1 tjrob137 | Most general case of the Twin Paradox | Wednesday 25 September 2019 |

2 tjrob137 | Re :Most general case of the Twin Paradox | Friday 27 September 2019 |

3 tjrob137 | Re :Most general case of the Twin Paradox | Friday 27 September 2019 |

4 tjrob137 | Re :Most general case of the Twin Paradox | Friday 27 September 2019 |

5 tjrob137 | Re :Most general case of the Twin Paradox | Saturday 28 September 2019 |

6 tjrob137 | Re :Most general case of the Twin Paradox | Saturday 28 September 2019 |

7 tjrob137 | Re :Most general case of the Twin Paradox | Monday 30 September 2019 |

8 tjrob137 | Re :Most general case of the Twin Paradox | Monday 30 September 2019 |

9 tjrob137 | Re :Most general case of the Twin Paradox | Tuesday 1 October 2019 |

10 tjrob137 | Re :Most general case of the Twin Paradox | Wednesday 2 October 2019 |

11 tjrob137 | Re :Most general case of the Twin Paradox | Wednesday 2 October 2019 |

12 Edgar L. Owen | Re :Most general case of the Twin Paradox | Thursday 3 October 2019 |

13 Paul B. Andersen | Re :Most general case of the Twin Paradox | Thursday 3 October 2019 |

14 tjrob137 | Re :Most general case of the Twin Paradox | Friday 4 October 2019 |

15 tjrob137 | Re :Most general case of the Twin Paradox | Monday 7 October 2019 |

Most general case of the Twin Paradox

210 posts by 21 authors

https://groups.google.com/g/sci.physics.relativity/c/ovXhGHJ5t78/m/8GXMOQ8GBQAJ

keywords = SR, Moving clocks

On 9/23/19 5:39 PM, Edgar L. Owen wrote:

> |
Disregarding gravitation:
Twins A and B separate from any possible point in the universe, each take any separate path, then meet again at any possible location. In general when they meet their clocks will read different elapsed proper times. Is it true that in all possible cases each twin's elapsed proper time will be equal to the sum of its Lorentz time dilation due to its velocity RELATIVE to the eventual meeting point? Where the Lorentz time dilation is of their own proper time RELATIVE to a clock at the eventual meeting point (i.e. the coordinate time of their clock as measured relative to a clock at the eventual meeting point)? |

This is true IF AND ONLY IF there exists an inertial frame in which the clock at the eventual meeting point is always at rest, and that frame is used to calculate all of the "time dilations".

Speed "relative to a point" is meaningless. Speed can only be measured relative to some coordinate system, and is only simple if that coordinate system is an inertial frame. So it makes no sense for a point to be at rest in an inertial frame. But it does make sense for a clock. So the "time dilations" are relative to that inertial frame of the clock, not the eventual meeting point.

Tom Roberts

On 9/25/19 11:33 AM, Edgar L. Owen wrote:

> | 'Velocity relative to a point' is most certainly meaningful. |

You OBVIOUSLY do not understand that those words mean.

Velocity can only be measured relative to a coordinate system, and a single point does not define one. Even an object is not sufficient -- it is much more than a point (a worldline, trajectory, or path through spacetime), but also does not define a coordinate system.

Tom Roberts

On 9/23/19 10:02 PM, Edgar L. Owen wrote:

> | OK, now here's what I don't understand. To me it seems that the different elapsed proper times when the twins compare clocks upon meeting is clearly something actual since both twins agree on it. |

Yes. "actual" = "real".

> | Both proper time rates are now the same. The elapsed disparity is not due to coordinate times. |

I have no idea what you are trying to say here.

> | That seems in clear contradiction to Tom and rotchm claiming 'clocks always run at the same rate'. |

It is not -- it's quite clear that YOU do not understand this.

The clocks traveled DIFFERENT paths through spacetime. Different paths can have different path lengths. For timelike paths, such as those of clocks, the path length is elapsed proper time.

So the reunited clocks show different elapsed proper times BECAUSE THEY TRAVELED DIFFERENT PATHS THROUGH SPACETIME. Not because they "tick at different rates".

- If clocks did tick at different rates, that would mean that the laws of physics governing their ticking were different -- that would violate the first postulate. No such violation has ever been observed.

In terms of some specific inertial frame, the moving clock does indeed tick slower than an identical clock at rest in the frame. But the phrase "a clock's tick rate" IS NOT THAT -- the phrase makes no mention of any frame, and is discussing only the clock ITSELF; for that, the first postulate applies and the clock MUST tick at its usual rate.

- To measure a clock's tick rate you MUST use its (local) rest frame. Using any other frame cannot possibly get the tick rate of the clock, it gets "the tick rate of that clock measured in this frame", which is NOT THE SAME AT ALL.

> | For the two clocks to show different elapsed times they can only have been running at different proper time rates during their separation. |

Not true. They might have traveled different paths through spacetime that have different path lengths (= elapsed proper time).

> | [... more nonsense displaying colossal ignorance of physics] |

You seem to be implicitly using one particular frame to discuss everything. That's OK, but you MUST recognize that such a procedure does NOT give "the moving clock's tick rate", it gives THE MOVING CLOCK'S TICK RATE MEASURED IN THAT FRAME.

Tom Roberts

On 9/24/19 3:08 AM, Engr. Ravi wrote:

> | For the two clocks to show different elapsed times they can only have been running at different proper time rates during their separation. |

You keep claiming this, even after many discussions about what you are missing in the assumptions you implicitly make AND NEVER ACKNOWLEDGE. That implies your imagination is FAR too narrow to understand very basic physics.

Find another hobby more suited to your abilities.

Tom Roberts

On 9/27/19 4:03 PM, Edgar L. Owen wrote:

> | When you say the two clocks travel different paths through ST of course, but one could just as well claim the earth clock was the one that traveled the longer path. |

No, you cannot do that, because it is simply not true. One can MEASURE the path length of a path (calculate it in a gedanken). This calculation is an invariant -- it depends on the path but is independent of coordinates (observer).

- For a timelike path, the path length is the elapsed proper time of a clock that followed the path. In an experiment this is observed directly; in a gedanken, or to compare theory with experiment, it is calculated via an integral of the metric over the path.

> | you are implicitly assuming an ST stationary wrt earth |

NONSENSE:

A) that makes no sense -- spacetime and "stationary" are
incommensurate concepts.

B) the calculation is an INVARIANT.

This is not "gospel", but it _IS_ the way SR and GR are formulated.

> | I hope now you understand the issue |

I do. But YOU don't.

Tom Roberts

On 9/27/19 5:29 PM, RichD wrote:

> | One cannot measure a clock's tick rate within its local frame, since it is THE REFERENCE, by definition; there is nothing to measure it against. |

Not true. One simply puts a time standard next to the clock and measures its tick rate using the standard. How do you think clocks are calibrated in the real world? -- by comparing them to a time standard.

Of course the time standard is itself a clock....

> | Once one (arbitrarily) designates a device as "the clock", |

This is not "arbitrary" -- we designate certain devices as clock because they behave the way clock must behave, and because they do so when compared to other devices already designated as clocks.

Tom Roberts

On 9/28/19 2:40 PM, Edgar L. Owen wrote:

> | On Saturday, September 28, 2019 at 3:04:16 PM UTC-4, tjrob137 wrote: |

>> | On 9/27/19 5:29 PM, RichD wrote: |

>>> | One cannot measure a clock's tick rate within its local frame, since it is THE REFERENCE, by definition; there is nothing to measure it against. |

>> |
Not true. One simply puts a time standard next to the clock and measures its tick rate using the standard. How do you think clocks are calibrated in the real world? -- by comparing them to a time standard. |

> |
NO, this just measures the accuracy of individual clocks, not the intrinsic tick rate of time in a frame. |

This most certainly does measure the tick rate OF THE CLOCK. Because that is what those words mean.

- Tine is what clocks measure. But time does not "tick", it flows equably into the future.

> | That can only be determined by comparing to tick rates of an identical clock in another frame... |

Nonsense. If you attempt to describe how to compare tick rates of clocks in different inertial frames, you will either: a) screw it up, or b) find it compares signals rather than tick rates, or c) find it has systematic errors so large that it is useless.

Relativity only works if clocks always display their own elapsed proper times [#]. Anything else would violate Einstein's first postulate. But no measurements inconsistent with this have ever been observed.

- [#] Calculated by integrating the metric over the clock's path through spacetime. Measured by a physical clock.

> | Comparing clocks in the same frame to claim they always are running at the same rate is irrelevant |

No. Clocks always run at their usual rate, regardless of whichever frame they are at rest in. Note the difference between my "usual" and your "the same".

> | Time is not something that refers to clocks but to frames. |

Tine is what clocks measure. Clocks need not be at rest in some frame.

Tom Roberts

On 9/28/19 2:01 PM, Edgar L. Owen wrote:

> | Fundamental method for calculating TD for any clock in a gravity free universe: [...] |

This is fundamentally WRONG.

You clearly do not understand basic physics. You cannot hope to write such things without understanding them FIRST. Get a good book on relativity and STUDY. You are just wasting your time posting nonsense to the net.

Tom Roberts

On 9/30/19 4:08 PM, Edgar L. Owen wrote:

> | Standard twin case such that twin B returns age 10 when twin A is 60. |

I have to guess what you are trying to discuss, because your description is WOEFULLY INADEQUATE.

> | But calculating from the frame in which B was initially at rest when it left earth, it's A that traveled the same distance at the same velocity as B did relative to the earth frame. So an observer in this frame, call him C, calculates that it is A, not B, that is only 10 years old when A and B meet. |

No. You did not actually do the calculation, YOU GUESSED. And you guessed WRONG.

When C does the calculation CORRECTLY, she obtains the same result: at their reunion B is 10 and A is 60. Indeed ANY observer at rest in ANY inertial frame calculates this answer. Moreover, observers not at rest in any inertial frame ALSO calculate the same answer (but the calculation is complicated).

- Hint: the elapsed proper time over any given path is invariant.

> | [... further nonsense omitted] |

You REALLY need to learn something about the subject before attempting to write about it. Get a good book and STUDY. You're just wasting your time posting nonsense to the net.

Tom Roberts

Re: Crank Ken Shito on his "book" On 10/1/19 1:48 PM, Edgar L. Owen wrote:

> | OK, think I've got it. |

Obviously not.

> | Assume earth frame time is measured 6x slower in C frame. Then it will also take 6x longer in C's proper time for the twins to complete their separation. Therefore both C frame and earth frame see twins meeting when their ages are the same. |

This is completely unintelligible. You MUST describe the physical situation you are discussing.

It is QUITE CLEAR that you do not understand this. Get a good book and STUDY. You are just wasting your time posting nonsense to the net.

Tom Roberts

Re: Crank Ken Shito on his "book" On 10/2/19 9:06 AM, Edgar L. Owen wrote:

> | This is a general principle. |

Apparently this is a general principle: Edgar L. Owen does not understand basic physics and only writes garbled nonsense.

> | MY general principle gives the CORRECT ANSWER |

TO WHAT QUESTION????

As I keep saying: you MUST describe the physical situation you are discussing.

Given your demonstrated lack of understanding of basic physics, it is highly likely that your "principle" is completely useless, even if you were to describe what it is about.

Tom Roberts

Re: Crank Ken Shito on his "book" All,

1. Suppose twin B leaves Earth at birth and returns 60 years later aged 10.

2. Now suppose Earth is being observed by an observer in frame C, in which Earth time appears to be running 3x slower.

3. In that frame it will obviously take 3x longer (3x as much of frame C proper time) for the twins to reunite.

4. So in that frame from separation to reunion will take 180 years.

5. But when they reunite twin A will still be 60, and twin B will still be 10.

6. There is only 1 ACTUAL separation and reunion so the ages of the reunited twins must be 60 and 10 in all frames. It just takes 3x longer to happen in a frame in which it seems to happen 3x slower.

This should be obvious to everyone.

Edgar

Re: Crank Ken Shito on his "book" Den 03.10.2019 02.22, skrev Edgar L. Owen:

> |
Rotchm,
It's really rather sad and amusing you are unable to even understand what I'm saying or that it's so obviously correct > and would instead rather regress to creative writing criticisms... I'm hoping at least Paul now understands it and that it's correct since he suddenly went dark when I explained it to > him. We'll see if he is man enough to admit I'm right on this one without letting his ego get in the way? Edgar |

Do you think I am monitoring this forum all the time? :-D

|Edgar L. Owen wrote: |> It's very simple. |> Assume coordinate time in the earth frame |> is being measured as n times slower in some other frame.

And the earth frame is supposedly where twin A is stationary.

But why would the coordinate time be measured to run slower in some other frame? If a frame is moving relative to you, and you compare your own clock to the coordinate time in that frame as it passes you, the coordinate time in that other frame will _always_ appear to run _faster_ than your clock.

If you meant that twin A's clock was measured to (apparently) run n times slower in some other frame, why didn't you say so?

|> Then obviously it will take n times as much proper time in |> that other frame for the twins to meet.

And what is this other frame? Twin B isn't stationary in any inertial frame during the whole journey, so it can't be B's rest frame. Is it some arbitrary moving frame where twin A's clock is measured to run slow? And what do you mean with 'proper time in that other frame'? It can't be the proper time of twin B since he isn't stationary in 'that other frame'. Is it the proper time of some stationary clock in that frame? But that clock will never meet twin A, so between which events do you think that clock should show the proper time?

But if you despite its impossibility meant that it was B's clock that would show n times as much proper time to meet A, then B would be the elder twin. We know the opposite is true.

There is now way this can be interpreted to make sense.

|> |> This is a general principle that holds for all inertial frames. |> Thus in any inertial frame the twins will always meet when they are the same ages.

What I now have understood is the that you weren't trolling when you wrote the above.

So I can only repeat: You are even more ignorant of basic physics than I thought.

When I now suddenly go dark it is because I understand that you will never understand.

-- Paul

Re: Crank Ken Shito on his "book" On 10/3/19 8:53 AM, Edgar L. Owen wrote:

> | 1. Suppose twin B leaves Earth at birth and returns 60 years later aged 10. |

Presumably your "60 years" is in the earth frame, and the "10 years" is the elapsed proper time of twin B between leaving and returning. This is kind of implied by your words, but really needs to be made explicit.

> |
2. Now suppose Earth is being observed by an observer in frame C, in which Earth time appears to be > running 3x slower. 3. In that frame it will obviously take 3x longer (3x as much of frame C proper time) for the twins > to reunite. 4. So in that frame from separation to reunion will take 180 years. 5. But when they reunite twin A will still be 60, and twin B will still be 10. |

Yes. Note that if the observer in frame C passes the earth when twin B leaves, B's return to earth will occur many lightyears away from the observer in frame C.

> | 6. There is only 1 ACTUAL separation and reunion so the ages of the reunited twins must be 60 and 10 > in all frames. It just takes 3x longer to happen in a frame in which it seems to happen 3x slower. |

Yes. Finally you have specified the physical situation with enough detail so it can be understood. This seems to be unrelated to the nonsense you have been writing previously.

> | Yes proper time is invariant in all frames AND I'm giving the reason why it is. |

Yes, proper time is indeed invariant, but you aren't giving any reason for it, you are just stating that it is so and giving an example.

The proper time of a clock is invariant, in the sense that it does not matter from which frame the clock is observed, because the clock always displays a single value. Any number of observers, with any properties whatsoever, can observe the clock and agree that at a given point along its worldline it displays a given value.

Mathematically this is modeled by calculating the elapsed proper time along a given path via an integral of the metric over the path; the integral and the metric are manifestly invariant, in that they do not depend on coordinates in any way.

Tom Roberts

Re: Crank Ken Shito on his "book" On 10/4/19 1:40 PM, Edgar L. Owen wrote:

> | The principle that I�ve been stating [...] |

is useless. The ACTUAL principle is:

- The elapsed proper time over a given path through spacetime is invariant (independent of coordinates used to calculate it).

Your embroidery is useless, confusing, and wholly contained in the above statement. As I said before, you don't "explain" anything, you just give an example.

Tom Roberts

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