Graphical Derivation of the CADO Equation

[Neddy Bate]

I will just address this one claim that Charlie must receive the information first-hand, since it is particularly egregious:

You've missed the point of the clock blowing up. In that case, Charlie's statement that Alice's clock reached 10 is false. No matter how much he stares at a local clock, no matter how well that clock is (was) synced up to Alice's, he is wrong.

Charlie does not state that Alice's clock has currently reached 10. Charlie states that Alice's clock would currently display 10 if it were properly functioning and Einstein-synchronised to all of the other clocks in the stay-home frame.

This is the whole point in equations such as t = γ(t' + (vx' / c²)). They can be used to calculate the time on distant and relatively moving clocks which are functioning properly and synchronised according to the standard configuration of SR. Of course the question of whether or not any particular clock has been vandalized before it reaches a particular time is not addressed by the equation, nor can it be. I don't see where you think you are going with this non-argument.

It's a clear demonstration that Charlie (at that time) has no information about Alice's clock around the time it would have been showing 10. So Charlie cannot make any definitive statements (yet) about the state of Alice's clock.

Information cannot travel faster than c in either SR or GR, and I never claimed otherwise. I don't see where you think you are going with this non-argument.

You've forgotten to describe a critical part of your scenario: the information traveling from Charlie's assistent to Charlie. Demonstrate that it does, and that when it reached Charlie, he'll still be of the opinion that Alice's clock ticked backwards.

Okay, let's do that, if it will make you happy. We will give Charlie's assistant his own television broadcast station, complete with transmitter. We will also give Charlie a DVR on the train which he can set to record his assistant's broadcast at a certain time based on the clocks on the train.

Likewise, we will give Alice her own television broadcast station, complete with transmitter. We will also give Charlie a DVR on the ground which he can set to record Alice's broadcast at a certain time based on the clocks on the ground.

In the traveling frame, the distance between Charlie and his assistant is 17.32 light years. So, at the time when both of their clocks display 20, his assistant can point the television camera at Alice's clock displaying 10, with his own clock displaying 20 in the same picture. He broadcasts that picture live on his own personal television network. Charlie sets his DVR to record the broadcast 17.32 years later when the information is due to arrive.

In the meantime, Charlie jumps off the train, and lands on the ground right next to a home-frame-synched clock which displays 40. He quickly sets the ground-based DVR to record Alice's current television broadcast which will arrive 34.64 years later because the distance between himself and Alice is 34.64 light years in the stay-home frame. In Alice's current broadcast, she is showing her own clock displaying 40, and also a train-clock passing her and displaying 80, just for good measure.

Charlie quickly jumps back on the train, while his own clock and his assistant's clock both still display 20. He waits 17.32 years and then turns on his TV. He sees his assistant's broadcast from 17.32 years earlier, and sees Alice's clock displaying 10, with his assistant's clock displaying 20, as expected.

Charlie jumps off the train and takes another super-fast train back to where he left his DVR on the ground. Assuming at least 34.64 years have passed on the ground clocks, he can watch the recording that he scheduled. So he watches it, and sees Alice's clock displaying 40, with a train clock near her displaying 80, as expected.

So Charlie concludes that his calculations were correct. Alice's clock went from 10 to 40 when he jumped off the train, and it went from 40 to 10 when he jumped back on the train. He even has the DVR recordings to show to everyone.

But Charlie does need to receive information about Alice's clock. You continually forget about that part. I wonder why?

Because it does not change the outcome, as I demonstrated above. It only adds a huge delay of time between the time when Charlie makes his calculations, and when he actually verifies them first-hand.

 

[NotEinstein]

I will just address this one claim that Charlie must receive the information first-hand, since it is particularly egregious:



Charlie does not state that Alice's clock has currently reached 10. Charlie states that Alice's clock would currently display 10 if it were properly functioning and Einstein-synchronised to all of the other clocks in the stay-home frame.

This is the whole point in equations such as t = γ(t' + (vx' / c²)). They can be used to calculate the time on distant and relatively moving clocks which are functioning properly and synchronised according to the standard configuration of SR. Of course the question of whether or not any particular clock has been vandalized before it reaches a particular time is not addressed by the equation, nor can it be. I don't see where you think you are going with this non-argument.

Great! So you finally agree with me that the conclusions Charlie draws from his calculations aren't (necessarily) real! I'm glad you've finally seen the light. I guess my "non-argument" wasn't that "non" after all, because it certainly managed to convince you!

Information cannot travel faster than c in either SR or GR, and I never claimed otherwise.

And neither have I?

I don't see where you think you are going with this non-argument.

You have once again missed the point. At the moment Charlie calculates Alice's clock to say 10, he can't know that. And as you've just admitted, he indeed doesn't. I guess my "non-argument" wasn't that "non" after all, because I certainly managed to convince you!

Okay, let's do that, if it will make you happy. We will give Charlie's assistant his own television broadcast station, complete with transmitter. We will also give Charlie a DVR on the train which he can set to record his assistant's broadcast at a certain time based on the clocks on the train.

Likewise, we will give Alice her own television broadcast station, complete with transmitter. We will also give Charlie a DVR on the ground which he can set to record Alice's broadcast at a certain time based on the clocks on the ground.

Okay... I don't see why you'd need all that equipement, but sure, let's see where you're going with this...

In the traveling frame, the distance between Charlie and his assistant is 17.32 light years. So, at the time when both of their clocks display 20, his assistant can point the television camera at Alice's clock displaying 10, with his own clock displaying 20 in the same picture. He broadcasts that picture live on his own personal television network. Charlie sets his DVR to record the broadcast 17.32 years later when the information is due to arrive.

In the meantime, Charlie jumps off the train, and lands on the ground right next to a home-frame-synched clock which displays 40. He quickly sets the ground-based DVR to record Alice's current television broadcast which will arrive 34.64 years later because the distance between himself and Alice is 34.64 light years in the stay-home frame. In Alice's current broadcast, she is showing her own clock displaying 40, and also a train-clock passing her and displaying 80, just for good measure.

Charlie quickly jumps back on the train, while his own clock and his assistant's clock both still display 20. He waits 17.32 years and then turns on his TV. He sees his assistant's broadcast from 17.32 years earlier, and sees Alice's clock displaying 10, with his assistant's clock displaying 20, as expected.

Charlie jumps off the train and takes another super-fast train back to where he left his DVR on the ground. Assuming at least 34.64 years have passed on the ground clocks, he can watch the recording that he scheduled. So he watches it, and sees Alice's clock displaying 40, with a train clock near her displaying 80, as expected.

So Charlie concludes that his calculations were correct. Alice's clock went from 10 to 40 when he jumped off the train, and it went from 40 to 10 when he jumped back on the train. He even has the DVR recordings to show to everyone.

You haven't demonstrated squat. You've just put a (very verbose) story around your scenario. Where are the calculations of the traveling information? Where is the Minkowski diagram? All you've done boils down to more of what you said earlier.

Because it does not change the outcome, as I demonstrated above.

You only demonstrated that you can write a story.

It only adds a huge delay of time between the time when Charlie makes his calculations, and when he actually verifies them first-hand.

Well, now that you claim to have it all figured out, draw the Minkowski diagram that Charlie would, after this entire ordeal. What does Alice's worldline does in it? And have Alice make one too. What does her worldline does in it? Remember, there can only be one truth/reality: the two Minkowski diagrams must be identical, except they can only be different by a single boost.

 

[NotEinstein]

Well, now that you claim to have it all figured out, draw the Minkowski diagram that Charlie would, after this entire ordeal. What does Alice's worldline does in it? And have Alice make one too. What does her worldline does in it? Remember, there can only be one truth/reality: the two Minkowski diagrams must be identical, except they can only be different by a single boost.

And before you accuse me of just flailing around, focus on this question specifically:

In the meantime, Charlie jumps off the train, and lands on the ground right next to a home-frame-synched clock which displays 40. He quickly sets the ground-based DVR to record Alice's current television broadcast which will arrive 34.64 years later because the distance between himself and Alice is 34.64 light years in the stay-home frame. In Alice's current broadcast, she is showing her own clock displaying 40, and also a train-clock passing her and displaying 80, just for good measure.

So the broadcast that is going to be recorded, will have Alice's clock display 40 as she jumps onto the train.

(And in what post did you calculate that 80? I don't remember.)

Charlie quickly jumps back on the train, while his own clock and his assistant's clock both still display 20. He waits 17.32 years and then turns on his TV. He sees his assistant's broadcast from 17.32 years earlier, and sees Alice's clock displaying 10, with his assistant's clock displaying 20, as expected.

So the broadcast that is going to be seen, has Alice's clock displaying 10 as she jumps onto the train.

Alice jumping onto the train is one event. You and I agreed earlier that it will have one unique set of (x, t) associated with it when expressed in the coordinates of a specific reference frame. What value does t have in Alice's frame (so the value displaying on the clock right next to her in the home-staying frame) when she jumps onto the train? Because right now, you have a video recording of two values: 10 and 40. How do you resolve this conflict?

Before you say that that's not a problem, let me introduce another bomb. Alice is standing on a pressure place, connected to her clock. If she jumps off of it when the clock reads 10, it won't explode. Any other reading, it detonates. Does the bomb explode when Alice jumps off of it, onto the train?

And if you're now going to say that the two broadcasts are not of the same event (namely, the moment Alice jumps onto the train), then it's not surprising the times don't match up: you're compare two different events.

 

[NotEinstein]

*clicked reply instead of edit*

 

[phyti]

I've got another question. Imagine there is a triplet: Alice, Bob, and Charlie. Alice and Bob

complete stop (with respect to Alice and Bob). In other words, v=0v=0. Afterwards, according to your formula, all will agree how old Alice is, because with v=0v=0your equation reduces to: CADOT=CADOHCADOT=CADOH.
But now, both Bob and Charlie start moving (both in the same direction that Charlie was moving in earlier) with the same speed (using the same acceleration profile, if that's relevant). Even though they are in the same reference frame all the time, Bob and Charlie will now disagree on how old Alice is, due to Charlie starting with L=DL=Dwhile Bob starts with L=0L=0. That seems weird to me; I'm pretty sure SR predicts that Bob and Charlie would agree on Alice's age, because in SR time dilation is a function only of speed, not distance.

How do you resolve this conflict?

With the A-frame as a reference, C moves in x at .6c. At time A5, B moves in x at .6c.
The red axis provides simultaneous events for A and C. At A5, C is 1 t-unit behind B. The blue lines represent C sending a light signal to A for a clock image (A3.2), and assigning the event to C4, which is part of the clock synch convention. At A10, A and B assign different times for A. When A and B both read t8, the assigned A events are 1 unit different. B had a different speed profile than C, which results in different assignments.

The C-frame reference is on the right.

 

[Neddy Bate]

Alice jumping onto the train is one event. You and I agreed earlier that it will have one unique set of (x, t) associated with it when expressed in the coordinates of a specific reference frame. What value does t have in Alice's frame (so the value displaying on the clock right next to her in the home-staying frame) when she jumps onto the train? Because right now, you have a video recording of two values: 10 and 40. How do you resolve this conflict?

In my above 'story' it is only Charlie who jumps anyplace. When his clock displays 20, he quickly sets his train-DVR to start recording in 17.32 years, then he jumps off the train, quickly sets the ground-DVR to start recording in 34.64 years, and jumps right back on the train.

In the end there are two videos. One shows Alice with her clock displaying 10, and one shows Alice with her clock displaying 40. I didn't necessarily have her jumping anyplace in my scenario, but we can add that if you want.

Let's say she jumps on the train when her clock displays 10, and momentarily stands next to Charlie's assistant while he broadcasts his video. Then she immediately jumps back off the train. The video broadcast by Charlie's assistant will show her jumping on the train and jumping back off, with her clock displaying 10 while his displays 20.

When her clock displays 40, Alice broadcasts her own video. Let's say that she also decides to jump on and off the train at that time. Her video broadcast will show her jumping on the train and jumping back off, with her clock displaying 40. The train assistant that she lands next to will have a clock displaying 80.

I don't know if I showed the calculation for 80 before, so here it is:
t' = γ(t - (vx / c²))
where:
γ = 2
t = 40
v = 0.866
x = 0
t' = 2(40 - (0.866 * 0))
t' = 80

Before you say that that's not a problem, let me introduce another bomb. Alice is standing on a pressure place, connected to her clock. If she jumps off of it when the clock reads 10, it won't explode. Any other reading, it detonates. Does the bomb explode when Alice jumps off of it, onto the train?

It won't explode the first time she jumps, because her clock displays 10 at that time. The second time she jumps, it will explode.

And if you're now going to say that the two broadcasts are not of the same event (namely, the moment Alice jumps onto the train), then it's not surprising the times don't match up: you're compare two different events.

Yes, of course it is two different events. The event where Alice's clock displays 10 is a totally different event than the event where her clock displays 40.

But the two most important things to remember are these:
1. The event where Alice's clock displays 10 is simultaneous (in the train frame) with Charlie jumping off/on the train.
2. The event where Alice's clock displays 40 is simultaneous (in the ground frame) with Charlie jumping off/on the train.

 

[Neddy Bate]

Where are the calculations of the traveling information?

The distance between Alice and Charlie (as measured by the train frame) when Charlie's clock displays 20:
d' = vt'
d' = 0.866 * 20
d' = 17.32 light years
So the time required for the train assistant's broadcast to reach Charlie's train-DVR is 17.32 years (as measured by the train frame).

The distance between Alice and Charlie (as measured by the ground frame) when Alice's clock displays 40:
d = vt
d = 0.866 * 40
d = 34.64 light years
So the time required for Alice's broadcast to reach Charlie's ground-DVR is 34.64 years (as measured by the ground frame).

What did you think the distances were?
How long did you think it would take light to travel those distances?
Did your really think you could make SR fail by just letting light signals travel from Alice to Charlie? Strange.

 

[Neddy Bate]

Dear NotEinstein,

I think these might be the calculations you are looking for. These calculations are all going to be done in Alice's stay-home frame.

At the time Alice's clock displays 10, the light from her clock which is carrying the "showing 10" information is 10-10=0 light years away from Alice, as that light is just being emitted toward Charlie. But at that time, Charlie is a distance of 10*0.866=8.66 light years away, and his clock displays 10/2=5. Remember, this is all from the stay-home frame.

By the time Alice's clock displays 40, the light from her clock which is carrying the "showing 10" information is 40-10=30 light years away from Alice. And Charlie is a distance of 40*0.866=34.64 light years away from Alice, while his clock displays 40/2=20.

By the time Alice's clock displays 74.64, the light from her clock which is carrying the "showing 10" information is 74.64-10=64.64 light years away from Alice. And Charlie is also a distance of 74.64*0.866=64.64 light years away from Alice, while his clock displays 74.64/2=37.32. This demonstrates the event when the light from Alice's clock which is carrying the "showing 10" information arrives at Charlie, because both Charlie and the light are in the same place; both are 64.64 light years away from Alice.

Does that help?

 

[NotEinstein]

With the A-frame as a reference, C moves in x at .6c. At time A5, B moves in x at .6c.
The red axis provides simultaneous events for A and C. At A5, C is 1 t-unit behind B. The blue lines represent C sending a light signal to A for a clock image (A3.2), and assigning the event to C4, which is part of the clock synch convention. At A10, A and B assign different times for A. When A and B both read t8, the assigned A events are 1 unit different. B had a different speed profile than C, which results in different assignments.

The C-frame reference is on the right.
View attachment 2163

I'm not sure your Minkowski diagram matches my scenario? I don't see the clock-sync of C with A right before B starts moving? Also, B and C should have the same speed profile (after the quite irrelevant set-up phase).​

 

[NotEinstein]

In my above 'story' it is only Charlie who jumps anyplace. When his clock displays 20, he quickly sets his train-DVR to start recording in 17.32 years, then he jumps off the train, quickly sets the ground-DVR to start recording in 34.64 years, and jumps right back on the train.

In the end there are two videos. One shows Alice with her clock displaying 10, and one shows Alice with her clock displaying 40. I didn't necessarily have her jumping anyplace in my scenario, but we can add that if you want.

Let's say she jumps on the train when her clock displays 10, and momentarily stands next to Charlie's assistant while he broadcasts his video. Then she immediately jumps back off the train.

And once again you have to modify the scenario in order to make ends meet.

The video broadcast by Charlie's assistant will show her jumping on the train and jumping back off, with her clock displaying 10 while his displays 20.

When her clock displays 40, Alice broadcasts her own video. Let's say that she also decides to jump on and off the train at that time. Her video broadcast will show her jumping on the train and jumping back off, with her clock displaying 40. The train assistant that she lands next to will have a clock displaying 80.

I don't know if I showed the calculation for 80 before, so here it is:
t' = γ(t - (vx / c²))
where:
γ = 2
t = 40
v = 0.866
x = 0
t' = 2(40 - (0.866 * 0))
t' = 80



It won't explode the first time she jumps, because her clock displays 10 at that time. The second time she jumps, it will explode.

There is no second jump; that's something you just came up with.

Yes, of course it is two different events. The event where Alice's clock displays 10 is a totally different event than the event where her clock displays 40.

So then what does your VCR-scenario prove? That two different recordings made in two different reference frames of two different events are different?

But the two most important things to remember are these:
1. The event where Alice's clock displays 10 is simultaneous (in the train frame) with Charlie jumping off/on the train.
2. The event where Alice's clock displays 40 is simultaneous (in the ground frame) with Charlie jumping off/on the train.

Right, and now where is your backwards ticking clock? Because that was what you were trying to prove, remember?

The distance between Alice and Charlie (as measured by the train frame) when Charlie's clock displays 20:
d' = vt'
d' = 0.866 * 20
d' = 17.32 light years
So the time required for the train assistant's broadcast to reach Charlie's train-DVR is 17.32 years (as measured by the train frame).

The distance between Alice and Charlie (as measured by the ground frame) when Alice's clock displays 40:
d = vt
d = 0.866 * 40
d = 34.64 light years
So the time required for Alice's broadcast to reach Charlie's ground-DVR is 34.64 years (as measured by the ground frame).

Right, so Charlie's DVR's receive Alice's clock reading 10 long before they receive Alice's clock reading 40. Where's the backwards ticking?

What did you think the distances were?

It's your scenario; it's your task to work it through to the end.

How long did you think it would take light to travel those distances?

Well, light travels at c in a vacuum, so...

Did your really think you could make SR fail by just letting light signals travel from Alice to Charlie?

Nope, and I never made that claim. But did you really think you could make clocks tick backwards in SR?

Strange.

Indeed.

I think these might be the calculations you are looking for.

Actually, I'm still waiting for that Minkowski diagram...

These calculations are all going to be done in Alice's stay-home frame.

At the time Alice's clock displays 10, the light from her clock which is carrying the "showing 10" information is 10-10=0 light years away from Alice, as that light is just being emitted toward Charlie. But at that time, Charlie is a distance of 10*0.866=8.66 light years away, and his clock displays 10/2=5. Remember, this is all from the stay-home frame.

By the time Alice's clock displays 40, the light from her clock which is carrying the "showing 10" information is 40-10=30 light years away from Alice. And Charlie is a distance of 40*0.866=34.64 light years away from Alice, while his clock displays 40/2=20.

By the time Alice's clock displays 74.64, the light from her clock which is carrying the "showing 10" information is 74.64-10=64.64 light years away from Alice. And Charlie is also a distance of 74.64*0.866=64.64 light years away from Alice, while his clock displays 74.64/2=37.32. This demonstrates the event when the light from Alice's clock which is carrying the "showing 10" information arrives at Charlie, because both Charlie and the light are in the same place; both are 64.64 light years away from Alice.

Right, and now, instead of the chosen value of "10", make it a variable. Then check whether Charlie ever receives light "out of order"; i.e. will he ever observe Alice's clock ticking backwards? Oh, what do you know, look at that, that never happens!

Does that help?

I already knew the outcome of this before you posted it, but I hope it helps you realize clocks don't tick backwards in SR.

 

[Neddy Bate]

And once again you have to modify the scenario in order to make ends meet.

There is no second jump; that's something you just came up with.

So why did you ask me whether she jumps at t=10 or t=40, if you had already decided she only jumps at t=10? Oh, that's right, because you thought there would be a contradiction there, because you did not even realize that her clock displaying two different times, t=10 and t=40, could not be one single event.

If you have decided that she only jumps on and off the train at t=10, then the pressure pad does not explode. The train assistant still broadcasts her time displaying 10 when she jumps on/off the train, which is recorded 17.32 years later by Charlie's train-DVR. And Alice herself still broadcasts her own clock when it is displaying 40, as recorded 34.46 years later by Charlie's ground-DVR.

The only difference now is that you are saying she is not allowed to jump on/off the train when her clock displays 40, because it did not result in the contradiction you had hoped for. Yet, the result is still unchanged. Charlie still concludes her clock displayed t=40 at the moment when he was on the ground, and Charlie still concludes her clock displayed t=10 at the moment he had jumped back on the train.

So more wasted time caused by you.

So then what does your VCR-scenario prove? That two different recordings made in two different reference frames of two different events are different?

It shows that Charlie cannot help but conclude that Alice's clock displayed t=40 at the moment when he was on the ground, and conclude that her clock displayed t=10 at the moment he had jumped back on the train.

All of the DVR stuff was not required. I only added it because you were complaining, for some unknown reason, that Charlie did not receive the information in the same instant that he calculated it. Once given enough time to receive the information, unsurprisingly, we find that Charlie's calculations were correct, and sufficient in and of themselves.

So more wasted time caused by you.

Right, and now where is your backwards ticking clock? Because that was what you were trying to prove, remember?

Charlie concludes that Alice's clock displayed t=40 at the moment when he was on the ground, and Charlie concludes her clock displayed t=10 at the moment he had jumped back on the train.

Right, so Charlie's DVR's receive Alice's clock reading 10 long before they receive Alice's clock reading 40. Where's the backwards ticking?

Charlie concludes that Alice's clock displayed t=40 at the moment when he was on the ground, and Charlie concludes her clock displayed t=10 at the moment he had jumped back on the train.

Right, and now, instead of the chosen value of "10", make it a variable. Then check whether Charlie ever receives light "out of order"; i.e. will he ever observe Alice's clock ticking backwards? Oh, what do you know, look at that, that never happens!

Well then, it is a good thing that I never claimed Charlie ever receives light "out of order".

...clocks don't tick backwards in SR.

Charlie concludes that Alice's clock displayed t=40 at the moment when he was on the ground, and Charlie concludes her clock displayed t=10 at the moment he had jumped back on the train. Since you claim that is impossible, the onus is on you to explain to Charlie which of his conclusions are false. You have the floor.

 

[NotEinstein]

So why did you ask me whether she jumps at t=10 or t=40, if you had already decided she only jumps at t=10? Oh, that's right, because you thought there would be a contradiction there,

No, I didn't realize you were introducing another scenario into the mix. It's quite spectacular to see you changing up the scenario time after time without addressing the points I raise about your previous one.

because you did not even realize that her clock displaying two different times, t=10 and t=40, could not be one single event.

Actually, if you read back my posts, it's me that explicitly pointing that out. You got that exactly the wrong way around.

If you have decided that she only jumps on and off the train at t=10, then the pressure pad does not explode. The train assistant still broadcasts her time displaying 10 when she jumps on/off the train, which is recorded 17.32 years later by Charlie's train-DVR. And Alice herself still broadcasts her own clock when it is displaying 40, as recorded 34.46 years later by Charlie's ground-DVR.

The only difference now is that you are saying she is not allowed to jump on/off the train when her clock displays 40, because it did not result in the contradiction you had hoped for. Yet, the result is still unchanged. Charlie still concludes her clock displayed t=40 at the moment when he was on the ground, and Charlie still concludes her clock displayed t=10 at the moment he had jumped back on the train.

But, as I said before, the full recordings of both DVR's also show no sign of clocks ticking backwards.

So more wasted time caused by you.

Says the person introducing new irrelevant scenario's every other post...

It shows that Charlie cannot help but conclude that Alice's clock displayed t=40 at the moment when he was on the ground, and conclude that her clock displayed t=10 at the moment he had jumped back on the train.

All of the DVR stuff was not required. I only added it because you were complaining, for some unknown reason, that Charlie did not receive the information in the same instant that he calculated it.

And he still doesn't. In fact, take your scenario. When does the DVR-displaying-10 receive that information? Does that give Charlie enough time to act on it, before her clock reaches 40?

And does Charlie (not a DVR, but Charlie) ever see Alice's clock ticking backwards?

Once given enough time to receive the information, unsurprisingly, we find that Charlie's calculations were correct,

And I never claimed otherwise.

and sufficient in and of themselves.

Sufficient for what?

So more wasted time caused by you.

Once again, you got that the wrong way around. This entire scenario turns out to be to demonstrate that Charlie's calculations are correct, something I've never doubted.

Charlie concludes that Alice's clock displayed t=40 at the moment when he was on the ground, and Charlie concludes her clock displayed t=10 at the moment he had jumped back on the train.

Yeah, and when does Charlie reach that conclusion? Oh right, long after Alice's clock (according to Alice) passes 40, so it's too late to act on that information.

And I see you've dodged the question I asked you. How very intellectually honest. You have failed to show a clock ticking backwards; you have only showed that space-like events can be re-ordered by doing boosts, and you've failed to show that that re-ordering is physically meaningful.

Well then, it is a good thing that I never claimed Charlie ever receives light "out of order".

So you agree that Charlie never ever sees Alice's clock ticking backwards? Great! I'm glad you finally agree!

Charlie concludes that Alice's clock displayed t=40 at the moment when he was on the ground, and Charlie concludes her clock displayed t=10 at the moment he had jumped back on the train. Since you claim that is impossible,

That's a misrepresentation: I only claim that clocks can't tick backwards.

the onus is on you to explain to Charlie which of his conclusions are false.

The "conclusion" that Alice's clock ticked backwards is false.

You have the floor.

And back to you.

 

[NotEinstein]

By the way, I love your selective quoting. Perhaps I should start doing the same; just ignore "the difficult" parts of your posts? Seems like a good and intellectually honest strategy!

 

[Neddy Bate]

But, as I said before, the full recordings of both DVR's also show no sign of clocks ticking backwards.

True. But there are two recordings. One is timed to be simultaneous with Charlie landing on the ground after jumping off the train. The other is timed to be simultaneous with Charlie jumping back on the train.

Note that Charlie jumps back on the train after he lands on the ground. Yet that is the recording which shows the earlier time on Alice's clock.

And he still doesn't. In fact, take your scenario. When does the DVR-displaying-10 receive that information? Does that give Charlie enough time to act on it, before her clock reaches 40?

The question is not, and has never been, whether Charlie has time to "act on it." The question is what time is currently being displayed on Alice's clock.

Do you think I would have to be able to "act on it" if I wanted to know what time it is on the opposite side of the earth from me right now? No, I would just have to do a calculation based on the time zones of the earth.

And does Charlie (not a DVR, but Charlie) ever see Alice's clock ticking backwards?

No, not even the VCR sees that. But there are two recordings. One is timed to be simultaneous with Charlie landing on the ground after jumping off the train. The other is timed to be simultaneous with Charlie jumping back on the train.

Note that Charlie jumps back on the train after he lands on the ground. Yet that is the recording which shows the earlier time on Alice's clock.

Sufficient for what?

Sufficient to know what time is currently being displayed on Alice's clock, assuming it is properly functioning, and synchronised according to the standard configuration of SR.

Yeah, and when does Charlie reach that conclusion? Oh right, long after Alice's clock (according to Alice) passes 40, so it's too late to act on that information.

The question is not, and has never been, whether Charlie has time to "act on it." The question is what time is currently being displayed on Alice's clock.

You have failed to show a clock ticking backwards; you have only showed that space-like events can be re-ordered by doing boosts, and you've failed to show that that re-ordering is physically meaningful.

The physical meaning is simply, "Alice's clock displays this time right now, even though I won't be able to see that information until later."

So you agree that Charlie never ever sees Alice's clock ticking backwards? Great! I'm glad you finally agree!

I never said he sees Alice's clock go backwards. I said he concludes it must because it displays 40 before it displays 10.

Charlie concludes that Alice's clock displayed t=40 at the moment when he was on the ground, and Charlie concludes her clock displayed t=10 at the moment he had jumped back on the train. Since you claim that is impossible,

That's a misrepresentation: I only claim that clocks can't tick backwards.

Okay, then we are making progress. You and I agree that Charlie concludes that Alice's clock displayed t=40 at the moment when he was on the ground, and Charlie concludes her clock displayed t=10 at the moment he had jumped back on the train. Now all you have to do is explain how a clock can go from displaying t=40 to displaying t=10 without going backwards. How?

 

[NotEinstein]

True. But there are two recordings. One is timed to be simultaneous with Charlie landing on the ground after jumping off the train. The other is timed to be simultaneous with Charlie jumping back on the train.

Note that Charlie jumps back on the train after he lands on the ground. Yet that is the recording which shows the earlier time on Alice's clock.

Please address my point instead of merely repeating your previous one.

The question is not, and has never been, whether Charlie has time to "act on it."

False; it was brought up as a side-discussion with the "physically meaningful" part. You know, the part of our discussion you've chosen to ignore?

The question is what time is currently being displayed on Alice's clock.

No, the big discussion is whether or not Alice's clock ticks backwards.

Do you think I would have to be able to "act on it" if I wanted to know what time it is on the opposite side of the earth from me right now? No, I would just have to do a calculation based on the time zones of the earth.

False; you've forgotten about the bomb-issue. You don't know what time it is on the opposite side of the earth.

I find it funny: you are claiming I'm wasting time, but here you are, selectively forgetting whole parts of our discussion so we have to repeat it.

No, not even the VCR sees that.

Cool, then this part of the discussion is done. Alice's clock doesn't tick backwards.

But there are two recordings. One is timed to be simultaneous with Charlie landing on the ground after jumping off the train. The other is timed to be simultaneous with Charlie jumping back on the train.

Note that Charlie jumps back on the train after he lands on the ground. Yet that is the recording which shows the earlier time on Alice's clock.

Irrelevant to the issue of Alice's clock ticking backwards.

Sufficient to know what time is currently being displayed on Alice's clock, assuming it is properly functioning,

So about that assumption... seems like my bombs were quite important after all!

and synchronised according to the standard configuration of SR.

Yep, and I don't believe I've ever claimed otherwise.

The question is not, and has never been, whether Charlie has time to "act on it." The question is what time is currently being displayed on Alice's clock.

No, the issue is whether Alice's clock ever ticks backwards. And you just admitted that it doesn't (at least, in this scenario, according to Charlie and his VCR's).

The physical meaning is simply, "Alice's clock displays this time right now, even though I won't be able to see that information until later."

Please give the definition of "physical meaning" you are using.

I never said he sees Alice's clock go backwards.

Great!

I said he concludes it must because it displays 40 before it displays 10.

So he *calculates* Alice's clock ticking backwards? Tell me, have you by now figured out what the second law of thermodynamics is about? You know, that thing I brought up that you totally ignored as well? What does Charlie calculate entropy does in a closed system next to Alice, that (according to Alice) has its entropy steadily increasing during this entire scenario?

Okay, then we are making progress.

Indeed, after all these posts, you are finally catching on!

You and I agree that Charlie concludes that Alice's clock displayed t=40 at the moment when he was on the ground, and Charlie concludes her clock displayed t=10 at the moment he had jumped back on the train.

Correct, but only if you replace the word "concludes" with "calculates".

Now all you have to do is explain how a clock can go from displaying t=40 to displaying t=10 without going backwards. How?

I've already addressed this question many posts ago; go back and look for statements about the re-ordering of space-like separated events. You have to explain how you think it is possible for time to go backwards.

 

[Neddy Bate]

False; you've forgotten about the bomb-issue. You don't know what time it is on the opposite side of the earth.

But I do know what time it currently is on the opposite side of the earth. I may not know what time it is on a particular clock which might have been set incorrectly, or vandalized. That does not change the current time on the opposite side of the earth! Please, try to be reasonable.

So about that assumption... seems like my bombs were quite important after all!

Not at all. There is no limit to the number of clocks we can have in a thought experiment. You can blow Alice's clock up, but Charlie's calculation is still correct for some other clock which happens to be co-located with Alice, properly functioning & synchronised. That is all he was ever tasked to calculate, at least by reasonable people.

So he *calculates* Alice's clock ticking backwards? Tell me, have you by now figured out what the second law of thermodynamics is about? You know, that thing I brought up that you totally ignored as well? What does Charlie calculate entropy does in a closed system next to Alice, that (according to Alice) has its entropy steadily increasing during this entire scenario?

Yes, he calculates that Alice's clock displayed 40 before it displayed 10, as you agreed. If you are worried about entropy, then you tell me what you think happens.

Correct, but only if you replace the word "concludes" with "calculates".

What's the difference? He is interested in the current time on Alice's clock. He does a calculation, and concludes that is what time the clock should be displaying at that time. He is correct in that conclusion, unless the clock was vandalized by an unreasonable person, in which case he is still correct for some other clock which was not vandalized.

So why can't he conclude that his calculation is what the clock actually displays? Are you trying to argue that every single clock might be vandalized? It would be unreasonable to expect people to work out SR problems with that possibility on the table.

I've already addressed this question many posts ago; go back and look for statements about the re-ordering of space-like separated events.

Oh, so the clock goes from 40 to 10 because there was a re-ordering of space-like separated events, but it did not go backwards? Most reasonable people would call it going backwards.

You have to explain how you think it is possible for time to go backwards.

When a clock displays 40 before it displays 10, that is not the normal "forward" direction of time that we are all used to. If you want to call it a re-ordering of space-like separated events, then that is up to you. Most reasonable folks would agree it is a clock going backward, but then again, not everyone is reasonable.

 

[NotEinstein]

But I do know what time it currently is on the opposite side of the earth.

No, you don't. You don't know for certain that the opposite side of the earth still exists. Think big bombs.

I may not know what time it is on a particular clock which might have been set incorrectly, or vandalized. That does not change the current time on the opposite side of the earth! Please, try to be reasonable.

I am merely pointing out that you are incorrect. If you call that unreasonable, that's on you, not me.

Not at all. There is no limit to the number of clocks we can have in a thought experiment. You can blow Alice's clock up, but Charlie's calculation is still correct for some other clock which happens to be co-located with Alice, properly functioning & synchronised.

*sigh* You've once again lost sight of what we were discussing. Sure, you can have your clocks, but the statement that Charlie knows what some particular distant clock shows at that moment is wrong. Of course, under the assumption that the clock functions properly and isn't vandalized, Charlie's calculation will turn out to be correct, but until the information travels from Alice to Charlie, he cannot know.

That is all he was ever tasked to calculate,

Erm, you tasked him with that, I never did.

at least by reasonable people.

Reasonable people would address the points raised in the discussion, not some imaginary other things that were never brought up or in doubt.

Yes, he calculates that Alice's clock displayed 40 before it displayed 10, as you agreed. If you are worried about entropy, then you tell me what you think happens.

*sigh* And again you've forgotten parts of our previous discussion. I've already told you what "I think happens". How about you answer the question I asked you, instead of dodging it once again?

What's the difference?

Calculates is less ambiguous. You could call the result of the calculation a conclusion (I'm fine with that), but you could also call the result he reaches after assuming the clock isn't vandalized a conclusion. That latter one is a lot less certain.

He is interested in the current time on Alice's clock. He does a calculation, and concludes that is what time the clock should be displaying at that time. He is correct in that conclusion, unless the clock was vandalized

I see that you are using it in the former sense; good. All I was doing was avoiding the ambiguity I just explicitly pointed out.

by an unreasonable person,

What does Dave have to do with this?

in which case he is still correct for some other clock which was not vandalized.

Which is besides the point.

In fact, clocks don't even need to be vandalized. Alice can also simply walk off right before go-time. Is Alice now unreasonable too, for proving that Charlie's conclusion cannot be considered a certainty by him until after he receives information about the event?

So why can't he conclude that his calculation is what the clock actually displays?

Because, as I pointed out umpth-teen times, and you've already agreed to multiple times: he may be incorrect.

Are you trying to argue that every single clock might be vandalized?

Nope, and I don't know where you got that idea from?

It would be unreasonable to expect people to work out SR problems with that possibility on the table.

No, it would be unreasonable to just blindly assume things that can obviously happen don't happen. That would be dumb.

Oh, so the clock goes from 40 to 10 because there was a re-ordering of space-like separated events, but it did not go backwards?

Please point out in your Minkowski diagram where on the worldline of Alice's clock it goes backwards. Or on Charlie's worldline. Or on any of the DVR's worldlines.

Most reasonable people would call it going backwards.

Most reasonable people would have given their definition of "physically meaningful", so this discussion can actually take place.

When a clock displays 40 before it displays 10, that is not the normal "forward" direction of time that we are all used to.

Yep, that normal direction is called "the arrow of time", and it's quite an important concept in physics. Look it up, you might learn something about it being one-way and all...

If you want to call it a re-ordering of space-like separated events, then that is up to you.

No, that's not what I want to do. Why do you try to put word in my mouth?

Most reasonable folks would agree it is a clock going backward,

Your new pet-word, "reasonable"?

Anyway, no. Charlie never sees a clock going backwards, and neither does Alice, nor Bob, or even Charlie's assistant or his DVR's. In fact, every single observer agrees with that; you can demonstrate that clearly in your Minkowski diagram.

What you are talking about is the results of the calculation, and its interpretation. Since events that are space-like separated can be arbitrarily ordered by doing particular boosts, there's no physical meaning ascribed to the order of events in the space-like region.

but then again, not everyone is reasonable.

Oh, I fully agree. Some people use misrepresentations, misattributions, miss the point (repeatedly), put words in other people's mouths, selectively respond, etc. Quite unreasonable behavior on a science forum, I agree!

 

[Neddy Bate]

No, you don't. You don't know for certain that the opposite side of the earth still exists. Think big bombs.

Because of the convention of time zones that we have established on earth, it could be argued that the time on the opposite side of the earth is still whatever time it would have been had it not been blown up. Time does not cease to exist because of big bombs.

*sigh* You've once again lost sight of what we were discussing. Sure, you can have your clocks, but the statement that Charlie knows what some particular distant clock shows at that moment is wrong. Of course, under the assumption that the clock functions properly and isn't vandalized, Charlie's calculation will turn out to be correct, but until the information travels from Alice to Charlie, he cannot know.

In an SR thought experiment, Charlie should not be tasked with "knowing" something that SR says he cannot possibly know. That is why Charlie is only tasked with calculating the current time in Alice's location, which is usually represented in SR by a standard configuration clock.

Charlie does not say, "Alice's actual clock displays 10 right now" he says, "The event (t=10, x=0) is simultaneous with the event (t'=20, x'=0) according to definition of simultaneity used it the train frame." Which is why your nit-picking is way off topic.

*sigh* And again you've forgotten parts of our previous discussion. I've already told you what "I think happens". How about you answer the question I asked you, instead of dodging it once again?

I am not concerned with entropy. I don't know much about the block universe theory, but I think it says the universe is a fixed block where time only appears to pass.

Calculates is less ambiguous. You could call the result of the calculation a conclusion (I'm fine with that), but you could also call the result he reaches after assuming the clock isn't vandalized a conclusion. That latter one is a lot less certain.

How about simply, Charlie says, "The event (t=10, x=0) is simultaneous with the event (t'=20, x'=0) according to definition of simultaneity used it the train frame," which only assumes that time and space are related as SR says they are? Seems pretty unambiguous to me.

In fact, clocks don't even need to be vandalized. Alice can also simply walk off right before go-time.

Even if she does, Charlie can still say, "The event (t=10, x=0) is simultaneous with the event (t'=20, x'=0) according to definition of simultaneity used it the train frame," and he would be correct.

Because, as I pointed out umpth-teen times, and you've already agreed to multiple times: he may be incorrect.

He would not be incorrect to say, "The event (t=10, x=0) is simultaneous with the event (t'=20, x'=0) according to definition of simultaneity used it the train frame."

Please point out in your Minkowski diagram where on the worldline of Alice's clock it goes backwards. Or on Charlie's worldline. Or on any of the DVR's worldlines.

I thought one of the advantages of Minkowski diagrams was that you could draw a plane of simultaneity for one frame, and easily see where it intersects the time axis of the other frame. When Charlie jumps off/on the train, his plane of simultaneity shifts to a different angle. It should not be difficult to follow the two different planes of simultaneity to where they intersect the time axis, and see where the conclusion of time going from 40 to 10 comes from. Or you can just do the Lorentz transformations, which are the non-graphical equivalent.

Yep, that normal direction is called "the arrow of time", and it's quite an important concept in physics. Look it up, you might learn something about it being one-way and all...

Okay, so when t=40 is followed by t=10 we should just say that time always goes forward? Is that it?

Anyway, no. Charlie never sees a clock going backwards, and neither does Alice, nor Bob, or even Charlie's assistant or his DVR's. In fact, every single observer agrees with that; you can demonstrate that clearly in your Minkowski diagram.

Does your Minkowsi diagram have the different planes of simultaneity drawn in?

What you are talking about is the results of the calculation, and its interpretation. Since events that are space-like separated can be arbitrarily ordered by doing particular boosts, there's no physical meaning ascribed to the order of events in the space-like region.

I'm not sure what you are really saying, other than perhaps it is okay for t=10 to come after t=40 because it does not matter.

My interpretation of it is more along the lines of, when Charlie is at rest in the stay-home frame, he must use the definition of simultaneity used it the stay-home frame. And, when Charlie is at rest in the train frame, he must use the definition of simultaneity used it the train frame. This results in t=40 being followed by t=10 when he changes frames.

The only meaning I ascribe to it is that the event with coordinates (t=40, x=0) is simultaneous (in the ground frame) with Charlie landing on the ground. And the event with coordinates (t=10, x=0) is simultaneous (in the train frame) with Charlie jumping and landing on the train. I can't see how Charlie can get around concluding that time must have gone backwards there, when he jumped.

Maybe you are having trouble because we chose to use idealized instantaneous accelerations. That makes the t=40 immediately become t=10. To me, that is still backwards, but maybe it would help if you thought of it as a finite acceleration over an arbitrarily small period of time. This means Charlie concludes that the time at x=0 sweeps continuously from t=40 back to t=10. If that is not a clock running backwards, then I don't know what is.

 

[Neddy Bate]

I thought one of the advantages of Minkowski diagrams was that you could draw a plane of simultaneity for one frame, and easily see where it intersects the time axis of the other frame. When Charlie jumps off/on the train, his plane of simultaneity shifts to a different angle. It should not be difficult to follow the two different planes of simultaneity to where they intersect the time axis, and see where the conclusion of time going from 40 to 10 comes from. Or you can just do the Lorentz transformations, which are the non-graphical equivalent.

Indeed.

Glad that is settled.

 

[NotEinstein]

Because of the convention of time zones that we have established on earth, it could be argued that the time on the opposite side of the earth is still whatever time it would have been had it not been blown up.

(Emphasis mine.)
Yes, of course, under the assumption that it hasn't blown up, you can. I've already said that multiple times. But because you don't know (yet), it's always under that assumption. Without that assumption, you don't know.

Time does not cease to exist because of big bombs.

And I've never claimed otherwise.

In an SR thought experiment, Charlie should not be tasked with "knowing" something that SR says he cannot possibly know.

He's not. He just can't know it yet.

That is why Charlie is only tasked with calculating the current time in Alice's location, which is usually represented in SR by a standard configuration clock.

Correct. "Calculate", not "know".

Charlie does not say, "Alice's actual clock displays 10 right now" he says, "The event (t=10, x=0) is simultaneous with the event (t'=20, x'=0) according to definition of simultaneity used it the train frame." Which is why your nit-picking is way off topic.

I can't help it that you use the incorrect words, and have trouble correcting yourself.

I am not concerned with entropy.

Perhaps you really should read about the "arrow of time".

I don't know much about the block universe theory,

What is the "block universe theory"? *Wikipedia-moment* Oh, okay, not sure why you bring that up? Especially after just accusing me of pulling the thread off-topic!

but I think it says the universe is a fixed block where time only appears to pass.

Sooo, another question dodge? Well, at least you are consistent.

Look, if you refuse to be intellectually honest, that's fine with me. It just makes you look bad, that's all.

How about simply, Charlie says, "The event (t=10, x=0) is simultaneous with the event (t'=20, x'=0) according to definition of simultaneity used it the train frame," which only assumes that time and space are related as SR says they are? Seems pretty unambiguous to me.

Yep, that's fine too. I'm glad you agree that using the word "conclusion" (without specifying it) is not unambiguous.

Even if she does, Charlie can still say, "The event (t=10, x=0) is simultaneous with the event (t'=20, x'=0) according to definition of simultaneity used it the train frame," and he would be correct.

True; I'm glad you agree with my that Charlie doesn't know.

He would not be incorrect to say, "The event (t=10, x=0) is simultaneous with the event (t'=20, x'=0) according to definition of simultaneity used it the train frame."

I'm glad you agree with me Charlie could be incorrect if he said he knows.

I thought one of the advantages of Minkowski diagrams was that you could draw a plane of simultaneity for one frame, and easily see where it intersects the time axis of the other frame. When Charlie jumps off/on the train, his plane of simultaneity shifts to a different angle. It should not be difficult to follow the two different planes of simultaneity to where they intersect the time axis, and see where the conclusion of time going from 40 to 10 comes from. Or you can just do the Lorentz transformations, which are the non-graphical equivalent.

Yep, you can see the result of the calculations. But you've dodged my question, because you can also see no clock (worldline) ever goes backwards.

Okay, so when t=40 is followed by t=10 we should just say that time always goes forward? Is that it?

*sigh* More intellectual dishonesty.

No, re-ordering events doesn't make time tick backwards or forwards. I've been saying that for pages now. Haven't you been paying attention to anything I said?

And I see you haven't even bothered looking up the "arrow of time", because it explicitly disagrees with what you just wrote. So what you just wrote is another of your imaginations, not mine.

Does your Minkowsi diagram have the different planes of simultaneity drawn in?

Obviously, yes. Does your Minkowski diagram have worldlines going back in time?

I'm not sure what you are really saying, other than perhaps it is okay for t=10 to come after t=40 because it does not matter.

My interpretation of it is more along the lines of, when Charlie is at rest in the stay-home frame, he must use the definition of simultaneity used it the stay-home frame. And, when Charlie is at rest in the train frame, he must use the definition of simultaneity used it the train frame. This results in t=40 being followed by t=10 when he changes frames.

It's okay for t=40 being followed by t=10 in the space-like region, as long as it's in order when the events cross through the light-like line into the time-like region. I've explained this to you before; you have once again forgotten things previously discussed.

The only meaning I ascribe to it is that the event with coordinates (t=40, x=0) is simultaneous (in the ground frame) with Charlie landing on the ground. And the event with coordinates (t=10, x=0) is simultaneous (in the train frame) with Charlie jumping and landing on the train. I can't see how Charlie can get around concluding that time must have gone backwards there, when he jumped.

And I can't get around concluding that that interpretation has to be wrong, because time doesn't tick backwards. But we've been over this before. It seems you have some mental blockage when it comes to this, and you are unwilling and/or unable to consider any other point of view. Perhaps this entire discussion should be moved to the pseudo-science section because of that?

(And I see you're back to using the word "concluding" again?)

Maybe you are having trouble because we chose to use idealized instantaneous accelerations.

Nope, as we've established before (after much struggle from you), all large enough accelerations cause this effect.

That makes the t=40 immediately become t=10. To me, that is still backwards, but maybe it would help if you thought of it as a finite acceleration over an arbitrarily small period of time. This means Charlie concludes that the time at x=0 sweeps continuously from t=40 back to t=10. If that is not a clock running backwards, then I don't know what is.

Please point me to the worldline associated to that clock in the Minkowski diagram, and demonstrate how it is going backwards.

Indeed.

Glad that is settled.

Let's see, Alice's worldline is in cyan. Seems to be ticking forwards quite nicely, yes. Indeed, that settles it! Alice's clock is always ticking forwards; I'm glad you finally agree!