Let us understand each other (a moment of optimism).
The immense, the incomparable, the fantastic relativist theorist
(that's me) wrote:
"If two different observers travel an identical path in equal
observable times,
then their proper times will necessarily be equal.
Let's set up a Galilean frame of reference, in which a Galilean mobile moves from left to right on the x axis....
Let us pose another body, but this time in uniformly accelerated...
motion, and whose speed will be specially chosen so that To=12.915.
This means that in R, if they leave together, they arrive together
(even if they do not have the same speed between them).
Their own times will be equal.
It's even a very speech full of abstract religiosity.
W dniu 13.04.2024 o 13:49, Python pisze:
Le 13/04/2024 à 08:36, Richard "Hachel" Lengrand a écrit :
Let us understand each other (a moment of optimism).
The immense, the incomparable, the fantastic relativist theorist
(that's me) wrote:
"If two different observers travel an identical path in equal
observable times,
then their proper times will necessarily be equal.
[note : I posted a more detailed answer on fr.sci.physique]
This claim is logically ill-founded to begin with :
Oh, stinker Python is opening its muzzle again,
and trying again to pretend he knows something.
Tell me, poor stinker, have you already learnt
what a function is? Are you still trying to
determine its properties applying a French
definition of a different word?
Le 13/04/2024 à 08:36, Richard "Hachel" Lengrand a écrit :
Let us understand each other (a moment of optimism).
The immense, the incomparable, the fantastic relativist theorist
(that's me) wrote:
"If two different observers travel an identical path in equal
observable times,
then their proper times will necessarily be equal.
[note : I posted a more detailed answer on fr.sci.physique]
This claim is logically ill-founded to begin with :
Le 13/04/2024 à 16:01, Maciej Wozniak a écrit :
W dniu 13.04.2024 o 13:49, Python pisze:
Le 13/04/2024 à 08:36, Richard "Hachel" Lengrand a écrit :
Let us understand each other (a moment of optimism).
The immense, the incomparable, the fantastic relativist theorist
(that's me) wrote:
"If two different observers travel an identical path in equal
observable times,
then their proper times will necessarily be equal.
[note : I posted a more detailed answer on fr.sci.physique]
This claim is logically ill-founded to begin with :
Oh, stinker Python is opening its muzzle again,
and trying again to pretend he knows something.
Tell me, poor stinker, have you already learnt
what a function is? Are you still trying to
determine its properties applying a French
definition of a different word?
Woz, this thread is not about your own confusions about
your own confusions.
Woz, this thread is not about your own confusions about
your own confusions.
"If two different observers travel an identical path in equal observable times,
then their proper times will necessarily be equal.
Let's set up a Galilean frame of reference, in which a Galilean mobile moves from left to right on the x axis.
When it goes to O, in R, we click.
When it passes x, let's say 12 al, we click.
For example, we obtain To=12.915.
Or a speed of Vo=x/To=0.9291c
Tr=To.sqrt(1-Vo²/c²)=4.776
Let us pose another body, but this time in uniformly accelerated motion,
and whose speed will be specially chosen so that To=12.915.
This means that in R, if they leave together, they arrive together (even
if they do not have the same speed between them).
Le 13/04/2024 à 17:45, Maciej Wozniak a écrit :
[same bullshit]
Woz, this thread is not about your own confusions about
your own confusions.
Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times,
"Observable"?
WHO is doing the observing? Presumably, an observer at earth, yes?
then their proper times will necessarily be equal.
This is not possible.
Let's set up a Galilean frame of reference, in which a Galilean
mobile moves from left to right on the x axis.
When it goes to O, in R, we click.
When it passes x, let's say 12 al, we click.
For example, we obtain To=12.915.
Or a speed of Vo=x/To=0.9291c
Tr=To.sqrt(1-Vo²/c²)=4.776
This is assuming a constant velocity of v = 0.9291c, yes. The proper time of the trip is 4.7746 years, using c = 299792458 m/s and one year equal to 3600*24*365.25 = 31557600 seconds.
AAnd in the meantime in the real world, forbidden
by your bunch of idiots "improper" clocks keep
measuring t'=t, just like all serious clocks
always did.
Maciej Wozniak wrote:
AAnd in the meantime in the real world, forbidden
by your bunch of idiots "improper" clocks keep
measuring t'=t, just like all serious clocks
always did.
The "improper" clocks in orbit keep proper time
only on earth.
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
[... enormous amount of gibberish ignored.]
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course.
My response differs from others because I interpret your context
differently.
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course.
My response differs from others because I interpret your context
differently.
Since you mention "proper times", your context must be relativity;
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course. My response differs from others because I interpret your context differently.
Since you mention "proper times", your context must be relativity; it
does not matter whether SR or GR, because "travel an identical path"
means they travel along a single worldline through spacetime -- i.e.
they are always co-located and co-moving, so of course their elapsed
proper times are equal (counting from any event on their worldline).
Your "in equal observable times" is redundant. For any observer this
directly follows from them following the same worldline through
spacetime.
Note this is essentially the first time I agree with Hachel. I doubt
that he understands why what he wrote is actually correct, because he followed it with a bunch of obfuscatory nonsense.
Yes, of course.
My response differs from others because I interpret your context
differently.
Since you mention "proper times", your context must be relativity; it
does not matter whether SR or GR, because "travel an identical path"
means they travel along a single worldline through spacetime -- i.e.
they are always co-located and co-moving, so of course their elapsed
proper times are equal (counting from any event on their worldline).
Your "in equal observable times" is redundant. For any
observer this directly follows from them following the
same worldline through spacetime.
Note this is essentially the first time I agree with Hachel. I doubt
that he understands why what he wrote is actually correct, because he followed it with a bunch of obfuscatory nonsense.
[... enormous amount of gibberish ignored.]
Le 20/04/2024 18:52, Tom Roberts a crit :
Yes, of course.
My response differs from others because I interpret your context
differently.
Since you mention "proper times", your context must be relativity; it
does not matter whether SR or GR, because "travel an identical path"
means they travel along a single worldline through spacetime -- i.e.
they are always co-located and co-moving, so of course their elapsed
proper times are equal (counting from any event on their worldline).
Your "in equal observable times" is redundant. For any
observer this directly follows from them following the
same worldline through spacetime.
Note this is essentially the first time I agree with Hachel. I doubt
that he understands why what he wrote is actually correct, because he
followed it with a bunch of obfuscatory nonsense.
[... enormous amount of gibberish ignored.]
J'ai rien compris.
On 04/20/2024 09:52 AM, Tom Roberts wrote:
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course.
My response differs from others because I interpret your context
differently.
Since you mention "proper times", your context must be relativity; it
does not matter whether SR or GR, because "travel an identical path"
means they travel along a single worldline through spacetime -- i.e.
they are always co-located and co-moving, so of course their elapsed
proper times are equal (counting from any event on their worldline).
Your "in equal observable times" is redundant. For any
observer this directly follows from them following the
same worldline through spacetime.
Note this is essentially the first time I agree with Hachel. I doubt
that he understands why what he wrote is actually correct, because he
followed it with a bunch of obfuscatory nonsense.
[... enormous amount of gibberish ignored.]
Tom Roberts
Can you help further explain for the rest of us why
this isn't necessarily the usual interpretation or
why it sort of doesn't arrive at the same results of
some of the usual thought experiments like the traveling twins?
Le 21/04/2024 à 15:53, Ross Finlayson a écrit :
On 04/20/2024 09:52 AM, Tom Roberts wrote:
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course.
My response differs from others because I interpret your context
differently.
Since you mention "proper times", your context must be relativity; it
does not matter whether SR or GR, because "travel an identical path"
means they travel along a single worldline through spacetime -- i.e.
they are always co-located and co-moving, so of course their elapsed
proper times are equal (counting from any event on their worldline).
Your "in equal observable times" is redundant. For any
observer this directly follows from them following the
same worldline through spacetime.
Note this is essentially the first time I agree with Hachel. I doubt
that he understands why what he wrote is actually correct, because he
followed it with a bunch of obfuscatory nonsense.
[... enormous amount of gibberish ignored.]
Tom Roberts
Can you help further explain for the rest of us why
this isn't necessarily the usual interpretation or
why it sort of doesn't arrive at the same results of
some of the usual thought experiments like the traveling twins?
In the twins scenario, twins does not share the same
space-time path.
Le 21/04/2024 à 15:53, Ross Finlayson a écrit :
On 04/20/2024 09:52 AM, Tom Roberts wrote:
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course.
My response differs from others because I interpret your context
differently.
Since you mention "proper times", your context must be relativity; it
does not matter whether SR or GR, because "travel an identical path"
means they travel along a single worldline through spacetime -- i.e.
they are always co-located and co-moving, so of course their elapsed
proper times are equal (counting from any event on their worldline).
Your "in equal observable times" is redundant. For any
observer this directly follows from them following the
same worldline through spacetime.
Note this is essentially the first time I agree with Hachel. I doubt
that he understands why what he wrote is actually correct, because he
followed it with a bunch of obfuscatory nonsense.
[... enormous amount of gibberish ignored.]
Tom Roberts
Can you help further explain for the rest of us why
this isn't necessarily the usual interpretation or
why it sort of doesn't arrive at the same results of
some of the usual thought experiments like the traveling twins?
In the twins scenario, twins does not share the same
space-time path.
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course.
My response differs from others because I interpret your context
differently.
You interpret path as a path in space-time while Hachel is considering
path in space only (so it is true in only ONE reference frame)
For instance, he claims that if two travelers would go from Earth
to Tau Ceti, one at constant velocity w.r.t Earth the other one
with a constant acceleration both will measure the same (proper)
time for their trips.
Le 21/04/2024 à 16:48, Python a écrit :
Le 21/04/2024 à 15:53, Ross Finlayson a écrit :
On 04/20/2024 09:52 AM, Tom Roberts wrote:
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course.
My response differs from others because I interpret your context
differently.
Since you mention "proper times", your context must be relativity; it
does not matter whether SR or GR, because "travel an identical path"
means they travel along a single worldline through spacetime -- i.e.
they are always co-located and co-moving, so of course their elapsed
proper times are equal (counting from any event on their worldline).
Your "in equal observable times" is redundant. For any
observer this directly follows from them following the
same worldline through spacetime.
Note this is essentially the first time I agree with Hachel. I doubt
that he understands why what he wrote is actually correct, because he
followed it with a bunch of obfuscatory nonsense.
[... enormous amount of gibberish ignored.]
Tom Roberts
Can you help further explain for the rest of us why
this isn't necessarily the usual interpretation or
why it sort of doesn't arrive at the same results of
some of the usual thought experiments like the traveling twins?
In the twins scenario, twins does not share the same
space-time path.
We may not have the same space-time path, and have the same proper time,
the same path, and the same improper time (for other observers).
Note that "space-time path", I don't understand the geometric concept
very well.
[snip babbling]
Le 20/04/2024 à 19:07, Python a écrit :
On 4/13/24 1:36 AM, Richard Hachel wrote:
"If two different observers travel an identical path in equal
observable times, then their proper times will necessarily be equal.
Yes, of course.
My response differs from others because I interpret your context
differently.
You interpret path as a path in space-time while Hachel is considering
path in space only (so it is true in only ONE reference frame)
For instance, he claims that if two travelers would go from Earth
to Tau Ceti, one at constant velocity w.r.t Earth the other one
with a constant acceleration both will measure the same (proper)
time for their trips.
Yes, this is entirely correct.
If two different protagonists use different frames of reference to join
Tau Ceti (12al), and if we notice that in the terrestrial frame of
reference the travel times are similar and the path is similar (direct
path), then the two proper times will be similar.
We may not have the same space-time path, and have the same proper time,
the same path, and the same improper time (for other observers).
To have the same improper time is meaningless because in a given
frame of reference there is a single time elapsed between any
two events. And there are only two events : travelers' depart
and travelers' meeting again.
Moreover to have the same path (in the spatial sense) is a
property that is only true in a single frame of reference.
It makes no sense (outside of a trivial one in Galilean Relativity)
to claim that a condition depending on a given chosen frame could
imply a property (equality of proper times) that does not depend
on that.
Note that "space-time path", I don't understand the geometric concept
very well.
It's not a big deal! It is the set of events "traveler is at this
position at that time".
[snip babbling]
Your claim is directly violating the principle of Relativity. End of
Story.
The equality of travel times in any frame is always true.
Le 21/04/2024 à 23:19, Python a écrit :
We may not have the same space-time path, and have the same proper
time, the same path, and the same improper time (for other observers).
To have the same improper time is meaningless because in a given
frame of reference there is a single time elapsed between any
two events. And there are only two events : travelers' depart
and travelers' meeting again.
Moreover to have the same path (in the spatial sense) is a
property that is only true in a single frame of reference.
It makes no sense (outside of a trivial one in Galilean Relativity)
to claim that a condition depending on a given chosen frame could
imply a property (equality of proper times) that does not depend
on that.
Note that "space-time path", I don't understand the geometric concept
very well.
It's not a big deal! It is the set of events "traveler is at this
position at that time".
[snip babbling]
Your claim is directly violating the principle of Relativity. End of
Story.
Having the same improper time is on the contrary an important
relativistic concept, as important as having the same proper time.
Le 21/04/2024 à 23:39, Python a écrit :
The equality of travel times in any frame is always true.
? ? ?
Le 21/04/2024 à 23:47, Python a écrit :
Oh come on!
There are only two events involved! One for when the travelers
departs and another one when they meet again.
Hence a SINGLE time elapsed between these two events, as seen in any
frame of reference.
You seemed to have, at least, understood this recently. It appears
you don't. You are insufferable.
[snip babbing]
It goes without saying that it defines the time between the joint
departure event of the two rockets and the joint arrival of the two
rockets.
We have entered into tautology there.
Le 21/04/2024 à 23:31, Richard Hachel a écrit :
[snip babbling]
1. You must only use one reference frame during the entire trip.
"use" a reference frame... sigh...
2. the accelerated twin must go into “native” state, i.e. at rest.
At rest with respect to what? Absolute space again! You are
insufferable.
Oh come on!
There are only two events involved! One for when the travelers
departs and another one when they meet again.
Hence a SINGLE time elapsed between these two events, as seen in any
frame of reference.
You seemed to have, at least, understood this recently. It appears
you don't. You are insufferable.
[snip babbling]
1. You must only use one reference frame during the entire trip.
2. the accelerated twin must go into “native” state, i.e. at rest.
Sysop: | Keyop |
---|---|
Location: | Huddersfield, West Yorkshire, UK |
Users: | 365 |
Nodes: | 16 (2 / 14) |
Uptime: | 20:03:06 |
Calls: | 7,787 |
Calls today: | 2 |
Files: | 12,917 |
Messages: | 5,750,729 |