Acceleration is not important in the twin paradox

[xyzt]

xyzt, would you say this statement is accurate:

 

Adding the proper time measured by Clock B between events AB and BC to the proper time measured by Clock C between events BC and AC will yield a number that is equivalent to the proper time experienced by an observer that travels from event AB to event BC and then instantly accelerates to travel back to event AC?

OK, I understand your notation. The above is:

1. true only if B and C have the same speed wrt A

2. false if B and C have different speeds wrt A

 

I showed this in my writeup.

[Delta1212]

OK, I understand your notation. The above is:

1. true only if B and C have the same speed wrt A

2. false if B and C have different speeds wrt A

 

I showed this in my writeup.

I believe this is true if the accelerating observer travels back at the same speed at which he leaves, but that doesn't seem like a requirement of the twin paradox.

[xyzt]

I believe this is true if the accelerating observer travels back at the same speed at which he leaves, but that doesn't seem like a requirement of the twin paradox.

In the most general form of the paradox, equal speed profiles are not a requirement.

[Delta1212]

In the most general form of the paradox, equal speed profiles are not a requirement.

Let me try doing some math for that then.

 

Let's use Clock A's frame of reference. Let's say that A observes Clock B traveling at 0.8c and Clock C traveling at 0.5c. Let's also say that B passes C at a point 10 ly from A.

 

A will measure B traveling between events AB and BC over the course of 12.5 years. A will also measure C traveling from between BC and AC over the course of 20 years, so A's elapsed proper time between AB and AC will be 32.5 years.

 

B's proper time between AB and BC is

 

12.5 * sqrt(1-(0.8/c)^2) = 7.5 years

 

C sets it's time to 7.5 years at BC and then measures a further proper time between BC and AC of

 

20 * sqrt(1-(0.5/c)^2) = 17.32 years

 

7.5 + 17.32 = 24.82 years

 

24.82 < 32.5

 

Which agrees with your assertion that, if u < v (which it is here), tau_C will be less than tau_A.

 

However, if I reverse the speeds, but keep the distance the same:

 

A will see B travel at 0.5c and cover the 10 ly to BC in 20 years. A will see C travel at 0.8c and cover the 10 ly from BC to AC in 12.5 years.

 

B's proper time recorded between AB and BC will be

 

20 * sqrt(1-(0.5/c)^2) = 17.32 years

 

C will then set it's time to 17.32 and cover the distance from BC to AC in an additional

 

12.5 * sqrt(1-(0.8/c)^2) = 7.5 years

 

17.32 + 7.5 = 24.82

 

24.82 < 32.5

 

Here, v<u but tau_C is still less than tau_A. One of us must have made a mistake somewhere, and as someone who wouldn't have understood any of the math I just typed if I'd read it this morning, I'd appreciate being corrected on any mistakes I make so I can avoid them going forward.

[Iggy]

OK, I understand your notation. The above is:

1. true only if B and C have the same speed wrt A

2. false if B and C have different speeds wrt A

 

I showed this in my writeup.

 

That is not correct.

 

It is true for any speed (different or the same) that B and C might have relative to A so long as it is less than c.

 

You can easily prove this by figuring the proper time of A from the perspective of B and C, adding them, and seeing if it is the same as A would figure from his perspective.

 

If B measures [math]\tau_B[/math] between AC and BC then he will calculate [math]\tau_A[/math] between those events as follows,

 

[math]\tau_A = \tau_B \sqrt{1 - {v_B}^2 } + \tau_B \frac{{v_B}^2}{\sqrt{1-{v_B}^2}}[/math]

 

The second term accounts for the relativity of simultaneity (since A is not at BC) and is taken from the lorentz transforms where tv has substituted the distance, x, and this of course assumes light units.

 

If C measures [math]\tau_C[/math] between BC and CA then he will calculate [math]\tau_A[/math] between those events as follows,

 

[math]\tau_A = \tau_C \sqrt{1 - {v_C}^2 } + \tau_C \frac{{v_C}^2}{\sqrt{1-{v_C}^2}}[/math]

 

 

The question is then, if we add these together do we get the correct equation from A's perspective. Let's see...

 

[math]\tau_A \stackrel{?}{=} \tau_B \sqrt{1 - {v_B}^2 } + \tau_B \frac{{v_B}^2}{\sqrt{1-{v_B}^2}} + \tau_C \sqrt{1 - {v_C}^2 } + \tau_C \frac{{v_C}^2}{\sqrt{1-{v_C}^2}}[/math]

 

Pull [math]\tau_B[/math] and [math]\tau_C[/math] out,

 

[math]\tau_A \stackrel{?}{=} \tau_B \left( \sqrt{1 - {v_B}^2 } + \frac{{v_B}^2}{\sqrt{1-{v_B}^2}} \right) + \tau_C \left( \sqrt{1 - {v_C}^2 } + \frac{{v_C}^2}{\sqrt{1-{v_C}^2}} \right)[/math]

 

Common denominator,

 

[math]\tau_A \stackrel{?}{=} \tau_B \left( \frac{{v_B}^2 + 1 - {v_B}^2}{\sqrt{1-{v_B}^2}} \right) + \tau_C \left( \frac{{v_C}^2 + 1 - {v_C}^2}{\sqrt{1-{v_C}^2}} \right)[/math]

 

add,

 

[math]\tau_A \stackrel{?}{=} \tau_B \left( \frac{1}{\sqrt{1-{v_B}^2}} \right) + \tau_C \left( \frac{1}{\sqrt{1-{v_C}^2}} \right)[/math]

 

Now that is indeed the proper equation from A's perspective. It is therefore true for any velocity of B and C. The proper times of B and C, each multiplied by the gamma factor equals the proper time of A, just like your good old fashion twin paradox if the two legs of the trip had a different velocity.

Edited May 7, 2013 by Iggy

[xyzt]

 

If B measures [math]\tau_B[/math] between AC and BC then he will calculate [math]\tau_A[/math] between those events

 

This seems to make no sense: according to "Delta" AC is the event of C

passing through A and BC is the event of B passing through C so

[math]\tau_B[/math] seems to have no sense for the problem. Nor does

your attempt at applying corrections through RoS, the standard way of

calculating proper time is to integrate [math]\sqrt{1-v^2}dt[/math], nor

RoS "corrections" needed .

Let me try doing some math for that then.

 

Let's use Clock A's frame of reference. Let's say that A observes Clock B traveling at 0.8c and Clock C traveling at 0.5c. Let's also say that B passes C at a point 10 ly from A.

 

A will measure B traveling between events AB and BC over the course of 12.5 years. A will also measure C traveling from between BC and AC over the course of 20 years, so A's elapsed proper time between AB and AC will be 32.5 years.

 

B's proper time between AB and BC is

 

12.5 * sqrt(1-(0.8/c)^2) = 7.5 years

This doesn't look right, proper time is frame invariant, so A and B

need to measure the time between the same two events (AB,BC) as being

the same:

 

[math](cd \tau)^2=(cdt)^2-dx^2=(cdt')^2-dx'^2=.....[/math]

 

You are applying a "gamma correction" when going from a frame comoving with A to a frame comoving with B, this is not correct. The times measured by A are proper times, not coordinate times , so no "gamma correction" is allowed.

Edited May 7, 2013 by xyzt

[Delta1212]

 

This seems to make no sense: according to "Delta" AC is the event of C

passing through A and BC is the event of B passing through C so

[math]\tau_B[/math] seems to have no sense for the problem. Nor does

your attempt at applying corrections through RoS, the standard way of

calculating proper time is to integrate [math]\sqrt{1-v^2}dt[/math], nor

RoS "corrections" needed .

This doesn't look right, proper time is frame invariant, so A and B

need to measure the time between the same two events (AB,BC) as being

the same:

 

[math](cd \tau)^2=(cdt)^2-dx^2=(cdt')^2-dx'^2=.....[/math]

 

You are applying a "gamma correction" when going from a frame comoving with A to a frame comoving with B, this is not correct.

B is moving with respect to A. They should not measure the same time between AB and BC.

 

Edit to respond to your edit: Hold on. Let me go back and look at what I did.

Edited May 7, 2013 by Delta1212

[Iggy]

This seems to make no sense: according to "Delta" AC is the event of C passing through A and BC is the event of B passing through C so [math]\tau_B[/math] seems to have no sense for the problem.

 

Delta is correct. AC (what I said) is the same as AB. A and B start the experiment collocated.

 

Nor does your attempt at applying corrections through RoS,

 

One must account for the relativity of simultaneity when an event is considered in two different frames. In A's frame, BC is simultaneous with one event along A's world line, but it is simultaneous with two different events along A's world line in the frames of B and C. If one doesn't account for this when calculating the proper time from 3 different perspectives then one will most assuredly get the wrong answer.

 

the standard way of calculating proper time is to integrate [math]\sqrt{1-v^2}dt[/math], nor RoS "corrections" needed .

 

The simplification to inertial motion is explained here.

 

The standard time dilation equation is given here.

Edited May 7, 2013 by Iggy

[Delta1212]

 

This seems to make no sense: according to "Delta" AC is the event of C

passing through A and BC is the event of B passing through C so

[math]\tau_B[/math] seems to have no sense for the problem. Nor does

your attempt at applying corrections through RoS, the standard way of

calculating proper time is to integrate [math]\sqrt{1-v^2}dt[/math], nor

RoS "corrections" needed .

This doesn't look right, proper time is frame invariant, so A and B

need to measure the time between the same two events (AB,BC) as being

the same:

 

[math](cd \tau)^2=(cdt)^2-dx^2=(cdt')^2-dx'^2=.....[/math]

 

You are applying a "gamma correction" when going from a frame comoving with A to a frame comoving with B, this is not correct. The times measured by A are proper times, not coordinate times , so no "gamma correction" is allowed.

Ok, here's a point I need clarification on. As far as I am aware, proper time is the time recorded by an observer between two events, both of which the observer was at. You can determine the proper time measured by A between AB and AC. You can measure the proper time recorded by B between AB and BC. You can measure the proper time recorded by C between BC and AC. You cannot record the proper time measured by A between AB and BC because A was not at BC.

 

If this is the case, then A cannot measure proper times between AB and BC or between BC and AC, meaning that the measurements A makes of the times the other clocks take to travel between these events are coordinate times and a gamma correction must be applied.

 

I really do need someone to check what I just said.

Edited May 7, 2013 by Delta1212

[xyzt]

Delta is correct. AC (what I said) is the same as AB. A and B start the experiment collocated.

 

 

One must account for the relativity of simultaneity when an event is considered in two different frames. In A's frame, BC is simultaneous with one event along A's world line, but it is simultaneous with two different events along A's world line in the frames of B and C. If one doesn't account for this when calculating the proper time from 3 different perspectives then one will most assuredly get the wrong answer.

 

 

The simplification to inertial motion is explained here.

 

The standard time dilation equation is given here.

I don't need the basic stuff on time dilation.

The calculations are very simple, in A frame B and C meet at [math]t=\frac{D}{v_B+v_C}[/math]

The total elapsed proper time in frame A for:

 

1. B to reach the BC event is [math]\tau_B=\int_0^t{\sqrt{1-v_B^2}dt}=\frac{D}{v_B+v_C}\sqrt{1-v_B^2}[/math]

2. C to reach the BC event is [math]\tau_C=\int_0^t{\sqrt{1-v_c^2}dt}=\frac{D}{v_B+v_C}\sqrt{1-v_c^2}[/math]

 

Very simple.

[Iggy]

If this is the case, then A cannot measure proper times between AB and BC or between BC and AC, meaning that the measurements A makes of the times the other clocks take to travel between these events are coordinate times and a gamma correction must be applied.

 

I really do need someone to check what I just said because.

That is correct. We can use Md's velocity of v = 13/15c (.8666c) and 4 total years of proper time. Half of that time B travels away until bumping into C. That means A has 2 years of proper time between AC and BC. To calculate the proper time of B it is like you say,

 

[math]\tau_A = \tau_B \frac{1}{\sqrt{1-v^2}}[/math]

 

rearrange

 

[math]\tau_B = \tau_A \sqrt{1-v^2}[/math]

 

[math]\tau_B = (2) \sqrt{1-(0.8666)^2 } = 2 \cdot 0.5 = 1[/math]

 

A therefore calculates that the proper time of B is 1 year.

 

We can also calculate the proper time of A during half of the trip using the perspective and proper time of B. This is the equation that Xyzt disagrees with, but let's see if it works,

 

[math]\tau_B = \tau_A \frac{1}{\sqrt{1 - v^2 }} - \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math]

 

rearrange,

 

[math]\tau_A = \tau_B \sqrt{1 - v^2 } + \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math]

 

[math]\tau_A = 1 \sqrt{1 - (0.8666)^2 } + 1 \frac{(0.8666)^2}{\sqrt{1-( \mbox{0.8666})^2 }}[/math]

 

[math]\tau_A = 1 (0.5) + 1 (1.5) = 2[/math]

 

So, B, from his frame, calculates that A measures a proper time of 2 years over the span of half of his trip. And this, we know, is correct.

 

 

---

 

 

I don't need the basic stuff on time dilation.

The calculations are very simple, in A frame B and C meet at [math]t=\frac{D}{v_B+v_C}[/math]

The total elapsed proper time in frame A for:

 

1. B to reach the BC event is [math]\tau_B=\int_0^t{\sqrt{1-v_B^2}dt}=\frac{D}{v_B+v_C}\sqrt{1-v_B^2}[/math]

2. C to reach the BC event is [math]\tau_C=\int_0^t{\sqrt{1-v_c^2}dt}=\frac{D}{v_B+v_C}\sqrt{1-v_c^2}[/math]

 

Very simple.

Calculating things in the coordinate system of A is hardly a good way of disagreeing with my method of calculating things in the coordinate system of B and C, is it?

 

If you disagree with my method of calculating things from B and C's perspective then I welcome you to do it differently. It would help to use numbers like I have done above.

Edited May 7, 2013 by Iggy

[md65536]

If this is the case, then A cannot measure proper times between AB and BC or between BC and AC, meaning that the measurements A makes of the times the other clocks take to travel between these events are coordinate times and a gamma correction must be applied.

If you mean that for A to measure using its own clock the proper time experienced by B's clock, then yes a gamma factor is applied. Observer A can consistently observe, measure, predict, or calculate the proper time elapsed on another clock that passes through the events, which is why we would say "All observers agree on proper time". If you measure using a clock that's actually experiencing the proper time, you would accept its measurement without adjustment, regardless of the clock rate relative to your local clock. So in the examples, B ages 1 year between AB and BC according to its own clock, regardless of what the observer's local clock measures or the value of gamma during that time.

 

Instead of "the proper time at A between AB and BC", you might say "the proper time between AB and an event at A that is simultaneous according to A with BC". In these examples since B recedes at the same relative speed that C approaches, such an event is halfway between AB and AC.

 

So, everyone agrees on the proper time measured by a particular clock's path between 2 events. In general you can't agree on the time that a second, distant clock ages during that time, because the simultaneity of events between the two clocks depends on observer. However you could choose the second clock's frame of reference to determine simultaneity, and then figure out the proper time that it ticks "during" the first clock's proper time. However, if they start or end separated, observers (including the first clock) would generally disagree that the two clocks' proper time measurements were simultaneous.

 

Also just for pedantry if we say "the proper time between AB and AC" we're implying (in this thread at least) "the proper time measured by A" or ie. an inertial clock passing through the 2 events, because any clock on any path between the two events would measure a proper time (which is exactly what the traveling twin of the typical twin paradox does).

 

Sorry if this is mostly basic and rambling, I'm just trying to work through my understanding of it.

 

The calculations are very simple, in A frame B and C meet at [math]t=\frac{D}{v_B+v_C}[/math]

Why do you add the velocities?

This calculation implies that if v_B = 0.1c and v_C = 0.9c, they would meet at the same time (according to A) as if v_B = 0.9c and v_C = 0.1c. That's not true. Nevermind, I just figured out what D represents.

Edited May 7, 2013 by md65536

[michel123456]

1. So far, i haven't seen Md or Iggy or Delta take some understanding on xyzt's point that when the proper time of B's clock is transfered to C's clock then the paradox falls apart. There is no paradox anymore and xyzt is right.

 

2. Also I have not seen from Md Iggy and delta some understanding that in order to keep the paradox, when C takes the reading of B's clock, the traveler must say "what a crap, this reading is wrong, mine is correct" and C to continue with it's own reading, because B and C times do not correspond to each other: there is a jump as if B & C met in a place of spacetime where observations all around differ, the one observing on Earth the 2000 Olympics in Sidney while the other observes the 2004 in Athens.

 

When B & C correspond, both observing Sidney, then we go back to point 1 and there is no paradox.

In this case C must say, "my wrong, I am dreaming, I do not observe Athens (although he does), lets keep B's time and observe Sidney this is bogus."

Edited May 7, 2013 by michel123456

[Delta1212]

1. So far, i haven't seen Md or Iggy or Delta take some understanding on xyzt's point that when the proper time of B's clock is transfered to C's clock then the paradox falls apart. There is no paradox anymore and xyzt is right.

 

2. Also I have not seen from Md Iggy and delta some understanding that in order to keep the paradox, when C takes the reading of B's clock, the traveler must say "what a crap, this reading is wrong, mine is correct" and C to continue with it's own reading, because B and C times do not correspond to each other: there is a jump as if B & C met in a place of spacetime where observations all around differ, the one observing on Earth the 2000 Olympics in Sidney while the other observes the 2004 in Athens.

 

When B & C correspond, both observing Sidney, then we go back to point 1 and there is no paradox.

In this case C must say, "my wrong, I am dreaming, I do not observe Athens (although he does), lets keep B's time and observe Sidney this is bogus."

2. This doesn't make sense, especially- Oh, I just realized why xyzt thinks setting B and C to the same speed makes B irrelevant.

 

Ok, here's why I disagree with both of these points anyway:

 

This isn't an example of C measuring a discrepancy in elapsed proper time since the beginning of the experiment with A in the way that the twin paradox is generally formulated as a discrepancy between the time experienced by one twin at rest with another twin that takes a non inertial path. Rather, it is a method of measuring the time experienced along that path without sending someone the full length of the trip.

 

Here, say we have a pair of twins all set to do the experiment, but they can't agree on which is to stay home and which is to leave. They are about to give up in despair when a friend from NASA comes to them and says "I have a probe returning from deep space to Earth that will be passing through the area you would have turned your ship around. If we send a clock out to meet it, we can record the length of time the clock measures over the course of the first leg of the trip that you would have taken. Then we can use the probe's internal clock to measure the time experienced on the second leg of the trip."

 

"Nothing will be returning having aged less than those who stated behind, but we can at least experimentally verify the length of time the traveling twin would have experienced so you can compare it with the amount both of you actually age while staying here on Earth."

 

C doesn't say "hey, I'm right not you" upon meeting B, because C isn't accepting B's time as the correct time at that moment, it is accepting that as the proper time of the first leg of a hypothetical twin experiment. C didn't start at the same location as the at rest partner, nor travel between that partner and the "turnaround point" where B passes C. C cannot, therefore, measure a proper time between those two events on its own clock, but it can see the proper time that B measured on that leg (which B actually traveled) and say "Ah, now I know how much time would be experienced by the twin on his trip away. As I am about to travel the path such a twin would have taken on the way back, I can add B's time to the time I experience between this moment and the one when I pass A and determine how much time such a twin would have experienced had he traveled away from A (with the velocity of B) to the point at which BC took place, and then turned around and traveled back to A (with the velocity of C)."

 

If you mean that for A to measure using its own clock the proper time experienced by B's clock, then yes a gamma factor is applied. Observer A can consistently observe, measure, predict, or calculate the proper time elapsed on another clock that passes through the events, which is why we would say "All observers agree on proper time". If you measure using a clock that's actually experiencing the proper time, you would accept its measurement without adjustment, regardless of the clock rate relative to your local clock. So in the examples, B ages 1 year between AB and BC according to its own clock, regardless of what the observer's local clock measures or the value of gamma during that time.

 

Instead of "the proper time at A between AB and BC", you might say "the proper time between AB and an event at A that is simultaneous according to A with BC". In these examples since B recedes at the same relative speed that C approaches, such an event is halfway between AB and AC.

 

So, everyone agrees on the proper time measured by a particular clock's path between 2 events. In general you can't agree on the time that a second, distant clock ages during that time, because the simultaneity of events between the two clocks depends on observer. However you could choose the second clock's frame of reference to determine simultaneity, and then figure out the proper time that it ticks "during" the first clock's proper time. However, if they start or end separated, observers (including the first clock) would generally disagree that the two clocks' proper time measurements were simultaneous.

 

Also just for pedantry if we say "the proper time between AB and AC" we're implying (in this thread at least) "the proper time measured by A" or ie. an inertial clock passing through the 2 events, because any clock on any path between the two events would measure a proper time (which is exactly what the traveling twin of the typical twin paradox does).

 

Sorry if this is mostly basic and rambling, I'm just trying to work through my understanding of it.

 

No, it's good. I did already understand most of that, but it's nice to have confirmation that I'm not misinterpreting what proper time is. I already basically grasped the concept of it (that everyone would agree on the time a particular clock measured between events it was present at), but I didn't know it was called that or exactly how to calculate it from measurements taken in another frame until yesterday. Edited May 7, 2013 by Delta1212

[xyzt]

 

 

 

---

 

 

 

Calculating things in the coordinate system of A is hardly a good way of disagreeing with my method of calculating things in the coordinate system of B and C, is it?

 

If you disagree with my method of calculating things from B and C's perspective then I welcome you to do it differently. It would help to use numbers like I have done above.

I simply pointed out how I do the calculations, complete with the derivation from scratch.

You haven't answered my question, how do you arrive to your formulas? Where is the derivation?

Edited May 7, 2013 by xyzt

[md65536]

1. So far, i haven't seen Md or Iggy or Delta take some understanding on xyzt's point that when the proper time of B's clock is transfered to C's clock then the paradox falls apart. There is no paradox anymore and xyzt is right.

 

2. Also I have not seen from Md Iggy and delta some understanding that in order to keep the paradox, when C takes the reading of B's clock, the traveler must say "what a crap, this reading is wrong, mine is correct" and C to continue with it's own reading, because B and C times do not correspond to each other: there is a jump as if B & C met in a place of spacetime where observations all around differ, the one observing on Earth the 2000 Olympics in Sidney while the other observes the 2004 in Athens.

 

When B & C correspond, both observing Sidney, then we go back to point 1 and there is no paradox.

In this case C must say, "my wrong, I am dreaming, I do not observe Athens (although he does), lets keep B's time and observe Sidney this is bogus."

This is all incorrect.

 

Nobody says "My measurements are wrong because another observer measured something different." If B and C understand SR, they'll accept that their different measurements are consistent.

 

In your example, B and C aren't "seeing" the olympics but rather saying "the olympics are going on now (simultaneous)". So they both might be observing Earth appearing as it was a couple years before the 2000 olympics, and B says "This light signal has taken a couple years to reach me" while C says "This light signal has taken a couple + 4 years to reach me (4 years by Earth's clock that is)". Both are correct.

 

 

The clock sync at BC is no different than the clock sync at AB (or if there is a difference, eg setting it to a value other than 0, the experiment can be changed to make it exactly the same and it won't affect the proper time measured on any leg of the journey*). Why doesn't A and B setting their clocks to 0 at AB destroy the paradox? What was their time before they set their clocks?!?! Isn't that a jump in time!!!? If you understand the answers to these, then you'll see that the sync at BC is no different and doesn't affect the paradox, but affects only the way C measures it.

 

 

* What I mean by that is, whether C sets its clock to 1 and measures a proper time of 1 ending with a clock that says "2" when it reaches AC, or whether it sets its clock to 0 and measures a proper time of 1 with a clock that says "1" when it reaches AC, the proper time of that leg of the trip is still 1.

Edited May 7, 2013 by md65536

[Iggy]

I simply pointed out how I do the calculations, complete with the derivation from scratch.

You haven't answered my question, how do you arrive to your formulas? Where is the derivation?

Like I said when I first used it, it's a straightforward application of the Lorentz transformations. I can give a thought experiment explaining if you're interested, but I think you're more trying to debate me.

 

Would anyone like to fix some of the (-1)'s xyzt is throwing around? I fixed what I could, but can't get them all if you know what I mean.

[xyzt]

Like I said when I first used it, it's a straightforward application of the Lorentz transformations. I can give a thought experiment explaining if you're interested, but I think you're more trying to debate me.

 

 

 

 

How about you showed the derivation of your formulas? This is the second time I ask you.

I have shown you the way it is done in standard textbooks, so how about you explained how you arrived to the formulas that you posted?

[md65536]

This seems to make no sense: according to "Delta" AC is the event of C

passing through A and BC is the event of B passing through C so

[math]\tau_B[/math] seems to have no sense for the problem. Nor does

your attempt at applying corrections through RoS, the standard way of

calculating proper time is to integrate [math]\sqrt{1-v^2}dt[/math], nor

RoS "corrections" needed .

This doesn't look right, proper time is frame invariant, so A and B

need to measure the time between the same two events (AB,BC) as being

the same:

Is this a fair analogy of your counter-argument?:

 

Say that observer C is an astronaut, and B is a passing meteor, and that C is 40 years old at event AC. Then everyone will agree on the proper time measured by C, ie. that it is 40 years old at AC, and that is true regardless of when the meteor passed it, so meteor B is irrelevant.

 

 

I get the impression that you believe that a clock can only measure one proper time at a time? Is this fair? Ie. if you're measuring proper time since birth, the proper time on its 40th birthday is 40 years, and the proper time between its 38th and 39th birthday is ... ???

a) Can't be measured

b) 39 years

c) 1 year (and the time between the passing meteor event BC and AC can be say 1 year even though C has also recorded a proper time of 40 years since another event)

d) Other?

Edited May 7, 2013 by md65536

[Iggy]

How about you showed the derivation of your formulas? This is the second time I ask you.

I have shown you the way it is done in standard textbooks, so how about you explained how you arrived to the formulas that you posted?

 

I used the standard time dilation formula which is derived from the Lorentz transforms at the link I gave you: hyperphysics time dilation I'm not going to copy it over.

 

The other term I used is, like I said, a straightforward application of the Lorentz transforms to account for the relativity of simultaneity. The t formula for switching coordinate systems is,

 

[math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math]

 

You can get it here. All I did was set t'=0 (imagining the origin of the coordinate system be event BC). And, like I said when I gave the term, I subbed x' = vt', since that locates A's world line and reveals the deficit of proper time from the shift of simultaneity along that world line.

 

[math]\tau_B = \frac{v^2}{\sqrt{1-v^2}}[/math]

 

I've shown that the equation works, and now explained how I got it. If you have a different way of switching coordinate systems then that's fine, I'd love to see it, but I'm not going to debate the efficacy of something as common as the Lorentz transformations.

[xyzt]

 

I used the standard time dilation formula which is derived from the Lorentz transforms at the link I gave you: hyperphysics time dilation I'm not going to copy it over.

 

The other term I used is, like I said, a straightforward application of the Lorentz transforms to account for the relativity of simultaneity. The t formula for switching coordinate systems is,

 

[math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math]

 

You can get it here. All I did was set t'=0 (imagining the origin of the coordinate system be event BC). And, like I said when I gave the term, I subbed x' = vt', since that locates A's world line and reveals the deficit of proper time from the shift of simultaneity along that world line.

 

[math]\tau_B = \frac{v^2}{\sqrt{1-v^2}}[/math]

 

I've shown that the equation works, and now explained how I got it. If you have a different way of switching coordinate systems then that's fine, I'd love to see it, but I'm not going to debate the efficacy of something as common as the Lorentz transformations.

You have just showed two disjoined formulas without being able to connect them into your final formula that you are advocating. How do you derive this:

If B measures [math]\tau_B[/math] between AC and BC then he will calculate [math]\tau_A[/math] between those events as follows,

 

[math]\tau_A = \tau_B \sqrt{1 - {v_B}^2 } + \tau_B \frac{{v_B}^2}{\sqrt{1-{v_B}^2}}[/math]

 

There is no "coordinate switching" in my approach, all calculations are done in A frame. This is the third time I am telling you this.

Edited May 8, 2013 by xyzt

[Iggy]

You have just showed two disjoined formulas without being able to connect them into your final formula that you are advocating. How do you derive this:

If B measures [math]\tau_B[/math] between AC and BC then he will calculate [math]\tau_A[/math] between those events as follows,

 

[math]\tau_A = \tau_B \sqrt{1 - {v_B}^2 } + \tau_B \frac{{v_B}^2}{\sqrt{1-{v_B}^2}}[/math]

 

I just explained the derivation of both terms. They are added together because we are looking for the proper time of A along half of his trip and they each find part of the length of that line. The time dilation formula finds the first segment and the shift of simultaneity finds the second. If I didn't think you were trying to debate me I would make a minkowski diagram and show you visually what is happening.

 

There is no "coordinate switching" in my approach, all calculations are done in A frame. This is the third time I am telling you this.

 

You think I don't know this?

 

I solved the proper time of A from the frames of B and C. You haven't done that. All you did was find their proper time from A's frame. That is just A's time multiplied by the gamma factor. That's all you've done.

 

If you want to disagree with what I'm doing then at least have a different method of doing it. Or show something that I've done wrong. I already showed with numbers that the equation you're bickering about works.

 

Once again you're just objecting because somebody is doing something that you haven't considered or that you are unable to do. I'm not going to encourage that argument any more.

[xyzt]

I just explained the derivation of both terms. They are added together because we are looking for the proper time of A along half of his trip and they each find part of the length of that line. The time dilation formula finds the first segment and the shift of simultaneity finds the second. If I didn't think you were trying to debate me I would make a minkowski diagram and show you visually what is happenin

Please do so, there is no derivation for the second term that you are adding in.

 

<quote>You think I don't know this?</quote>

 

Then stop claiming that I am switching frames, would you?

Edited May 8, 2013 by xyzt

[Delta1212]

Actually, you said that B and C begin moving simultaneously from A's frame along a segment measuring length d (also from A's frame). You then measured the proper time that A would measure if it traveled d distance at C's speed (in other words, A's proper time as measured from C's frame). Great, everyone agrees on proper time.

 

Except that in frame C, C did not reach distance d from A and B simultaneous to the moment that they separated, because C is in a different location and moving at a different velocity to A. In C's frame, it is not d distance away at the start of the experiment, so while it does measure A approaching it at the same velocity that A measures C approaching, they do not agree on the time that the experiment started, and therefore on the distance that needs to be traveled.

 

You use d/v as the time that C measures A to take while finding A's proper time, but d is the length of the distance traveled since the start of the experiment as measured in frame A, not in frame C.

[xyzt]

Actually, you said that B and C begin moving simultaneously from A's frame along a segment measuring length d (also from A's frame). You then measured the proper time that A would measure if it traveled d distance at C's speed (in other words, A's proper time as measured from C's frame). Great, everyone agrees on proper time.

 

Except that in frame C, C did not reach distance d from A and B simultaneous to the moment that they separated, because C is in a different location and moving at a different velocity to A. In C's frame, it is not d distance away at the start of the experiment, so while it does measure A approaching it at the same velocity that A measures C approaching, they do not agree on the time that the experiment started, and therefore on the distance that needs to be traveled.

 

You use d/v as the time that C measures A to take while finding A's proper time, but d is the length of the distance traveled since the start of the experiment as measured in frame A, not in frame C.

Frame C has no impact on the solution, all proper times are calculated in one frame, frame A. You (and Iggy) need to come with grips with this fact.