Special relativity: Can you explain the paradox?

[vuquta]

if A and B are both at the origin at T=0 and A instantly accelerates to v at T=0 then you just switch frames immediately to the new frame.

In that frame it is B that is moving. Therefore it is B's clock that is ticking slower in that frame.

At T=1 B stops moving in that frame.

In that frame it is therefore B that ages less.

 

I suspect that you are just arguing with people to try to get them to do your homework for you.

I wouldnt have posted this answer but it was trivial and I couldnt resist the urge to point out how obviously wrong you are.

LOL, you are funny.

 

Now, I will demonstrate where you are wrong.

 

Conclusions of A:

 

1) A takes off from the common frame of A and B.

2) A is then stationary for a time period.

3) Eventually B joins the frame of A.

4) Since B was the moving frame, A concludes the clock of B is time dilated.

 

Conclusions of B:

 

1) A takes off from the common frame of A and B.

2) B is then stationary for a time period.

3) Eventually B joins the frame of A.

4) Since A was the moving frame, B concludes the clock of A is time dilated.

 

 

This is very basic stuff.

 

Actually, who is "younger" depends upon which frame you are considering it from.

 

Here's the space-time diagrams for the situation as seen from both the staring inertial rest frame of A and B and from their final inertial rest frame.

 

 

From the original frame, A (the blue world-line) ends up younger than B,(the red World-lines)

 

However, when you switch to the final rest frame, B is younger than A.

 

This is the above from Janus is correct answer.

Edited September 14, 2010 by vuquta

[granpa]

I have put you on my ignore list.

I will no longer read or respond to your posts.

[vuquta]

I have put you on my ignore list.

I will no longer read or respond to your posts.

 

What?

 

I thought you functioned by logic.

 

My answer was simply logical. I am sorry it offended you. What is your problem with logic?

 

Otherwise, refute it.

[swansont]

LOL, you are funny.

 

!

Moderator Note

This will stop right now.

 

 

———

 

 

 

This is the above from Janus is correct answer.

 

granpa and Janus came up with the same answer. How can one be right and the other wrong?

[between3and26characterslon]

Special relativity says that a clock on a spaceship moving relative to a stationary observer appears to be running more slowly than if it was stationary. So if this is true, what would you get in the following scenario?

 

Stationary observer Andy and moving observer Brian are both initially stationary and in the same location, and each has an atomic clocks measuring time in days, and displaying what they can see on a large monitor visible to both. In addition both clocks emit an electronic "pip" whenever one day passes to the next, and each records the number of pips made since reset. Day = 0, and the pip counter is set to 0.

 

Brian rapidly accelerates towards a distant star, and it takes him one second to reach around half the speed of light. After 100 days of travelling Brian stops in 1 second and accelerates to the same speed back towards Andy (in one second), and arrives back at Andy in a further 100 days.

Before they got out their clocks to compare what they showed, they each gave their opinion as to what they were expecting them to show.

 

 

Brian reports that Brian's clock shows that 200 days has elapsed, and he reported having received 200 pips from it.

Brian reports that Andy's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it.

 

Andy reports that Andy's clock shows that 200 days has elapsed, and he reported having received 200 pips from it.

Andy reports that Brian's clock appeared to be running at half the speed on both the outbound and return journeys, and he had correspondingly received 100 pips from it.

 

When they both look at their clocks, and read their pip counters, what do they see?

 

Thanks

 

 

Having read all the replies I can say it has nothing to do with dopler effect, the problem is you are not playing by the rules, you have false axioms in your question. You are assuming that as Brian moves away from Andy that this is no different to Andy moving away from Brian or both moving away from each other. This is wrong.

 

This leads you to conclude that they will both see each other's clocks running slower and therefore they will both be younger than each other, hence the paradox. This is also wrong.

 

Let us reset the experiment using the correct axioms.

 

 

 

1) Andy and Brian must find themselves an inertial frame. That is, as I'm sure you know, Where a body is sufficiently far removed from any force so as to not be subject to any force.

 

2) We will now refer to this inertial frame as a system of coordinates and, using Einstein's notation, call it coordinate system K made up of coordinates (x, y, z)

 

3) In this inertial frame (coordinate system K) Newton's law of inertia is true. That is a body at rest with respect to K will remain at rest with respect to K until acted upon by a force and a body traveling with uniform translation (i.e. same speed in a straight line, i.e. constant velocity) with respect to K will continue with constant velocity with respect to K until acted upon by a force.

 

4) Both Andy and Brian are at rest (motionless) with respect to K.

 

5) Andy and Brian both have clocks which are both precisely accurate and synchronous whith each other. They both have measuring rods s which are identical to each other (you'll see why later).

 

6) Brian is then subjected to a force and therefore accelerated in the direction of x of coordinate system K. He has no velocity or acceleration in the direction of y or z.

 

7) It is now true for coordinate system K that Brian is in motion (i.e in reality) and Andy is at rest (i.e in reality) with respect to K.

 

8) Brian is now in a coordinate system K' which is moving with respect to K.

 

9) In K time=t and distance=s.

 

10) in K' time=t' and distance=s'.

 

11) The Lorentz transformation will tell you what t' and s' are in K' when measured from K.

 

12) It is important to understand that the axes (x, y, z) of K are parralell to and positive in the same direction as (x', y', z') of K'.

 

13) When Andy, who is in K, looks at Brian he sees Brians clock running slower and his measuring rod will be shorter.

 

14) This means that when viewed from K, t'> t (seconds t' are longer than seconds t) and s'< s (distance s' is less than distance s)

 

15) You're now thinking "what's the difference between K' moving relative to K or K moving relative to K' ?"

 

 

Here's the bit I struggled with, I am not an expert (physicist or mathematician) and I'm sure people will disagree but, it must be true within coordinate system K, the values of t, x and s in K are negative with respect to K' .

 

 

16) See 12, If (x, y, z) coincide with and are positive in the same direction as (x', y', z') then K' is moving with (+)velocity with respect to K and K is moving with (-)velocity with respect to K'

 

17) Therefore when viewed from K', t< t' (seconds t are shorter than seconds t') and s> s' (distance s is longer than distance s')

 

18) Andy will see Brian's clock go slower and his measuring rod get shorter, Brian will see Andy's clock go faster and his measuring rod get longer (lucky ol' Andy).

 

19) However much time and distance change in one system of coordinates they change equally and oppositely in the other.

 

20) I hope that all makes sense and more than that I hope it's correct but it makes sense to me.

Edited September 14, 2010 by between3and26characterslon

[Mike-from-the-Bronx]

I stand corrected. I'm the one who got lost and confused. As has been pointed out, two events that are not simultaneous in one reference frame can be simultaneous in another. However the point I was trying to make still holds. Two event cannot switch chronological order.

 

And again I jump the gun. That's not right either.

 

I'm just trying to say that all observers must agree on who is younger once "A" and "B" are in the same reference frame again

Edited September 14, 2010 by Mike-from-the-Bronx

[losfomot]

OK, let's see.

 

I have two A and B in the same frame clocks synched.

 

 

A takes off and instantly acquires v.

 

After some time B takes of exactly the same way.

 

They will end up in the same frame.

 

Which is younger?

Actually, who is "younger" depends upon which frame you are considering it from.

 

Here's the space-time diagrams for the situation as seen from both the staring inertial rest frame of A and B and from their final inertial rest frame.

 

 

From the original frame, A (the blue world-line) ends up younger than B,(the red World-lines)

 

However, when you switch to the final rest frame, B is younger than A.

 

So if A and B synchronized their clocks in the starting rest frame (before anyone took off anywhere, say at time = -1), then A 'takes off', accelerating instantly to some velocity (the final rest frame)... and then sometime later (in your diagram it appears to be 1 second later?) B takes off, accelerating instantly to A's frame (the final rest frame)... they compare clocks in the final rest frame and B's is behind A's?

 

This doesn't make a lot of sense to me.

[Sisyphus]

I stand corrected. I'm the one who got lost and confused. As has been pointed out, two events that are not simultaneous in one reference frame can be simultaneous in another. However the point I was trying to make still holds. Two event cannot switch chronological order.

 

And again I jump the gun. That's not right either.

 

I'm just trying to say that all observers must agree on who is younger once "A" and "B" are in the same reference frame again

 

So wait, are you correcting yourself within this post? The last sentence is right. The "two events cannot switch chronological order" part is not right.

[between3and26characterslon]

So if A and B synchronized their clocks in the starting rest frame (before anyone took off anywhere, say at time = -1), then A 'takes off', accelerating instantly to some velocity (the final rest frame)... and then sometime later (in your diagram it appears to be 1 second later?) B takes off, accelerating instantly to A's frame (the final rest frame)... they compare clocks in the final rest frame and B's is behind A's?

 

This doesn't make a lot of sense to me.

 

If they both start of at rest with respect to coordinate system K and A takes off at v with respect to K and continues with velocity v with respect to K then B, who leaves one second later, will have to travel faster than v to catch up with A, their clocks will be synchronous.

 

If A accelerates to v with respect to K and then decelerates with respect to K eventually coming to rest again with respect to K and B does the same thing, but one second behind, their clocks will be synchronous.

[losfomot]

If they both start of at rest with respect to coordinate system K and A takes off at v with respect to K and continues with velocity v with respect to K then B, who leaves one second later, will have to travel faster than v to catch up with A, their clocks will be synchronous.

 

If A accelerates to v with respect to K and then decelerates with respect to K eventually coming to rest again with respect to K and B does the same thing, but one second behind, their clocks will be synchronous.

 

Although I think you may be right about both of those situations (not positive about the first one), neither applies in this instance. There was never any mention of B 'catching up to' A... They start in the same rest frame and end in the same rest frame, they undergo the same acceleration (apparently instant) just at different times.

[between3and26characterslon]

Although I think you may be right about both of those situations (not positive about the first one), neither applies in this instance. There was never any mention of B 'catching up to' A... They start in the same rest frame and end in the same rest frame, they undergo the same acceleration (apparently instant) just at different times.

 

 

So if A remains at constant v with respect to K and B remains at constant v with respect to K albeit 1 second time t and vt(I think??) distance behind A then A will be younger than B by more than 1 second of time t'

Edited September 14, 2010 by between3and26characterslon

[losfomot]

So if A remains at constant v with respect to K and B remains at constant v with respect to K albeit 1 second time t and vt(I think??) distance behind A then A will be younger than B by more than 1 second of time t'

I don't think that is possible no matter how fast they were going. edit - <scratch that

 

There was no quantity for v given, so I don't think you can work out how much the clocks will differ, only who's clock will be ahead or behind.

Edited September 14, 2010 by losfomot

[between3and26characterslon]

I don't think that is possible no matter how fast they were going. edit - <scratch that

 

There was no quantity for v given, so I don't think you can work out how much the clocks will differ, only who's clock will be ahead or behind.

 

No, your right. My bad, I was assuming that v=c

 

If v did =c and B left 1 second after A set off (that is one second t with respect to K) then both A and B would agree that A is 1 second t or less than 1 second t' younger than B. That is it will be t=1 seconds or t'<1 seconds before B sets off.

 

But as v is not known then the time difference can only be given as a mathmetical expression and not as a numerical value.

[Janus]

 

granpa and Janus came up with the same answer. How can one be right and the other wrong?

 

I was wondering the same thing. At the most, one might be able to say that granpa's answer was incomplete because it only gave the answer for one frame, but by the same token, my answer could be considered incomplete because I didn't mention that there was a frame in which A and B never differ in age.

[vuquta]

!

Moderator Note

This will stop right now.

 

 

———

 

 

 

 

granpa and Janus came up with the same answer. How can one be right and the other wrong?

Grandpa said,

 

if A and B are both at the origin at T=0 and A instantly accelerates to v at T=0 then you just switch frames immediately to the new frame.

In that frame it is B that is moving. Therefore it is B's clock that is ticking slower in that frame.

At T=1 B stops moving in that frame.

In that frame it is therefore B that ages less.

 

Where exactly did he sat from the view of each frame, the other is younger? What I read is that he claimed B is always younger since he never stated a case in which A is younger in which Janus clearly stated.

 

OK,

 

then once A and B are in the frame and reciprocal time dilation is the correction conclusions to A and B, they perform Einstein's clock sync method and can decide which is older and which is younger.

 

A says B is younger.

 

B says A is younger.

 

 

What does the clock sync say?

[swansont]

Grandpa said,

 

if A and B are both at the origin at T=0 and A instantly accelerates to v at T=0 then you just switch frames immediately to the new frame.

In that frame it is B that is moving. Therefore it is B's clock that is ticking slower in that frame.

At T=1 B stops moving in that frame.

In that frame it is therefore B that ages less.

 

Where exactly did he sat from the view of each frame, the other is younger? What I read is that he claimed B is always younger since he never stated a case in which A is younger in which Janus clearly stated.

 

Granpa said "In the new frame both will agree that B is younger." That is not wrong, and it is not claiming that B is always younger. Even in the quote you provided, he repeatedly uses in that frame to be very clear about in which frame the conclusion holds, and to indicate that the conclusion does not hold in general.

[between3and26characterslon]

Ok let me explain something here because I think people are missing something quite fundemental (apologies if you're not)

 

(let me just say I'm not fantastically intelligent or knowledgeable but I am currently reading a book, it's called Relativity and it's written by a chap called Albert Einstein, if you don't know who he is google him) sarcasm!

 

Anyway suppose you are on a small boat out in the middle of the Pacific ocean and there is no land visable. The water is moving at speed and in one direction under the boat. The wind is moving at a different speed and in a different direction to the water. Another boat comes past at a different speed and direction to both the wind and the water. You're at full throttle and moving at yet a different speed and direction.

 

Questions

1) How fast are you going?

2) What direction are you going in?

 

There is no meaning in these questions unless you construct a frame of reference with which to measure against, this is what we do and we call it longditude and lattitude. This frame is a pratically rigid structure and is considered to be absolutely at rest and all other velocities (water, wind, boats) are measured with respect to this frame.

 

As there is nothing in nature that is absolutley at rest we have to construct an imaginary frame which, for the purposes of our experiment, we consider to be absolutely at rest.

 

Going back to our friends A and B they are not at rest relative to each other, they are both at rest relative to a practically rigid structure, an imaginary framework of axes (x, y, z). This is a system of coordinates and we call it coordinate system (K)

 

When A accelerates and aquires velocity (v) he does so relative to (K). It can now be considered that he is in his own system of coordinates and it is called coordinate system (K'). So now we consider that (K) is at rest and (K') is moving with velocity (v) relative to (K)

 

(K) has time=t, distance=s and coordinates (x, y, z)

 

(K') has time=t', distance=s' and coordinates (x', y', z')

 

All the axes (x, y, z) and (x', y', z') are parralell (ie x is parralell to x', y to y' and z to z') and (+)positive in the same direction. This means that when using the Lorentz transformation you are seeing how (K') changes with respect to/as viewed from (K)

 

If (K') is moving with velocity (v) in a (+)positive direction along the (x) axis of (K) then

(K) is moving with velocity (v) in a (-)negative direction along the (x') axis of (K')

 

This means when using the Lorentz transformation in (K) to measure changes in (K') (v) has a (+)value

But when using the Lorentz transformation in (K') to measure changes in (K) (v) has a (-)value to indicate it is in the opposite direction.

 

When A has (v) with respect to (K) he experiences time at a slower rate than (K) ie (t')<(t) (t' seconds are longer than t seconds)

If B leaves 1 second (t) or <1 second (t') later and attains the same velocity as A then both A and B will consider A is younger

There will be a difference between how much A and B think A is younger

 

If B leaves 1 second (t) after A then A has not experienced 1 full second (t') yet

 

(A) is younger because he is the first one to experience time at a slower rate. He will remain younger for as long as they both have constant (v). If A stops (relative to K) and B catches up to A and stops (relative to K) they will both be the same age again.

[vuquta]

Granpa said "In the new frame both will agree that B is younger." That is not wrong, and it is not claiming that B is always younger. Even in the quote you provided, he repeatedly uses in that frame to be very clear about in which frame the conclusion holds, and to indicate that the conclusion does not hold in general.

 

There are two lines of thnking in this thread.

 

One by the OP and one I proposed.

 

For the OP, I provided the math to prove the accelerating twin will have the slower clock.

 

Then someone said, the twons paradox will always produce and answer.

 

I then provided one in which you cannot decide the outcome. That is the one grandpa was talking about.

 

Ok let me explain something here because I think people are missing something quite fundemental (apologies if you're not)

 

(let me just say I'm not fantastically intelligent or knowledgeable but I am currently reading a book, it's called Relativity and it's written by a chap called Albert Einstein, if you don't know who he is google him) sarcasm!

 

Anyway suppose you are on a small boat out in the middle of the Pacific ocean and there is no land visable. The water is moving at speed and in one direction under the boat. The wind is moving at a different speed and in a different direction to the water. Another boat comes past at a different speed and direction to both the wind and the water. You're at full throttle and moving at yet a different speed and direction.

 

Questions

1) How fast are you going?

2) What direction are you going in?

 

There is no meaning in these questions unless you construct a frame of reference with which to measure against, this is what we do and we call it longditude and lattitude. This frame is a pratically rigid structure and is considered to be absolutely at rest and all other velocities (water, wind, boats) are measured with respect to this frame.

 

As there is nothing in nature that is absolutley at rest we have to construct an imaginary frame which, for the purposes of our experiment, we consider to be absolutely at rest.

 

Going back to our friends A and B they are not at rest relative to each other, they are both at rest relative to a practically rigid structure, an imaginary framework of axes (x, y, z). This is a system of coordinates and we call it coordinate system (K)

 

When A accelerates and aquires velocity (v) he does so relative to (K). It can now be considered that he is in his own system of coordinates and it is called coordinate system (K'). So now we consider that (K) is at rest and (K') is moving with velocity (v) relative to (K)

 

(K) has time=t, distance=s and coordinates (x, y, z)

 

(K') has time=t', distance=s' and coordinates (x', y', z')

 

All the axes (x, y, z) and (x', y', z') are parralell (ie x is parralell to x', y to y' and z to z') and (+)positive in the same direction. This means that when using the Lorentz transformation you are seeing how (K') changes with respect to/as viewed from (K)

 

If (K') is moving with velocity (v) in a (+)positive direction along the (x) axis of (K) then

(K) is moving with velocity (v) in a (-)negative direction along the (x') axis of (K')

 

This means when using the Lorentz transformation in (K) to measure changes in (K') (v) has a (+)value

But when using the Lorentz transformation in (K') to measure changes in (K) (v) has a (-)value to indicate it is in the opposite direction.

 

When A has (v) with respect to (K) he experiences time at a slower rate than (K) ie (t')<(t) (t' seconds are longer than t seconds)

If B leaves 1 second (t) or <1 second (t') later and attains the same velocity as A then both A and B will consider A is younger

There will be a difference between how much A and B think A is younger

 

If B leaves 1 second (t) after A then A has not experienced 1 full second (t') yet

 

(A) is younger because he is the first one to experience time at a slower rate. He will remain younger for as long as they both have constant (v). If A stops (relative to K) and B catches up to A and stops (relative to K) they will both be the same age again.

 

1) This "other" third frame, how is its motion related to the A and B?

 

2) If B leaves 1 second (t) after A then A has not experienced 1 full second (t') yet

 

From the view of A, if B elapses 1 second, then A elapses 1/γ seconds.

 

This is called reciprocal time dilation and is a consequence of SR.

[swansont]

There are two lines of thnking in this thread.

 

One by the OP and one I proposed.

 

For the OP, I provided the math to prove the accelerating twin will have the slower clock.

 

Then someone said, the twons paradox will always produce and answer.

 

I then provided one in which you cannot decide the outcome. That is the one grandpa was talking about.

 

 

 

And yet, when Janus provided the same answer to that same problem, you declared him to be right.

[between3and26characterslon]

There are two lines of thnking in this thread.

 

One by the OP and one I proposed.

 

For the OP, I provided the math to prove the accelerating twin will have the slower clock.

 

Then someone said, the twons paradox will always produce and answer.

 

I then provided one in which you cannot decide the outcome. That is the one grandpa was talking about.

 

 

 

1) This "other" third frame, how is its motion related to the A and B?

 

2) If B leaves 1 second (t) after A then A has not experienced 1 full second (t') yet

 

From the view of A, if B elapses 1 second, then A elapses 1/γ seconds.

 

This is called reciprocal time dilation and is a consequence of SR.

 

1) For the purpose of relativity you have to construct an imaginary rest frame and call it (K)

A is in a frame called (K')

B is in a frame called (K'')

 

(K') and (K'') are both in frame (K)

 

(K) has time=t, distance=s and (x,y,z)

(K') has time=t', distance=s' and (x',y',z')

(K'') has time=t'', distance=s'' and (x'',y'',z'')

 

The lorentz transformation will tell you how (K') and (K'') will appear as viewed from (K) when you know their (v) relative to (K)

 

When you know how (K') and (K'') behave relative to (K)

you can work out how t' and t'' behave relative to t

 

then by knowing how (K') and (K'') behave relative to (K) you can work out how (K') and (K'') behave relative to each other and so how t' behaves relative to t''

 

Lets go back to the boats in the ocean

boat 1 = (K')

boat 2 = (K'')

and longditude and lattitude(L&L)= (K)

 

you measure boat 1's (v) relative to (L&L) that is you measure (K') relative to (K)

you measure boat 2's (v) relative to (L&L) that is you measure (K'') relative to (K)

so now you can work out boat 1's (v) relative to boat 2's, That is (K')'s (v) relative to (K'')

 

It is only by knowing how A and B are behaving relative to (K) that you can know how A and B are behaving relative to each other.

 

2) "From the view of A, if B elapses 1 second, then A elapses 1/γ seconds."

If B leaves 1 second later then whose second does he measure, his seconds are different to A's. does he leave one of A's seconds t'=1 or does he leave one of his seconds t''=1 bearing in mind (t'=t'' is false)

 

He leaves 1 second t as measure in (K) becasue it is (K) we are measuring everything from

 

whilst B is motionless relative to (K) he will have the same time as (K)

whilst A in is motion relative to (K) he will have less time than (K)

 

So B leaves one second (measured from K) after A but A thinks "hang on, that's not been a full second by my clock"

 

So I think we are agreeing here if 1/γ seconds<1 second t

 

They will both agree A is younger than B but they will disagree by how much A is younger. For arguements sake A will think he is less than 1 second younger, B will think A is more than one second younger and the difference will be exactly 1 second t (as measured in K)

 

I will go back and read this thread again just to make sure we're on the same page (so to speak) but I think it applies to everything I've read so far that you need to construct an imaginary frame of refference and measure everything relative to that frame. That is entirely the point of relativity, if you do not create this frame of refference then you are disregarding the principle of relativity and the law of inertia which are the two axioms which you assume to be true for relativity to work otherwise it's like trying to measure something using an elastic tape measure.

Edited September 16, 2010 by between3and26characterslon

[vuquta]

And yet, when Janus provided the same answer to that same problem, you declared him to be right.

 

No, Janus said each frame will conclude the other is time dilated.

 

If grandpa said that, or at least meant to but didn't, then fine.

 

It is not worth debating since the fact remains, both will view the other as time dilated.

 

1) For the purpose of relativity you have to construct an imaginary rest frame and call it (K)

A is in a frame called (K')

B is in a frame called (K'')

 

(K') and (K'') are both in frame (K)

 

(K) has time=t, distance=s and (x,y,z)

(K') has time=t', distance=s' and (x',y',z')

(K'') has time=t'', distance=s'' and (x'',y'',z'')

 

The lorentz transformation will tell you how (K') and (K'') will appear as viewed from (K) when you know their (v) relative to (K)

 

When you know how (K') and (K'') behave relative to (K)

you can work out how t' and t'' behave relative to t

 

then by knowing how (K') and (K'') behave relative to (K) you can work out how (K') and (K'') behave relative to each other and so how t' behaves relative to t''

 

Lets go back to the boats in the ocean

boat 1 = (K')

boat 2 = (K'')

and longditude and lattitude(L&L)= (K)

 

you measure boat 1's (v) relative to (L&L) that is you measure (K') relative to (K)

you measure boat 2's (v) relative to (L&L) that is you measure (K'') relative to (K)

so now you can work out boat 1's (v) relative to boat 2's, That is (K')'s (v) relative to (K'')

 

It is only by knowing how A and B are behaving relative to (K) that you can know how A and B are behaving relative to each other.

 

2) "From the view of A, if B elapses 1 second, then A elapses 1/γ seconds."

If B leaves 1 second later then whose second does he measure, his seconds are different to A's. does he leave one of A's seconds t'=1 or does he leave one of his seconds t''=1 bearing in mind (t'=t'' is false)

 

He leaves 1 second t as measure in (K) becasue it is (K) we are measuring everything from

 

whilst B is motionless relative to (K) he will have the same time as (K)

whilst A in is motion relative to (K) he will have less time than (K)

 

So B leaves one second (measured from K) after A but A thinks "hang on, that's not been a full second by my clock"

 

So I think we are agreeing here if 1/γ seconds<1 second t

 

They will both agree A is younger than B but they will disagree by how much A is younger. For arguements sake A will think he is less than 1 second younger, B will think A is more than one second younger and the difference will be exactly 1 second t (as measured in K)

 

I will go back and read this thread again just to make sure we're on the same page (so to speak) but I think it applies to everything I've read so far that you need to construct an imaginary frame of refference and measure everything relative to that frame. That is entirely the point of relativity, if you do not create this frame of refference then you are disregarding the principle of relativity and the law of inertia which are the two axioms which you assume to be true for relativity to work otherwise it's like trying to measure something using an elastic tape measure.

 

I want to make sure I understand your argument.

 

You are holding back the origina frame as an imaginery frame.

 

Let's say A B C are all synched observers.

 

A takes off, then after some time B takes off and joins the frame of A.

 

Is this what you are saying?

 

If so, what does A conclude about B. You did not mention that.

 

So, we have, at the instant B rejoins A,

 

1) A concludes B and C are the same age since we are using instant acceleration.

2) B concludes A < C and A < B and B = C since we are using instant acceleration.

3) C concludes A < C and A < B and B = C since we are using instant acceleration.

 

 

But, this is only if I understood you correctly.

[between3and26characterslon]

No, Janus said each frame will conclude the other is time dilated.

 

If grandpa said that, or at least meant to but didn't, then fine.

 

It is not worth debating since the fact remains, both will view the other as time dilated.

 

 

 

I want to make sure I understand your argument.

 

You are holding back the origina frame as an imaginery frame.

 

Let's say A B C are all synched observers.

 

A takes off, then after some time B takes off and joins the frame of A.

 

Is this what you are saying?

 

If so, what does A conclude about B. You did not mention that.

 

So, we have, at the instant B rejoins A,

 

1) A concludes B and C are the same age since we are using instant acceleration.

2) B concludes A < C and A < B and B = C since we are using instant acceleration.

3) C concludes A < C and A < B and B = C since we are using instant acceleration.

 

 

But, this is only if I understood you correctly.

 

1,2 and 3) are all correct if B does not accellerate, C is acting as the rest frame and B and C are motionless relative to each other.

 

I think the problem is you're thinking if A has (v) relative to B then B has (v) relative to A and I'm saying this is wrong. This leads you to the opinion that the will both see each other as time dilated (as you said above) and this is wrong (as I understand it)

 

To simplify the problem without altering the effects

 

A accelerates (instantly) to (v) travels at (v) for one second and then decelerates (instantly) to rest again.

 

 

Here's the logic

 

A and B are motionless relative to each other (the principle of relativity says nothing is absolutely motionless BUT two things can be absolutely motionless relative to each other)

 

Now construct a frame around A and B which is absolutely at rest relative to A and B (constructing this frame is just like using a ruler to measure something and it acts as a datum)

 

Newtons law of inertia, a body at rest will remain at rest until acted upon by a force (it is essential you understand what this implies)

 

A and B are at rest therefore no forces are acting upon them

 

Apply a force to A and A only. A and A only accelerates. A and A only has (v) relative to the rest frame

 

There is no force acting upon B so B is still motionless (law of inertia)

 

Therefore A is absolutely in motion and B is absolutely at rest (within this frame of refference)

 

A's time will slow down relative to the rest frame (the rest frame is still at rest because no force has been applied to it)

 

B's time will be the same as the rest frame's because he is motionless relative to it (still no force applied to B)

 

If A is experiencing time at a slower rate than B then for whose one second does he travel at (v) for

 

He travels at (v) for one second of the rest frame (we measure against the rest frame)

 

One second in the rest frame = one second of B's time (B is motionless relative to the rest frame), BUT

One second in the rest frame < one second of A's time (A is moving relative to the rest frame)

 

So if B experiences one full second, A (whose time is slower), has not experienced one full second yet

and if A experiences one full second, B (whose time is faster), has experienced more than one second

 

They will both agree A is younger

They will both agree B is older

 

They will disagree by how much A is younger

They will disagree by how much B is older

 

Somebody posted the ladder paradox (may have been you??) but the paradox does not exist. It comes from a misunderstanding of the theory of relativity. In this paradox the ladder is both longer and shorter than the garrage and the garrage is both longer and shorter than the ladder.

 

Nonsense

 

If you apply a force to the ladder then the ladder IS moving.

 

As viewed from the garrage the ladder will appear shorter.

As viewed from the ladder the garrage will appear longer.

 

If the ladder remains at rest (ie no force applied) and you accelerate the garrage instead then

 

As viewed form the ladder the garrage will be shorter and

As viewed from the garrage the ladder will be longer

[swansont]

 

Somebody posted the ladder paradox (may have been you??) but the paradox does not exist. It comes from a misunderstanding of the theory of relativity. In this paradox the ladder is both longer and shorter than the garrage and the garrage is both longer and shorter than the ladder.

 

Nonsense

 

If you apply a force to the ladder then the ladder IS moving.

 

As viewed from the garrage the ladder will appear shorter.

As viewed from the ladder the garrage will appear longer.

 

If the ladder remains at rest (ie no force applied) and you accelerate the garrage instead then

 

As viewed form the ladder the garrage will be shorter and

As viewed from the garrage the ladder will be longer

 

I posted it, and like most of these so-called paradoxes, they really aren't — it's a paradox only if one assumes simultaneity is absolute. The reason I mentioned it was to rebut the notion that the order of events couldn't differ in different frames. The ladder "paradox" is a classic example of that very thing; the order of the events, i.e. the ladder entering the barn and the doors shutting and opening, depends on which frame the observer is in.

 

No forces need be involved, though. You can look at the situation after the ladder is moving relative to the barn, and there are no accelerations to worry about.

[losfomot]

It confuses me when an expert replies to a comment but does not refute facts stated in the comment that are clearly wrong. It seems like the expert is agreeing. I start to question my understanding (which is great when I have things wrong, but unproductive when I have it right) Above is a case in point.

 

If you apply a force to the ladder then the ladder IS moving.

 

As viewed from the garrage the ladder will appear shorter.

As viewed from the ladder the garrage will appear longer.

 

If the ladder remains at rest (ie no force applied) and you accelerate the garrage instead then

 

As viewed form the ladder the garrage will be shorter and

As viewed from the garrage the ladder will be longer

 

To me this is clearly wrong. It doesn't matter who is moving, if they are moving relative to each other (toward each other), then both should be shorter to the other. If I am wrong here, please let me know and excuse this post. If I am right, why not point out the shortcoming?

 

You may have, kind of, here:

 

No forces need be involved, though. You can look at the situation after the ladder is moving relative to the barn, and there are no accelerations to worry about.

 

But, I'm not sure.... and I can easily see others being confused.

[Janus]

1,2 and 3) are all correct if B does not accellerate, C is acting as the rest frame and B and C are motionless relative to each other.

 

I think the problem is you're thinking if A has (v) relative to B then B has (v) relative to A and I'm saying this is wrong. This leads you to the opinion that the will both see each other as time dilated (as you said above) and this is wrong (as I understand it)

 

Sorry, but your understanding is wrong.

 

To simplify the problem without altering the effects

 

A accelerates (instantly) to (v) travels at (v) for one second and then decelerates (instantly) to rest again.

 

 

Here's the logic

 

A and B are motionless relative to each other (the principle of relativity says nothing is absolutely motionless BUT two things can be absolutely motionless relative to each other)

 

Now construct a frame around A and B which is absolutely at rest relative to A and B (constructing this frame is just like using a ruler to measure something and it acts as a datum)

 

Newtons law of inertia, a body at rest will remain at rest until acted upon by a force (it is essential you understand what this implies)

 

A and B are at rest therefore no forces are acting upon them

 

Apply a force to A and A only. A and A only accelerates. A and A only has (v) relative to the rest frame

 

There is no force acting upon B so B is still motionless (law of inertia)

 

Therefore A is absolutely in motion and B is absolutely at rest (within this frame of refference)

 

A's time will slow down relative to the rest frame (the rest frame is still at rest because no force has been applied to it)

 

B's time will be the same as the rest frame's because he is motionless relative to it (still no force applied to B)

 

If A is experiencing time at a slower rate than B then for whose one second does he travel at (v) for

 

He travels at (v) for one second of the rest frame (we measure against the rest frame)

 

One second in the rest frame = one second of B's time (B is motionless relative to the rest frame), BUT

One second in the rest frame < one second of A's time (A is moving relative to the rest frame)

 

So if B experiences one full second, A (whose time is slower), has not experienced one full second yet

and if A experiences one full second, B (whose time is faster), has experienced more than one second

 

They will both agree A is younger

They will both agree B is older

 

They will disagree by how much A is younger

They will disagree by how much B is older

 

Again, wrong. In SR, as long as A and B have a constant relative motion with respect to each other it does not matter how that motion came to be. As far as time dilation, length contraction, etc are concerned all that counts is the relative motion. A Will see B as time dilated, and B will see A as time dilated. It is only during the actual acceleration that this symmetry is broken (how the symmetry is broken depends on both the direction of the acceleration and the distance between A and B with respect to the acceleration direction).

 

Somebody posted the ladder paradox (may have been you??) but the paradox does not exist. It comes from a misunderstanding of the theory of relativity. In this paradox the ladder is both longer and shorter than the garrage and the garrage is both longer and shorter than the ladder.

 

Nonsense

 

If you apply a force to the ladder then the ladder IS moving.

 

As viewed from the garrage the ladder will appear shorter.

As viewed from the ladder the garrage will appear longer.

 

If the ladder remains at rest (ie no force applied) and you accelerate the garrage instead then

 

As viewed form the ladder the garrage will be shorter and

As viewed from the garrage the ladder will be longer

 

You are violating the Principle of Relativity. If there is no absolute rest, then there is no way that you can say that it the ladder that is moving.

 

Let's put it this way, just because you applied a force to the ladder does not always mean that it is moving. A force can also bring a moving object to a halt. The only way that could say that the force accelerated or decelerated the ladder would be to assume that there is an state of absolute rest by which to measure motion.

 

For instance, once the ladder finishes accelerating, I can establish a new inertial reference frame in which the ladder is at rest and the garage is moving. Now according to you, in this frame the Garage will appear longer. I apply a force to the ladder so that matches the speed of the garage. By your argument above, the ladder should length contract due to its acceleration, making it even shorter than the garage. However, it is now in the same rest frame as the garage, and so as seen from the garage and ladder should be the same length. This is a contradiction.

 

So the only way for this to work as you say is for there to be a absolute rest frame by which to judge the various motions and accelerations. The principle of Relativity forbids this however.