Twin paradox (Pete and MacM)

[Pete]

I do wish you and others here would take the wax out of yours ears. I have not claimed events simultaneous in one frame are simultaneous in any other frame.

Oh, good.

Then when you said:

And finally. It IS my view that when "A" stops at 10 hours and B stops at 4.359 hours they do IN FACT stop simultaneously in real time. I know you will argue that point and say "Prove it" but that frankly is not only nonsense which requires relativity to be wrong in first principles.

You meant that they stopped simultaneously (after 36000 seconds) in A's frame only?

And you acknowledge that in B's frame they don't stop simultaneously?
That in B's frame, clcok B stops after 15692 seconds and clock A stops after 82590 seconds?

 

[MacM]

Oh, good.

Then when you said:

You meant that they stopped simultaneously (after 36000 seconds) in A's frame only?

And you acknowledge that in B's frame they don't stop simultaneously?
That in B's frame, clcok B stops after 15692 seconds and clock A stops after 82590 seconds?

Well, you at least got half of it right. We agree that "A" stops at 36,000 seconds and "B" simultaneously stops at 15,692 seconds.

Now since "B" sees "A" reading only 6,840 seconds when his clock stops which is to indicate the end of the test ant that "A" has stopped, it defies "B" view of Relativity.

But more importantly is the fact that you talk outright nonsense when you claim "A" runs to 82,590 seconds. It has stopped and the only thing running is "B's" view of "A" but only upto 36,000 seconds.

What you mean to say is that "B" runs to 82,590 seconds to see "A" reach 36,000 seconds. Which is rather hard to justify since "B" stopped at 15,692 seconds and is no longer running.

 

[Pete]

But more importantly is the fact that you talk outright nonsense when you claim "A" runs to 82,590 seconds. It has stopped and the only thing running is "B's" view of "A" but only upto 36,000 seconds.

What you mean to say is that "B" runs to 82,590 seconds to see "A" reach 36,000 seconds. Which is rather hard to justify since "B" stopped at 15,692 seconds and is no longer running.

I said what I mean to say. If you choose to misinterpret, there's not much I can do.


Why do you think B must be running in order to "see" A run?
Do you imagine that B's perception of A has some effect on A?

 

[Pete]

Well, you at least got half of it right. We agree that "A" stops at 36,000 seconds and "B" simultaneously stops at 15,692 seconds.

Now since "B" sees "A" reading only 6,840 seconds when his clock stops which is to indicate the end of the test ant that "A" has stopped, it defies "B" view of Relativity.

In B's frame:
Why do you think that the test ends when "B" stops?
The test doesn't end until both clocks have stopped.

 

[MacM]

I said what I mean to say. If you choose to misinterpret, there's not much I can do.


Why do you think B must be running in order to "see" A run?
Do you imagine that B's perception of A has some effect on A?

No. But you seem to think so. Now reverse directions and have a velocity toward clock "A". Tell me relative velocity affects are not simply "delayed information" affects.

 

[MacM]

Mac, I think you need to carefully examine the concepts involved here.

You stated that you accept that simultaneity alter due to perspective. It's time to clearly state exactly what you mean.

Think about the scenario of two objects (A and B) firing signals toward each other.
It is clear that when A and B have equal and opposite velocities (our mean velocity frame), that if the signals are emitted simultaneously they are received simultaneously.

However, in this diagram:
<img src="/attachment.php?attachmentid=3349&stc=1">

It is clear that in A's frame, when the signals are emitted simultaneously, they are not received simultaneously. Similarly, if the signals are received simultaneously, they are not emitted simultaneously.

Now... think carefully about what that means...

It might help if you think of the signals as powerful lasers that will destroy their targets.

In one frame, A and B are destroyed simultaneously - these are real events!
In other frames, one is destroyed before the other...

So which one is really destroyed first? Does this question have an absolute answer? Or does the question lack information - is a reference frame needed to provide a meaningful answer?

I have my computer back now and will begin to respond to your diagrams and text.

A referance frame does not alter the simultaneity of the events. It provides a mathematical means of determining the reality of the simultaneity from the distorted "Perception" of the moving frame.

 

[MacM]

Now, we can look at the same situation in B's frame.
Here is the situation when B emits its flash:

<img src="/attachment.php?attachmentid=3358&stc=1">​

Note that initial distances are adjusted for length contraction, and that the emission of light from A is not marked because we don't yet know when that happened.
Rolling forward 0.87 seconds, B's light flash reaches E... and now we know where A's flash is as well!

<img src="/attachment.php?attachmentid=3359&stc=1">​

This means that if we want to, we can work backward and determine that A emitted that flash 0.29 seconds earlier, or 0.58 seconds after B emitted its flash.

Rolling forward another 0.87 seconds, we find that the two flashes reach their receivers simultaneously (interesting!!).

<img src="/attachment.php?attachmentid=3360&stc=1">​

So in B's frame, B receives a flash 1.73 seconds after emitting, and A receives a flash 1.15 seconds after emitting..

How did you go?

For reader convenience and efficient reuse, I have also combined all diagrams into a simgle image: Combined diagrams

You cannot merely adjust distance (Length Contraction) for "B's" view only. It is either 0.5 light seconds distance for both or it is 0.43 light seconds for both.

Reciprocity holds.

 

[Silas]

You cannot merely adjust distance (Length Contraction) for "B's" view only. It is either 0.5 light seconds distance for both or it is 0.43 light seconds for both.

Reciprocity holds.

Er, no, the length contraction always happens to the other object, not to the reference frame object. If I'm riding in a 40ft long cabin in a spaceship at 0.95c, I see the cabin as 40ft, and myself as approximately 18 inches from front to back (I'm quite fat). It is in this way that I'm still able to see that if I were to measure the speed of light within my cabin it is still 3x10⁸m/s. This is where relativity comes from.

I'm afraid I don't have the math to refute your point about the energy, but I'm pretty sure that relativity does not require that the energy applied to the one object is relative to the other object. The energy required to accelerate an object to 0.95c applies to 0.95c compared to the comparatively at-rest Universe.

If the energy to accelerate a car to 60mph perpendicular to the quasar mass is x, and then we turn a corner and happen to be accelerating towards the quasar mass, according to you this would involve having to expend 3.2x in energy to drive in that direction. This would surely imply "spooky action at a distance", which I'm sure neither you believe nor Einstein believed in. In any case, 3.2 times the energy is the total multiplier of the amount of energy to accelerate an object from rest to 0.95c according to Newton. To increase velocity from 0.95c to 0.95c + 60mph does not take 3.2 times the energy required to accelerate from 0 to 60mph.

I will say, however, that it has been interesting discussing these issues with you, Dan. I've certainly learnt a great deal about relativity during it!

 

[Yuriy]

Please, see a new thread "One Lecture on Twins Paradox"

 

[Persol]

I'm afraid I don't have the math to refute your point about the energy

He seems to have dropped it when everyone pointed out that it was wrong.

 

[Pete]

I do wish you and others here would take the wax out of yours ears. I have not claimed events simultaneous in one frame are simultaneous in any other frame.

A referance frame does not alter the simultaneity of the events.

I do wish you would make up your mind, Mac.


Do you think that what is simultaneous in one frame is simultaneous in all frames, or not?

 

[Pete]

Reciprocity holds.

You keep saying so, but can't prove it...

Can you say AD HOC FIAT???

 

[Pete]

You cannot merely adjust distance (Length Contraction) for "B's" view only. It is either 0.5 light seconds distance for both or it is 0.43 light seconds for both.

Are you sure?

Don't forget that D, E, and A are all stationary with respect to each other. They could be rigidly connected.

Imagine that they are connected by a long rod, and think it through again.

 

[Pete]

It provides a mathematical means of determining the reality of the simultaneity from the distorted "Perception" of the moving frame.

In this diagram, who's perception do you think is reperesented, and exactly how do you think it is distorted?
<img src="/attachment.php?attachmentid=3349&stc=1">

It isn't A's perception... All that A has perceived so far is that it emitted a light flash. A has yet to receive any information from B.

So who's perception do you think is shown? Is it a perception-based diagram at all??

 

[MacM]

Er, no, the length contraction always happens to the other object, not to the reference frame object. If I'm riding in a 40ft long cabin in a spaceship at 0.95c, I see the cabin as 40ft, and myself as approximately 18 inches from front to back (I'm quite fat). It is in this way that I'm still able to see that if I were to measure the speed of light within my cabin it is still 3x10⁸m/s. This is where relativity comes from.

Certainly the observer at rest always sees no affects of Relativity but sees the afrfects of the other that is moving relative to him.

The point you seem to have missed is that Reciprocity is a matter of swapping the observers views. Relativity mandates that all motion is relative. that is each observer sees himself as at rest. When you compare those views Reciprocity eliminates any actual physical differentials.

I'm afraid I don't have the math to refute your point about the energy, but I'm pretty sure that relativity does not require that the energy applied to the one object is relative to the other object. The energy required to accelerate an object to 0.95c applies to 0.95c compared to the comparatively at-rest Universe.

If the energy to accelerate a car to 60mph perpendicular to the quasar mass is x, and then we turn a corner and happen to be accelerating towards the quasar mass, according to you this would involve having to expend 3.2x in energy to drive in that direction. This would surely imply "spooky action at a distance", which I'm sure neither you believe nor Einstein believed in. In any case, 3.2 times the energy is the total multiplier of the amount of energy to accelerate an object from rest to 0.95c according to Newton. To increase velocity from 0.95c to 0.95c + 60mph does not take 3.2 times the energy required to accelerate from 0 to 60mph.

James has added technical detail where I was being pretty general. But his mathematics shows an even greater affect than I was claiming.

I will say, however, that it has been interesting discussing these issues with you, Dan. I've certainly learnt a great deal about relativity during it!

Well, it is a pleasure to have a dialog which is based on the issues and not repeated personal attacks. Some here however, seem ill equipped to actually discuss the matter and so choose to jpretend to be superior but they aren't. There failures clearly show that.

 

[MacM]

Er, no, the length contraction always happens to the other object, not to the reference frame object. If I'm riding in a 40ft long cabin in a spaceship at 0.95c, I see the cabin as 40ft, and myself as approximately 18 inches from front to back (I'm quite fat). It is in this way that I'm still able to see that if I were to measure the speed of light within my cabin it is still 3x10⁸m/s. This is where relativity comes from.

Certainly the observer at rest always sees no affects of Relativity but sees the afrfects of the other that is moving relative to him.

The point you seem to have missed is that Reciprocity is a matter of swapping the observers views. Relativity mandates that all motion is relative. that is each observer sees himself as at rest. When you compare those views Reciprocity eliminates any actual physical differentials.

I'm afraid I don't have the math to refute your point about the energy, but I'm pretty sure that relativity does not require that the energy applied to the one object is relative to the other object. The energy required to accelerate an object to 0.95c applies to 0.95c compared to the comparatively at-rest Universe.

If the energy to accelerate a car to 60mph perpendicular to the quasar mass is x, and then we turn a corner and happen to be accelerating towards the quasar mass, according to you this would involve having to expend 3.2x in energy to drive in that direction. This would surely imply "spooky action at a distance", which I'm sure neither you believe nor Einstein believed in. In any case, 3.2 times the energy is the total multiplier of the amount of energy to accelerate an object from rest to 0.95c according to Newton. To increase velocity from 0.95c to 0.95c + 60mph does not take 3.2 times the energy required to accelerate from 0 to 60mph.

James has added technical detail where I was being pretty general. But his mathematics shows an even greater affect than I was claiming.

I will say, however, that it has been interesting discussing these issues with you, Dan. I've certainly learnt a great deal about relativity during it!

Well, it is a pleasure to have a dialog which is based on the issues and not repeated personal attacks. Some here however, seem ill equipped to actually discuss the matter and so choose to pretend to be superior but they aren't. There failures clearly show that.

 

[MacM]

In B's frame:
Why do you think that the test ends when "B" stops?
The test doesn't end until both clocks have stopped.

You just agreed that "A" stops at 36,000 seconds and that "B" simultaneously stops at 15,692 seconds.

How do you believe tha tthe test continues until 82,590 seconds?

What actually happens is "B" sees his clcok stop at 15,692 seconds (which means it never records 82,590 seconds or even 15,692 seconds. It is stopped. He however sees "A" clock read 6,840 seconds, which is what it should read according to him for his recorded stopped time; however, he notices that "A" keeps running in volation of his expected time according to Relativity.

When the clocks are brought back together "A" reads 36,000 seconds and "B" reads 15,692 seconds. What happened to B's" view of "A" as being 6,480 seconds for his 15,692 seconds.? It was never recorded since it was not the reality of clock "A". It was a perception by "B" but not reality.

 

[MacM]

He seems to have dropped it when everyone pointed out that it was wrong.

Don't you ever get tired of being wrong? I haven't dropped anything. the only thing wrong was the numerical value I was claiming. James fixed that.

The bottom line is you cannot claim increased energy as a function of relative velocity and then simply disregard energy required to accelerate in a frame which includes Quasar ejecta moving at 0.95c.

Relative velocity is relative velocity. You can't claim the Quasar material is moving and we are not. Either position is valid in Relativity. It is "Relative Velocity" which is used to compute energy.

As James has pointed out I believe it was about 124/1 times the energy to go from 0.95c to 0.96c vs going from 0 - 0.1c. My examples are valid physics. If Relativity were reality then we would see pertabations in our physics of F = ma even at sub-luminal velocities. We see NONE. Guess what that means?

 

[Persol]

If Relativity were reality then we would see pertabations in our physics of F = ma even at sub-luminal velocities. We see NONE. Guess what that means?

What a dumbass. At the velocities we travel the change is minimal.

You stated that we should have measured some change in F with regard to the quasar. This is impossible, as we are not in the quasar's frame of reference, but a frame extremely similar to whatever we are measuring.

 

[MacM]

I do wish you would make up your mind, Mac.


Do you think that what is simultaneous in one frame is simultaneous in all frames, or not?

In a different frame? NO. But you are avoiding the fact that by reversing the observer at rest's view the problem is negated. There is no net affect in physical reality.

i.e. - Both clocks slow equally. Both masses increase equally. Lorentz Contraction shrinks distance equally.

This is called reciprocity and is mandated by Relativity itself.