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. 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. AG* 

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.

*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. 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 *

You 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*


*Start with this video:  *

*I Never Understood How To Intuitively Solve The Twin's Paradox ... Until 
Now!* <https://www.youtube.com/watch?v=3V00tAfcHCI>

*John K Clark    See what's on my new list at  Extropolis 
<https://groups.google.com/g/extropolis>*

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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