because your concept of simultaneity (assuming you try to define it similar to an inertial frame) is changing moment to moment. This change in what events are considered simultaneous radically changes the way signals from a distance are interpreted.

In this case, there is a simple way to see the validity of what I suggest. Imagine the pole observer sending out a signal every second. This signal is an expanding spherical wave front. The revolving observers get this signal every 1/γ seconds, by their clocks. The pole's clocks appear fast, and light from the pole appears blueshifted, at all times, by both revolving observers. Correspondingly, the pole observer sees the revolving clocks slow and red shifted.
qualiaoftao said:
What I mean by "see" or "perceive" would be when the non-inertial twins path of travel converged, they would directly observe the other's clock and physical condition (aging). Here I am assuming they would detect a difference but it appears by your comments and the comments of other, my assumption is incorrect.

The revolving clocks would see the same time on their clocks each time they passed each other. If they ever slowed down and joined the pole clock, they would see that they were younger (their clocks were behind).

[edit: typo noted by gwellsjr corrected.]
 
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  • #21
The revolving clocks would see the same time on their clocks each time they passed each other. ...

Thank you for taking the time and effort to explain this to me, it is much appreciated.

But how would this be the case? From the various simple examples of a stationary observer and an observer on a train traveling at a constant velocity, each would observe the others' vertical light pulse clocks as running slower. Since the path of light traveled by the other would appear to them as a diagonal path rather than perpendicular and thus travel farther.

If there are two trains traveling in opposite directions (each at a constant rate of velocity), I assume the observation would be that the time of each pulse would be slower than in the instance of the stationary observer example.

Am I misunderstanding these simple examples?

Now, if both trains are on the same set of tracks and traveling directly toward each other, the results should be the same as the above example, correct?

If that is correct, let's assume a new set of tracks that is very long... but happens to have some curvature. The fact the track does curve and the trains maintain the same constant velocity negates the previously observed time dilation with the straight tracks?
 
  • #22
MikeLizzi said:
ghwellsjr said:
I keep hearing this but nobody ever comes through with such a transformation. Why don't you be the first to show the OP one of the infinite ways there are of doing this. And then we'll let the OP explain it to me.
Yep. I get frustrated with folks who do that too. I put a summary on my website (because I'm terrible with latex).

http://www.relativitysimulation.com/Documents%20/AccelerationTransformation%20in%20SR.html

The hard part is going to be when you ask me to solve a problem with it.
Your link won't work in this case because it is for a constant acceleration. In this case the acceleration is not constant--it keeps changing direction. But at least your link gives a taste of what you're in for when you deviate from an inertial frame.

But the real question is why bother with a non-inertial frame? Is it just so you can "say" it applies to the accelerating observer so that he can do what? It does not help us learn what he sees or learn what he understands about what is going on. It doesn't affect what he sees and it doesn't help him understand what is going on. I already said that what he sees is not related to any frame of reference or any theory, although we can use SR and any frame of reference to help us understand what he sees. So why not use the easiest frame to calculate and understand what is going on? It's so easy to do this in an inertial frame in which the twins start out at rest and in which the pole remains at rest. There is no other frame that will increase our understanding of what is happening or what the twins experience, including how they see the other clocks progressing. All frames include all the same information.

I really think the concern over non-inertial frames is a result of not understanding that observers don't own any frame even though we talk about an observer's frame, meaning one in which he is always at rest, presumably at the spatial origin. But all the other objects, clocks and observers also "reside" in this observer's "rest" frame and we should always describe and analyze any scenario (including all observers, objects and clocks) from just one frame of our choice before we transform the entire scenario into some other frame, if we are so inclined and like to torture ourselves.
 
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  • #23
qualiaoftao said:
Thank you for taking the time and effort to explain this to me, it is much appreciated.

But how would this be the case? From the various simple examples of a stationary observer and an observer on a train traveling at a constant velocity, each would observe the others' vertical light pulse clocks as running slower. Since the path of light traveled by the other would appear to them as a diagonal path rather than perpendicular and thus travel farther.

If there are two trains traveling in opposite directions (each at a constant rate of velocity), I assume the observation would be that the time of each pulse would be slower than in the instance of the stationary observer example.

Am I misunderstanding these simple examples?

Now, if both trains are on the same set of tracks and traveling directly toward each other, the results should be the same as the above example, correct?

If that is correct, let's assume a new set of tracks that is very long... but happens to have some curvature. The fact the track does curve and the trains maintain the same constant velocity negates the previously observed time dilation with the straight tracks?

I will try to foster your physical intuition to short circuit your confusion. The key thing about the twin 'paradox' is a genuine asymmetry - one twin is inertial the other is not. Suppose each twin started off in a rocket going opposite directions, turned around after an hour as they measure it, and met back at the same point. Since there is perfect symmetry now, what could possibly pick out which has aged less? Do you think it is possible, after they are back together, for each to see the other as younger? In your scenario, also, there is perfect symmetry between the revolving clocks.

As for the mechanism, have you seen the doppler explanation of the twin 'paradox'? The twin that turns around immediately (visually) sees the resting twin's clock start going fast (and look blueshifted). The resting twin only sees the turn around twin's clock start going fast (and look blueshifted) quite a while later (when light from the turnaround reaches him). Thus the twin that doesn't turn around sees the other twin's clock running slow for much longer. In any truly symmetric case (like your revolving observers), each sees identical variation in the other clock, and necessarily, the clocks agree when they are adjacent.
 
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  • #24
PAllen said:
...
In your scenario, also, there is perfect symmetry between the revolving clocks.
...
In any truly symmetric case (like your revolving observers), each sees identical variation in the other clock, and necessarily, the clocks agree when they are adjacent.

But in my example is there not asymmetry at all times since their acceleration is in a different direction at all times? I ask this because the wikipedia entry on the twins paradox states that it is the turn around point where the asymmetry comes into play which accounts for the difference in time. In my example, are they not constantly in the process of "turning around"?

I looked at the "Spacetime Symmetries" wikipedia entry but it is well beyond my comprehension.
 
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  • #25
I visited the Stanford page on Symmetry and Symmetry Breaking and think now that my example does have symmetry, equal but opposite would indeed be symmetrical.

I have a better understanding, thanks!

I will work more on the Doppler effect to better visualise what the twins would observe during their flight.
 
  • #26
Has anyone ever brought an atomic clock back from orbit and compared it with one on the ground?
 
  • #27
This should help you. There’s absolutely no time dilation/length contraction between the two circling observers in your example because they’re both doing exactly the same thing. It’s just Doppler shift. If the two objects are moving in opposite directions round the circle then when they’re moving away from each other the light gets red shifted and a longer wave length means it will look slower because each wave represents a certain amount of time. When they start they’re moving directly away from each other so we start with the maximum red shift when eases off as the go around the circle and their velocity relative to each other decreases. At the half way point there isn’t any shift because they’re parallel but each sees a delayed image of the other. Now they’re starting to head towards each other so they’re blue shifted and at an increasing amount as their velocity relative to each other increases. It’s the exact opposite of the red shift from before so they see each other speed up at an increasing rate until they fly past each other and their clocks match up again. When they're going round the circle they're moving around a curved path in three dimensions but in four they're moving in a straight line because they're not accelerating.

This is an interesting example because both observers are actually time dilated relative to an observer at the centre of the circle when they compare watches afterwards because they were traveling in a circle. Curvature and acceleration are the same thing so it’s equivalent to observers that accelerate away from each other and then meet back up. If they accelerate away from each other then in three dimensions they're moving in a straight line but a curved line in four dimensions because a constant velocity is a straight line in four dimensions and acceleration is a curved line. Gravity works the same way.
 
  • #28
plbeale said:
Has anyone ever brought an atomic clock back from orbit and compared it with one on the ground?

http://metaresearch.org/cosmology/gps-relativity.asp

I think that's what you're looking for
 
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  • #29
law said:
At the half way point there isn’t any shift because they’re parallel but each sees a delayed image of the other.
Several of us have made this same mistake on this thread by it actually happens before the half way point. When they've each gone half way, they meet up again on the other side. Just a minor point that doesn't change the analysis at all.
 
  • #30
qualiaoftao said:
While attempting to wrap my head around the classical Twin Paradox, I asked myself the following. So far I have not been able to reach a well reasoned answer... which is no big surprise. :)

Twins (A and B) are oriented in space such that A could be said to be standing on B but facing in the opposite direction. Both twins have a jetpack and clock. They are both tethered to a pole which is positioned 1ly/2∏ away. On this pole is mounted a third clock.

All three clock are syncronized and the twins ignite their jetpacks and begin to accelerate in the opposite direction around the 1ly circumference path. They both finally reach and maintain a constant velocity of .5c in the first revolution and continue as this velocity for 100 years.

Can one say the twins have equivalent accelerated reference frames? Maybe equivalent but opposite? I am unsure as to what constitutes a reference frame.

From the perspective of the pole, the twin's clocks should be slower. But what of the twins with respect to each other? Shouldn't they both see the other's clock as slower and the other twin age at a slower rate?
Well I guess you have shown that "moving clock is a slow clock" leads to serious difficulties. Well done. This is not sarcasm. Clearly the twins must see the clocks to run identically, and they should see the other's clock as slower.Let's say you see a moving clock. So it's a slow clock. Well that's ok. But it's also ok to decide that the clock is actually still, and you are moving, which means that the clock runs normally.
 
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  • #31
jartsa said:
Well I guess you have shown that "moving clock is a slow clock" leads to serious difficulties. Well done. This is not sarcasm. Clearly the twins must see the clocks to run identically, and they should see the other's clock as slower.


Let's say you see a moving clock. So it's a slow clock. Well that's ok. But it's also ok to decide that the clock is actually still, and you are moving, which means that the clock runs normally.

The two revolving clocks are both non-inertial, and can readily detect this with local experiments (accelerometers). Thus, neither can use laws in their inertial form to interpret the clock of the other (including the simple rule that clock moving relative to me is running slow), and the Lorentz transform cannot be used between coordinates in which each stays stationary.

See post #20 and #23.
 
  • #32
PAllen said:
The two revolving clocks are both non-inertial, and can readily detect this with local experiments (accelerometers). Thus, neither can use laws in their inertial form to interpret the clock of the other (including the simple rule that clock moving relative to me is running slow), and the Lorentz transform cannot be used between coordinates in which each stays stationary.

See post #20 and #23.

Well I guess you can use the rule that a clock moving relative to you runs slowly, anytime you never meet with the clock again.

You can't use the rule if you are also doing any kind of checking that the clock really did run slow.

Fine rule indeed. This is sarcasm.

I have a theory about the source of all the confusion: the rule that a clock moving relative to you runs slowly.
 
  • #33
jartsa said:
Well I guess you can use the rule that a clock moving relative to you runs slowly, anytime you never meet with the clock again.

You can't use the rule if you are also doing any kind of checking that the clock really did run slow.

Fine rule indeed. This is sarcasm.

I have a theory about the source of all the confusion: the rule that a clock moving relative to you runs slowly.
The rule is that a clock moving relative to an inertial Frame of Reference runs slowly. Any number of clocks can be moving or be stationary according to that FoR. It doesn't matter what the relative speed is between them but unless they start out together and end up together, the rates of the clocks will change depending on your selected FoR.

I wouldn't propose or even suggest your own rule on this forum, it's a good way to get yourself banned. This forum is for learning the established theory of relativity. If you don't want to learn and instead promote your own private theory, you are not abiding by the rules you agreed to when you signed up. Just a friendly word of caution...

By the way, if you choose to learn relativity, your confusion will vanish.
 
  • #34
ghwellsjr said:
The rule is that a clock moving relative to an inertial Frame of Reference runs slowly. Any number of clocks can be moving or be stationary according to that FoR. It doesn't matter what the relative speed is between them but unless they start out together and end up together, the rates of the clocks will change depending on your selected FoR.

I wouldn't propose or even suggest your own rule on this forum, it's a good way to get yourself banned. This forum is for learning the established theory of relativity. If you don't want to learn and instead promote your own private theory, you are not abiding by the rules you agreed to when you signed up. Just a friendly word of caution...

By the way, if you choose to learn relativity, your confusion will vanish.
Here is my own rule: If you think a clock just seems to be moving, because you are moving yourself, then you must stick to that idea, until you feel an acceleration.

If you thought you were standing still before an acceleration, then after an acceleration you must be not standing still.
 
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  • #35
jartsa said:
Here is my own rule: If you think a clock just seems to be moving, because you are moving yourself, then you must stick to that idea, until you feel an acceleration.

If you thought you were standing still before an acceleration, then after an acceleration you must be not standing still.
That's not such a bad rule--I don't think it's the sort of rule that will get you banned--but it's not precise. Here's the way you should think of it:

It doesn't matter whether you think you or the clock is moving, you must choose a Frame of Reference and say whose moving in that frame. If you choose a frame in which you are at rest, then the clock is moving (and experiencing time dilation). If you choose a frame where the clock is at rest, then you are moving and experiencing time dilation (although you won't be able to discern that). You could also pick a frame that was kind of half way between the motion of you and the clock and then you both would be moving and experiencing time dilation although not as much as in the first two cases.

Any time you accelerate from a position of rest within a Frame of Reference, then of course you will be moving in that Frame of Reference.

If you just always think in terms of a Frame of Reference, then you'll have no problem telling who's moving and who's not. Many times people assume a FoR without actually explicity saying so, like when they say "from my point of view" or "in my frame" but the problem is when the use that terminology and then at the same time switch to "in the clock's frame" and imply that you are not in the clock's frame and it is not in your frame. Everything and everybody is in every frame you want to consider and you don't want to share parameters between two frames without clearly saying which frame they apply to.
 

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