Understanding Time Dilation: How Fast Do You Have to Go?

[Sam Woole]

Rather than use your letters ABCDE I’d prefer to use abcdefghijklm. Each of my letters represents 5 minutes of Alice’s travel time.

B           T
o           e
b           d
|           |
abcdefghijklm
A--B--C--D--E

Ted’s     Ted’s     Ted’s
own       view of   view of
clock     Alice’s   Bob’s
clock     clock
17:20(m)  12:00(a)  12:00(a)
17:30(m)  12:15(d)  12:10(a)
17:40(m)  12:30(g)  12:20(a)
17:50(m)  12:45(j)  12:30(a)
18:00(m)  13:00(m)  12:40(a)

Each row of this table represents a ray of light traveling from Bob to Ted. For example in the second row, a ray of light leaves Bob, at (a), at 12:10 Bob-time, passes Alice, at (d), at 12:15 Alice-time, and arrives at Ted, at (m), at 17:30 Ted-time. If you ignore Alice’s column and just look at the rays leaving Bob and arriving at Ted, each ray arrives at Ted(m) at a Ted-time that is 5h20m later than the Bob-time that it left Bob(a).
The first diagram I have attached to this post may make things clearer. Each spot represents a 5-minute interval. The yellow lines are light rays.
The conclusion to draw from this table is that if you travel away from somebody and observe an apparent 2/3 “red-shift” change of clock rates (Alice observing Bob), then it must follow that if you travel towards somebody at the same speed, you must observe an apparent 3/2 “blue-shift” change of clock rates (Ted observing Alice).
For the return journey my table shows, not EDCBA (mjgda) but mkigeca. That was my choice.

Alice’s   Alice’s
own       view of
clock     Bob’s clock
13:00(m)  12:40(a)
13:10(k)  12:55(a)
13:20(i)  13:10(a)
13:30(g)  13:25(a)
13:40(e)  13:40(a)
13:50(c)  13:55(a)
14:00(a)  14:10(a)

The reason for the 3/2 rate is because the speed of Bob towards Alice, during the return, is exactly the same as was the speed of Alice towards Ted on the outward journey. Einstein’s postulates imply that any effects you can measure should depend only on the relative velocity and on nothing else. So the same rate of 3/2 applies in both cases.
The second diagram I have attached to this post shows Alice’s whole journey.

I would say I have learned why and how the Doppler effect was applied. But I have also found more difficulties.

The first was, when Alice moved toward Ted, the 3/2 rate was applied to the moving clock of Alice, not to the stationary clock of Ted; while Alice moved toward Bob, the 3/2 rate was applied to the stationary clock of Bob. I could not understand this unbalanced treatment.

The second was, when Alice was moving away from Ted, the stationary observer Ted did the observation, applying the 2/3 rate to Alice's clock. This was the reverse of Alice moving away from Bob, the stationary observer. So what will happen if we let Bob observe Alice's journey both ways and apply the 2/3 and 3/2 to Alice's clock?

When I tried, I got the following table, supposing the starting time be 00 minutes, and eliminating the 3rd observer.

At start both clocks show 00 minutes:
00.......00, at start. After every 15 minutes on Bob's clock, he applies 2/3 to Alice's clock,

Bob's view of Alice clock...Bob's view of his own.
10........15,
20........30,
30........45,
40........60, In Bob's view, Alice has traveled 60 minutes, and she returns at this moment. Now Bob will observer every 10 minutes, and apply the 3/2rate.
55........70,
70........80,
85........90,
100.......100,
115.......110,
130.......120.

My table showed, after 2 hours of travel by Alice from Bob's view, Alice's clock should have accumulated 10 minutes more than his. If my table were sound, we could deduce that the earthbound twin got younger than his traveled brother.

Let me repeat my difficulties. When Alice was moving toward Ted, why the 3/2 rate was not applied to Ted's clock? Second, can Bob do the observation?

 

[Sherlock]

First off, your assertion that the clocks cannot be compared during the acceleration phases is not correct.

Thanks for doing that explanation. I've printed it out and will refer to it as I read the books I've gotten ("Introduction to the Theory of Relativity" by Peter Gabriel Bergmann;"Einstein's Theory of Relativity" by Max Born; and a college physics text by Halliday, Resnick and Walker that has a chapter on it. I got these particular books because they were very cheap.)

This is much more complicated than I thought it was going to be. :-)

Regarding the assertion that clocks can't be compared during acceleration phases -- I forget where I got that. I was Googling a lot last night and read something like that somewhere. It wasn't from a quack site (at least I don't think it was). I also remember reading somewhere that a clock's tick rate is independent of any acceleration that the clock might experience.

 

[Janus]

Thanks for doing that explanation. I've printed it out and will refer to it as I read the books I've gotten ("Introduction to the Theory of Relativity" by Peter Gabriel Bergmann;"Einstein's Theory of Relativity" by Max Born; and a college physics text by Halliday, Resnick and Walker that has a chapter on it. I got these particular books because they were very cheap.)
This is much more complicated than I thought it was going to be. :-)
Regarding the assertion that clocks can't be compared during acceleration phases -- I forget where I got that. I was Googling a lot last night and read something like that somewhere. It wasn't from a quack site (at least I don't think it was).

What might have been meant is that the clocks cannot be compared in an absolute way. IOW, If you compare clocks from the perspective of each clock they might both determine that is the other clock that is running slow and there is no way to say that one of them is absolutely right and the other wrong

I also remember reading somewhere that a clock's tick rate is independent of any acceleration that the clock might experience.

That is correct, a clocks rate is independent of any acceleration it experiences. What is not indedependent of that acceleration is how that clock measures the time rate of other clocks separated from it by a distance measured along the line of acceleration.

 

[DrGreg]

I would say I have learned why and how the Doppler effect was applied. But I have also found more difficulties.
The first was, when Alice moved toward Ted, the 3/2 rate was applied to the moving clock of Alice, not to the stationary clock of Ted; while Alice moved toward Bob, the 3/2 rate was applied to the stationary clock of Bob. I could not understand this unbalanced treatment.

The basic principle of Relativity is that the concepts of "stationary" and "moving" are relative terms. Something is stationary relative to something else, or something is moving relative to something else.

When Alice moved towards Ted, Alice was moving relative to Ted with a Doppler shift, relative to Ted, of 3/2.

When Alice moves towards Bob, it is also true to say that Bob is moving relative to Alice. From Alice's point of view, Bob is moving towards her. And the speed at which he is moving is exactly the same speed that, before, Alice was moving towards Ted. That is why, in both cases, the 3/2 Doppler shift applies.

This is the crux of what Relativity is all about. If you are sat in a moving train, you can think of yourself as being stationary and the rail tracks as moving relative to you. As long as the train moves at a constant speed in a straight line, any experiments you carry out inside the train carriage will give exactly the same results as if the train were stopped.

If A moves at speed v relative to B, then B moves at speed v relative to A. And if A observes B with a Doppler shift of 3/2, then B observes A with a Doppler shift of 3/2.

The second was, when Alice was moving away from Ted, the stationary observer Ted did the observation, applying the 2/3 rate to Alice's clock. This was the reverse of Alice moving away from Bob, the stationary observer. So what will happen if we let Bob observe Alice's journey both ways and apply the 2/3 and 3/2 to Alice's clock?

In my original post https://www.physicsforums.com/showpost.php?p=783491&postcount=68", the last table I gave was for Bob's view of Alice. (I never gave a table for Ted's view of Alice on her return journey.)

When I tried, I got the following table, supposing the starting time be 00 minutes, and eliminating the 3rd observer.
At start both clocks show 00 minutes:
00.......00,
at start. After every 15 minutes on Bob's clock, he applies 2/3 to Alice's clock,
Bob's view of Alice clock...Bob's view of his own.
10........15,
20........30,
30........45,
40........60,

Yes, that is correct.

In Bob's view, Alice has traveled 60 minutes, and she returns at this moment. Now Bob will observer every 10 minutes, and apply the 3/2 rate.
55........70,
70........80,
85........90,
100.......100,
115.......110,
130.......120.

No. Bob has only observed 40 minutes of Alice's journey so far. We must keep going with the 2/3 rate until Bob has seen the whole hour of Alice's journey:

50.......75
60.......90

Now, Bob sees Alice return and the factor 3/2 applies:

60.......90
75.......100
90.......110
105........120
120........130

Another diagram attached...

 

[jtbell]

[at the 60-minute mark, according to Bob's clock] Bob has only observed 40 minutes of Alice's journey so far. We must keep going with the 2/3 rate until Bob has seen the whole hour of Alice's journey:

To make this more explicit, in Bob's reference frame, when Alice turns around she is 25 light-minutes away. It takes 25 minutes for light from this event to travel back to Bob, and that's when he actually sees her turn around. In the meantime Bob continues to watch Alice's clock running at the 2/3 rate.

 

[Sam Woole]

No. Bob has only observed 40 minutes of Alice's journey so far. We must keep going with the 2/3 rate until Bob has seen the whole hour of Alice's journey:
50.......75
60.......90
Now, Bob sees Alice return and the factor 3/2 applies:
60.......90
75.......100
90.......110
105........120
120........130
Another diagram attached...

Here I might have to disagree with you. Alice turned around when she saw 13:00 on her clock. If Bob observed Alice's clock from (a), he would see 12:40. Therefore, I believe Alice's return trip should begin when Alice saw 60 on her clock, namely when Bob saw 40 on Alice's clock.

If you insist, you have to explain 2 things. First, where was the 20 minutes delay? When Bob saw 40 on Alice's clock, Alice has traveled 60 minutes already due to the 20 minutes delay.

Secondly, why the equal row disappeared. Your table on post #86 had an equal row at (e), 13:40 = 13:40, and mine had an equal row 100=100.
This new table above of yours had no equal row.

Your table on post #86:
Alice’s... Alice’s
own... view of
clock..... Bob’s clock
13:00(m)... 12:40(a)
13:10(k)... 12:55(a)
13:20(i)... 13:10(a)
13:30(g)... 13:25(a)
13:40(e)... 13:40(a)
13:50(c)... 13:55(a)
14:00(a)... 14:10(a)

 

[jtbell]

I believe Alice's return trip should begin when Alice saw 60 on her clock, namely when Bob saw 40 on Alice's clock.

Imagine that in order to initiate the turnaround, Alice has to push a button on her ship's control panel, and the button is right next to her clock. From her point of view, she pushes the button when her clock reads 60 minutes after departure. Do you really believe that Bob is going to see Alice pushing the button when her clock (which is right next to the button!) reads 40?

 

[Sam Woole]

Thanks for doing that explanation. I've printed it out and will refer to it as I read the books I've gotten ("Introduction to the Theory of Relativity" by Peter Gabriel Bergmann;"Einstein's Theory of Relativity" by Max Born; and a college physics text by Halliday, Resnick and Walker that has a chapter on it. I got these particular books because they were very cheap.)

Since you have the book by Halliday, Resnick and Walker, would you please pay attention to the language used in the chapter on Relativity. By this I meant, try to find out whether there is a lie in the said chapter.

 

[Sherlock]

Since you have the book by Halliday, Resnick and Walker, would you please pay attention to the language used in the chapter on Relativity. By this I meant, try to find out whether there is a lie in the said chapter.

What did you read that causes you to say this? I'll look for it.

 

[Sherlock]

... a clock's rate is independent of any acceleration it experiences.

I don't think I understand what this means.

A and B are on Earth with identical, synchronized clocks. A accelerates away and eventually assumes a uniform speed wrt B and earth.

B sees A's clock as ticking at a slower rate than his. How is this independent of the acceleration that brought A to the relative speed that is causing the time dilation?

 

[Janus]

I don't think I understand what this means.
A and B are on Earth with identical, synchronized clocks. A accelerates away and eventually assumes a uniform speed wrt B and earth.
B sees A's clock as ticking at a slower rate than his. How is this independent of the acceleration that brought A to the relative speed that is causing the time dilation?

In that acceleration itself does not cause any time dilation effect on a clock outside of that caused by a change in relative speed. Remember, you can have acceleration without any change of relative speed, as acceleration is a change of velocity and velocity consists of both speed and direction.

For example, you can put a clock on a centrifuge and spin it at a constant RPM, and will be both constantly accelerating and maintaining a constant speed. In this case, an observer watching the centrifuge from a frame at rest with respect to the centrifuge axis will only measure a time dilation for the clock that is due to its constant relative speed.

 

[jtbell]

Another way to put it: time dilation does not depend directly on acceleration. It depends only indirectly on acceleration insofar as acceleration produces a change in an object's speed. It is not necessary to know an object's "acceleration history" in order to calculate the amount of time dilation, only the object's current speed.

 

[DrGreg]

Sam,

You must remember that each row in my tables does not show two things happening at the same time. Each row represents a ray of light traveling from Bob to Alice in one table, or from Alice to Bob in the other table. There is always a delay between light being sent and being received.

First, where was the 20 minutes delay? When Bob saw 40 on Alice's clock, Alice has traveled 60 minutes already due to the 20 minutes delay.

The “20 minute delay” is for light sent from Bob at (a) to Alice at (m). There is also a “20 minute delay” for light sent from Alice at (i) (not (m)) to Bob at (a). When Bob sees 40 on Alice’s clock, Bob is seeing Alice when she was at (i). As Alice traveled from (i) to (m) she was still traveling away from Bob, so when Bob sees Alice traveling from (i) to (m) the delay is still increasing and the Doppler rate of 2/3 still applies. The fact that Bob’s own clock has gone past 60 is irrelevant – he is still seeing light which left Alice before her clock reached 60 so the delay from Alice to Bob is still increasing, not decreasing.

Secondly, why the equal row disappeared. Your table on post #86 had an equal row at (e), 13:40 = 13:40, and mine had an equal row 100=100. This new table above of yours had no equal row.

This is because these are different tables. The table with 13:40(e)…13:40(a) that you quoted was Alice’s view of Bob. The table in my last message was Bob’s view of Alice.

It is wrong to say that the event of Alice’s clock showing 13:40 occurs “at the same time” as the event of Bob’s clock showing 13:40. The table with 13:40(e)…13:40(a) shows that light leaving (a) at 13:40 on Bob’s clock arrives at (e) at 13:40 on Alice’s clock. If these two events occurred at the same time, light would have traveled instantly from (a) to (e) at infinite speed. This is impossible.

In Relativity, the concept of “at the same time” is a relative concept, which is determined by a convention. Using that convention, two different observers can disagree whether two separated events occur “at the same time”.

However, in my argument, I make no assumptions about simultaneity. Clocks are compared only by sending light from one clock to another, which takes time.

It might avoid confusion if I redesign the experiment so that Alice and Bob do not synchronize their clocks at the start. Let’s say that we start with Bob’s clock showing 12:00 but Alice’s clock showing 15:00.

The two tables now look like this.


Alice’s view of Bob

Bob...abcdefghijklm.[/color]Alice[/color]
12:00 a[/color]a ...15:00[/color]
12:10 a[/color]>>>d ...15:15[/color]
12:20 a[/color]>>>>>>g ...15:30[/color]
12:30 a[/color]>>>>>>>>>j ...15:45[/color]
12:40 a[/color]>>>>>>>>>>>>m 16:00[/color]
12:40 a[/color]>>>>>>>>>>>>m 16:00[/color]
12:55 a[/color]>>>>>>>>>>k ..16:10[/color]
13:10 a[/color]>>>>>>>>i ...16:20[/color]
13:25 a[/color]>>>>>>g ...16:30[/color]
13:40 a[/color]>>>>h ...16:40[/color]
13:55 a[/color]>>c ...16:50[/color]
14:10 a[/color]a ...17:00[/color]
Bob...abcdefghijklm.[/color]Alice[/color]


Bob’s view of Alice

Bob...abcdefghijklm.[/color]Alice[/color]
12:00 a[/color]a ...15:00[/color]
12:15 a[/color]<<c ...15:10[/color]
12:30 a[/color]<<<<e ...15:20[/color]
12:45 a[/color]<<<<<<g ...15:30[/color]
13:00 a[/color]<<<<<<<<i ...15:40[/color]
13:15 a[/color]<<<<<<<<<<k ..15:50[/color]
13:30 a[/color]<<<<<<<<<<<<m 16:00[/color]
13:30 a[/color]<<<<<<<<<<<<m 16:00[/color]
13:40 a[/color]<<<<<<<<<j ...16:15[/color]
13:50 a[/color]<<<<<<g ...16:30[/color]
14:00 a[/color]<<<d ...16:45[/color]
14:10 a[/color]a ...17:00[/color]
Bob...abcdefghijklm.[/color]Alice[/color]


The only times that we can sensibly compare Bob’s clock directly with Alice’s clock is when Bob and Alice are at the same place (a). At the beginning there is a difference of 3 hours but at the end the difference is 3 hours 10 minutes. We cannot say what the difference is in between.

 

[Sam Woole]

Sam,
You must remember that each row in my tables does not show two things happening at the same time. Each row represents a ray of light traveling from Bob to Alice in one table, or from Alice to Bob in the other table. There is always a delay between light being sent and being received.
The “20 minute delay” is for light sent from Bob at (a) to Alice at (m). There is also a “20 minute delay” for light sent from Alice at (i) (not (m)) to Bob at (a). When Bob sees 40 on Alice’s clock, Bob is seeing Alice when she was at (i). As Alice traveled from (i) to (m) she was still traveling away from Bob, so when Bob sees Alice traveling from (i) to (m) the delay is still increasing and the Doppler rate of 2/3 still applies. The fact that Bob’s own clock has gone past 60 is irrelevant – he is still seeing light which left Alice before her clock reached 60 so the delay from Alice to Bob is still increasing, not decreasing.
This is because these are different tables. The table with 13:40(e)…13:40(a) that you quoted was Alice’s view of Bob. The table in my last message was Bob’s view of Alice.
It is wrong to say that the event of Alice’s clock showing 13:40 occurs “at the same time” as the event of Bob’s clock showing 13:40. The table with 13:40(e)…13:40(a) shows that light leaving (a) at 13:40 on Bob’s clock arrives at (e) at 13:40 on Alice’s clock. If these two events occurred at the same time, light would have traveled instantly from (a) to (e) at infinite speed. This is impossible.
In Relativity, the concept of “at the same time” is a relative concept, which is determined by a convention. Using that convention, two different observers can disagree whether two separated events occur “at the same time”.
However, in my argument, I make no assumptions about simultaneity. Clocks are compared only by sending light from one clock to another, which takes time.
It might avoid confusion if I redesign the experiment so that Alice and Bob do not synchronize their clocks at the start. Let’s say that we start with Bob’s clock showing 12:00 but Alice’s clock showing 15:00.
The two tables now look like this.
Alice’s view of Bob
Bob...abcdefghijklm.[/color]Alice[/color]
12:00 a[/color]a ...15:00[/color]
12:10 a[/color]>>>d ...15:15[/color]
12:20 a[/color]>>>>>>g ...15:30[/color]
12:30 a[/color]>>>>>>>>>j ...15:45[/color]
12:40 a[/color]>>>>>>>>>>>>m 16:00[/color]
12:40 a[/color]>>>>>>>>>>>>m 16:00[/color]
12:55 a[/color]>>>>>>>>>>k ..16:10[/color]
13:10 a[/color]>>>>>>>>i ...16:20[/color]
13:25 a[/color]>>>>>>g ...16:30[/color]
13:40 a[/color]>>>>h ...16:40[/color]
13:55 a[/color]>>c ...16:50[/color]
14:10 a[/color]a ...17:00[/color]
Bob...abcdefghijklm.[/color]Alice[/color]
Bob’s view of Alice
Bob...abcdefghijklm.[/color]Alice[/color]
12:00 a[/color]a ...15:00[/color]
12:15 a[/color]<<c ...15:10[/color]
12:30 a[/color]<<<<e ...15:20[/color]
12:45 a[/color]<<<<<<g ...15:30[/color]
13:00 a[/color]<<<<<<<<i ...15:40[/color]
13:15 a[/color]<<<<<<<<<<k ..15:50[/color]
13:30 a[/color]<<<<<<<<<<<<m 16:00[/color]
13:30 a[/color]<<<<<<<<<<<<m 16:00[/color]
13:40 a[/color]<<<<<<<<<j ...16:15[/color]
13:50 a[/color]<<<<<<g ...16:30[/color]
14:00 a[/color]<<<d ...16:45[/color]
14:10 a[/color]a ...17:00[/color]
Bob...abcdefghijklm.[/color]Alice[/color]
The only times that we can sensibly compare Bob’s clock directly with Alice’s clock is when Bob and Alice are at the same place (a). At the beginning there is a difference of 3 hours but at the end the difference is 3 hours 10 minutes. We cannot say what the difference is in between.

I do not think your tables are honest. Your first table showed the turn around took place when the moving clock has accumulated 60 minutes, 60 vs 40 (40 was the illusion in the telescope. Actual number was 60 due to 20 minutes delay). Your second table showed the turn around took place when the moving clock has accumulated 90 minutes, 90 vs 60 (60 is the illusion in the telescope. Actual number is 80 due to 20 minutes delay.). This was the point I could not understand. I did not find any solution to it in your post.

My understanding of your original tables was, it was a two hour return journey, one hour each way. I do not understand how could the turn around take place when one clock has accumulated 90 minutes.

I did not challenge the truthfulness of the equal row, nor the definition of simultaneity. What I did challenge was your consistency. You changed rules of the game by prolonging the one way journey to 90 minutes which made the equal row disappear, and as a result the turn around took place 30 minutes later. To be consistent, we should have kept that equal row, we should have the turn around to take place when the moving clock has accumulated 60 minutes, meaning the journey's one end.

Although you changed the time on Alice clock to be 15:00 hours at start, but the equal row was still there in the first table above, 13:40 and 16:40, accumulation of 100 minutes on each clock, or 1:40 hours, meaning light is instantaneous. How could it be so?

I had an inkling that people cannot make the time dilation idea stand unless they contradict themselves, such as the equal row. I believe it was all an illusion. You said so in the beginning; you knew it was.

 

[JesseM]

IYour second table showed the turn around took place when the moving clock has accumulated 90 minutes, 90 vs 60 (60 is the illusion in the telescope.

You're misunderstanding the second table. The second table shows that Alice's clock read 16:00 at the time of the turnaround, but the light from that event didn't reach Bob until his clock read 13:30. The first table, on the other hand, says that Alice's clock read 16:00 at the time of the turnaround, and that at that moment she was seeing light from Bob's clock which read 12:40. So in both tables it's true that the turnaround happened at 16:00 according to Alice's clock, so both tables say the moving clock accumulated 60 minutes.

 

[DrGreg]

I do not think your tables are honest. Your first table showed the turn around took place when the moving clock has accumulated 60 minutes, 60 vs 40 (40 was the illusion in the telescope. Actual number was 60 due to 20 minutes delay). Your second table showed the turn around took place when the moving clock has accumulated 90 minutes, 90 vs 60 (60 is the illusion in the telescope. Actual number is 80 due to 20 minutes delay.). This was the point I could not understand. I did not find any solution to it in your post.

The turn around occurs when Alice reaches (m), which is at 16:00 according to her own clock. This is something everybody agrees upon. Alice knows it because she is there and looks at her own clock when it happens. Bob does not know that it has happened immediately but has to wait until light has traveled from (m) to (a). When this has happened, Bob can see that Alice has turned round and he can see that her clock was showing 16:00 when it happened.

If you accept that the Doppler factor of 2/3 applies to both Bob's view of Alice and Alice's view of Bob, then you have to accept that the red rows of both tables are correct.

My understanding of your original tables was, it was a two hour return journey, one hour each way.

...according to Alice.

I do not understand how could the turn around take place when one clock has accumulated 90 minutes.

It doesn't. Bob sees the turn around when his clock shows 13:30, but the turn around has already taken place some time earlier. It takes time for light to travel from Alice at (m) to Bob at (a).

I did not challenge the truthfulness of the equal row, nor the definition of simultaneity. What I did challenge was your consistency. You changed rules of the game by prolonging the one way journey to 90 minutes which made the equal row disappear, and as a result the turn around took place 30 minutes later.

Alice's journey appears to Bob to be 90 minutes because of the delay in light traveling from Alice to Bob. The 2/3 factor applies: a journey of 60 minutes by Alice's clock looks to Bob like a journey of 60 / (2/3) = 90 minutes. Alice's journey appears to Alice to be 60 minutes because she sees no delay on her own clock. The return trip journey is not symmetrical -- it is Alice who turns round, not Bob. The event of her turning round is experienced directly by Alice but can only be observed remotely, after a delay, by Bob.

Although you changed the time on Alice clock to be 15:00 hours at start, but the equal row was still there in the first table above, 13:40 and 16:40, accumulation of 100 minutes on each clock, or 1:40 hours, meaning light is instantaneous. How could it be so?

I answered that in my last post. Once clocks have been separated you cannot apply the time on one clock to events that occur at the position of the other clock.

Think of a clock like the trip-meter on a car. If you zero the trip-meters on two cars, and then the two cars follow each other along the same road, at the end both trip-meters show the same mileage. But if the two cars follow different routes, when they later meet, the mileage may be different.

For example suppose you and I meet in New York and zero the trip-meters on our cars. You drive directly to Los Angeles. I drive to New Orleans first, then I drive on to Los Angeles. When you and I meet again in Los Angeles, I have driven further than you. Our trip-meters show different numbers even though we are both in the same place.

Clocks behave the same way. If two clocks follow two different routes through space-time, they may show different accumulated times when they are brought back together again.

I had an inkling that people cannot make the time dilation idea stand unless they contradict themselves, such as the equal row. I believe it was all an illusion. You said so in the beginning; you knew it was.

This is what I said in post #68:

From Alice’s point of view, it looks as though Bob’s clock is ticking at 2/3 of the rate of her own clock. Of course, this is an illusion – it is nothing more or less than the Doppler effect, caused by the delay of the light signals.

The illusion would be if Alice believed that Bob's clock really was ticking at 2/3 the rate. It is still true that what she sees (after a delay) is Bob's clock's image which ticks at 2/3 the rate of the image of her own clock.

 

[JesseM]

Alice's journey appears to Bob to be 90 minutes because of the delay in light traveling from Alice to Bob.

Just so Sam doesn't get confused here and think it's all an illusion due to light-signal delays, it should be pointed out that that's not the only reason it appears to be 90 minutes--there is also genuine time dilation of Alice's clock happening in Bob's frame, the relativistic doppler shift equation takes this into account (that's why it's different from the non-relativistic doppler shift equation). DrGreg, what was the relative velocity between Alice and Bob in this problem?

 

[jtbell]

what was the relative velocity between Alice and Bob in this problem?

Inverting the relativistic Dopper effect equation gives me
\frac{v}{c} = \frac {1-R^2}{1+R^2}
where R = 2/3, so v = 0.3846c, or \gamma = 1.0833.

 

[Sam Woole]

You're misunderstanding the second table. The second table shows that Alice's clock read 16:00 at the time of the turnaround, but the light from that event didn't reach Bob until his clock read 13:30. The first table, on the other hand, says that Alice's clock read 16:00 at the time of the turnaround, and that at that moment she was seeing light from Bob's clock which read 12:40. So in both tables it's true that the turnaround happened at 16:00 according to Alice's clock, so both tables say the moving clock accumulated 60 minutes.

No, I was not misunderstanding the second table.
Let me quote your words: "The first table, on the other hand, says that Alice's clock read 16:00 at the time of the turnaround, and that at that moment she was seeing light from Bob's clock which read 12:40." Yes, but at this juncture what would Bob see on Alice's clock? Of course he would not see 16:00, but 15:40 on Alice's clock (16:00 minus 20 minutes delay). I think this positively means that, the turnaround took place when Bob saw 15:40 on Alice's clock. Therefore from my viewpoint, the second table should be revised accordingly.

 

[JesseM]

No, I was not misunderstanding the second table.
Let me quote your words: "The first table, on the other hand, says that Alice's clock read 16:00 at the time of the turnaround, and that at that moment she was seeing light from Bob's clock which read 12:40." Yes, but at this juncture what would Bob see on Alice's clock?

What do you mean by "at this juncture"? Do you mean, "at the same time Alice's clock read 16:00"? The problem is, again, that different frames define simultaneity differently, so they will have different answers to what Bob's clock read at the same time that Alice's read 16:00. Since \gamma = 1.0833 in this example, then in Alice's frame Bob's clock is running slower than hers by a factor of 1.0833, so after her clock has accumulated 60 minutes, Bob's clock has only accumulated 60/1.0833 = 55.386 minutes, meaning that his clock reads 12:55.386 at that moment. In Bob's frame, it is Alice's clock that is running slow by a factor of 1.0833, so when her clock has accumulated 60 minutes, his clock has accumulated 60*1.0833 = 64.998 minutes, meaning in his frame, his clock reads 13:04.998 minutes at the same moment her clock has accumulated 60 minutes.

Of course he would not see 16:00, but 15:40 on Alice's clock (16:00 minus 20 minutes delay).

Delay times don't change the readings on a clock right next to an event! If Alice pushes the button to turn around, and a clock on her dashboard right next to that button reads 16:00, then all observers, no matter how far away, will see her clock reading 16:00 when she pushes the button when they look through their telescope. Imagine that Alice wrote down the time on a piece of paper after pushing the button--do you imagine that different observers would see her write down different numbers depending on the delay time?

Maybe you're not talking about what time Bob would see on Alice's clock at the same moment he saw her turn around, but what time Bob would see on Alice's clock at the actual moment she was turning around, even though the light from this event would take a while to reach him. If so, again, asking what Bob was seeing "at the same moment" Alice was turning around depends on your reference frame. In Alice's frame, this question would be equivalent to "what was Bob seeing on Alice's clock at the moment his clock read 12:55.386?" but in Bob's frame this question would be equivalent to "what was Bob seeing on Alice's clock at the moment his clock read 13:04.998?" And it's not true in either case that Bob would simply see Alice's clock read 20 minutes earlier than 16:00. I can only assume you're basing that on the fact that Alice saw Bob's clock read 12:40 at the moment her clock read 16:00, but as I said in my earlier post to DrGreg, the relativistic Doppler shift equation is not based only on light-speed delays, it's also based on time dilation. In fact, when Bob's clock reads 12:40, he is not at a distance of 20 light-minutes from Alice, in either frame. Since their relative velocity is 0.3846c, then after 40 minutes have passed in his frame, they should be a distance of 40*0.3846 = 15.384 light-minutes apart in Bob's frame. And since she is continuing to move away at 0.3846c, the light will take longer than 15.384 minutes to catch up with her in his frame--you can find the time by solving c*t = 0.3846c*t + 15.384, which gives t = 24.998 minutes for the light to reach her in his own frame. So, the light will reach her 40 + 24.998 = 64.998 minutes after they departed in his frame. But since her clock is only ticking at 1/1.0833 the normal rate in his frame, her clock will only have elapsed 64.998/1.0833 = 60 minutes when the light from this event reaches her.

If you look at it from the point of view of her frame, it's also not true that the light took 20 minutes to reach her. From her point of view, it was Bob's clock that was ticking at 1/1.083333 the normal rate, so when his clock had elapsed 40 minutes, hers had elapsed 40*1.083333 = 43.33332 minutes. Since he was moving away at 0.384615c in her frame, at this moment he would be at a distance of 43.33332*0.384615c = 16.66664 light-minutes away. And since she's at rest in her own frame, the light will take 16.66664 minutes to reach her, so it'll reach her after 43.33332 + 16.66664 = 59.99996 minutes have elapsed on her own clock (it would be exactly 60 if I hadn't rounded off the numbers).

So, in both frames you conclude that the light from Bob's clock reading 12:40 reached Alice when her clock read 16:00, but in neither frame did the light take 20 minutes to travel between Bob and Alice.

 

[Sam Woole]

The turn around occurs when Alice reaches (m), which is at 16:00 according to her own clock. This is something everybody agrees upon. Alice knows it because she is there and looks at her own clock when it happens. Bob does not know that it has happened immediately but has to wait until light has traveled from (m) to (a). When this has happened, Bob can see that Alice has turned round and he can see that her clock was showing 16:00 when it happened..

Now I think I have got it. Here you pointed out that Bob could not apply the 3/2 rate until light from Alice has reached him, that would be 13:30 Bob's time. JesseM also pointed this out to me but I did not take it. I am sorry to both of you. Before now I have insisted that the 3/2 rate be applied as soon as Alice's turnaround took place. I admit I was wrong. Before I accept your second table as correct, I still have one more question. For Bob to see 15:50 (second table) so that he may apply the 2/3 rate, it meant to me that light from Alice after her turnaround has already arrived at Bob, before 16:00. Would you explain why this 15:50 does not mean so?

This is what I said in post #68:
The illusion would be if Alice believed that Bob's clock really was ticking at 2/3 the rate. It is still true that what she sees (after a delay) is Bob's clock's image which ticks at 2/3 the rate of the image of her own clock.

So far I must say thanks again for your patience toward a slow student like me. Here comes my final difficulty. My understanding of your comments about illusion and delay was that, when Alice saw 15 minutes on her clock, Bob also saw 15 minutes on his clock, the same accumulation on both clocks. My understanding must be right because you also gave number 5 as the first delay and double (10) for the second delay, etc. Such a sequence of events means to me, the two clocks will accumulate identical numbers in the end. When the tables showed otherwise (10 minutes more), it was only an illusion created by mathematical works; it was not the true picture. But it was claimed in Relativity to be the true picture, why? It is as if mathematics will remotely affect the workings of mechanical devices and even life forms, as some believe so. This remoteness implies instantaneity. I think it is really hard to believe so. When one twin applies the 3/2 rate many light minutes (or years) away from earth, his brother's clock or body flesh on Earth is instantaneously affected. Is this possible?

 

[JesseM]

Now I think I have got it. Here you pointed out that Bob could not apply the 3/2 rate until light from Alice has reached him, that would be 13:30 Bob's time. JesseM also pointed this out to me but I did not take it. I am sorry to both of you. Before now I have insisted that the 3/2 rate be applied as soon as Alice's turnaround took place. I admit I was wrong. Before I accept your second table as correct, I still have one more question. For Bob to see 15:50 (second table) so that he may apply the 2/3 rate, it meant to me that light from Alice after her turnaround has already arrived at Bob, before 16:00.

Why do you say that? He sees Alice's clock reading 15:50 before he sees her clock read 16:00, which is when she turns around. 15:50 is ten minutes before 16:00, no? And he sees her clock read 15:50 at 13:15 on his clock, which is 75 minutes after she departed at 12:00 according to his clock. 2/3 of 75 is 50, and 15:50 is 50 minutes after she departed at 15:00 according to her own clock.

My understanding of your comments about illusion and delay was that, when Alice saw 15 minutes on her clock, Bob also saw 15 minutes on his clock, the same accumulation on both clocks.

No, DrGreg specifically pointed out that there is no objective way to compare what two clocks read "at the same time" if they are at different locations in space:

In Relativity, the concept of “at the same time” is a relative concept, which is determined by a convention. Using that convention, two different observers can disagree whether two separated events occur “at the same time”.

Once clocks have been separated you cannot apply the time on one clock to events that occur at the position of the other clock.

In Bob's frame, when his own clock has accumulated 15 minutes, Alice's clock has only accumulated 15/1.0833 = 13.847 minutes. Likewise, in Alice's frame during the outbound phase of the trip (note that Alice switches frames when she turns around), when her clock has accumulated 15 minutes, Bob's has only accumulated 13.847 minutes.

But it was claimed in Relativity to be the true picture, why? It is as if mathematics will remotely affect the workings of mechanical devices and even life forms, as some believe so. This remoteness implies instantaneity. I think it is really hard to believe so. When one twin applies the 3/2 rate many light minutes (or years) away from earth, his brother's clock or body flesh on Earth is instantaneously affected. Is this possible?

I like to think of it in geometric terms. Imagine drawing two dots on a piece of paper, and then drawing two paths between them, one a straight line and the other with a bend in it. No matter what coordinate system you use to calculate the lengths of the paths, you will get the same answer for the two lengths, and you will also always find that the straight path is shorter than the non-straight one. Does this mean one path is instantaneously affecting another? No it's just geometry. Similarly, Alice and Bob take two different paths through spacetime between two events (the event of them departing from a single location in space, and the event of them reuiniting at a single location in space), and you can use different space-time coordinate systems to calculate the time elapsed on each path, but all coordinate systems will give the same answer, and you'll always find that the straight path between the points has a shorter amount of time than paths involving changes in direction/speed.

 

[DrGreg]

DrGreg, what was the relative velocity between Alice and Bob in this problem?

Inverting the relativistic Dopper effect equation gives me
\frac{v}{c} = \frac {1-R^2}{1+R^2}
where R = 2/3, so v = 0.3846c, or \gamma = 1.0833.

Correct.

For the benefit of jtbell and JesseM, I'd like to say that in my presentation to Sam Woole I have tried to avoid quoting any formulas or other results from relativity theory. So I've never invoked any definition of (relative) simultaneity, I haven't discussed time dilation (between frames) or length contraction (between frames), I haven't used any Lorentz transformations or Doppler equations. I wanted to keep the argument as simple as possible and avoid quoting any result that Sam might have difficulty accepting. That's also why I gave an explicit numerical example rather than an algebraic proof.

I believe that in my argument I have used only two assumptions:
- that Alice’s speed relative to Bob is the same as Bob’s speed relative to Alice
- that Alice’s Doppler shift relative to Bob is the same as Bob’s Doppler shift relative to Alice.
Everything else is a logical consequence of those assumptions. (And, in fact, with hindsight, I think I could reword the argument to avoid the first assumption. But that would have complicated the presentation.)


In fact you can deduce the speed from the tables that I quoted in post #103:


12:40 a[/color]>>>>>>>>>>>>m 16:00[/color]
13:30 a[/color]<<<<<<<<<<<<m 16:00[/color]

This shows that it takes light 50 minutes to make the round trip from (a) (Bob) to (m) (Alice's furthest position) and back again, so Alice's total distance traveled is 50 light-minutes, according to Bob. And we also know that her total journey takes 130 minutes according to Bob. So her constant speed must be (50 light-minutes) / (130 minutes) = ⁵/₁₃ c, relative to Bob.

And \gamma (which is the same for both halves of the journey) is the total journey time according to Bob divided by the the total journey time according to Alice, 130/120 = 13/12.

 

[DrGreg]

Sam,

I think your questions to me in post #111 have been answered by Jesse in post #112, but if you don't think so, please ask again.

 

[jtbell]

I've never invoked any definition of (relative) simultaneity, I haven't discussed time dilation (between frames) or length contraction (between frames), I haven't used any Lorentz transformations or Doppler equations.

Of course, the assumption that the frequency ratios that Bob and Alice observe are reciprocals (3/2 and 2/3) means that you've implicitly assumed the relativistic Doppler effect. In the non-relativistic Doppler effect, one of them would use the ratio 1 + v/c = 18/13 and the other one, 1 - v/c = 8/13, which are obviously not reciprocals of each other. As their relative speed v decreases (becomes less relativistic), the non-relativistic ratios become closer and closer to the relativistic ones, and more and more nearly reciprocals of each other.

 

[Sam Woole]

Why do you say that? He sees Alice's clock reading 15:50 before he sees her clock read 16:00, which is when she turns around. 15:50 is ten minutes before 16:00, no? And he sees her clock read 15:50 at 13:15 on his clock, which is 75 minutes after she departed at 12:00 according to his clock. 2/3 of 75 is 50, and 15:50 is 50 minutes after she departed at 15:00 according to her own clock.

I am sorry again. I realized my mistake a few minutes after I shut down my computer. You are right.

No, DrGreg specifically pointed out that there is no objective way to compare what two clocks read "at the same time" if they are at different locations in space: In Bob's frame, when his own clock has accumulated 15 minutes, Alice's clock has only accumulated 15/1.0833 = 13.847 minutes. Likewise, in Alice's frame during the outbound phase of the trip (note that Alice switches frames when she turns around), when her clock has accumulated 15 minutes, Bob's has only accumulated 13.847 minutes.
I like to think of it in geometric terms. Imagine drawing two dots on a piece of paper, and then drawing two paths between them, one a straight line and the other with a bend in it. No matter what coordinate system you use to calculate the lengths of the paths, you will get the same answer for the two lengths, and you will also always find that the straight path is shorter than the non-straight one. Does this mean one path is instantaneously affecting another? No it's just geometry. Similarly, Alice and Bob take two different paths through spacetime between two events (the event of them departing from a single location in space, and the event of them reuiniting at a single location in space), and you can use different space-time coordinate systems to calculate the time elapsed on each path, but all coordinate systems will give the same answer, and you'll always find that the straight path between the points has a shorter amount of time than paths involving changes in direction/speed.

On this subject, I think I have not moved over to your point of view. When I mentioned readings on clocks, I was referring to a mechanical device. When you gave numbers such as 15/1.0833 = 13.847, and the geometry, you were referring to math, which I agree totally. But when the math shows 13.847 minutes, will the clock show 13.847 minutes. This was my difficulty, or my inability to believe.

As to whether there is an objective way to compare two clocks, I believe DrGreg's demonstration itself was the objective way. It showed, when Alice saw her own clock to read 15 minutes, she saw 10 minutes on Bob's clock thru the telescope. DrGreg pointed out, there was a 5 minute delay. This is to say, Bob's clock did accumulate 15otherwise Alice could not have seen 10. If Bob looked into his telescope, of course he would find the same number 10 on Alice' clock, from which Bob concludes that Alice's clock has accumulated the same number of minutes as his, 15. To me, this should be the objective way to compare clocks.

Did you see my difficulty here? Your math produced 13.847and 15 in one reference frame, while my mechanical method did 15 for both frames. Will the 13.847 produced in Alice's frame influence the workings of a clock in Bob's frame, remotely?

 

[DrGreg]

Of course, the assumption that the frequency ratios that Bob and Alice observe are reciprocals (3/2 and 2/3) means that you've implicitly assumed the relativistic Doppler effect. In the non-relativistic Doppler effect, one of them would use the ratio 1 + v/c = 18/13 and the other one, 1 - v/c = 8/13, which are obviously not reciprocals of each other. As their relative speed v decreases (becomes less relativistic), the non-relativistic ratios become closer and closer to the relativistic ones, and more and more nearly reciprocals of each other.

For what it's worth, long ago, in my original post https://www.physicsforums.com/showpost.php?p=783491&postcount=68", I gave an argument to show that the blue and red shifts must be reciprocal (assuming that shifts depend only on relative velocity).

Note I've now added another paragraph in post #113.

 

[DrGreg]

Would you explain why this 15:50 does not mean so?

I think JesseM has answered this in post #112.

(By the way, I don’t think JesseM’s post #110 will help you very much. Everything he says is correct but he is quoting some results from Relativity theory that you are probably not familiar with.)

When one twin applies the 3/2 rate many light minutes (or years) away from earth, his brother’s clock or body flesh on Earth is instantaneously affected. Is this possible?

No it isn’t. When Alice starts applying the 3/2 rate, it makes no difference to any time that has already accumulated on her own clock, or to any time that she has already seen accumulate on her image of Bob’s clock. What changes is the relative rate of accumulation after that point. For every extra 10 minutes added to her clock, she sees an extra 15 minutes added to her image of Bob’s clock. This change is due to her change of motion relative to Bob. Nothing has happened to Bob.

You are still thinking in terms of simultaneity, which is confusing you.

Let me give an analogy.

Suppose you and I stand next to each other. You walk 100 metres north. I walk 100 metres northeast. We have both walked 100 metres forward but we are not in the same place.

From your point of view, you are 100m north of the start, but I am only 70.7m north of the start. You have walked further forward (north) than me.

From my point of view, I am 100m northeast of the start, but you are only 70.7m northeast of the start. I have walked further forward (northeast) than you.

Now I turn and walk 100m northwest. You continue to walk 41.4m north, until we both meet again. The point where we meet is 141.4m north of the start.

You have now walked 141.4m (north). I have walked a total of 200m (100m NE + 100m NW). We have walked different distances but are at the same place.

In this analogy, “distance walked forwards” is the analogue of time. We each have our own distance that we have walked. There is no “absolute distance” that can be applied to everybody. 100m N is not the same as 100m NE. If you specify that two places are 100m apart, that does not specify their location – you have to specify a direction as well.

The same goes for time. You cannot say that two events occur 10 minutes apart – you have to specify a “direction in spacetime”, which means you have to specify the motion of the clock that will measure the time. Different clocks (moving at different speeds) will measure different times, in the same way that in the first half of my analogy we disagreed over who had walked “forward” the furthest. (I thought I had walked 29.3m further forward than you. You thought you had walked 29.3m further forward than me.)

In the analogy, the person who changed direction (me) walked further, in total, than the other. Spacetime is slightly different – the person who changes direction (Alice) takes less time, in total, than the other.

 

[DrGreg]

As to whether there is an objective way to compare two clocks, I believe DrGreg's demonstration itself was the objective way. It showed, when Alice saw her own clock to read 15 minutes, she saw 10 minutes on Bob's clock thru the telescope. DrGreg pointed out, there was a 5 minute delay.

This apparent "delay" is a comparison between a time on Alice's clock and another time on Bob's clock. Times on different clocks cannot be directly compared. There is no absolute time, only time relative to an observer.

This is to say, Bob's clock did accumulate 15 otherwise Alice could not have seen 10.

But you cannot say that the moment when Bob's clock has accumulated 15 minutes occurs "at the same time as" the moment when Alice's clock has accumulated 15 minutes. These events cannot be compared absolutely and different observers will disagree which of these events occurred first.

If Bob looked into his telescope, of course he would find the same number 10 on Alice' clock, from which Bob concludes that Alice's clock has accumulated the same number of minutes as his, 15. To me, this should be the objective way to compare clocks.

There is no "objective way" that everyone can agree on. In the same way that, in my analogy in post #118, there is no "objective way" to say who has traveled further forward, because everyone disagrees about which direction is "forward".

 

[JesseM]

For what it's worth, long ago, in my original post https://www.physicsforums.com/showpost.php?p=783491&postcount=68", I gave an argument to show that the blue and red shifts must be reciprocal (assuming that shifts depend only on relative velocity).

Which section of post #68 are you referring to? I may be thinking about this wrong, but it doesn't make sense to me that assuming the shifts depend only on relative velocity is enough to conclude the shifts must be reciprocal...imagine that 2 observers in a Newtonian universe are shooting pellets at each other at a rate of 1 pellet per second, and that both shoot the pellets at the same constant velocity in their rest frame. Wouldn't it be true that the frequency of incoming pellets depends only on the two observers' relative velocity, yet the frequency of incoming pellets when they are moving apart at velocity v is not the reciprocal of the frequency when they are moving towards each other at velocity v?