Question about relation acceleration en special relativity

[Rettich]

Suppose we are in deep space in a rocket. We don't accelerate, we experience no significant gravity forces and we don't rotate. Ten (10) lightyears apart from us there is an other rocket (called B) wich is at a stationary distant and direction to us. We receive a clocksignal from B.

We want to go to B so we accelerate towards it.

Now the following two statements concerning the speed of the clocksignal we see from B during acceleration:

1- Due to non-relativistic-Dopler effect we see the clocksignal from B going faster (what I'm saying is that if there were no relativistic effects, then on arrival at B, its clock would have gained 10 years due to its original distance)

2- As a consequence of our speed towards B we see a growing slowing down of the clocksignal from B, due to special relativity

 

Are both statements true?

Do these two statements give a complete picture of the effect we see, during acceleration, concerning the speed of the clockssignal we see from B?

[xyzt]

Suppose we are in deep space in a rocket. We don't accelerate, we experience no significant gravity forces and we don't rotate. Ten (10) lightyears apart from us there is an other rocket (called B) wich is at a stationary distant and direction to us. We receive a clocksignal from B.

We want to go to B so we accelerate towards it.

Now the following two statements concerning the speed of the clocksignal we see from B during acceleration:

1- Due to non-relativistic-Dopler effect we see the clocksignal from B going faster (what I'm saying is that if there were no relativistic effects, then on arrival at B, its clock would have gained 10 years due to its original distance)

2- As a consequence of our speed towards B we see a growing slowing down of the clocksignal from B, due to special relativity

 

Are both statements true?

Do these two statements give a complete picture of the effect we see, during acceleration, concerning the speed of the clockssignal we see from B?

We see a relativistic Doppler effect , so 1 is incorrect. See here.

[Rettich]

Thanks for your answer, I will think about the reference you gave.

[Alan McDougall]

Suppose we are in deep space in a rocket. We don't accelerate, we experience no significant gravity forces and we don't rotate. Ten (10) lightyears apart from us there is an other rocket (called B) wich is at a stationary distant and direction to us. We receive a clocksignal from B.

We want to go to B so we accelerate towards it.

Now the following two statements concerning the speed of the clocksignal we see from B during acceleration:

1- Due to non-relativistic-Dopler effect we see the clocksignal from B going faster (what I'm saying is that if there were no relativistic effects, then on arrival at B, its clock would have gained 10 years due to its original distance)

2- As a consequence of our speed towards B we see a growing slowing down of the clocksignal from B, due to special relativity

 

Are both statements true?

Do these two statements give a complete picture of the effect we see, during acceleration, concerning the speed of the clockssignal we see from B?

 

I statement 2 is incorrect , clock signal from B should be relatively faster because it is stationary. When we move or accelerate toward something, time immediately begins to go "slower in out personal time frame, than that of the destination target. This happens but imperceptibly, even by you walking toward a friend in the mall, if he is standing still relative to you, time is moving slower for you and faster for him. There simply at the fundamental level nothing such as standard time, the flow of time even differs from your head to your feet, because of the effect of gravity, thus your feet are always a little younger than your head.

[michel123456]

you don't need acceleration during the entire trip. A can accelrate for an hour or two and continue to travel towards B at constant velocity. if A keeps accelerating during the entire trip then the relative velocity will be huge when they meet. just think of an acceleration of 1m/sec^2 for (at least) 10 years...

[Rettich]

The link xyzt gave helped me to understand the twin paradox, thanks.

[Alan McDougall]

you don't need acceleration during the entire trip. A can accelrate for an hour or two and continue to travel towards B at constant velocity. if A keeps accelerating during the entire trip then the relative velocity will be huge when they meet. just think of an acceleration of 1m/sec^2 for (at least) 10 years...

 

You are right all you need is to move towards the object for the effect to come into effect at B

[md65536]

1- Due to non-relativistic-Dopler effect we see the clocksignal from B going faster (what I'm saying is that if there were no relativistic effects, then on arrival at B, its clock would have gained 10 years due to its original distance)

This is true but is attributed to the changing delay of light, not to time. The original signals were delayed 10 years so B's clock appeared to be 10 years behind, and on arrival it appears not delayed at all. If by "see" you're talking about appearance, then it's true. If you're talking about measurements, ie. the "correct time" at B when accounting for the delay of light, then it's not true.

 

Even taking into consideration relativistic Doppler effect, B's clock will appear to tick faster as you approach it, due to the decreasing delay of light which overwhelms the time dilation effect at speeds less than c.

[phyti]

1- Due to non-relativistic-Dopler effect we see the clocksignal from B going faster (what I'm saying is that if there were no relativistic effects, then on arrival at B, its clock would have gained 10 years due to its original distance)

2- As a consequence of our speed towards B we see a growing slowing down of the clocksignal from B, due to special relativity

 

Are both statements true?

Do these two statements give a complete picture of the effect we see, during acceleration, concerning the speed of the clockssignal we see from B?

 

The B clock will have ticked 10 yr while your clock ticked less, depending on your speed.

As a result of this you will interpret the distance as less than 10 ly.

You will see an increase in frequency of B signals.

All this per SR.

[Jerry Wickey]

There may be a more explanatory way to ask this question. This may require a real expert to answer.

 

Rocket ship A and B are traveling toward each other at some relativistically significant portion of the speed of light in intergalactic space. They are unaffected by any significant gravitational forces and they are so far from any other object that essentially, they have only each other for spacial reference.

They have no way of knowing who is going faster. They can't even tell if one is standing still and the other is doing all the moving. This is just a simple way of saying all motion is relative. It doesn't really matter which is moving. Each observes the other as the one moving closer each from their own respective frames of reference.

 

With a very powerful telescope each can see the clock placed on the front of the other rocket ship.

 

Rocket ship A observers the clock on rocket ship B moving at a slower rate than his own because B is moving at relativistic speed with reference to A's frame of reference. A observes time dilation of B. In the 1 day travel time it takes B to arrive at A, B's clock will show only one hour has passed.

 

Paradoxically, Rocket ship B simultaneously observers the clock on rocket ship A moving at a slower rate than his own because A is moving at relativistic speed with reference to B's frame of reference. B observes time dilation of A. In the 1 day travel time it takes A to arrive at B, A's clock will show only one hour has passed.

 

As they pass each other, which clock will show more time passed?

 

 

[Janus]

There may be a more explanatory way to ask this question. This may require a real expert to answer.

 

Rocket ship A and B are traveling toward each other at some relativistically significant portion of the speed of light in intergalactic space. They are unaffected by any significant gravitational forces and they are so far from any other object that essentially, they have only each other for spacial reference.

They have no way of knowing who is going faster. They can't even tell if one is standing still and the other is doing all the moving. This is just a simple way of saying all motion is relative. It doesn't really matter which is moving. Each observes the other as the one moving closer each from their own respective frames of reference.

 

With a very powerful telescope each can see the clock placed on the front of the other rocket ship.

 

Rocket ship A observers the clock on rocket ship B moving at a slower rate than his own because B is moving at relativistic speed with reference to A's frame of reference. A observes time dilation of B. In the 1 day travel time it takes B to arrive at A, B's clock will show only one hour has passed.

 

Paradoxically, Rocket ship B simultaneously observers the clock on rocket ship A moving at a slower rate than his own because A is moving at relativistic speed with reference to B's frame of reference. B observes time dilation of A. In the 1 day travel time it takes A to arrive at B, A's clock will show only one hour has passed.

 

As they pass each other, which clock will show more time passed?

 

 

Your question assumes that each ship will agree that their clocks read the same time when they were 1 day apart. This is not the case. IOW, while both ships will agree what time was on their respective clocks at the moment they meet, they will not agree as to what time was on their clocks when they were separated.

[Jerry Wickey]

Not so, They can establish simultaneous clocks before beginning their journey toward each other. This is how.

 

Before beginning their journey toward each other, both ships send a light signal to the other. Each bounces it back. This tells both exactly how far apart they are. They do this twice to establish that the distance between them is not changing over time. At this point neither is moving relative to the other and they could also have transmitted the reading from their respective clocks.

 

Since both know and agree how far apart they are before they begin to move, both know what each other's clock read and how far, how long ago it read that. Both know what the other's clock reads simultaneous to their own before they begin the journey toward each other.

 

As their journeys begin, both observe the other experiencing time dilation. How is this paradox resolved?

[Janus]

Not so, They can establish simultaneous clocks before beginning their journey toward each other. This is how.

 

Before beginning their journey toward each other, both ships send a light signal to the other. Each bounces it back. This tells both exactly how far apart they are. They do this twice to establish that the distance between them is not changing over time. At this point neither is moving relative to the other and they could also have transmitted the reading from their respective clocks.

 

Since both know and agree how far apart they are before they begin to move, both know what each other's clock read and how far, how long ago it read that. Both know what the other's clock reads simultaneous to their own before they begin the journey toward each other.

 

As their journeys begin, both observe the other experiencing time dilation. How is this paradox resolved?

Once they begin their journey towards each other, any agreement as to their respective clock readings goes away. This is due to the relativity of simultaneity. As long as they are at rest with respect to each other they are in the same inertial frame and agree on simultaneity. However, the instant they accelerate towards each other, they are no longer in the same inertial frame and no longer agree on simultaneity. Each ship will determine that the others clock will have advanced ahead of his while he was accelerating. The end result will be that even when all is done, each ship will say that the other clock advanced ahead of his during acceleration and then ran slow during the trip, and that the sum of these times will end up equaling the same amount of time that passed on his clock from start of acceleration to their meeting.

[DimaMazin]

just think of an acceleration of 1m/sec^2 for (at least) 10 years...

The speed will be more than "c". Every subsequent 1 m/c² of acceleration takes more energy than previous,therefore such acceleration can continuously be during 10 years never.

Edited February 16, 2014 by DimaMazin

[michel123456]

The speed will be more than "c". Every subsequent 1 m/c² of acceleration takes more energy than previous,therefore such acceleration can continuously be during 10 years never.

The astronaut in the accelerating rocket can constantly have the same thrust all the time. The astronaut will never feel any "relativistic force" that will prevent him to send his super-fuel to the thrusters, nor will he feel the rocket getting heavier because of a relativistic mass increase.

 

For him things will remain exactly as usual.

 

The only thing is that all observers at rest will observe him gaining mass and accelerating less and less.

I suppose that the astronaut in the rocket, looking through the window at the planet he started from will observe the planet accelerating less and less, and gaining mass.

 

Otherwise I suppose the rest of the universe will look as usual.

 

After a hundred years, I still miss what would be the ending observation from the observer at rest at the launch pad.

[Jerry Wickey]

michel is right.

 

Accelerations doesn't change anything. As each accelerates closer and closer to the speed of light, time is slowing for them, mass in increasing and distance is shortened in the direction of their travel, but for them all their rules, scales and clocks still read exactly as they expect. They measure their mass the same and their fuel consumption the same.

 

This in spite of the fact that an observer at rest observes the mass of their fuel reserves increasing.

 

But more importantly, both observe the other rocket which is speeding toward them experience the effect of time dilation. Both observe the other passing through time more slowly. This effect does not go away when they slow down to greet each other as Janus poses. Each observes the other passing through time more slowly throughout their whole journey until they come together.

 

At which time each says of the other, you are younger than I.

 

How does relativity resolve the paradox?

[michel123456]

michel is right.

 

Accelerations doesn't change anything. As each accelerates closer and closer to the speed of light, time is slowing for them, mass in increasing and distance is shortened in the direction of their travel, but for them all their rules, scales and clocks still read exactly as they expect. They measure their mass the same and their fuel consumption the same.

 

This in spite of the fact that an observer at rest observes the mass of their fuel reserves increasing.

 

But more importantly, both observe the other rocket which is speeding toward them experience the effect of time dilation. Both observe the other passing through time more slowly. This effect does not go away when they slow down to greet each other as Janus poses. Each observes the other passing through time more slowly throughout their whole journey until they come together.

 

At which time each says of the other, you are younger than I.

 

How does relativity resolve the paradox?

The bold part is the thing I cannot understand (or do not want to understand, in all honesty).

 

To me everything must be symmetrical: what you build can be unbuild. What you gain must be lost. I don't understand how some effect becomes one-way. It becomes paradoxal.

Edited February 16, 2014 by michel123456

[Janus]

michel is right.

 

Accelerations doesn't change anything. As each accelerates closer and closer to the speed of light, time is slowing for them, mass in increasing and distance is shortened in the direction of their travel, but for them all their rules, scales and clocks still read exactly as they expect. They measure their mass the same and their fuel consumption the same.

 

This in spite of the fact that an observer at rest observes the mass of their fuel reserves increasing.

 

But more importantly, both observe the other rocket which is speeding toward them experience the effect of time dilation. Both observe the other passing through time more slowly. This effect does not go away when they slow down to greet each other as Janus poses. Each observes the other passing through time more slowly throughout their whole journey until they come together.

 

At which time each says of the other, you are younger than I.

 

How does relativity resolve the paradox?

I never said that any thing goes away when they slow down. I never even said anything about them slowing down. I said that when they accelerate at the beginning of the trip, they will measure a jump forward in the other ship's time(Relativity of Simultaneity), and that jump forward will offset any slowing of the clock that takes place during the trip.

 

In this case, the ship clocks read the same time both before they just their trip and when they pass each other because both ships undergo symmetrical accelerations at the start of their trips.

 

If on the other hand, only one of the Ships under goes acceleration, then the following happens:

 

For the ship that undergoes acceleration:

 

His acceleration causes him to determine that the other ship's clock runs fast during the acceleration and then run slow during the rest of the trip. Because the difference in the two ship's velocity is entirely due to his acceleration, the time the other ship's clock gains by running fast will exceed the time it losses by running slow, and when they pass each other, the other clock will show more time then his.

 

For the ship that does not accelerate:

 

Since he has no period of acceleration himself, there is no period when the other ship's clock runs fast and thus for him, the other Ship's clock always runs slow for the whole trip and when they meet will show less time than his.

 

This is how Relativity deals with this. You have to remember that the simple time dilation for SR only works from within inertial frames, and when a ship is accelerating it is no longer in an inertial frame and you can not use the simple time dilation rule to determine how other clocks behave according to it.

 

When you are in an accelerating frame, the rules of Relativity say that clocks in the direction of your acceleration run fast by a factor that depends on both your acceleration and their distance from you in that direction as measured by you. Clocks that are in the other direction will run slow by the same factor.

 

If on the other hand, you are in an inertial frame watching a clock that is accelerating, that clock's time dilation will be only due to its relative motion to you at any given moment.

 

To give a better look at how thing work out in the proposed scenario, we can look at space-time diagrams.

 

This first diagram shows the events as they occur according to the original pre-acceleration frame of the ships. This is also the same frame as someone who never undergoes acceleration during this setup.

 

 

The numbers mark out equal time periods and the yellow lines light signals traveling from ship to ship.

 

Note that at the moment just before acceleration each ship receives a signal from the other ship that left 8 time units ago. Thus at this point each ship knows that it is 8 ticks at the other ship just like at his own.

 

Also note that the the ticks take place further apart for each ship after they start moving towards each other. This shows the time dilation of each ship as seen in this frame. Since each ship shows the same time dilation, both ships meet with their clocks reading the same time.

 

In addition, consider the light signals each ship sees. Before the ships start on their trips, they will receive signals at a rate of one per time period. Immediately after they start moving, they start to see signals arriving at rate of over 2 a time period. Note that for a while, they are still receiving signals that left the other ship before that ship started moving. It isn't until sometime during the 13th time period that they start seeing signals arriving from after the other ship started moving and the signals start arriving at a rate of 3 signals per time period.

 

Now let's see how things look from the frame of on of the ships after it starts moving. Here is the space-time diagram for that:

 

 

 

Here we note that each ship still receives a signal form the other ship that left at zero time for the other ship and arrives at time period 8 of his own, but these signals no longer tell him that it is time period 8 on the other ship when the signal arrives, Instead, he determines that is sometime during time period 11 own the other ship when it is time period 8 own his own. From this point on, the other ship's clock runs slower than his, ticking away some 4.2 periods for his 7 until they meet up, each reading ~15.

 

The light signals each receives matches up with the other space-time diagram. Each receives signals at a little over 2 a time period until the 13th time period and then start getting them at a rate of 3 per period.

 

The space-time diagram for the other ship would be the mirror image of this diagram.

 

There is no paradox and all frames agree that the ships read the same time upon meeting, even though according to each ship, the other ship's clock ran slower while they were traveling.

 

Edited February 16, 2014 by Janus

[Jerry Wickey]

That is very helpful. Thank you. Your diagrams is very helpful, but it is fails to explain the paradox.

 

I understand the vertical axis to be time and the horizontal to be distance through space. Light always moves on this graph at 45degrees, because of course, it is moving as fast through space as it is through time. Your graphs presume a sudden acceleration to a then constant velocity. This impossibility posses no difficulty or error in our examination of the circumstances.

 

Everything is limited to the speed of light and so if we move very fast through space and remained moving at the same speed through time, the sum of our vectors through space and time would exceed the speed of light. Therefore, if we move very fast through space the sum of over velocity vectors sqroot (space_distance ^2 + time_passed ^2) / time must always be c. This is the Lorentz.

 

You are exactly correct in the observation that neither rockets perceive the movement of the other for some time, but the diagram has two errors. First the rate at which the ticks occur changes after the rockets begin to move (see my marks on your graph) and in the second graph the rocket on the right is not stationary but is moving away from the other. This makes your graph appear to explain the paradox, but does not.

 

If the rocket on the right remains stationary in space, it is still moving through time. While the rocket on the left is moving through space and time. Because it can not exceed the speed of light, its movement through time must slow to accommodate its rapid movement through space. Because of this, the rocket on the left perceives light ticks emanating from the rocket on the right at a more rapid rate. He perceives time having quickened for the rocket on the right. He understands, his own time as slowed.

 

When they meet, whether they come to a stop or not doesn't matter. More time will have passed for the rocket on the right as for the rocket on the left. eight ticks for the rocket on the right while the rocket on the left will have passed nine clicks. They both have moved through the same amount of time, but because the rocket on the left has also moved through considerable space, his movement through time must be slowed otherwise the sqroot of the sum of the squares of his distance and time would exceed the speed of light.

 

This I understand. Relativity is why the clocks on GPS satellite run slower. Actually general relativity has more to do with that than special, but the point is if an astronaut ever visits one of those satellites it's clock will not have caught up to his as you claim "each reading ~15"

 

The only reason the same number of clicks appear is because the distance you put between them is adjusted, but neither for the passage of time nor for their respective movement through time. They seem to be adjusted just to make them match up.

 

Here is a graph showing equal distances between clicks. i.e the speed of light for all. As one rocket moves through space it can not move through time as quickly, because of course the rockets aggregate speed would exceed light.

 

The paradox of the relativity of motion is still unexplained. It does't matter for the rocket on the left this he is moving. From his perspective it is just as true to say the rocket on the right is moving instead and his clock should be slowed showing the passage of less time, when they pass each other. This is equally true for the other.

 

I would really like to understand this from someone who truly understands relativity.

 

[Janus]

There are no mistakes in the diagrams. They were made by a software program specifically designed to draw S-T diagrams.

 

Just to clarify how I drew it up:

 

First I drew the two lines which represented the two ships before they started moving as shown in the first diagram and placed the circles at regular intervals.

I then added the two lines that converge on each other which represent the ships after they move toward each other.

I then chose own of these lines and told the software to match velocities. It then redraws the diagram with this line going straight up and all the other lines adjusted to this being the rest frame.

 

Now I placed circles on this line at equal intervals to the ones I did for the first line. This way, the time between ticks as measured in each frame are equal.

 

I repeated the process for the last line (matching velocities and marking out equal time intervals)

 

Lastly, I added 45 degree lines for the light signals.

 

By choosing either the green or dark blue lines and having the software match velocities with it, you get the first diagram.

By choosing the lighter blue line and have the software math velocities with it, you get the second diagram.

 

If you have a problem with the way either of these diagrams appear, then you have a problem with how Relativity explains this scenario.

 

For example: For this diagram, I chose 0.8c (the value from the OP) for the relative velocity of the ships relative to each other as measured by the ships.

Using the Relativistic addition of velocities, this means that the ships are moving at 0.5c relative to the rest frame of the first diagram. Time dilation at 0.5c is 0.866, so the vertical distance between ticks for the Red and lgt blue line should be 1.155 times longer than that for the Green and Drk Blue lines. This shows to be the case in the diagram.

 

In the second diagram, after the ship is moving, the relative velocity between ships is 0.8 c. The time dilation for 0.8c is 0.6. This means that the vertical distance between ticks for the red line should be 1.667 times that for the Drk blue line in this diagram, and they are.

 

Now consider Relativistic Doppler shift.

 

For the first part of its inward leg, the ship on the right sees light coming from the other ship while they have a relative velocity of 0.5c

 

The formula for Relativistic Doppler shift gives as answer of 1.732, which means the light should cross the Drk blue line at a rate of 1.732 per time period for this part. This is what the diagrams show.

 

After this, the ship sees light arriving from a ship with a relative velocity of 0.8c, this gives a Doppler shift factor of 3, again agreeing with what is shown on the diagram.

 

I didn't adjust anything to make it work out, They work out because Relativity says that they do.

 

Your "corrected" diagram is wrong because it shows one ship as treating both the before and after segments for one of the ships as being the same inertial frame. This cannot be true, These are two separate inertial frames. In SR, the ship can treat either of these frames as "being at rest" but not both simultaneously.

 

As far as the GPS satellite goes. The thing to remember here is that even disregarding the GR effect and only considering the motion of the satellite, the satellite is still in an non inertial frame. It travels in a circle around the Earth, and circular motion is accelerated motion. So you are dealing with the same type of situation as I described when just one ship accelerated; you end up with a accumulative time difference for the satellite.

 

I have given you the correct answer as to how Relativity shows that there is no paradox in this scenario, no one else with a grasp of Relativity is going to give you any different answer, because it is the answer Relativity gives.

[DimaMazin]

After a hundred years, I still miss what would be the ending observation from the observer at rest at the launch pad.

t = 100 years = 3155692544.5 seconds

In your case: dp=dt_*m_*a

v=p/(gamma_*m)

v=tma/(gamma_*m)

v=ta(c²-v²)^1/2 / c

v²c²=t²a²(c²-v²)

v²=t²a²c²/(c²+t²a²)

v = tac/(c²+t²a²)^1/2

v = 298,448,718 m/s

a - initial acceleration

p - momentum created by invariable force

Edited February 17, 2014 by DimaMazin

[Jerry Wickey]

Janus seems very sure of his take on this.

 

But to clarify. Is Janus saying?

 

A) The space between two space ships is closing and it doesn't matter who is moving, both toward each other or one stationary and the other moving, in both cases each observe time dilation in the other, but that when they pass each other, both will have experienced the passage of the same amount of time.

 

B) A space traveler leaves earth, a significant source of gravity, moving at near light speed. Upon his return to earth he will have experienced the passage of less time than those having remained on earth. Earth will have aged more than the traveler.

 

C) A space traveler leaves a space station far from any significant source of gravity, moving at near light speed relative to the motion of the space station. Upon his return to the space station he will have experienced the passage of less time than those having remained on the space station. The space station will have aged more than the traveler.

 

Janus, Are you claiming all three of these statements are true? Or are you claiming that A and B are true and that C is false?

[Janus]

Janus seems very sure of his take on this.

 

But to clarify. Is Janus saying?

 

A) The space between two space ships is closing and it doesn't matter who is moving, both toward each other or one stationary and the other moving, in both cases each observe time dilation in the other, but that when they pass each other, both will have experienced the passage of the same amount of time.

 

B) A space traveler leaves earth, a significant source of gravity, moving at near light speed. Upon his return to earth he will have experienced the passage of less time than those having remained on earth. Earth will have aged more than the traveler.

 

C) A space traveler leaves a space station far from any significant source of gravity, moving at near light speed relative to the motion of the space station. Upon his return to the space station he will have experienced the passage of less time than those having remained on the space station. The space station will have aged more than the traveler.

 

Janus, Are you claiming all three of these statements are true? Or are you claiming that A and B are true and that C is false?

 

Let's deal with B and C first, as it is A that seems to be giving you the most trouble.

 

With B, I agree with the result given.

With C, I agree with the result given.

 

A is ambiguous.

 

The answer depends on which frame of reference you are working from.

 

So for example, If you are in a frame where from your perspective, the two ships are approaching each other from different directions at equal speeds, Then yes, both ships will age at the same rate. Thus if, according to you, when they are 2 light years apart, each ship's clock read 0, and each ship was traveling at 0.5 c, then each clock will read 1.732 yrs( time dilation factor of 0.866) when they meet 2 yrs later by your clock. To illustrate this better, let's assume that there are three space buoys in this scenario, at rest in this frame and stationed 1 ly apart ( as measured in this frame) with the middle one located where the ships meet. Each of the Ship's clocks are set to zero when they pass one of the buoys. For convenience sake, we will say that all three buoys also have clocks that are synchronized in this frame and read zero when the ships pass the end buoys

 

If you are in the frame of one of the ships, then this is what happens:

An outer buoy passes the ship and sets the ship clock to zero. Because of length contraction, the other two buoys are 0.866 and 1.732

light years distant and traveling towards you at 0.5c. It will take 1.732 years by your clock for the middle buoy to reach you.

Now, because of the Relativity of Simultaneity, None of the buoys will be synchronized to each other in your frame. As the buoy that passed you reads 0, the middle buoy's clock will read 0.5 years, and the furthest buoy will read 1 year. Both buoy's clocks will be time dilated and run 0.866 times as fast as yours. Thus when the middle buoy reaches you it will read 0.5+ 1.732(0.866)= 2 yrs. exactly what it reads according to the first frame we worked with. Note that the buoy's clock aged less than the ship's clock during the interval between the ship being at each buoy, but since the middle buoy was already ahead of the ships clock, it reads a later time when they meet.

 

Now consider the other ship. As we noted, when the first buoy passes our ship, the time at this buoy is 1 year. Since the other ship passes that buoy when it reads 0, it means that according to this frame, when the first buoy passes our ship, the other ship has already passed its buoy. In fact, since 1 year has passed on for the buoy since the ship passed it, then 1.155 yrs have passed on our ship's clock since then other ship passed its buoy. Since the buoy was 1.732 light years away when we passed our buoy and has been approaching at 0.5 c, this means that when the two passed each other, they were 1.732+ 1.155*.5 = 2.31 light years away from our ship when they passed each other.

 

If we use the relativistic addition of velocities, we find that the other ship's speed relative to us is 0.8c. Thus it has traveled 0.924 light years in the 1.155 yrs since it passed its outer buoy, and at the moment our ship passes its buoy is 2.31-.924= 1.386 light years from our ship. and its clock, having run at 0.6 the rate of ours for 1.155 years since it read zero will read 0.693 years. It will also be 0.52 light years from the middle buoy. It will take 5.2/(.8-.3)= 1.732 years (accounting for rounding errors) for the other ship to catch up to the middle buoy. This means that it arrives at the middle buoy at the same time the middle buoy arrives at our ship. It its clock will advance by 1.0392 yrs from the moment the first buoy passes our ship and we meet up with it at the middle buoy, and will read 1.0392+0693= 1.732 yr when we meet. Again exactly as in the first frame.

 

Again note, that while the other ship's clock only aged 1.0392 yrs during the period that our ship's clock aged 1.732 yrs, at the moment our clock read 0, the other ship's clock already read 0.693 years, and thus when we meet up the clocks read the same time.

 

If we were in the other ship, things would be reversed.

 

What needs to be taken away here is that in this particular scenario, there is no absolute answer as to whose clock aged slower than the other, this is all frame dependent.

 

What makes this different than the other two cases is that in both, we start and end the scenario at the same point, and one of the participants undergoes acceleration.

 

If both were to undergo equal accelerations, for example if you had two ships that each accelerated in opposite directions and then reversed course and came back together, when they meet they will have aged the same amount. (however if you were to ask someone on either ship which ship had aged more at any point of the trip other then when they were together, you would get different answers.)

 

When dealing with Relativity you have to account for all the relativistic effects: Time dilation, length contraction, and the relativity of simultaneity. When you forget to account for any of them is when you are going run into apparent but nonexistent paradoxes.

 

When your clock reads 0 you

[DimaMazin]

t = 100 years = 3155692544.5 seconds

In your case: dp=dt_*m_*a

v=p/(gamma_*m)

v=tma/(gamma_*m)

v=ta(c²-v²)^1/2 / c

v²c²=t²a²(c²-v²)

v²=t²a²c²/(c²+t²a²)

v = tac/(c²+t²a²)^1/2

v = 298,448,718 m/s

a - initial acceleration

p - momentum created by invariable force

My calculation is correct!