> If you recall, you recently posted some links to the TP and accused me of not reading them. Well, I certainly intended to read them and I explained why. But then I got involved in other discussions here and put that temporarily aside. But now those links are, from my pov, lost in a myriad of discussion threads. So, please be so kind as to post them again here. TY, AG
fgt
>>Start with this video:
> I've encountered that fellow before. He speaks too fast,
> It seems to me that GR solves the problem,
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>>Start with this video:> I've encountered that fellow before. He speaks too fast,Try it again. Any YouTube video can be slowed down or sped up with no change in pitch with just a few clicks of a mouse button, and he provides the clearest explanation of how to resolve the twin paradox that I know of, so if he can't give you an intuitive understanding of it then I'm not going to be able to either.> It seems to me that GR solves the problem,You need to have a good feel for Special Relativity before you jump to General Relativity because it is far more complex and even less intuitive.
On Saturday, September 6, 2025 at 5:39:18 AM UTC-6 John Clark wrote:
I've encountered that fellow before. He speaks too fast, way too fast. It seems to me that GR solves the problem, and without frame jumping. Just imagine several changes in velocity, each spread out, so not instantaneous. During those changes, time for the traveler slows compared to rest frame on Earth, so when he returns to meet his twin, he is younger. Do you see anything wrong with this analysis? AG
fgt
No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
This result has nothing to do with accelerations experience by the traveler. This a common specious "explanation" trying to connect it to general relativity.
This is most easily seen by the triplet version of the paradox. In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth. When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.
Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.
Brent
On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:
No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG
This result has nothing to do with accelerations experience by the traveler. This a common specious "explanation" trying to connect it to general relativity.
Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AGThis is most easily seen by the triplet version of the paradox. In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth. When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.
Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin.
And I don't like the handing off of clock readings. AG
Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.
Brent
If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AG
On 9/6/2025 8:01 AM, Alan Grayson wrote:
On Saturday, September 6, 2025 at 5:39:18 AM UTC-6 John Clark wrote:
On Sat, Sep 6, 2025 at 1:10 AM Alan Grayson <agrays...@gmail.com> wrote:
> If you recall, you recently posted some links to the TP and accused me of not reading them. Well, I certainly intended to read them and I explained why. But then I got involved in other discussions here and put that temporarily aside. But now those links are, from my pov, lost in a myriad of discussion threads. So, please be so kind as to post them again here. TY, AG
I've encountered that fellow before. He speaks too fast, way too fast. It seems to me that GR solves the problem, and without frame jumping. Just imagine several changes in velocity, each spread out, so not instantaneous. During those changes, time for the traveler slows compared to rest frame on Earth, so when he returns to meet his twin, he is younger. Do you see anything wrong with this analysis? AG
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On 9/6/2025 11:20 PM, Alan Grayson wrote:
On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:
No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG
This result has nothing to do with accelerations experience by the traveler. This a common specious "explanation" trying to connect it to general relativity.
Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AGThis is most easily seen by the triplet version of the paradox. In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth. When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.
Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin.That's flat wrong.
And I don't like the handing off of clock readings. AG
Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.
Brent
If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AGYou may not like it, but it shows the twin paradox has nothing to do with general relativity or acceleration. It's simply the geometric fact that some paths are shorter than others. There is no reason to require that the clocks are set (not synched) equal at rest.
I've not "misstated" anything. You have apparently not understood the twins paradox.
Brent
On 9/6/2025 11:20 PM, Alan Grayson wrote:
On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:
No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AG
This result has nothing to do with accelerations experience by the traveler. This a common specious "explanation" trying to connect it to general relativity.
Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AGThis is most easily seen by the triplet version of the paradox. In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth. When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin.
That's flat wrong.
And I don't like the handing off of clock readings. AG
Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.
Brent
If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AG
mmm
On Sunday, September 7, 2025 at 11:17:42 AM UTC-6 Brent Meeker wrote:
On 9/6/2025 11:20 PM, Alan Grayson wrote:
On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:
No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AGThis result has nothing to do with accelerations experience by the traveler. This a common specious "explanation" trying to connect it to general relativity.Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AGThis is most easily seen by the triplet version of the paradox. In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth. When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin.That's flat wrong.
In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AG
And I don't like the handing off of clock readings. AGNotice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.
Brent
If there's no acceleration, then you've misstated the TP, where both twins start at rest, juxtaposed so their clocks are synched, and the traveler leaves and returns at rest to compare clocks. AGYou may not like it, but it shows the twin paradox has nothing to do with general relativity or acceleration. It's simply the geometric fact that some paths are shorter than others. There is no reason to require that the clocks are set (not synched) equal at rest. I've not "misstated" anything. You have apparently not understood the twins paradox.
Brent
On 9/6/2025 8:01 AM, Alan Grayson wrote:
On Saturday, September 6, 2025 at 5:39:18 AM UTC-6 John Clark wrote:
On Sat, Sep 6, 2025 at 1:10 AM Alan Grayson <agrays...@gmail.com> wrote:
> If you recall, you recently posted some links to the TP and accused me of not reading them. Well, I certainly intended to read them and I explained why. But then I got involved in other discussions here and put that temporarily aside. But now those links are, from my pov, lost in a myriad of discussion threads. So, please be so kind as to post them again here. TY, AG
I've encountered that fellow before. He speaks too fast, way too fast. It seems to me that GR solves the problem, and without frame jumping. Just imagine several changes in velocity, each spread out, so not instantaneous. During those changes, time for the traveler slows compared to rest frame on Earth, so when he returns to meet his twin, he is younger. Do you see anything wrong with this analysis? AG
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A complicated explanation of the triplet paradox. Length contraction is consistent, but it's not necessary to understand the effect. AG will reject it because he doesn't "believe in" handing off clock readings.
Brent
On 9/7/2025 12:12 PM, Alan Grayson wrote:
On Sunday, September 7, 2025 at 11:17:42 AM UTC-6 Brent Meeker wrote:
On 9/6/2025 11:20 PM, Alan Grayson wrote:
On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:
No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
Your solution is essentially no different than the two I explained, one using GR and other SR. All proposed solutions rely on some accepted result from relativity theory. AGThis result has nothing to do with accelerations experience by the traveler. This a common specious "explanation" trying to connect it to general relativity.Why "specious"? I used an established result in GR to show the traveling twin is younger when he returns and compares clock readings with the stationary twin. The fact that you prefer your solution to mine, doesn't mean using GR is wrong. AGThis is most easily seen by the triplet version of the paradox. In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth. When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.Without acceleration, there can be no comparison of clock readings when the inbound twin compares his clock readings with stationary twin.That's flat wrong.
In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AGActually he can. All he has to do it slingshot around a distant planet in order to head back to Earth:
You may object that he accelerated in turning around. But general relativity teaches us that force free motion in a gravitational field is geodesic and there is no acceleration.
On Sunday, September 7, 2025 at 2:29:11 PM UTC-6 Brent Meeker wrote:
On 9/7/2025 12:12 PM, Alan Grayson wrote:
In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AGActually he can. All he has to do it slingshot around a distant planet in order to head back to Earth:
But that's NOT how the TP is defined! AG
You may object that he accelerated in turning around. But general relativity teaches us that force free motion in a gravitational field is geodesic and there is no acceleration.
So the traveling twin turns around without acceleration? AG
On 9/7/2025 5:52 PM, Alan Grayson wrote:
On Sunday, September 7, 2025 at 2:29:11 PM UTC-6 Brent Meeker wrote:
On 9/7/2025 12:12 PM, Alan Grayson wrote:
In reality, there surely IS acceleration, even though it might not be necessary to use it to solve the problem. Can the traveling twin return without acceleration? Of course not! AGActually he can. All he has to do it slingshot around a distant planet in order to head back to Earth:
But that's NOT how the TP is defined! AG
What is this "defined"? It's not defined anywhere.
It's just a thought experiment that was paradoxical in Newtonian mechanics. Every version I've shown you is paradoxical in Newtonian mechanics in exactly the same way. If you'd open you eyes and mind, you'd see that they give an intuitive grasp on why they all give the same answer in relativity and so resolve the same paradox
You may object that he accelerated in turning around. But general relativity teaches us that force free motion in a gravitational field is geodesic and there is no acceleration.
So the traveling twin turns around without acceleration? AGRead my last sentence above over again a few times.
Brent
On Sunday, September 7, 2025 at 2:38:40 PM UTC-6 Brent Meeker wrote:
A complicated explanation of the triplet paradox. Length contraction is consistent, but it's not necessary to understand the effect. AG will reject it because he doesn't "believe in" handing off clock readings.
Brent
No, that's not it. Rather, I am uncomfortable with de-facto frame-jumping because I am unsure what happens to time when this is included in a solution. And if the twins are at rest and juxtaposed as the scenario begins -- which, BTW, is how the TP is habitually DEFINED -- the traveling twin MUST accelerate to begin his journey. But in the final analysis it's "your way or the highway", meaning that alternate solutions are unacceptable for you.
So, if there is acceleration, there is also gravity by applying the Equivalence Principle,
and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force,
>> Here is another video on the twin paradox by the same guy that I recommended before, he explains it in a slightly different way but it's still crystal clear at least in my mind. The guy is really good.
> Thanks. I'll view it, but I am satisfied I understand its resolution despite what Brent says. He thinks there is a unique solution, his, but that's not true. AG
> I'm not sure the impossibility of absolute simultaneity solves the problem,
On
>> Here is another video on the twin paradox by the same guy that I recommended before, he explains it in a slightly different way but it's still crystal clear at least in my mind. The guy is really good.> Thanks. I'll view it, but I am satisfied I understand its resolution despite what Brent says. He thinks there is a unique solution, his, but that's not true. AGYou really should look at that video because it shows a way to state the twin "paradox" such that no acceleration is involved, but even in that case it can be resolved and he demonstrates it's not a paradox at all, it's just a strange situation. People call it a "paradox" because, although they remember time dilation and length contraction, they forget a third equally important thing special relativity tells us about the universe, the impossibility of absolute simultaneously except for the case of two events occurring at the same place and at the same time. So even if you can run faster than me I can still beat you to the finish line if I hear the starting gun before you do.I'm not sure the impossibility of absolute simultaneity solves the problem, if it's caused by the fallacious assumption of symmetry. I haven't yet viewed the video you just posted, but I will. AG
kf9
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On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:> I'm not sure the impossibility of absolute simultaneity solves the problem,Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult.
Where did I use the word "force"? AG
OTOH, during periods when a force is applied to accelerate (not by any gravity field), his clock will slow down, compared to its rate while he's in geodesic motion. So for the traveler to return to Earth younger than his twin, his slower clock while accelerating, must be large enough overall, to cause his clock to fall behind his stationary twin. The traveler could apply a continous but changing acceleration, say by traveling in a circle, but whether his clock will slow enough to make him age less than his stationary twin I don't know. AG
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Second, the traveler and the stay home person each see the other's clock as running slow.
On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:
On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:
> I'm not sure the impossibility of absolute simultaneity solves the problem,
Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult.
I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger.
Do you agree that I've correctly stated the paradox? AG
John K Clark See what's on my new list at Extropolis
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On 9/8/2025 11:19 AM, Alan Grayson wrote:
On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:
On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:
> I'm not sure the impossibility of absolute simultaneity solves the problem,
Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult.
I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger.No that's wrong. The stay at home twin has a clock that indicates a longer interval than the traveling twins clock. They agree that the traveling twin is younger.
Brent
On 9/7/2025 5:44 PM, Alan Grayson wrote:
On Sunday, September 7, 2025 at 2:38:40 PM UTC-6 Brent Meeker wrote:
A complicated explanation of the triplet paradox. Length contraction is consistent, but it's not necessary to understand the effect. AG will reject it because he doesn't "believe in" handing off clock readings.
Brent
No, that's not it. Rather, I am uncomfortable with de-facto frame-jumping because I am unsure what happens to time when this is included in a solution. And if the twins are at rest and juxtaposed as the scenario begins -- which, BTW, is how the TP is habitually DEFINED -- the traveling twin MUST accelerate to begin his journey. But in the final analysis it's "your way or the highway", meaning that alternate solutions are unacceptable for you.Not at all. You think it depends on acceleration. Fine, then here's an alternate version with acceleration. The twins each accelerates exactly the same level for exactly the same duration. But Red is still younger than Blue for exactly the same reason; his path is longer in space and therefore shorter in spacetime.
So, if there is acceleration, there is also gravity by applying the Equivalence Principle,So did you apply gravitational time dilation to each twin above?
and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force,Above you seem to think the equivalence principle means acceleration implies gravity
Brent
It's your way or the highway. Or shall we say a touch of arrogance? Haven't you ever heard of the EP? Let me remind you. Gravity is locally equivalent to acceleration, so when the traveling twin accelerates, it's equivalent to being in a gravity field, where clock rates are slower compared to rest frames. AG
But in the turnaround the time in the gravity field can be arbitrarily short compared to the the coasting phase before and after.
Where did I use the word "force"? AG" During the turnaround the motion is NOT force free."
Brent
On 9/7/2025 5:44 PM, Alan Grayson wrote:
On Sunday, September 7, 2025 at 2:38:40 PM UTC-6 Brent Meeker wrote:
A complicated explanation of the triplet paradox. Length contraction is consistent, but it's not necessary to understand the effect. AG will reject it because he doesn't "believe in" handing off clock readings.
Brent
No, that's not it. Rather, I am uncomfortable with de-facto frame-jumping because I am unsure what happens to time when this is included in a solution. And if the twins are at rest and juxtaposed as the scenario begins -- which, BTW, is how the TP is habitually DEFINED -- the traveling twin MUST accelerate to begin his journey. But in the final analysis it's "your way or the highway", meaning that alternate solutions are unacceptable for you.Not at all. You think it depends on acceleration. Fine, then here's an alternate version with acceleration. The twins each accelerates exactly the same level for exactly the same duration.
But Red is still younger than Blue for exactly the same reason; his path is longer in space and therefore shorter in spacetime.
> I'm not sure the impossibility of absolute simultaneity solves the problem,Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult.
I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox?
> I'm not sure the impossibility of absolute simultaneity solves the problem,Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult.I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox?Yes, it's impossible for all observers to agree that two events are simultaneous unless they happen at the same place. But I'm sure that is a little too succinct to convince you, that's why you need to watch the video, it's only a little longer. And neither acceleration nor General Relativity is the key to resolving the twin "paradox", as far as this matter is concerned you'd do best to forget about them.The twins ARE at the same place when the thought experiment begins, and since they're juxtaposed when the thought experiment begins, their clocks are synchronized without anything to do with absolute simultaneity! And second, using a spacetime diagram, the paths are unequal, and that's because only one path represents the accelerating twin who is traveling. If you don't believe me, look at (ds)^2 for each twin to confirm my claim. Take notice of the second order differentials. AGCORRECTION: The spacetime path lengths are INVARIANT, but along the path of the stationary twin, the second order differentials are zero since that twin is NOT accelerating, but those second order differentials for the traveling twin are NON-ZERO since he's accelerating. Now look at the proper times along both paths and you'll see that it's greater for stationary twin because the path lengths are invariant. Hence, stationary twin ages more than traveling twin! Note also that this analysis uses SR. If the video is using a simultaneity argument, then the video is wrong. AG
No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
This result has nothing to do with accelerations experience by the traveler.
This a common specious "explanation" trying to connect it to general relativity. This is most easily seen by the triplet version of the paradox. In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth. When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.
Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.
Brent
On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
If the spacetime distance between two points IS invariant, how can you say one path has a different length than another? BTW, considering acceleration does NOT necessarily mean invoking GR, and the invariance is a result of SR. AG
On Sunday, September 7, 2025 at 7:49:01 PM UTC-6 Brent Meeker wrote:
...
Later I posted why my GR model doesn't work. There's no obvious way for the twins to compare clock and determine their relative ages. It might depend on the paths taken, and I don't see how to do a calculation for any particular path for the traveling twin. Nonetheless, your denial of acceleration is mistaken. In your diagram with two spacetime paths, the proper times differ because along one path all the spatial derivatives are zero, unlike along the other path of the traveling twin.
This is your de-facto admission that differences in accelerations is the key to solving the paradox.
Your solution is ostensibly simpler because you fail to state exactly why the proper times are different along the two paths. AG
and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force,Above you seem to think the equivalence principle means acceleration implies gravity
Brent
When you're accelerating, it seems as if you're in a local gravitational field; that is, you cannot distinguish your acceleration from local gravity field. If that's not what the EP is, what's your take? AG
On Saturday, September 6, 2025 at 5:56:36 PM UTC-6 Brent Meeker wrote:
No. You're over complicating the problem. It's as simple as the fact that two different thru spacetime are different lengths. Because the spatial coordinate distance, X, appears with a minus sign relative to the coordinate time, T, the proper time, S (which is what a clock measures). So the rocket, which takes the longer spatial path, experiences less proper time lapse.
If the spacetime distance between two points in invariant, how can you say one path has a different length than another?
BTW, considering acceleration does NOT necessarily mean invoking GR, and the invariance is a result of SR. AG
This result has nothing to do with accelerations experience by the traveler.
I disagree since it is what indicates lack of symmetry, which causes the paradox.
Moreover, referring to acceleration does not necessarily invoke GR. AGThis a common specious "explanation" trying to connect it to general relativity. This is most easily seen by the triplet version of the paradox. In this version the one triplet stays home, one travelers away from Earth, and one who has been far away returns to Earth. When the outbound triplet passes the inbound triplet he hands off the time to the inbound one, so together they measure the same path as the turnaround twin.
Notice that I have also avoided any acceleration at the beginning and end, so no triplet ever accelerates.
Brent
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Brent
On 9/9/2025 12:59 AM, Alan Grayson wrote:
On Sunday, September 7, 2025 at 7:49:01 PM UTC-6 Brent Meeker wrote:
...
Later I posted why my GR model doesn't work. There's no obvious way for the twins to compare clock and determine their relative ages. It might depend on the paths taken, and I don't see how to do a calculation for any particular path for the traveling twin. Nonetheless, your denial of acceleration is mistaken. In your diagram with two spacetime paths, the proper times differ because along one path all the spatial derivatives are zero, unlike along the other path of the traveling twin.
Spatial derivatives of what?? Red and Blue each accelerate the same level for the same duration.
This is your de-facto admission that differences in accelerations is the key to solving the paradox.There are no differences in accelerations. Look at the damn diagram!
Your solution is ostensibly simpler because you fail to state exactly why the proper times are different along the two paths. AGThe proper times are different because one path is longer than the other (in Mikowski 4-space metric). And I explicitly note the times.and clocks in gravitational fields slow down, and this applies solely to the traveling twin. Notice, I never used or applied the concept of force,Above you seem to think the equivalence principle means acceleration implies gravity
Brent
When you're accelerating, it seems as if you're in a local gravitational field; that is, you cannot distinguish your acceleration from local gravity field. If that's not what the EP is, what's your take? AGMy take is that it's true but completely irrelevant to the twins paradox.
Even in the usual story the acceleration level and duration for the traveling twin can be made arbitrarily small compared to the geometric effect simply by choosing a more distant turnaround point.
Brent
You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done.
AG
Brent
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On Tuesday, September 9, 2025 at 5:39:58 PM UTC-6 Brent Meeker wrote:
...
This is your de-facto admission that differences in accelerations is the key to solving the paradox.There are no differences in accelerations. Look at the damn diagram!
Your diagram has no relation to the TP as originally stated.
You say no relation, but it produces exactly the same paradox for exactly the same reason.
You claim both spaceships have the same acceleration. How this relates to the stationary twin who never changes his location you provide no clue!
On 9/9/2025 9:57 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 5:18:14 PM UTC-6 Brent Meeker wrote:
On 9/8/2025 8:45 PM, Alan Grayson wrote:
On Monday, September 8, 2025 at 9:35:09 PM UTC-6 Brent Meeker wrote:
On 9/8/2025 11:19 AM, Alan Grayson wrote:
On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:
On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:
> I'm not sure the impossibility of absolute simultaneity solves the problem,
Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult.
I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger.No that's wrong. The stay at home twin has a clock that indicates a longer interval than the traveling twins clock. They agree that the traveling twin is younger.
Brent
Can't you understand English? I was stating the paradox and its cause. With an accurate analysis, the traveling twin is younger. Also, FWIW, for the traveling twin to return for the clock comparison, some acceleration is necessary, although it can be minimized if the comparison is done by fly-by. a AGBut notice that the acceleration is entirely incidental, as illustrated by the case in which Red and Blue each accelerates the same amount. IT'S JUST GEOMETRY. ONE PATH IS LONGER THAN THE OTHER.
In the original statement of the "paradox', the traveling twin must accelerate to return so the clocks can be compared. Please explain how this can happen without acceleration.I've shown two different ways without acceleration and I've also shown the paradox with equal accelerations by both twins. Why can't you just accept that it's geometry; that one path is longer than the other.
Brent
On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:
On 9/9/2025 9:57 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 5:18:14 PM UTC-6 Brent Meeker wrote:
On 9/8/2025 8:45 PM, Alan Grayson wrote:
On Monday, September 8, 2025 at 9:35:09 PM UTC-6 Brent Meeker wrote:
On 9/8/2025 11:19 AM, Alan Grayson wrote:
On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:
On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:
> I'm not sure the impossibility of absolute simultaneity solves the problem,
Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult.
I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger.No that's wrong. The stay at home twin has a clock that indicates a longer interval than the traveling twins clock. They agree that the traveling twin is younger.
Brent
Can't you understand English? I was stating the paradox and its cause. With an accurate analysis, the traveling twin is younger. Also, FWIW, for the traveling twin to return for the clock comparison, some acceleration is necessary, although it can be minimized if the comparison is done by fly-by. a AGBut notice that the acceleration is entirely incidental, as illustrated by the case in which Red and Blue each accelerates the same amount. IT'S JUST GEOMETRY. ONE PATH IS LONGER THAN THE OTHER.
In the original statement of the "paradox', the traveling twin must accelerate to return so the clocks can be compared. Please explain how this can happen without acceleration.I've shown two different ways without acceleration and I've also shown the paradox with equal accelerations by both twins. Why can't you just accept that it's geometry; that one path is longer than the other.
If both twins are accelerating, then you've redefined the TP.
If you have two paths in spacetime, starting at the same point and ending at the same point, or at a different point, how can you tell which is longer? AG
You seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done.But I've done more than that. I've done it while maintaining exactly the same paradox.
Brent
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On 9/9/2025 10:36 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:
On 9/9/2025 9:57 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 5:18:14 PM UTC-6 Brent Meeker wrote:
On 9/8/2025 8:45 PM, Alan Grayson wrote:
On Monday, September 8, 2025 at 9:35:09 PM UTC-6 Brent Meeker wrote:
On 9/8/2025 11:19 AM, Alan Grayson wrote:
On Monday, September 8, 2025 at 5:06:36 AM UTC-6 John Clark wrote:
On Mon, Sep 8, 2025 at 7:00 AM Alan Grayson <agrays...@gmail.com> wrote:
> I'm not sure the impossibility of absolute simultaneity solves the problem,
Watch the video! If you follow what he does step-by-step you will see that he is right. It's not difficult.
I'll definitely watch it, very soon, but a-priori the impossibility of absolute simultaneity can't solve the paradox because it's not its cause. Can you succinctly state the cause of the paradox? It's the application of time dilation in SR, under the mistaken assumption that the twins take symmetric paths; that their situations are symmetric. This results in the situation that when they meet and compare clock readings, each concludes the other is younger.No that's wrong. The stay at home twin has a clock that indicates a longer interval than the traveling twins clock. They agree that the traveling twin is younger.
Brent
Can't you understand English? I was stating the paradox and its cause. With an accurate analysis, the traveling twin is younger. Also, FWIW, for the traveling twin to return for the clock comparison, some acceleration is necessary, although it can be minimized if the comparison is done by fly-by. a AGBut notice that the acceleration is entirely incidental, as illustrated by the case in which Red and Blue each accelerates the same amount. IT'S JUST GEOMETRY. ONE PATH IS LONGER THAN THE OTHER.
In the original statement of the "paradox', the traveling twin must accelerate to return so the clocks can be compared. Please explain how this can happen without acceleration.I've shown two different ways without acceleration and I've also shown the paradox with equal accelerations by both twins. Why can't you just accept that it's geometry; that one path is longer than the other.
If both twins are accelerating, then you've redefined the TP.No. But Blue does not accelerate while Red travels out and back.
If you have two paths in spacetime, starting at the same point and ending at the same point, or at a different point, how can you tell which is longer? AGThese are paths in spacetime. They start and end at the same event, a point in 4-space.
The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals. You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2} where T is the coordinate time difference (in the same reference frame you measured distance).
BrentYou seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done.But I've done more than that. I've done it while maintaining exactly the same paradox.
Brent
On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:
On 9/9/2025 10:36 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:
...
If both twins are accelerating, then you've redefined the TP.
No. But Blue does not accelerate while Red travels out and back.
But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AG
But not "while Red travels out and back". Which IS what my diagram shows. You can imagine Blue as decelerating, landing on Earth, and then accelerating to join up again as Red comes back by. Next time read the whole sentence, not just up until a question occurs to you.
If you have two paths in spacetime, starting at the same point and ending at the same point, or at a different point, how can you tell which is longer? AGThese are paths in spacetime. They start and end at the same event, a point in 4-space.
If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AG
The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals. You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2} where T is the coordinate time difference (in the same reference frame you measured distance).
BrentYou seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done.But I've done more than that. I've done it while maintaining exactly the same paradox.
So you admit you're defying the laws of physics? AG
On 9/9/2025 11:52 PM, Alan Grayson wrote:
On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:
On 9/9/2025 10:36 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:
...
If both twins are accelerating, then you've redefined the TP.No. But Blue does not accelerate while Red travels out and back.
But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AGBut not "while Red travels out and back". Which IS what my diagram shows. You can imagine Blue as decelerating, landing on Earth, and then accelerating to join up again as Red comes back by. Next time read the whole sentence, not just up until a question occurs to you.
If you have two paths in spacetime, starting at the same point and ending at the same point, or at a different point, how can you tell which is longer? AGThese are paths in spacetime. They start and end at the same event, a point in 4-space.If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AGThe clock readings don't define events any more that odometer readings define distances.
The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals. You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2} where T is the coordinate time difference (in the same reference frame you measured distance).
BrentYou seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done.But I've done more than that. I've done it while maintaining exactly the same paradox.
So you admit you're defying the laws of physics? AGWhy? Because something "seemed to defy basic physics" to you? Can you explain how basic physics is defied?
Brent
On Wednesday, September 10, 2025 at 1:55:07 PM UTC-6 Brent Meeker wrote:
On 9/9/2025 11:52 PM, Alan Grayson wrote:
On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:
On 9/9/2025 10:36 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:
...
If both twins are accelerating, then you've redefined the TP.No. But Blue does not accelerate while Red travels out and back.
But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AGBut not "while Red travels out and back". Which IS what my diagram shows. You can imagine Blue as decelerating, landing on Earth, and then accelerating to join up again as Red comes back by. Next time read the whole sentence, not just up until a question occurs to you.
Thank you for replying, but please cease adding crap to it. I read it, and looked at your plot. It sure seemed as if both are accelerating. AGIf you have two paths in spacetime, starting at the same point and ending at the same point, or at a different point, how can you tell which is longer? AGThese are paths in spacetime. They start and end at the same event, a point in 4-space.If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AGThe clock readings don't define events any more that odometer readings define distances.
That's what I thought. But when traveling twin returns, it's to a different event because the time label on the coordinate differs from the event. So the return event is not the same as starting event when the twins are juxtaposed. AG
The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals. You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2} where T is the coordinate time difference (in the same reference frame you measured distance).
BrentYou seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done.But I've done more than that. I've done it while maintaining exactly the same paradox.
So you admit you're defying the laws of physics? AGWhy? Because something "seemed to defy basic physics" to you? Can you explain how basic physics is defied?
I now see you as a magician. In your special model of the problem, your traveling twin somehow turns around without acceleration. Or maybe you mean one of the triplets. In physics as I understand it, change in direction is acceleration and that's how the traveling twin returns. AG
On 9/10/2025 1:30 PM, Alan Grayson wrote:
On Wednesday, September 10, 2025 at 1:55:07 PM UTC-6 Brent Meeker wrote:
On 9/9/2025 11:52 PM, Alan Grayson wrote:
On Wednesday, September 10, 2025 at 12:08:39 AM UTC-6 Brent Meeker wrote:
On 9/9/2025 10:36 PM, Alan Grayson wrote:
On Tuesday, September 9, 2025 at 11:25:37 PM UTC-6 Brent Meeker wrote:
...
If both twins are accelerating, then you've redefined the TP.No. But Blue does not accelerate while Red travels out and back.
But earlier you claimed both Red and Blue have the same acceleration, and that's what your diagram shows. AGBut not "while Red travels out and back". Which IS what my diagram shows. You can imagine Blue as decelerating, landing on Earth, and then accelerating to join up again as Red comes back by. Next time read the whole sentence, not just up until a question occurs to you.
Thank you for replying, but please cease adding crap to it. I read it, and looked at your plot. It sure seemed as if both are accelerating. AGIf you have two paths in spacetime, starting at the same point and ending at the same point, or at a different point, how can you tell which is longer? AGThese are paths in spacetime. They start and end at the same event, a point in 4-space.If the paradox is resolved, then the clocks should read different values when finally compared. So the end point events are NOT the same. AGThe clock readings don't define events any more that odometer readings define distances.
That's what I thought. But when traveling twin returns, it's to a different event because the time label on the coordinate differs from the event. So the return event is not the same as starting event when the twins are juxtaposed. AGYou don't understand what an event is. It's a point in spacetime. It's independent of what coordinate labeling exists, just as a point on Earth is independent of what map you use. It's a physical thing. Have you never read a book on relativity?
The obvious way to tell which is longer in proper time is to carry an ideal clock along the two paths and compare the measured intervals. You could also measure the space distance X along the paths and and compute proper time S=\sqrt{T^2 - X^2} where T is the coordinate time difference (in the same reference frame you measured distance).
BrentYou seem to defying basic physics if this is your claim. I don't deny that the original problem can be restated in a way which avoids acceleration, and IMO this is what you've done.But I've done more than that. I've done it while maintaining exactly the same paradox.
So you admit you're defying the laws of physics? AGWhy? Because something "seemed to defy basic physics" to you? Can you explain how basic physics is defied?
I now see you as a magician. In your special model of the problem, your traveling twin somehow turns around without acceleration. Or maybe you mean one of the triplets. In physics as I understand it, change in direction is acceleration and that's how the traveling twin returns. AGThe point of the slingshot turnaround is that is shows the effect on the clock has nothing to do with acceleration. The whole trip is in free-fall. An accelerometer would read zero the whole time.
Brent
Every point in spacetime has a label, called cooordinates. So every event happens at some label.
If not that, then what? I see this in those spacetime diagrams. If their clocks disagree at the reunion, how do you define "the event"? AG
Wow, what a clever solution. If the clocks read differently when right beside one another there's no paradox. Somebody go wake up Albert.
Brent
> your traveling twin somehow turns around without acceleration
> your traveling twin somehow turns around without accelerationForget about acceleration, you're never going to understand the twin "paradox" if you become obsessed over that because, as I've mentioned before, there is a way to express it without any twin undergoing any acceleration.
But even in that case there is no paradox, the only reason people think there is one is because they forget there is no universal agreement about simultaneity.
> your traveling twin somehow turns around without accelerationForget about acceleration, you're never going to understand the twin "paradox" if you become obsessed over that because, as I've mentioned before, there is a way to express it without any twin undergoing any acceleration. But even in that case there is no paradox, the only reason people think there is one is because they forget there is no universal agreement about simultaneity.
> Those that make this claim cannot, it seems, explain how the traveling twin can start his journey without accelerating. AG
On Thu, Sep 11, 2025 at 7:39 AM Alan Grayson <agrays...@gmail.com> wrote:> Those that make this claim cannot, it seems, explain how the traveling twin can start his journey without accelerating. AGCannot explain? I've told you several times how you can see a demonstration of a version of the twin "paradox" with no acceleration involved at all.
If you could describe the model, I would be motivated to look at it
> I am not inclined to waste my time with it.
ygfi4b
On Thu, Sep 11, 2025 at 10:17 AM Alan Grayson <agrays...@gmail.com> wrote:If you could describe the model, I would be motivated to look at itI cannot describe it more succinctly or more clearly than has already been done in that video. That's why I believe my analogy between you and Galileo's critics was valid.
why don't you spend a few minutes describing the main differences between the usual TP and this different one?
On Thu, Sep 11, 2025 at 2:35 PM Alan Grayson <agrays...@gmail.com> wrote:why don't you spend a few minutes describing the main differences between the usual TP and this different one?Because if you can't be bothered to spend a few minutes watching the video then I can't be bothered to spend a few minutes making an inferior summary of that video. I can't easily provide snappy animations and illustrations needed to make things clearer.
On Wednesday, September 10, 2025 at 11:27:19 PM UTC-6 Brent Meeker wrote:
On 9/10/2025 6:41 PM, Alan Grayson wrote:
Every point in spacetime has a label, called cooordinates. So every event happens at some label.But the label is arbitrary. You draw a different map and give it a different label. But it's the same event. It doesn't become two different events because two different brought different clocks to it.
If not that, then what? I see this in those spacetime diagrams. If their clocks disagree at the reunion, how do you define "the event"? AGIt's the event of Red and Blue meeting again.
Brent
You have both accelerating in one of your diagrams, which is NOT how the TP is described, while also claiming there is no acceleration in explaining the TP. Is it any wonder that your presentation is unintelligible?
Real experts in relativity theory don’t find the Twin's Paradox fascinating or even interesting to discuss. But I sometimes see other physicists say misleading things that imply that it’s all about inertial v. non-inertial motion, i.e. acceleration. This is fine for the simplistic story of the one twin who stays on Earth, the inertial twin, and the non-inertial twin who travels far and returns at relativistic speeds. And remember the twins are just for dramatic effect. We're really talking about ideal clocks carried along the two paths. Clocks immune to the stress of space travel. Here’s the simple version:
And sometimes even physicists point to that turn-around at 2011 and say “It’s the acceleration there that makes the difference.” Well, sort of. If he’d just coasted inertially he wouldn’t have turned around and come back…or would he. In fact acceleration is just incidental. I some versions it's responsible for one path being longer than the other, but it's the path length difference that's essential, not how it's realized. This is illustrated in the slingshot version.
Suppose there’s a neutron star out there and our traveler just swings around it, using Its gravity, but no rocket thrust at all. Notice that the time in the gravitational field of the neutron star can be neglible compared to the travel time, so corrections due to gravitational time dilation can be ignored. The swing around the neutron star is entirely free-fall, zero acceleration in the general relativistic sense, no applied force.
But what about accelerating away from Earth and braking to a stop on return. Not necessary: I've shown Red flying by Earth on his way to the neutron star where he gets flung around by its gravity and flies by Earth going the other way. As he passes Earth outbound, he sets his clock to Earth time and as he passes Earth on the return he and Earth compare clocks and find the Twin Paradox is the same, a 2yr timedifference. So it’s not being non-inertial. It’s not some acceleration induced stress on the clock. Another way to see the same thing is The Triplet Paradox.
In the Triplet Paradox Red, with his clock, heads out setting his clock to equal Earth's as he passes. Four years later (ship's time) he passes Blue who is inbound. As they pass Blue sets his clock to match Red's, i.e. to 2011. Then when Blue passes Earth he compares to Earth's clock and finds exactly the same “paradoxical” disagreement.
The Triplet Paradox completely avoids even a change in direction. Proper time is just measured along three different inertial segments between the same two events.
And what if there’s acceleration, but each twin experiences exactly the same acceleration? Just not with the same intervals between them. In this case Red and Blue are coasting along together. We might suppose they are approaching Earth and Blue decides to stop there, but whether some planet is there or not, Blue fires his rockets and stops, while Red coasts on. But then years later, Red fires his rockets first to stop and then immediately after to go back. As he is reaching Blue, who has been stationary all this time, Blue fires his rockets and once again matches speed coasting along with Red. But their clocks register the same difference as in the other versions...even though each one experienced the same accelerations for the same durations.
By now you should be convinced that acceleration, per se, has nothing to do with the “paradox”. It’s just a matter of different paths between the same two events having different “lengths”, i.e. different intervals of proper time. So how should you looks at it? It’s (almost) the same as two cars who are driven from NYC to L.A. Blue takes a fairly direct route, while Red decides to visit the Alamo in San Antonio.
Their odometers measure the distance, not the as-the-crow-flies distance, but the proper distance along their paths. Is there any paradox that they measure different distances? No, they are just spatial versions of clocks. The tricky thing is that we have to distinguish between coordinate time (which are just labels) and proper time (which are physical measures). Einstein showed that spacetime has a minus sign in the metric, so more distance subtracts from the proper time. That’s why the twin who travels the greater spatial distance experiences less proper time distance. It has nothing to do with acceleration or not. Usually the inertial path (least spatial distance) is the longest proper time distance. The exceptions come when gravity bends spacetime and our mapping doesn’t take account of how it changes spacetime distances so that they don’t agree with naive space distances.
The misleading thing you will hear even from physicists (I’ve said it myself) is that time runs more slowly in a gravitational field. This is given as the explanation of Shapiro delay. But in the analogy above this is like saying odometers run fast near San Antonio. The right way to look at it is the clocks are ideal and there is just less time along a geodesic path thru a gravitational field. And paths are even shorter if you’re standing on a planet’s surface, because in that case your path isn’t geodesic. The force on the bottom of your feet is pushing you off the geodesic.
Moreover, in light cone diagrams, the labeled spacetime coordinates are referred to as "events" which are, or not, causally connected. So if I am confused, I suppose I should blame the "experts". AG
On Thursday, September 11, 2025 at 4:44:13 AM UTC-6 John Clark wrote:
> your traveling twin somehow turns around without acceleration
Forget about acceleration, you're never going to understand the twin "paradox" if you become obsessed over that because, as I've mentioned before, there is a way to express it without any twin undergoing any acceleration.
Firstly I'm not obsessed. If anyone is, it's those who claim acceleration plays no role in resolving the alleged paradox. Those that make this claim cannot, it seems, explain how the traveling twin can start his journey without accelerating. AG
But even in that case there is no paradox, the only reason people think there is one is because they forget there is no universal agreement about simultaneity.
If one appeals to acceleration, and thus asymmetry to resolve the TP, simultaneity plays no role since the twins sychronize their clocks when juxtaposed and at rest. AG
On Thursday, September 11, 2025 at 4:44:13 AM UTC-6 John Clark wrote:
> your traveling twin somehow turns around without acceleration
Forget about acceleration, you're never going to understand the twin "paradox" if you become obsessed over that because, as I've mentioned before, there is a way to express it without any twin undergoing any acceleration. But even in that case there is no paradox, the only reason people think there is one is because they forget there is no universal agreement about simultaneity.
Brent claims that using slingshot in a gravity field, he can avoid all acceleration.
So who is obsessed, and with what? How does the traveling twin separate from his twin if they're initially juxtaposed?
In free fall along a slingshot path? Not only has he changed the TP model to get the result he desires, but he is still unable to avoid acceleration. AG
John K Clark See what's on my new list at Extropolis
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On Thursday, September 11, 2025 at 1:16:02 PM UTC-6 John Clark wrote:
On Thu, Sep 11, 2025 at 2:35 PM Alan Grayson <agrays...@gmail.com> wrote:
why don't you spend a few minutes describing the main differences between the usual TP and this different one?
Because if you can't be bothered to spend a few minutes watching the video then I can't be bothered to spend a few minutes making an inferior summary of that video. I can't easily provide snappy animations and illustrations needed to make things clearer.
I did. I watched the first video and had a discussion with Brent about his alleged related solution, but I was finally turned off when he used the slingshot method for turnaround, but apparently couldn't explain how the traveling twin could start his adventure without accelerating.
If you ignore that issue, then voi-la, you've gotten the result you wanted, but with a different scenario than the original TP. AG
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On 9/11/2025 7:08 PM, Alan Grayson wrote:
On Thursday, September 11, 2025 at 7:58:36 PM UTC-6 Brent Meeker wrote:
On 9/11/2025 12:39 PM, Alan Grayson wrote:
On Thursday, September 11, 2025 at 1:16:02 PM UTC-6 John Clark wrote:
On Thu, Sep 11, 2025 at 2:35 PM Alan Grayson <agrays...@gmail.com> wrote:
why don't you spend a few minutes describing the main differences between the usual TP and this different one?
Because if you can't be bothered to spend a few minutes watching the video then I can't be bothered to spend a few minutes making an inferior summary of that video. I can't easily provide snappy animations and illustrations needed to make things clearer.
I did. I watched the first video and had a discussion with Brent about his alleged related solution, but I was finally turned off when he used the slingshot method for turnaround, but apparently couldn't explain how the traveling twin could start his adventure without accelerating.You never asked that, and it's obvious from the diagram.
Brent
I figured out that you could avoid acceleration entirely if the traveling twin is always in free fall and the clocks are synchronized when the traveler passes the resting twin. I don't recall you simply stating that, so I wouldn't have to decode your diagrams. What you left out is HOW the traveling twin gets into orbit.I left out how he combs his hair too.
Brent
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It's always the real prick the most oblivious to his condition.
All those moments will be lost in time, like tears in rain. (Roy Batty/Rutger Hauer)Le ven. 12 sept. 2025, 09:04, Alan Grayson <agrays...@gmail.com> a écrit :
On Thursday, September 11, 2025 at 10:41:59 PM UTC-6 Brent Meeker wrote:
On 9/11/2025 7:08 PM, Alan Grayson wrote:
On Thursday, September 11, 2025 at 7:58:36 PM UTC-6 Brent Meeker wrote:
On 9/11/2025 12:39 PM, Alan Grayson wrote:
On Thursday, September 11, 2025 at 1:16:02 PM UTC-6 John Clark wrote:
On Thu, Sep 11, 2025 at 2:35 PM Alan Grayson <agrays...@gmail.com> wrote:
why don't you spend a few minutes describing the main differences between the usual TP and this different one?
Because if you can't be bothered to spend a few minutes watching the video then I can't be bothered to spend a few minutes making an inferior summary of that video. I can't easily provide snappy animations and illustrations needed to make things clearer.
I did. I watched the first video and had a discussion with Brent about his alleged related solution, but I was finally turned off when he used the slingshot method for turnaround, but apparently couldn't explain how the traveling twin could start his adventure without accelerating.You never asked that, and it's obvious from the diagram.
Brent
I figured out that you could avoid acceleration entirely if the traveling twin is always in free fall and the clocks are synchronized when the traveler passes the resting twin. I don't recall you simply stating that, so I wouldn't have to decode your diagrams. What you left out is HOW the traveling twin gets into orbit.I left out how he combs his hair too.
Brent
And that you're an arrogant prick who is under the illusion that your plots are self-evident. I suppose you go mute and can't simply say that your TP models a traveling twin in orbit, and that it replicates the original problem perfectly (if it does). AG
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