Twin paradox (Pete and MacM)

[MacM]

GMonTag & Paul T,

I am not just ignoring your post but feel James R's is more productive. I may comment on your errors later.

MacM:

Some simple true-false questions for you. See if you can answer with just the word "true" or "false".

True or false?

1. Special relativity says simultaneity is relative.

True. And I use the calculated simultaneity to preprogram the timers to cause the clocks to physically stop simultaneously in both frames when B = 15,692. You have already agreed that number is common in both frames.

2. According to special relativity, 2 events which are simultaneous in one frame of reference are not simultaneous in a relatively-moving frame.

True. Which is why it is necessary to use Relativity to find the common denominator where data of the mathematics of such predictions can be compared, which is what I have done.

3. Special relativity says that the relativity of simultaneity is an effect which is totally independent of any delays caused by transmitting information.

True. But irrelavant to the issue of simultaneously stopped clocks in A and B's frames at the common time of 36,000 seconds for A and 15,692 seconds for B.

That's all for now. We need to establish whether you understand relativity, before we can decide whether you can refute it on its own grounds.

Are you brave enough to answer the questions? Only three words are required from you - one for each question.

Or will you do the MacM Texas Two Step?

No bravery required on my part. You have your "Three" word answers qualified as to their meaning which yo must address the real issue which is my qualifications.

Are my three words correct or not? Yes or No. Do you have ANY viable explanation or refutation of the qualified facts as to those answers? I think not.

 

[James R]

MacM:

I asked:

According to special relativity, 2 events which are simultaneous in one frame of reference are not simultaneous in a relatively-moving frame.

You answered:

True.

Next set of questions.

4. The reference frame in which clock B is stationary moves relative to the reference frame in which clock A is stationary. True or false?
5. In A's reference frame, clock A and clock B stop simultaneously. True or false?

Now, look at the first quote in this post again, and then answer the following question.

6. If clocks A and B stop simultaneously in A's reference frame, then relativity says that it is impossible for them to stop simultaneously in B's reference frame. True or false?

 

[MacM]

I claim that if time dilation is valid then after 15,692 seconds run time in B's frame and all clocks are stopped simultaneously that Relativity claims that A will only read 6,840 seconds.

Mac, Mac, Mac.... How many times do we have to tell you before it gets through your thick skull? When you are talking about predictions that relativity makes, and you claim that two events are simultaneous, you have to say in which reference frame. Otherwise your claim is meaningless.

My skull apparently isn't as thick as some others. My thought clarity seems to be superior given the facts of this case. I have infact relied upon Relativity to give me a precise number regarding simultaneity affects to preprogram timers such that they stop all clocks physically simultaneously in BOTH frames of referance.

As James has already finally admitted when B's accumulated time is 15,692 seconds it is common to both frames. That is when clocks are stopped. This is not a matter of not understanding Relativity. It is a matter of understanding Relativity and understanding how to use the predicted simultaneity so as to be able to compare stopped clocks displays against the displays predicted by Relativity. Relativity fails this test.

Do you agree that B's 15,692 seconds is a common number in both B's and A's frame of referance? Yes or No.

In relativity, simultaneity is frame dependant. I've told you this at least 5 times by now. God knows how many times JamesR has told you. Probably too many to count. And yet you still claim relativity predicts things that it does not.

What do you think I have said Relativity predicts that it does not. Talk is cheap.

In A's frame when A stops at 36,000 seconds does or does not B = 15,692 seconds? Yes or No.

In B's frame of referance when it accumulates 15,692 seconds is or is not that the same 15,692 seconds relative to A's frame? Yes or No.

If B accumulates only one number of 15,692 seconds which is common in both frames does or does not there now exist a requirement that to satisfy time in both frames of referance that A display both the 36,000 seconds and 6,840 seconds views? Yes or No.

If you argue that somehow there exists a dual reality (which is utter nonsense) that when B reaches 15,692 seconds it stops A at 6,840 seconds is that, or is it not, in conflict with the assumptions of Relativity that B will read 15,692 seconds when A reaches 36,000 seconds? Yes or No.

Remember 15,692 seconds displayed by B is a common prediction of Relativity in both frames. Therefore both A displays are predicted for that specific instant when B reads 15,692 seconds.

Is it not true that should B actually stop A at 6,840 seconds that MB (A's relavistic view of B) would claim that B reads only 2,892 seconds which is in conflict with B's true stopped reading of 15,692 seconds? Yes or No.

If all clocks stop simultaneously in B's frame at t'=15,692, then relativity does predict that A will read 6,840. But, since in your scenario, all the clocks don't stop simultaneously in B's frame, it's irrelevant.

There is your error. All clocks have stopped in Both frames OR THE PREDICTIONS OF RELATIVITY HAVE FAILED. Make up your mind is it simultaneity or time dilation which are invalid, or both?

A, B, and MA all stop simultaneously in A's frame. MA is not required to equal A in A's frame. MB is required to equal B, and it does.

A, B, and MA do not stop simultaneously in B's frame. In this frame MA is required to equal A, and it does. MB is not required to equal B, and it doesn't.

There is no conflict here.

Wrong you have not followed the simultaneous stopping of clocks required by preprogrammed timers in the test. The clocks are all stopped at the common accumulated time in A and B's frame where B = 15,692 seconds. In A's frame it has stopped at 36,000 seconds. In B's frame A is stopped at 6,840 seconds. From the common accumulated time of 15,692 seconds of B, frame A is shown to be required to display two different accumulated times to satisfy both frames from a common referance point of accumulated time.

I'm not arguing that B stops A at all. When B stops in B's frame at t'=15,692, A does read 6,840. But A doesn't stop then, because the clocks don't stop simultaneously in B's frame.

They certainly do or predictions of Relativity Simultaneity fail. You are left with only two choices.

1 - When A reaches 36,000 seconds and stops also stopping B and MA, B's stopped display reads 15,692 seconds and MA claims A should read 6,840 seconds. But A doesn't.

2 - When B reaches 15,692 seconds and stops and also stopping A and MB, A displays 6,840 seconds but MB predicts B must read 2,892 seconds and it doesn't

NOTE: Both frames stop at the common accumulated time in B's frame of 15,692 seconds. Both conditions of A therefore are invalid predictions of A's displayed time or of B's displayed time. ALL clocks are infact physically stopped at the same physical instant in both frames views.

MB does not equal B in B's frame, but that alright, because it isn't required to. MA only has to equal A in B's frame, and MB only has to equal B in A's frame. This is because MA is B's predictions of what A reads, and MB is A's predictions of what B reads. A's and B's predictions are only valid in their respective frames.

Crap. The start and stop times and duration of the test are equal and simultaneous. A's test time is 36,000 seconds. B's test time is 15,692 seconds that makes the stopping of the clocks simultaneous in BOTH frames.

You are trying to declare a duality of realities. It is not physically possible and only mathematically possible if you disregard the actual conditions timing wise of the test conditions.

All clocks only stop simultaneously in A's frame of referance.

False. You get it through your thick head. All clocks stop at the accumulated time of B of 15,692 seconds which is common (simultaneous) in BOTH frames.

Get this through your head Mac, there is *no* way to program the timers such that relativity will predict that the clocks stop simultaneously in both frames. The math simply will not work that way.

False. I have just shown you you can or that Relativity of Simultaneity fails and Relativity must be a falicy.

Achieving such simultaneity time dilation fails. Which one do you wish to blame for the failure of Relativity to be physically possible?

 

[MacM]

Only a dumb would set that kind of silly experiment. This seem to be the set up:

Put a clock on earth and another similar one in a spaceship. Preset both clocks to stop when they hit reading 36,000 seconds. Then let the spaceship move away from earth at relativistic velocity until the clock stop. Experiment done. Get the spaceship clock back to earth and compared their reading, and....voila they both show 36,000 seconds. No time dilation. Congratulation, mister...you had just proved that SR flawed...hahahahahaha


This 2-3 weeks time of waiting could be a nightmare for whoever supporting relativity here. What will you do after you get the book and discover that SR (time dilation concpet in particular) is seriously flawed even according to Einstein himself? Will you publish your finding in the scientific journal? Many seem to be unaware of the flawed. You will be remembered as a hero. OMG.

I am ignoring your post since it is just more personal attacks and distorted bullshit ignoring the issue and realities of the test description. You seem incapable of actually following the predictions of Relativity.

 

[1100f]

My skull apparently isn't as thick as some others. My thought clarity seems to be superior given the facts of this case. I have infact relied upon Relativity to give me a precise number regarding simultaneity affects to preprogram timers such that they stop all clocks physically simultaneously in BOTH frames of referance.

According to relativity you cannot have two events to be simultaneous in all frames unless they occur at the same place, since you prepare both clocks A and B to stop simultaneously in A's frame, when they are at different places, there is no way that they would stop simultaneously in B's frame. Since if you could make them stop simultaneously in both frames, you would get to the contradiction that in one frame A stops when it reads 36000 sec and in the other, it stops when it reads 6840 sec. since this is impossible, there is no way that the stopping of the clocks should be simultaneous in both reference frames.

What you are doing here is that you start with an assumption not supported by SR, get a contradiction and from this you conclude that SR is wrong. If you want to find flaws in SR, start with assumptions that are within the framework of SR.

As James has already finally admitted when B's accumulated time is 15,692 seconds it is common to both frames. That is when clocks are stopped. This is not a matter of not understanding Relativity. It is a matter of understanding Relativity and understanding how to use the predicted simultaneity so as to be able to compare stopped clocks displays against the displays predicted by Relativity. Relativity fails this test.

Do you agree that B's 15,692 seconds is a common number in both B's and A's frame of referance? Yes or No.

B's 15692 sec is a common number, it is not a common time since in A's frame, the time that has ellapsed is 36000 sec.

In A's frame when A stops at 36,000 seconds does or does not B = 15,692 seconds? Yes or No.

Yes.

In B's frame of referance when it accumulates 15,692 seconds is or is not that the same 15,692 seconds relative to A's frame? Yes or No.

No. In A's frame, when B reads 15692 sec, A reads 36000 sec.
In B's frame, when B reads 15692 sec, A reads 6840 sec.

If B accumulates only one number of 15,692 seconds which is common in both frames does or does not there now exist a requirement that to satisfy time in both frames of referance that A display both the 36,000 seconds and 6,840 seconds views? Yes or No.

No. Since in SR simulaneity is relative, it depends in which reference frame you are. In one frame you can read 36000 sec and in another one it can read 6840 sec.

If you argue that somehow there exists a dual reality (which is utter nonsense) that when B reaches 15,692 seconds it stops A at 6,840 seconds is that, or is it not, in conflict with the assumptions of Relativity that B will read 15,692 seconds when A reaches 36,000 seconds? Yes or No.

If you could start with the possibility that the clocks are predicted to stop simultaneously in both frames you would have this contradiction. Since from SR you cannot have this "dual reality", you cannot prepare your clocks to stop simultaneously in both frames.

Remember 15,692 seconds displayed by B is a common prediction of Relativity in both frames. Therefore both A displays are predicted for that specific instant when B reads 15,692 seconds.

Is it not true that should B actually stop A at 6,840 seconds that MB (A's relavistic view of B) would claim that B reads only 2,892 seconds which is in conflict with B's true stopped reading of 15,692 seconds? Yes or No.

According to SR, if one clock is prepared to stop when its reading is 36000, in all frame it will stop at 36000. If the second one is prepared to stop when it reads 15692, it will stop when it reads 15692. If both clocks stop simultaneously in one frame (in A's frame), in the second one, when one of the clocks stops (the A clock stops when it reads 36000), the second clock is reading 6840 sec and is still runing untill it reaches 15692 and then it will stop.

There is your error. All clocks have stopped in Both frames OR THE PREDICTIONS OF RELATIVITY HAVE FAILED. Make up your mind is it simultaneity or time dilation which are invalid, or both?

This is your error by not using SR to try to find flaws in SR.

Wrong you have not followed the simultaneous stopping of clocks required by preprogrammed timers in the test. The clocks are all stopped at the common accumulated time in A and B's frame where B = 15,692 seconds. In A's frame it has stopped at 36,000 seconds. In B's frame A is stopped at 6,840 seconds. From the common accumulated time of 15,692 seconds of B, frame A is shown to be required to display two different accumulated times to satisfy both frames from a common referance point of accumulated time.

You have not followed the rules of SR that say that transformation frome one reference from to another is performed by using the LOrentz transformation.

They certainly do or predictions of Relativity Simultaneity fail. You are left with only two choices.

1 - When A reaches 36,000 seconds and stops also stopping B and MA, B's stopped display reads 15,692 seconds and MA claims A should read 6,840 seconds. But A doesn't.

2 - When B reaches 15,692 seconds and stops and also stopping A and MB, A displays 6,840 seconds but MB predicts B must read 2,892 seconds and it doesn't

NOTE: Both frames stop at the common accumulated time in B's frame of 15,692 seconds. Both conditions of A therefore are invalid predictions of A's displayed time or of B's displayed time. ALL clocks are infact physically stopped at the same physical instant in both frames views.

I suggest that you use the Lorentz transformation

Crap. The start and stop times and duration of the test are equal and simultaneous. A's test time is 36,000 seconds. B's test time is 15,692 seconds that makes the stopping of the clocks simultaneous in BOTH frames.

You are trying to declare a duality of realities. It is not physically possible and only mathematically possible if you disregard the actual conditions timing wise of the test conditions.



False. You get it through your thick head. All clocks stop at the accumulated time of B of 15,692 seconds which is common (simultaneous) in BOTH frames.

Not according to relativity.

False. I have just shown you you can or that Relativity of Simultaneity fails and Relativity must be a falicy.

Achieving such simultaneity time dilation fails. Which one do you wish to blame for the failure of Relativity to be physically possible?

What you have shown is that if you prepare your clocks to stop simultaneously in both frames, then you get a contradiction. The conclusion from what you say is that you cannot prepare the clocks to stop simultaneously in both frames. But this is what SR claims.
Funny to see that you wanted to find flaws in SR, and instead, you prove again that you cannot find these flaws.

 

[MacM]

I asked:

“ According to special relativity, 2 events which are simultaneous in one frame of reference are not simultaneous in a relatively-moving frame. ”

You answered:

“ True. ”

Next set of questions.

4. The reference frame in which clock B is stationary moves relative to the reference frame in which clock A is stationary. True or false?

True.

5. In A's reference frame, clock A and clock B stop simultaneously. True or false?

True.

Now, look at the first quote in this post again, and then answer the following question.

6. If clocks A and B stop simultaneously in A's reference frame, then relativity says that it is impossible for them to stop simultaneously in B's reference frame. True or false?

True. And that should suggest that what Relativity claims is false since we agree that B = 15,692 seconds is a common time in both frames.

A stops at 36,000 seconds and B stops at 15,692 seconds and stops MA; which happens to be the same point in time in accordance with Relativity. You have already agreed that B's 15,692 seconds is a common time.

Do you agree that ALL three are stopped simultaneously? Yes or No.

Is 6,840 seconds recorded by MA the correct time that A should stop in B's FOR when B = 15,692? Yes or No.

Is that where "A" actually stopped? No. It stopped at 36,000 seconds (you can add the qualifier "In A's Frame).

If I place a timer in "A" with an OR logic gate to shut it down when:

1 - MA = 6,840 seconds OR

2 - B = 15,692 seconds OR

3 - A = 36,000 seconds;

Which input stops "A"? Remember "A" cannot be stopped twice. It is a physical clock and can only be stopped once.

Your answer please. Input 1, 2, or 3 ?

I believe the correct answer is that ALL three inputs are simultaneous. Leaving Relativity fiat declarations behind that translates into the following truthful statement:

"Relativity requires that When "A" = 36,000 seconds, B is required to equal 15,692 seconds and B requires that "A" equal 6,840 seconds.

Considering that ALL three clocks are started/stopped simultaneoulsy, this dictates that Relativity has the problem and not the description of the test.

Simply saying that "Relativity says these events are not simultaneous" does not make them unsimultaneous since precalibrated relavistic timers were used to correct simultaneity in the first place.

Your attempt to shift the discussion by referring to the other frame's view is unwarranted since I originally included that frame in my test using 6 clocks. Properly considered it shows that Relativity creates an impossible duality of reality.

Sorry your arguement is based on fiat and not facts.

 

[MacM]

1100f,

Rather than repeat long threads and arguements I direct your attention to:

[post=703476]My Response to James[/post]

What you are doing here is that you start with an assumption not supported by SR, get a contradiction and from this you conclude that SR is wrong. If you want to find flaws in SR, start with assumptions that are within the framework of SR.

What you seem to not realize is you are saying one must accept SR declarations as physical fact and in doing so you cannot prove the error. I prefer to stick with physical principles and follow the facts.

B's 15692 sec is a common number, it is not a common time since in A's frame, the time that has ellapsed is 36000 sec.

Certainly and therein lies the problem but you don't see it. From B's FOR it reads 15,692 when A reads only 6,840 but stopped clocks show that is not a possibility. The problem is not in the test description it is in the declarations of SRT.

No. In A's frame, when B reads 15692 sec, A reads 36000 sec.

In B's frame, when B reads 15692 sec, A reads 6840 sec.

I don't argue with this statement. Surprised? Don't be that is all that happens. It "Reads" different but it is "NOT" different at the local time level.

It is the local time level, the proper time, that is the reality, not the perception of moving observers.

When "A" stops at 36,000 seconds, B stops at 15,692 but B "reads "A" = 6,840 seconds. But that is not the case "A" is in physical reality stopped at 36,000 seconds.

Don't waste time saying by not in B's Frame, blah, blah. My test used 6 clocks showing all relavistic relationships and the physical impossibility persists. The only rational conclusion is that SRT declarations are being misapplied and should be understood as perceptions not realities.

This becomes even more evident when you realize that reciprocity requires that both clocks slow equally, such that there can be NO systemic measurable time differential between observers,

No. Since in SR simulaneity is relative, it depends in which reference frame you are. In one frame you can read 36000 sec and in another one it can read 6840 sec.

Again I agree, qualified as above.

This is your error by not using SR to try to find flaws in SR.

And it is the declarations of SRT that are the problem, not the test description.

I suggest that you use the Lorentz transformation

And I suggest one must keep their footing on solid physical ground and not let playing with numbers cause you to conclude nonsense.

What you have shown is that if you prepare your clocks to stop simultaneously in both frames, then you get a contradiction. The conclusion from what you say is that you cannot prepare the clocks to stop simultaneously in both frames. [/quote]

You cannot discard physical logic in favor of fiat.

Funny to see that you wanted to find flaws in SR, and instead, you prove again that you cannot find these flaws.

Not true. The flaws are clearly indicated. You just choose to claim that it shows the test is at fault and not SRT. You are accepting fiat over physical principles and logic.

 

[2inquisitive]

MacM, I hope you don't mind me posting another scenario concerning simultaneity in
your discussion. Maybe it will help everyone see what is happening in simultaneity.
I'm not going to get into much detail, just an example.
We will start with two observers located two hundred miles apart in TEXAS. Each has
a clock and they are synchronized with each other. Each has a telescope that they can see the other's clock with. Exactly halfway between them is a hydrogen bomb.
The clocks are setup to stop when the flash from the exploded bomb reaches it.
The bomb explodes and the clocks stop. No problem, each observer can look through
his telescope and confirm the others clock stopped exactly at the same time as his own. Now we will put another observer on the International Spacestation orbiting the
Earth. In standard Special Theory of Relativity speak, this observer is in the rest frame
and the Earth is passing below him. He sees the explosion and has two telescopes with
clocks attached to them that will stop when the flash from the bomb reaches a reflector at each location. This observer sees himself at rest and the Earth as the
moving frame, remember? His clocks will indicate that the flash did NOT reach each
ground-based clock at the same time. And he believes each ground-based observer
will read a different time on the others' clock. Which is reality?

 

[James R]

MacM,

Ok, let's recap before I respond to your post.

You have now agreed that relativity predicts that clocks A and B cannot stop simultaneously in the reference frames of both A and B. If they stop simultaneously in frame A, then relativity says they cannot stop simultaneously in frame B.

We now get to the real point of disagreement. You can no longer argue that relativity is in any way self-contradictory on the basis of all clocks not stopping simultaneously, because that would be inconsistent with your agreement that relativity does not allow all clocks to stop simultaneously in all frames. Therefore, your only course is to argue that relativity is not a correct description of reality. Since that is what you believe anyway, that should not be a problem for you.

At least we have made progress to the stage where you accept that relativity makes a particular prediction, even though you do not agree that that prediction represents "reality". So, I hope you will not, after this point, fall back on claims that relativity is inconsistent within itself. You have just agreed that relativity is consistent with itself, but you think it is inconsistent with reality. Can we take that as given from here on, or do you want to backpeddle?

Now, bearing that in mind, let's look at your post.

True. And that should suggest that what Relativity claims is false since we agree that B = 15,692 seconds is a common time in both frames.

I'm not sure what you mean by "a common time" here.

A stops at 36,000 seconds and B stops at 15,692 seconds and stops MA; which happens to be the same point in time in accordance with Relativity. You have already agreed that B's 15,692 seconds is a common time.

I have agreed to the following:

1. In A's frame, A stops at 36000, simultaneous with B stopping at 15692.
2. In B's frame, A stops at 36000, but not simultaneously with B stopping at 15692.
3. Clock MA stops at the same time as B, since it is with B, in either frame.
4. Clock MB stops at the same time as A, in either frame.

Do you agree that ALL three are stopped simultaneously? Yes or No.

In frame A, yes. In frame B, no.

Is 6,840 seconds recorded by MA the correct time that A should stop in B's FOR when B = 15,692? Yes or No.

No. A doesn't stop until it reaches 36000.

Is that where "A" actually stopped? No. It stopped at 36,000 seconds (you can add the qualifier "In A's Frame).

A stops when it displays 36000 in either frame. In A's frame, it displays 36000 at t=36000. In B's frame, it displays 36000 at t'=82590.

If I place a timer in "A" with an OR logic gate to shut it down when:

1 - MA = 6,840 seconds OR

2 - B = 15,692 seconds OR

3 - A = 36,000 seconds;

Which input stops "A"? Remember "A" cannot be stopped twice. It is a physical clock and can only be stopped once.

Correct. If your logic gate stops A on condition 1, you've changed the test, because MA travels with clock B. If A is set to stop when MA reads 6840, then in A's frame, this occurs at t=15692, so A would stop when it read 15692. So, you've changed the test, which was originally set up to stop A at t=36000.

Conditions 2 and 3 correspond to your original stop condition, since in A's frame, they happen simultaneously.

I believe the correct answer is that ALL three inputs are simultaneous.

No. In A's frame, inputs 2 and 3 are simultaneous, but 1 is not. In B's frame, inputs 1 and 2 are simultaneous, but 3 is not.

Leaving Relativity fiat declarations behind that translates into the following truthful statement:

"Relativity requires that When "A" = 36,000 seconds, B is required to equal 15,692 seconds and B requires that "A" equal 6,840 seconds.

How can you leave relativity fiat declarations behind when you're making statements about what relativity says? You HAVE to use relativity fiat declarations when you're working within the framework of relativity. Duh!

Your statement is wrong. Only two of the three events you mention (A=36000, B=15692, A=6840) occur simultaneously, in any frame, because A cannot ever have two readings at the same time. However, it is possible in A's frame to have the readings (A=36000, B=15692, MA = 6840) occur simultaneously. As I explained above, your monitor clocks do not do exactly what you think they do. They only work to display equivalent times correctly in one frame, not in both. Why? Because that is how you specified that they be set up. Moreover, relativity says it is impossible to set them up so they keep in step in both frames.

Considering that ALL three clocks are started/stopped simultaneoulsy, this dictates that Relativity has the problem and not the description of the test.

You have agreed that, according to relativity, it is impossible to stop all the clocks simultaneously in both frames. Relativity is a consistent picture, and your problem is a non-issue for it. Your only option is to dispute that relativity is real.

Simply saying that "Relativity says these events are not simultaneous" does not make them unsimultaneous since precalibrated relavistic timers were used to correct simultaneity in the first place.

This is your real argument.

Your real argument, when it comes down to it, is not that relativity is internally inconsistent, but that the relativity of simultaneity is not real. So, we are back to absolute time, which has been your underlying philosophy all along.

Your "precalibrated timers" used to "correct simultaneity" only work if relativity is wrong. If relativity is right, they do not do what you claim, as has been explained.

Where do we go from here, then?

We go back to the start. It is up to you to show that absolute time exists, and relativity is wrong.

You have utterly failed to show that relativity is inconsistent within itself. In fact, you have explicitly conceded that point, above, at least regarding the simultaneity issue.

So, can you show that relativity is not real?

 

[MacM]

James R,

Two clocks A and B are seperated in space at relative rest and are some acceptable distance apart with a start controller located midway between the clocks.

That controller sends a signal to each clock, when received it sets timers in each clock to a time based on an acceleration schedule for clock B such that from each clocks perspective once B has reached 0.9c and stopped accelerating both clocks start according to the precalculated time.

The test is designed such that the acceleration ends and timing starts when the moving clock is 9 light-hours seperation between clocks. This results in the 10 hours test before the clocks collide.

What does each clock display the instant they collide and certainly stop simultaneously.?

A ________

B ________

and what time does each clock observe the other clock as displaying at the instant they collide?


A ________

B ________

 

[MacM]

MacM, I hope you don't mind me posting another scenario concerning simultaneity in
your discussion. Maybe it will help everyone see what is happening in simultaneity.
I'm not going to get into much detail, just an example.
We will start with two observers located two hundred miles apart in TEXAS. Each has
a clock and they are synchronized with each other. Each has a telescope that they can see the other's clock with. Exactly halfway between them is a hydrogen bomb.
The clocks are setup to stop when the flash from the exploded bomb reaches it.
The bomb explodes and the clocks stop. No problem, each observer can look through
his telescope and confirm the others clock stopped exactly at the same time as his own. Now we will put another observer on the International Spacestation orbiting the
Earth. In standard Special Theory of Relativity speak, this observer is in the rest frame
and the Earth is passing below him. He sees the explosion and has two telescopes with
clocks attached to them that will stop when the flash from the bomb reaches a reflector at each location. This observer sees himself at rest and the Earth as the
moving frame, remember? His clocks will indicate that the flash did NOT reach each
ground-based clock at the same time. And he believes each ground-based observer
will read a different time on the others' clock. Which is reality?

Your contribution is appreciated and I have no qualms about your description.

My only comment would be the clocks at relative rest which measured the same time for the blast show a delayed time from the actual blast by 536 micro-seconds.

There view of the time of the event did not make it happen 536 micro-seconds later. Theirs is merely a perception of the time, not the actual time.

The same is true for the illusion of motion between the orbiting observer and the perception that the events were not simultaneous for the ground observers. The illusion that the time changed didn't change the time.

That is the entire arguement, not that simultaneity shift of views do not exist but only that such views do not alter the reality of the time of the event.

Thanks.

 

[James R]

MacM:

Seeing as you're introducing a new scenario, can I take it you now admit your original one is dead in the water?

Let's look at the new one, then.

Two clocks A and B are seperated in space at relative rest and are some acceptable distance apart with a start controller located midway between the clocks.

That controller sends a signal to each clock, when received it sets timers in each clock to a time based on an acceleration schedule for clock B such that from each clocks perspective once B has reached 0.9c and stopped accelerating both clocks start according to the precalculated time.

Let me make sure I have this right. A remains stationary. B, upon receiving the signal, accelerates towards A until it reaches a relative speed of 0.9c. The instant it reaches 0.9c, clocks A and B both start to tick. Is that right?

Now, we just need to know in whose frame they start to tick simultaneously to complete the specification of the test. When the clocks are started, clocks A and B are not in the same location, and they are travelling at 0.9c relative to one another. Relativity therefore says that they cannot start simultaneously in both frames, but only in one frame.

For now, I will assume they start simultaneously in A's frame, but not in B's, unless you want to change this. Remember, you have agreed that they cannot be started simultaneously in both frames if relativity is correct, and you have given no reason to suppose that relativity is not correct, so far.

The test is designed such that the acceleration ends and timing starts when the moving clock is 9 light-hours seperation between clocks. This results in the 10 hours test before the clocks collide.

In which frame is the 9 light-hours measured? A's frame, or B's frame? Because, as you are aware, length contraction effects in relativity say that the distance measured in two co-moving frames will be different. Again, unless you say something to the contrary, I will take it you are measuring the 9 light hours in A's frame, but not in B's.

What does each clock display the instant they collide and certainly stop simultaneously.? and what time does each clock observe the other clock as displaying at the instant they collide?

In frame A, given the above assumptions:

B needs to travel 9 light hours at speed 0.9c, which will take 10 hours.
Clock A will read 10 hours at the time of collision.

In frame B, given the assumptions:

A needs to travel 3.92 light hours at a speed 0.9c, which will take 4.35 hours.
Clock B will read 4.35 hours at the time of collision.

At the instant of the collision, both observers will agree that clock A reads 10 hours and clock B reads 4.35 hours.

In B's frame, clock A was started before clock B, which accounts for the longer time displayed on that clock, even though clock A runs slower than clock B as seen by B. Similarly, clock A sees clock B running slow, but A and B were started simultaneously, which accounts for the longer time displayed on clock A in this frame.

Can you show that this relativistic explanation is wrong?

Do you plan on addressing the issues I raised in my previous post?

 

[MacM]

Ok, let's recap before I respond to your post.

You have now agreed that relativity predicts that clocks A and B cannot stop simultaneously in the reference frames of both A and B. If they stop simultaneously in frame A, then relativity says they cannot stop simultaneously in frame B.

Correct. That is what Relativity says but that is perception and not reality. That perception changes with motion the reality of events do not.

We now get to the real point of disagreement. You can no longer argue that relativity is in any way self-contradictory on the basis of all clocks not stopping simultaneously, because that would be inconsistent with your agreement that relativity does not allow all clocks to stop simultaneously in all frames. Therefore, your only course is to argue that relativity is not a correct description of reality. Since that is what you believe anyway, that should not be a problem for you.

That sounds exactly right.

At least we have made progress to the stage where you accept that relativity makes a particular prediction, even though you do not agree that that prediction represents "reality". So, I hope you will not, after this point, fall back on claims that relativity is inconsistent within itself.

Actually, my views have not changed. Your understanding of my views have improved. I have never claimed relativity did not predict time dilation or was inconsistant in its mathematics but only that it is not real, it is an observational illusion.

You have just agreed that relativity is consistent with itself, but you think it is inconsistent with reality. Can we take that as given from here on, or do you want to backpeddle?

Repeating. I have not backpeddled previously.

Now, bearing that in mind, let's look at your post.

True. And that should suggest that what Relativity claims is false since we agree that B = 15,692 seconds is a common time in both frames.

I'm not sure what you mean by "a common time" here.

Now you are backpeddeling. You have already agreed that 15,692 is the same time in both frames. By accepting Relativity in my test when "A" predicts "B" reads 15,692, B does read 15,692. That time is the same in both frames.

A stops at 36,000 seconds and B stops at 15,692 seconds and stops MA; which happens to be the same point in time in accordance with Relativity. You have already agreed that B's 15,692 seconds is a common time.

Yes.

I have agreed to the following:

1. In A's frame, A stops at 36000, simultaneous with B stopping at 15692.

2. In B's frame, A stops at 36000, but not simultaneously with B stopping at 15692.

Agreed as to the observation but not as to actual stopped clocks. This is observation by a remote observer in motion and not the physical status of the clock.

3. Clock MA stops at the same time as B, since it is with B, in either frame.

4. Clock MB stops at the same time as A, in either frame.

Agreed.

Do you agree that ALL three are stopped simultaneously? Yes or No.

In frame A, yes. In frame B, no.

Here is where we disagree. Not as to the observation but as to the real status of clocks. The fact that B has seen A running slow and therefore is not seeing it stop in real time is observational and not A's actual physical status.

Perception vs Reality.

Is 6,840 seconds recorded by MA the correct time that A should stop in B's FOR when B = 15,692? Yes or No.

No. A doesn't stop until it reaches 36000.

I think you screwed up here. Have another look. If B reads 15,692 seconds then Relativity predicts that B sees A as being 6,840 seconds when B stops.

Remember B is stopping in B's FOR and B has a view of A in B's FOR. When B = 15,692, A must equal 6,840. I agree in this test that MA stops at the predicted time for B and that B sees A continue to run until it sees it stop at it actual stopped condition (which actually happened 29,160 seconds before B saw it happen).

But keep in mind if I put a timer in B to stop A when B stops at 15,692, since 36,000 and 15,692 exist simultaneously in both frames B cannot stop A at 6,840 since A is already at 36,000. Further if you do claim to stop A at 6,840 then A claims B is 2,892 not 15,692.

Is that where "A" actually stopped? No. It stopped at 36,000 seconds (you can add the qualifier "In A's Frame).

A stops when it displays 36000 in either frame. In A's frame, it displays 36000 at t=36000. In B's frame, it displays 36000 at t'=82590.

Agreed. Except B isn't timing to see the end of A. B stops at 15,692, MA stops at 6,840 stops; which is the time B claims A should read if stopped simultaneously between A and B.

You must remember that the functions are taking place simultaneously and Relativity is creating a conflict of "Physical", not "Observationsal" possibilities between clocks.

If I place a timer in "A" with an OR logic gate to shut it down when:

1 - MA = 6,840 seconds OR

2 - B = 15,692 seconds OR

3 - A = 36,000 seconds;

Which input stops "A"? Remember "A" cannot be stopped twice. It is a physical clock and can only be stopped once.

Correct. If your logic gate stops A on condition 1, you've changed the test, because MA travels with clock B. If A is set to stop when MA reads 6840, then in A's frame, this occurs at t=15692, so A would stop when it read 15692. So, you've changed the test, which was originally set up to stop A at t=36000.

Conditions 2 and 3 correspond to your original stop condition, since in A's frame, they happen simultaneously.

Wrong answer. MA reads 6,840 simultaneous with B = 15,692 and A = 36,000.

I believe the correct answer is that ALL three inputs are simultaneous.

No. In A's frame, inputs 2 and 3 are simultaneous, but 1 is not. In B's frame, inputs 1 and 2 are simultaneous, but 3 is not.

Wrong answer but I see what you are wanting to do. You are declaring MA a false signal claiming it doesn't happen in A's frame but my point is that it is happening in B's frame. It is B's view of A.

Now try linking your frames to function concurrently which is what happens in reality. That is to say A sees B and B sees A concurrently in reality. Try linking those signals to control the clocks. Got a problem don't you?

Go ahead put a timer in A that shuts it down when B sees 15,692 in B's FOR.

B's 15,692 is in B's FOR. There is your problem Do you see it yet?

Leaving Relativity fiat declarations behind that translates into the following truthful statement:

"Relativity requires that When "A" = 36,000 seconds, B is required to equal 15,692 seconds and B requires that "A" equal 6,840 seconds.

How can you leave relativity fiat declarations behind when you're making statements about what relativity says? You HAVE to use relativity fiat declarations when you're working within the framework of relativity. Duh!

Duh indeed. If such declarations present physical impossibilities, let me suggest you must leave them behind if you want to talk physics instead of mysticism.

You do believe in physical realities do you not or have you accepted voodoo for science?

Your statement is wrong. Only two of the three events you mention (A=36000, B=15692, A=6840) occur simultaneously, in any frame, because A cannot ever have two readings at the same time.

We certainly agree that A cannot have two different readings at the same time. We disagree that MA = 6,840 doesn't apply. It is inherent in that B =15,692 in B's FOR and when it reads 15,692 Relativity predicts A must read 6,840.

However, it is possible in A's frame to have the readings (A=36000, B=15692, MA = 6840) occur simultaneously. As I explained above, your monitor clocks do not do exactly what you think they do. They only work to display equivalent times correctly in one frame, not in both. Why? Because that is how you specified that they be set up. Moreover, relativity says it is impossible to set them up so they keep in step in both frames.

Please explain in a mutual relative motion scenario where Relativity is functioning concurrently on clocks how B can have two different accumulated times of 15,692 seconds.

The test is arranged such that A's test period is 36,000 seconds. B's test period is 15,692 seconds such that they run concurrently and stop simultaneously. If B's accumulated time is 15,692 seconds in its test period "A" MUST be at 6,840. seconds not 36,000 seconds.

I understand that "A" read 6,840 when B stopped but B sees it continue to run until it's 36,000 actual stop time because it was stopped in A's FOR.

However, what you are failing to do is to now place a timer in A to cause it to stop when B reads 15,692 (which wouldn't change anything) or to put a timer in A to stop it at 6,840 but which will show that B now only reads 2,892 not 15,692.

Considering that ALL three clocks are started/stopped simultaneoulsy, this dictates that Relativity has the problem and not the description of the test.

You have agreed that, according to relativity, it is impossible to stop all the clocks simultaneously in both frames. Relativity is a consistent picture, and your problem is a non-issue for it. Your only option is to dispute that relativity is real.

I have agreed that is what Relativity says and I agree that what I claim is Relativity is not real. It can only be perception. Not difficult to understand once you realize that reciprocity mandates no net time dilation between clocks.

Simply saying that "Relativity says these events are not simultaneous" does not make them unsimultaneous since precalibrated relavistic timers were used to correct simultaneity in the first place.

This is your real argument.

Your real argument, when it comes down to it, is not that relativity is internally inconsistent, but that the relativity of simultaneity is not real. So, we are back to absolute time, which has been your underlying philosophy all along.

Your "precalibrated timers" used to "correct simultaneity" only work if relativity is wrong. If relativity is right, they do not do what you claim, as has been explained.

False. They function if Relativity is right. You seem to not understand the manner in which I say All stop simultaneously. There is a differance between apearing to stop and actually being stopped. The physical principle of timers is valid. It is the concept of relativity which is shown invalid by their use.

Where do we go from here, then?

We go back to the start. It is up to you to show that absolute time exists, and relativity is wrong.

I have.

You have utterly failed to show that relativity is inconsistent within itself.

Never said it wasn't. Never tried to jprove that at all.

[quote\James R]In fact, you have explicitly conceded that point, above, at least regarding the simultaneity issue.

So, can you show that relativity is not real?[/quote]

I have.

 

[MacM]

Seeing as you're introducing a new scenario, can I take it you now admit your original one is dead in the water?

Absolutely not. The presentation stands without effective rebuttal. However, your resistance to acknowledging the conclusion requires another approach .

Let's look at the new one, then.

Two clocks A and B are seperated in space at relative rest and are some acceptable distance apart with a start controller located midway between the clocks.

That controller sends a signal to each clock, when received it sets timers in each clock to a time based on an acceleration schedule for clock B such that from each clocks perspective once B has reached 0.9c and stopped accelerating both clocks start according to the precalculated time.

Let me make sure I have this right. A remains stationary. B, upon receiving the signal, accelerates towards A until it reaches a relative speed of 0.9c. The instant it reaches 0.9c, clocks A and B both start to tick. Is that right?

Yes.

Now, we just need to know in whose frame they start to tick simultaneously to complete the specification of the test. When the clocks are started, clocks A and B are not in the same location, and they are travelling at 0.9c relative to one another. Relativity therefore says that they cannot start simultaneously in both frames, but only in one frame.

I didn't specify any simultaneous relative frames. I specified they start by their own FOR timer setting. That time is based on the amount of time that is calculated for B to reach his coasting velocity of 0.9c.

For now, I will assume they start simultaneously in A's frame, but not in B's, unless you want to change this. Remember, you have agreed that they cannot be started simultaneously in both frames if relativity is correct, and you have given no reason to suppose that relativity is not correct, so far.

Wrong. They are not simultaneous to any frame they start when calculated times for B to achieve 0.9c have accumulated from each of their respective views in their own FOR.

The test is designed such that the acceleration ends and timing starts when the moving clock is 9 light-hours seperation between clocks. This results in the 10 hours test before the clocks collide.

In which frame is the 9 light-hours measured? A's frame, or B's frame? Because, as you are aware, length contraction effects in relativity say that the distance measured in two co-moving frames will be different. Again, unless you say something to the contrary, I will take it you are measuring the 9 light hours in A's frame, but not in B's.

I hadn't considered contraction but you can go with "A" since the next step will be to reverse which clock is stationary and which clock accelerates.

In frame A, given the above assumptions:

B needs to travel 9 light hours at speed 0.9c, which will take 10 hours.
Clock A will read 10 hours at the time of collision.

In frame B, given the assumptions:

A needs to travel 3.92 light hours at a speed 0.9c, which will take 4.35 hours.
Clock B will read 4.35 hours at the time of collision.

Not disagreeing with your numbers at this time but noting that your presentation has "A" traveling and it was designated as at rest?

At the instant of the collision, both observers will agree that clock A reads 10 hours and clock B reads 4.35 hours.

In B's frame, clock A was started before clock B, which accounts for the longer time displayed on that clock, even though clock A runs slower than clock B as seen by B. Similarly, clock A sees clock B running slow, but A and B were started simultaneously, which accounts for the longer time displayed on clock A in this frame.

I didn't specify a FOR of referance for simultaneity. I specified the internal clock time to reach 0.9c. That must be correlated to the respective start time for each view.

Can you show that this relativistic explanation is wrong?

I won't do any calculations until we can agree on start times set into each clock based on the time required for B to reach 0.9c.

Do you plan on addressing the issues I raised in my previous post?

I already have.

 

[James R]

MacM,

You are backpeddling on statements you previously agreed to. For reference:

You have now agreed that relativity predicts that clocks A and B cannot stop simultaneously in the reference frames of both A and B. If they stop simultaneously in frame A, then relativity says they cannot stop simultaneously in frame B.

Correct. That is what Relativity says but that is perception and not reality.

We'll get back to this below.

Now you are backpeddeling. You have already agreed that 15,692 is the same time in both frames. By accepting Relativity in my test when "A" predicts "B" reads 15,692, B does read 15,692. That time is the same in both frames.

No.

B reads 15692 only when A reads 36000 in A's frame.
B reads 15692 only when A reads 6840 in B's frame.

B's display is "the same time" in both frames, but A's corresponding simultaneous times are different in each frame, due to the relativity of simultaneity.

A stops at 36,000 seconds and B stops at 15,692 seconds and stops MA; which happens to be the same point in time in accordance with Relativity.

Only in A's frame, not B's.

Agreed as to the observation but not as to actual stopped clocks. This is observation by a remote observer in motion and not the physical status of the clock.

This is where you go back on another previously-agreed statement. From above:

3. Special relativity says that the relativity of simultaneity is an effect which is totally independent of any delays caused by transmitting information.

True.

You cannot now claim that relativity says this effect is due to "observation" and not the "physical status" of the clock, since you previously agreed that the relativity of simultaneity is not an observational effect, but a real effect, according to the theory of relativity.

The most you can claim is that relativity is wrong, and that relativity's claim that this is a real effect is wrong. But you haven't provided any evidence to back up that claim.

(For reference, if needed

3. Clock MA stops at the same time as B, since it is with B, in either frame.
4. Clock MB stops at the same time as A, in either frame.

Agreed.

You: Do you agree that ALL three are stopped simultaneously? Yes or No.

Me: In frame A, yes. In frame B, no.

You: Here is where we disagree. Not as to the observation but as to the real status of clocks. The fact that B has seen A running slow and therefore is not seeing it stop in real time is observational and not A's actual physical status.

Yes, this is where we disagree. I say, and relativity says, this is A's actual status. Your job is to prove it isn't.

You: Is 6,840 seconds recorded by MA the correct time that A should stop in B's FOR when B = 15,692? Yes or No.

Me: No. A doesn't stop until it reaches 36000.

You: I think you screwed up here. Have another look. If B reads 15,692 seconds then Relativity predicts that B sees A as being 6,840 seconds when B stops.

I didn't screw up. You are correct that B sees A as being 6840 when B stops. But at this time, A hasn't stopped yet, according to B. Remember?

Remember B is stopping in B's FOR and B has a view of A in B's FOR. When B = 15,692, A must equal 6,840.

Agreed, in B's frame of reference only, not in A's.

I agree in this test that MA stops at the predicted time for B and that B sees A continue to run until it sees it stop at it actual stopped condition (which actually happened 29,160 seconds before B saw it happen).

Remember we factored out the signal delays. It is irrelevant when B "sees it happen". We're interested in the actual time it happened, measured in each reference frame. Relativity of simultaneity has nothing to do with signal delays. You agreed to that.

But keep in mind if I put a timer in B to stop A when B stops at 15,692, since 36,000 and 15,692 exist simultaneously in both frames B cannot stop A at 6,840 since A is already at 36,000.

36000 and 15692 are not simultaneous in both frames. You previously agreed that relativity says they cannot be.

Further if you do claim to stop A at 6,840 then A claims B is 2,892 not 15,692.

Right. And that would be a different test, wouldn't it?

Me: A stops when it displays 36000 in either frame. In A's frame, it displays 36000 at t=36000. In B's frame, it displays 36000 at t'=82590.

You: Agreed. Except B isn't timing to see the end of A. B stops at 15,692, MA stops at 6,840 stops; which is the time B claims A should read if stopped simultaneously between A and B.

6840 is the time B claims that A would read if it had stopped at the same time as B in B's frame. But A didn't stop at the same time as B in B's frame.

You must remember that the functions are taking place simultaneously and Relativity is creating a conflict of "Physical", not "Observationsal" possibilities between clocks.

According to relativity, the expression "the functions are taking place simultaneously" requires a reference frame. Otherwise it is meaningless. You previously agreed to that. See the quote at the top of this post.

Wrong answer. MA reads 6,840 simultaneous with B = 15,692 and A = 36,000.

I agree with that, in A's reference frame, but not in B's.

You keep failing to specify reference frames. That is sloppy. You keep trying to drag in the concept of absolute time by stealth, and I have to keep pulling you up on it. You can't simply assume, without proof, that what is simultaneous in one frame is simultaneous in all frames.

You need to prove your assumption, not just assert it over and over, or assume it.

Wrong answer but I see what you are wanting to do. You are declaring MA a false signal claiming it doesn't happen in A's frame but my point is that it is happening in B's frame. It is B's view of A.

I agree it gives an accurate reading of clock A in B's frame. It does not, however, given an accurate reading of clock A in A's frame. It was not set up to do that.

Now try linking your frames to function concurrently which is what happens in reality.

This is what you need to prove. This is your assumption of universal time.

Can you prove it?

That is to say A sees B and B sees A concurrently in reality.

And relativity says "concurrently" (another word for "simultaneously") is different in each frame, and that difference is real.

Try linking those signals to control the clocks. Got a problem don't you?

No. I've shown this is no problem for relativity. Relativity does not require absolute simultaneity. Only you require that.

Go ahead put a timer in A that shuts it down when B sees 15,692 in B's FOR.

Do you mean B sees 15692 on B's clock, or on clock MA? If B, then we have the same test as always. If MA, then you've changed the test again. You need to stick to one test at a time. Stop going off on tangents.

Duh indeed. If such declarations present physical impossibilities, let me suggest you must leave them behind if you want to talk physics instead of mysticism.

Fine. I'm quite happy to have that discussion.

Can we agree then that the relativistic explanation is internally consistent? In other words, do you agree that if relativity was correct, my presented explanation of this test would be correct?

If so, we can move on an discuss whether or not relativity is correct in the real world or not. If not, then we'll have to keep plugging away until you understand the relativistic explanation. (Yes, yes, I know you think you already understand it. Don't bother saying so.)

You do believe in physical realities do you not or have you accepted voodoo for science?

Back at you, MacM. Relativity is physical reality.

We certainly agree that A cannot have two different readings at the same time. We disagree that MA = 6,840 doesn't apply. It is inherent in that B =15,692 in B's FOR and when it reads 15,692 Relativity predicts A must read 6,840.

...in B's frame of reference, but not in A's.

You need to be clear about this, or we can't proceed.

Please explain in a mutual relative motion scenario where Relativity is functioning concurrently on clocks how B can have two different accumulated times of 15,692 seconds.

It can't. It only ever displays one reading at any time, according to each observer.

The test is arranged such that A's test period is 36,000 seconds. B's test period is 15,692 seconds such that they run concurrently and stop simultaneously.

... in A's frame, but not in B's.

If B's accumulated time is 15,692 seconds in its test period "A" MUST be at 6,840.

... in B's frame, but not in A's.

I understand that "A" read 6,840 when B stopped ...

in B's frame

but B sees it continue to run until it's 36,000 actual stop time because it was stopped in A's FOR.

Correct.

However, what you are failing to do is to now place a timer in A to cause it to stop when B reads 15,692 (which wouldn't change anything)...

if the timer is in A's frame, not in B's

... or to put a timer in A to stop it at 6,840 but which will show that B now only reads 2,892 not 15,692.

which would be changing the test.

Considering that ALL three clocks are started/stopped simultaneoulsy,...

in A's frame but not in B's

... this dictates that Relativity has the problem and not the description of the test.

It dictates that you have a problem of failing to adequately distinguish different frames, and of trying to sneak in absolute time everywhere.

I have agreed that is what Relativity says and I agree that what I claim is Relativity is not real. It can only be perception. Not difficult to understand once you realize that reciprocity mandates no net time dilation between clocks.

If "reciprocity" mandates no net time dilation in this test, then "reciprocity" is not in accordance with relativity. Remember, you previously agreed on what relativity says about the relativity of simultaneity. Explicitly, see the quote at the top of this post.

Simply saying that "Relativity says these events are not simultaneous" does not make them unsimultaneous since precalibrated relavistic timers were used to correct simultaneity in the first place.

This is just wrong. Your "monitors" do not act as you would like them to. They might work if absolute time existed, but unfortunately it doesn't.

False. They function if Relativity is right. You seem to not understand the manner in which I say All stop simultaneously. There is a differance between apearing to stop and actually being stopped. The physical principle of timers is valid. It is the concept of relativity which is shown invalid by their use.

In response, I again refer you to a previously agreed statement:

Me: 3. Special relativity says that the relativity of simultaneity is an effect which is totally independent of any delays caused by transmitting information.

You: True.

 

[James R]

MacM:

Regarding your new scenario.

I won't do any calculations until we can agree on start times set into each clock based on the time required for B to reach 0.9c.

Good. We can't jump into calculations until this is agreed, so you are taking exactly the right approach.

Please set out in detail how the clocks are to be started. If the start time relies on calculations, make sure you specify which frame any relevant observations which lead to inputs into the calculation are made in, because, as you have seen, such observations can be different in different frames. That is why we must be very careful.

I will wait until you specify all the details you feel are necessary. It's your test, after all. If I don't have enough information, I'll let you know.

 

[GMontag]

Mac, just curious...

Do you disagree with either or both of SR's postulates?

For reference the postulates are:

1. The speed of light is measured to be the same to all observers regardless of relative motion

2. The laws of physics can be written in such a way that they are the same for all observers regardless of relative motion.

 

[Quantum Quack]

Please excuse me for butting in here again but I wanted to ask the following question of James R.

James how does SR prove absolute time does not exist? Now I mean PROVE not postulate.?

What proof is there that absolute time does not exist?

Time dilation in itself is not proof. As atomic slowing does not mean that time is less absolute.

Just thought I'd ask this rather obvious question as you are asking MacM to provide proof of absolute time.

 

[Quantum Quack]

Say for arguments sake, one of the by-products of having velocity is the slowing of atomic rates thus the effect of time dilation. Why would this argue a case for the nulification of absolute time?
Is there another form of proof that absolute time is defunkt?

 

[MacM]

James R,

I will not dodge responding to your reply as to the first scenario but I happen to agree that we are making progress better in this last case. So I want to address it first and give you the opportunity to consider something interesting.

You stated making the assumptions you made (and I'm content to go with them and not debate a different presentation) that BOTH will agree that A = 10 and B = 4.35 hours.

Is that correct? Yes or No. Are you going to want to change that opinion keeping the assumptions you made. This is important since I am going to hold your feet to the fire and you will not substitute some other arguement when I expose the truth.

I await your answer.