The David Twin paradox - for STRICH
Dirk Van de moortel
Twin A is on planet at x = -L and uses coordinates x' and t'.
Twin B is on planet at x = L and uses coordinates x" and t".
In system K the twins have the same age 0 at time t = 0.
At event (x,t) = (-L,0) twin A suddenly moves with velocity v towards
x = 0 at his event (x',t') = (0,0).
At event (x,t) = (L,0) twin B suddenly moves with velocity -v towards
x = 0 at his event (x",t") = (0,0).
We use units where c = 1 (so -1 < v < 1) and abbreviate
g = 1/sqrt(1-v^2).
Lorentz transformation [LTA] between K and A is
{ x' = g ( x+L - v t )
{ t' = g ( t - v (x+L) )
and the inverse [LTAi]
{ x+L = g ( x' + v t' )
{ t = g ( t' + v x' )
Lorentz transformation [LTB] between K and B is
{ x" = g ( x-L + v t )
{ t" = g ( t + v (x-L) )
and the inverse [LTBi]
{ x-L = g ( x" - v t" )
{ t = g ( t" - v x" )
1) Looking from K:
Wordline (equation of motion) of origin of K:
x = 0
Wordline (equation of motion) of twin A:
x+L = v t
Wordline (equation of motion) of twin B:
x-L = -v t
Time of arrival at home at x = 0 for both twins:
t = L/v
Home arrival event H for both twins
(x,t) = (0,L/v)
Age of twin A at some time t = T is t'-coordinate of solution of
{ x' = 0
{ T = g ( t' + v x' )
giving
t' = 1/g T
Age of twin B at some time t = T is t"-coordinate of solution of
{ x" = 0
{ T = g ( t" - v x" )
giving
t" = 1/g T
Conclusion, in K, at any time t = T, the twins A and B have
the same age 1/g T.
2) Looking from A at event where A has age t' = 1/g T
i.o.w. at event ( x', t' ) = ( 0, 1/g T )
i.o.w. with [LTAi] at event ( x, t ) = ( -L + v T, T )
Line of A-simultaneity in K-coordinates:
x - (-L + v T) = 1/v (t - T)
Worldline of B in K-coordinates:
x-L = -v t
Point of A-simultaneity on worldline of B by solving
the previous 2 equations for x and t
{ x = ( (L- v T)/g^2 )/(1+v^2)
{ t = (T/g^2 + 2 L v)/(1+v^2)
Transforming to B-frame to find age t" of B and verifying
that x" = 0:
{ x" = 0
{ t" = ( 2 L v + T/g ) / ( g (1+v^2) )
So, when A has age t' = 1/g T, then according to A, his twin
brother B has age t" = ( 2 L v + T/g ) / ( g (1+v^2) ).
Let's see what happens when they meet at T = L/v:
t' = 1/g L/v
t" = 1/g L/v
which is the same age.
According to A, his brother B is older, but aging slower in
such a way that they have the same age when they meet.
3) Looking from B at event where B has age t" = 1/g T
i.o.w. at event ( x", t" ) = ( 0, 1/g T )
i.o.w. with [LTBi] at event ( x, t ) = ( L - v T, T )
Line of B-simultaneity in K-coordinates:
x - (L - v T) = -1/v (t - T)
Worldline of A in K-coordinates:
x+L = v t
Point of B-simultaneity on worldline of A by solving
the previous 2 equations for x and t
{ x = ( (-L+ v T)/g^2 )/(1+v^2)
{ t = (T/g^2 + 2 L v)/(1+v^2)
Transforming to A-frame to find age t' of A and verifying
that x' = 0:
{ x' = 0
{ t' = ( 2 L v + T/g ) / ( g (1+v^2) )
So, when B has age t" = 1/g T, then according to B, his twin
brother A has age t' = ( 2 L v + T/g ) / ( g (1+v^2) ).
Let's see what happens when they meet at T = L/v:
t" = 1/g L/v
t' = 1/g L/v
which is the same age.
According to B, his brother A is older, but aging slower in
such a way that they have the same age when they meet.
Dirk Vdm
PD
Which is exactly what I said, and you provided exactly the
mathematical presentation he asked for, and I'm sure it's going to go
clean over and to one side of his head.
PD
Dirk Van de moortel
0a9c2ec6-8bd6-46f9...@z72g2000hsb.googlegroups.com
To the Spaceside perhaps.
Maybe this will help then:
http://users.telenet.be/vdmoortel/dirk/Stuff/DavidTwinParadox.gsp.png
:-)
Dirk Vdm
Spaceman
LOL
So B is older than A AND A is older than B
simultaneously!~.
ROFLOL!
Freakin morons!
They actually think paradox parrotville is an OK place to be!
LOL
--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman
PD
wrote:
Simultaneously according to whom?
Simultaneity is frame-dependent.
:>)
Spaceman
Dirk and you doing the math.
:)
> Simultaneity is frame-dependent.
Then apparently you are both stuck in the same toilet frame
since you both get the same answer for both being older than the other.
:)
Sue...
SperM.hotmail.com> wrote:
> Intertial system K using coordinates x and t.
> Twin A is on planet at x = -L and uses coordinates x' and t'.
> Twin B is on planet at x = L and uses coordinates x" and t".
> In system K the twins have the same age 0 at time t = 0.
>
> At event (x,t) = (-L,0) twin A suddenly moves with velocity v towards
> x = 0 at his event (x',t') = (0,0).
> At event (x,t) = (L,0) twin B suddenly moves with velocity -v towards
> x = 0 at his event (x",t") = (0,0).
>
> We use units where c = 1 (so -1 < v < 1) and abbreviate
> g = 1/sqrt(1-v^2).
>
<<A Lorentz transformation or any other coordinate
transformation will convert electric or magnetic
fields into mixtures of electric and magnetic fields,
but no transformation mixes them with the
gravitational [inertial by equivalence] field. >>
http://scitation.aip.org/journals/doc/PHTOAD-ft/vol_58/iss_11/31_1.shtml
The rest has no basis.
See:
Relativistic particle dynamics
http://farside.ph.utexas.edu/teaching/em/lectures/node126.html
Sue...
[snip]
Dirk Van de moortel
W9ydnTmjfokw-j7V...@comcast.com
> PD wrote:
>> On Aug 13, 4:05 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
>> SperM.hotmail.com> wrote:
[snip]
>>> Conclusion, in K, at any time t = T, the twins A and B have
>>> the same age 1/g T.
[snip]
>>> According to A, his brother B is older, but aging slower in
>>> such a way that they have the same age when they meet.
[snip]
>>> According to B, his brother A is older, but aging slower in
>>> such a way that they have the same age when they meet.
>>>
>>> Dirk Vdm
>>
>> Which is exactly what I said, and you provided exactly the
>> mathematical presentation he asked for, and I'm sure it's going to go
>> clean over and to one side of his head.
>
> LOL
> So B is older than A AND A is older than B
> simultaneously!~.
> ROFLOL!
Yes, when Spaceman and his twin brother Strich stop cleaning
the floor and Spaceman looks at Strich from a distance through
a gap between his malfunctioning fingers, Strich looks smaller
than Spaceman, and when in turn Strich looks at Spaceman
through a gap between his malfunctioning fingers, Spaceman
looks smaller than Strich. And when they get together, they look
exactly the same throughs gaps between their malfunctioning
fingers.
Dirk Vdm
Dirk Van de moortel
fb891bc4-be4d-4f1a...@d77g2000hsb.googlegroups.com>> Inertial system K using coordinates x and t.
>> Twin A is on planet at x = -L and uses coordinates x' and t'.
>> Twin B is on planet at x = L and uses coordinates x" and t".
>> In system K the twins have the same age 0 at time t = 0.
>>
>> At event (x,t) = (-L,0) twin A suddenly moves with velocity v towards
>> x = 0 at his event (x',t') = (0,0).
>> At event (x,t) = (L,0) twin B suddenly moves with velocity -v towards
>> x = 0 at his event (x",t") = (0,0).
>>
>> We use units where c = 1 (so -1 < v < 1) and abbreviate
>> g = 1/sqrt(1-v^2).
>>
>> Lorentz transformation [LTA] between K and A is...
>
>
> <<A Lorentz transformation or any other coordinate
> transformation will convert electric or magnetic
> fields into mixtures of electric and magnetic fields,
> but no transformation mixes them with the
> gravitational [inertial by equivalence] field. >>
> http://scitation.aip.org/journals/doc/PHTOAD-ft/vol_58/iss_11/31_1.shtml
| "By the end of the 1970s, however, the SG models returned to
| the old design style for the most part, and current versions
| have returned to the 1967-1969 styling and construction with
| the large pickguard, which wraps around the pickups on the
| guitar body (though re-issues and variants of the small
| pickguard SG are still available)."
http://en.wikipedia.org/wiki/Gibson_SG
Dirk Vdm
stric...@gmail.com
Nice try Dirk. Only analyses (1) is correct as you used a frame K at
rest in the middle of twins A and B. Your analyses (2) and (3) still
used the same frame K, while purporting to analyze from A and B
respectively. Now either you are stupid and did the wrong analyses,
or you think we are stupid and such a glaring error will not be easily
seen. Have you forgotten your RELATIVITY?
Let me show you how to do a CORRECT second analyses from the A frame.
A is at coordinate (x, t) = (0,0) and B is at (x,t) = (D, 0) where
D=2L. More importantly, for A, velocity=0, and for B, velocity=V
where V can be simply V=v+v for small velocoties.
No wonder you posted this somewhere else... Was that a pathetic
attmept to win. I don't think you are that stupid.
Spaceman
> Yes, when Spaceman and his twin brother Strich stop cleaning
> the floor and Spaceman looks at Strich from a distance through
> a gap between his malfunctioning fingers, Strich looks smaller
> than Spaceman, and when in turn Strich looks at Spaceman
> through a gap between his malfunctioning fingers, Spaceman
> looks smaller than Strich. And when they get together, they look
> exactly the same throughs gaps between their malfunctioning
> fingers.
So you prove that you do not understand the limits of lightspeed
and observations using such limited lightspeed.
You poor thing.
Dirk Van de moortel
ec160f33-24c4-4049...@v57g2000hse.googlegroups.com
> On Aug 13, 5:05 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> SperM.hotmail.com> wrote:
>> Intertial system K using coordinates x and t.
>> Twin A is on planet at x = -L and uses coordinates x' and t'.
>> Twin B is on planet at x = L and uses coordinates x" and t".
>> In system K the twins have the same age 0 at time t = 0.
>>
>> At event (x,t) = (-L,0) twin A suddenly moves with velocity v towards
>> x = 0 at his event (x',t') = (0,0).
>> At event (x,t) = (L,0) twin B suddenly moves with velocity -v towards
>> x = 0 at his event (x",t") = (0,0).
>>
>> We use units where c = 1 (so -1 < v < 1) and abbreviate
>> g = 1/sqrt(1-v^2).
>>
>> Lorentz transformation [LTA] between K and A is
>> { x' = g ( x+L - v t )
>> { t' = g ( t - v (x+L) )
>> and the inverse [LTAi]
>> { x+L = g ( x' + v t' )
>> { t = g ( t' + v x' )
>>
>> Lorentz transformation [LTB] between K and B is
>> { x" = g ( x-L + v t )
>> { t" = g ( t + v (x-L) )
>> and the inverse [LTBi]
>> { x-L = g ( x" - v t" )
>> { t = g ( t" - v x" )
[snip]
>> Dirk Vdm
>
> Nice try Dirk. Only analyses (1) is correct as you used a frame K at
> rest in the middle of twins A and B. Your analyses (2) and (3) still
> used the same frame K, while purporting to analyze from A and B
> respectively.
Haven't you learned how to eliminate variables from systems
of equations in your highschool linear algebra courses?
> Now either you are stupid and did the wrong analyses,
> or you think we are stupid and such a glaring error will not be easily
> seen. Have you forgotten your RELATIVITY?
Analysis (2) used frames K and A with the Lorentz Transformation
between them.
Analysis (3) used frames K and B with the Lorentz Transformation
between them.
Although this is a standard technique, used as from page 3 in
introductory courses on special relativity, perhaps you haven't
heard of it. That would not surprise me.
Since according to linear algebra the Lorentz transfomations form
a group, these analyses are valid. But perhaps you think you have
some experimental evidence for the invalidity of linear algebra.
That would not surprise me either. Feel free to share it with us.
> Let me show you how to do a CORRECT second analyses from the A frame.
> A is at coordinate (x, t) = (0,0) and B is at (x,t) = (D, 0) where
> D=2L. More importantly, for A, velocity=0, and for B, velocity=V
> where V can be simply V=v+v for small velocoties.
So, feel free to provide the complete analysis, and compare
with mine. Don't forget to show every detail of your calculation,
so I can help you when you get stuck.
Don't forget that we are talking about arbitrary velocities and we
are working in special relativity, so make sure you take
V = 2 v/(1+v^2)
- you probably recognise this quantity in the expressions:
( 2 L v + T/g ) / ( g (1+v^2) )
In order not to get stuck, I advise you to use the same variables
As I have, namely (x',t') for A and (x",t") for B.
This way you can verify what happens if you manage to
eliminate variables x and t from [LTA] and [LTB] and compare
with you find.
Dirk Vdm
stric...@gmail.com
SperM.hotmail.com> wrote:
snip
> Dirk Vdm
CRAP
Dirk Van de moortel
Your lunch?
Or can't you make the exercise?
Dirk Vdm
PD
You seem to miss the fact that the point is not to convince you of
anything. You asked for a mathematical and qualitative accounting of
how SR resolves it, and it's been given to you.
You always have the right to close your eyes.
You always have the right to clamp your hands over your ears and
holler, "Na-na-na-na-na! Can't hear you! Don't believe you and you
CAN'T MAKE ME!"
This of course changes nothing.
PD
Spaceman
> On Aug 14, 10:13 am, strich....@gmail.com wrote:
>> On Aug 13, 5:05 pm, "Dirk Van de moortel"
>> <dirkvandemoor...@ThankS-NO-
>>
>> SperM.hotmail.com> wrote:
>>
>> snip
>>
>>> Dirk Vdm
>>
>> CRAP
>
> You seem to miss the fact that the point is not to convince you of
> anything. You asked for a mathematical and qualitative accounting of
> how SR resolves it, and it's been given to you.
I get it!
Both twins are older and younger than the other.
whoever is looking at the other is older and whoever is being
watched is the younger...
It is so perfectally logical.
NOT!
LOL
Dirk Van de moortel
Yes, when Spaceman and his twin brother Strich stop cleaning
the floor and Spaceman looks at Strich from a distance through
a gap between his malfunctioning fingers, Strich looks smaller
than Spaceman, and when in turn Strich looks at Spaceman
through a gap between his malfunctioning fingers, Spaceman
looks smaller than Strich. And when they get together, they look
exactly the same throughs gaps between their malfunctioning
fingers.
Dirk Vdm
Spaceman
LOL
The floor cleaners are you and PD,
Both of you have no clue about the limitations of the observations
made with light.
You love to ignore the light travel time to make up the younger
and older ages "as seen from" observer.
Dirk Van de moortel
Oops, I repeated myself. Sorry for that.
You will probably repeat yourself as well, and since the
floor must be sufficiently clean by now, there's no need
for me to include it here.
Dirk Vdm
PD
Dirk Van de moortel
I thought this one deserves a bit of (beyond the usual semi-automated) work:
"Reconstruction: Challenge, Reply, Contempt, Exercise, CRAP":
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrapReconstruction.html
Thanks for asking that first pertinent killer question ;-)
Dirk Vdm
stric...@gmail.com
SperM.hotmail.com> wrote:
> strich....@gmail.com <strich....@gmail.com> wrote in message
>
> ec160f33-24c4-4049-995e-20472e928...@v57g2000hse.googlegroups.com
>
> - Show quoted text -
Nice try hiding behind the power of group theory. But since you
invoke it, in any analyses using frame A, you can pick the simplest
wherein A has velocity zero (at rest) and B has V=V''~V' [let V'=v+v;
let V''=2v/(1+v^2); the formula does not impact the analyses as we are
talking about total velocity anyway]. "Since according to linear
algebra the Lorentz transfomations form a group", this analyses is
valid, and is simpler. Here velocity A is 0 and velocity B is V not
zero. Obviously, with velocity A being zero, it does not undergo time
dilation, while B undergoes time dilation proportional to V. Thus B
ages less. The converse analyses takes B at rest and A moving towards
it. Thus velocity B=0 and velocity A=-V, and again, it is obvious the
LTE yields no time dilation for B this time whereas A undergoes time
dilation proportional to V. Thus a contradiction. As for Dirk's
analyses, since he keeps on getting the wrong answers, the conclusions
are:
Dirk does not know LTE
Dirk does not linear algebra
Dirk does not know group theory
Dirk Van de moortel
As I already told you at
"Reconstruction: Challenge, Reply, Contempt, Exercise, CRAP":
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrapReconstruction.html
if you have experimental evidence for the failure of linear
algebra, feel free to share it with us.
And, quoting myself:
stric...@gmail.com
> if you have experimental evidence for the failure of linear
> algebra, feel free to share it with us.
>
> And, quoting myself:
> So, feel free to provide the complete analysis, and compare
> with mine. Don't forget to show every detail of your calculation,
> so I can help you when you get stuck.
>
> Don't forget that we are talking about arbitrary velocities and we
> are working in special relativity, so make sure you take
> V = 2 v/(1+v^2)
> - you probably recognise this quantity in the expressions:
> ( 2 L v + T/g ) / ( g (1+v^2) )
>
> In order not to get stuck, I advise you to use the same variables
> As I have, namely (x',t') for A and (x",t") for B.
> This way you can verify what happens if you manage to
> eliminate variables x and t from [LTA] and [LTB] and compare
> with you find.
>
>
The Ghost In The Machine
<dirkvand...@ThankS-NO-SperM.hotmail.com>
wrote
on Wed, 13 Aug 2008 23:05:56 +0200
<%0Iok.1$md...@newsfe27.ams2>:
> Intertial system K using coordinates x and t.
> Twin A is on planet at x = -L and uses coordinates x' and t'.
> Twin B is on planet at x = L and uses coordinates x" and t".
> In system K the twins have the same age 0 at time t = 0.
>
> At event (x,t) = (-L,0) twin A suddenly moves with velocity v towards
> x = 0 at his event (x',t') = (0,0).
> At event (x,t) = (L,0) twin B suddenly moves with velocity -v towards
> x = 0 at his event (x",t") = (0,0).
>
> We use units where c = 1 (so -1 < v < 1) and abbreviate
> g = 1/sqrt(1-v^2).
Pedant Point: Since A is moving towards x=0 with v, one
probably should restrict v to being positive. Not that
it otherwise matters. ;-)
Pedant Point #2: there are some issues as to how A and B
determine time zero; this is easily fixed by having
a light fire an event (0,-L) at the middle; when they see
the light, they go.
--
#191, ewi...@earthlink.net
Linux. Because life's too short for a buggy OS.
** Posted from http://www.teranews.com **
Androcles
"The Ghost In The Machine" <ew...@sirius.tg00suus7038.net> wrote in message
news:0q2kn5-...@sirius.tg00suus7038.net...
> In sci.physics.relativity, Dirk Van de moortel
> <dirkvand...@ThankS-NO-SperM.hotmail.com>
> wrote
> on Wed, 13 Aug 2008 23:05:56 +0200
> <%0Iok.1$md...@newsfe27.ams2>:
>> Intertial system K using coordinates x and t.
>> Twin A is on planet at x = -L and uses coordinates x' and t'.
>> Twin B is on planet at x = L and uses coordinates x" and t".
>> In system K the twins have the same age 0 at time t = 0.
>>
>> At event (x,t) = (-L,0) twin A suddenly moves with velocity v towards
>> x = 0 at his event (x',t') = (0,0).
>> At event (x,t) = (L,0) twin B suddenly moves with velocity -v towards
>> x = 0 at his event (x",t") = (0,0).
>>
>> We use units where c = 1 (so -1 < v < 1) and abbreviate
>> g = 1/sqrt(1-v^2).
>
> Pedant Point: Since A is moving towards x=0 with v, one
> probably should restrict v to being positive. Not that
> it otherwise matters. ;-)
Isn't the whole point to use a coordinate system and reverse v?
Not that Dork would see it that way, his entire drool revolves around
this one aspect of fucked-up relativity.
Dirk Van de moortel
> In sci.physics.relativity, Dirk Van de moortel
> <dirkvand...@ThankS-NO-SperM.hotmail.com>
> wrote
> on Wed, 13 Aug 2008 23:05:56 +0200
> <%0Iok.1$md...@newsfe27.ams2>:
>> Intertial system K using coordinates x and t.
>> Twin A is on planet at x = -L and uses coordinates x' and t'.
>> Twin B is on planet at x = L and uses coordinates x" and t".
>> In system K the twins have the same age 0 at time t = 0.
>>
>> At event (x,t) = (-L,0) twin A suddenly moves with velocity v towards
>> x = 0 at his event (x',t') = (0,0).
>> At event (x,t) = (L,0) twin B suddenly moves with velocity -v towards
>> x = 0 at his event (x",t") = (0,0).
>>
>> We use units where c = 1 (so -1 < v < 1) and abbreviate
>> g = 1/sqrt(1-v^2).
>
> Pedant Point: Since A is moving towards x=0 with v, one
> probably should restrict v to being positive.
Sure, if that would matter.
You should also restrict v to rational numbers if someone
would actually physically read the v-value on a gauge.
And you should restrict to 0 < v < 0.00000001 if they
have insufficient amounts of fuel, and bad helmets to
absorb the shock of the sudden acceleration.
> Not that
> it otherwise matters. ;-)
Right.
>
> Pedant Point #2: there are some issues as to how A and B
> determine time zero; this is easily fixed by having
> a light fire an event (0,-L) at the middle; when they see
> the light, they go.
Or by having a light fire an event (0,-L-k) at the middle, and
when they see the light, they wait for a time k and then they go.
Or by looking at their clocks.
I know you are the Champion of the Pendant Points, but you
forgot to make one about my first word :-)
Dirk Vdm
The Ghost In The Machine
<HzRpk.117738$bi3....@newsfe11.ams2>:
I wasn't feeling quite *that* pedantic. ;-)
--
#191, ewi...@earthlink.net
If your CPU can't stand the heat, get another fan.
PD
Dirk Van de moortel
And his Spaceman busier than "ever such" ;-)
Dirk Vdm
stric...@gmail.com
SperM.hotmail.com> wrote:
> PD <TheDraperFam...@gmail.com> wrote in message
>
> 3725ef59-88d7-4379-99e4-f742ef208...@k37g2000hsf.googlegroups.com
>
> > On Aug 14, 10:13 am, strich....@gmail.com wrote:
> >> On Aug 13, 5:05 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
>
> >> SperM.hotmail.com> wrote:
>
> >> snip
>
> >>> Dirk Vdm
>
> >> CRAP
>
> > Strich9 has certainly gotten very quiet....
>
> > PD
>
> And his Spaceman busier than "ever such" ;-)
>
> Dirk Vdm
...in any analyses using frame A, you can pick the simplest wherein A
has velocity zero (at rest) and B has V=V''~V' [let V'=v+v; let V''=2v/
(1+v^2); the formula does not impact the analyses as we are talking
about total velocity anyway]. "Since according to linear
algebra the Lorentz transfomations form a group", this analyses is
valid, and is simpler. Here velocity A is 0 and velocity B is V not
zero. Obviously, with velocity A being zero, it does not undergo time
dilation, while B undergoes time dilation proportional to V. Thus B
ages less. The converse analyses takes B at rest and A moving towards
it. Thus velocity B=0 and velocity A=-V, and again, it is obvious the
LTE yields no time dilation for B this time whereas A undergoes time
dilation proportional to V. Thus a contradiction.
Conclusion: As Dirk's analyses has the wrong answers, then:
Dirk Van de moortel
> On Aug 20, 2:01 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> SperM.hotmail.com> wrote:
>> PD <TheDraperFam...@gmail.com> wrote in message
>>
>> 3725ef59-88d7-4379-99e4-f742ef208...@k37g2000hsf.googlegroups.com
>>
>>> On Aug 14, 10:13 am, strich....@gmail.com wrote:
>>>> On Aug 13, 5:05 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
>>
>>>> SperM.hotmail.com> wrote:
>>
>>>> snip
>>
>>>>> Dirk Vdm
>>
>>>> CRAP
>>
>>> Strich9 has certainly gotten very quiet....
>>
>>> PD
>>
>> And his Spaceman busier than "ever such" ;-)
>>
>> Dirk Vdm
>
> ...in any analyses using frame A, you can pick the simplest wherein A
> has velocity zero (at rest) and B has V=V''~V' [let V'=v+v; let V''=2v/
> (1+v^2);
As I already told you at
"Reconstruction: Challenge, Reply, Contempt, Exercise, CRAP":
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrapReconstruction.html
if you have experimental evidence for the failure of linear
algebra, feel free to share it with us.
And, quoting myself:
So, feel free to provide the complete analysis, and compare
with mine. Don't forget to show every detail of your calculation,
so I can help you when you get stuck.
Don't forget that we are talking about arbitrary velocities and we
are working in special relativity, so make sure you take
V = 2 v/(1+v^2)
- you probably recognise this quantity in the expressions:
( 2 L v + T/g ) / ( g (1+v^2) )
In order not to get stuck, I advise you to use the same variables
iqgo...@gmail.com
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> strich....@gmail.com <strich....@gmail.com> wrote in message
>
>
>
>
>
>
> > On Aug 20, 2:01 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
> > SperM.hotmail.com> wrote:
> >> PD <TheDraperFam...@gmail.com> wrote in message
>
> >> 3725ef59-88d7-4379-99e4-f742ef208...@k37g2000hsf.googlegroups.com
>
> >>> On Aug 14, 10:13 am, strich....@gmail.com wrote:
> >>>> On Aug 13, 5:05 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
>
> >>>> SperM.hotmail.com> wrote:
>
> >>>> snip
>
> >>>>> Dirk Vdm
>
> >>>> CRAP
>
> >>> Strich9 has certainly gotten very quiet....
>
> >>> PD
>
> >> And his Spaceman busier than "ever such" ;-)
>
> >> Dirk Vdm
>
> > ...in any analyses using frame A, you can pick the simplest wherein A
> > has velocity zero (at rest) and B has V=V''~V' [let V'=v+v; let V''=2v/
> > (1+v^2);
>
> As I already told you at
> "Reconstruction: Challenge, Reply, Contempt, Exercise, CRAP":
> if you have experimental evidence for the failure of linear
> algebra, feel free to share it with us.
>
> And, quoting myself:
> So, feel free to provide the complete analysis, and compare
> with mine. Don't forget to show every detail of your calculation,
> so I can help you when you get stuck.
>
> Don't forget that we are talking about arbitrary velocities and we
> are working in special relativity, so make sure you take
> V = 2 v/(1+v^2)
> - you probably recognise this quantity in the expressions:
> ( 2 L v + T/g ) / ( g (1+v^2) )
>
> In order not to get stuck, I advise you to use the same variables
> As I have, namely (x',t') for A and (x",t") for B.
> This way you can verify what happens if you manage to
> eliminate variables x and t from [LTA] and [LTB] and compare
> with you find.
>
...in any analyses using frame A, you can pick the simplest wherein A
has velocity zero (at rest) and B has V=V''~V' [let V'=v+v; let
V''=2v/
(1+v^2); the formula does not impact the analyses as we are talking
about total velocity anyway]. "Since according to linear
algebra the Lorentz transfomations form a group", this analyses is
valid, and is simpler. Here velocity A is 0 and velocity B is V not
zero. Obviously, with velocity A being zero, it does not undergo
time
dilation, while B undergoes time dilation proportional to V. Thus B
ages less. The converse analyses takes B at rest and A moving
towards
it. Thus velocity B=0 and velocity A=-V, and again, it is obvious
the
LTE yields no time dilation for B this time whereas A undergoes time
dilation proportional to V. Thus a contradiction.
Clarification: Linear algebra did not fail. Relativity failed.
Dirk Van de moortel
f83b689b-158f-495a...@l42g2000hsc.googlegroups.com
Imbecile.
> the formula does not impact the analyses as we are talking
> about total velocity anyway]. "Since according to linear
> algebra the Lorentz transfomations form a group", this analyses is
> valid, and is simpler. Here velocity A is 0 and velocity B is V not
> zero. Obviously, with velocity A being zero, it does not undergo
> time dilation
No, A does not undergo time dilation w.r.t himself.
CONGRATULATIONS!
>, while B undergoes time dilation proportional to V.
As measured by A
> Thus B
> ages less.
Thus B ages slower than A as measured by A
> The converse analyses takes B at rest and A moving
> towards
> it. Thus velocity B=0 and velocity A=-V, and again, it is obvious
> the
> LTE yields no time dilation for B
No, B does not undergo time dilation w.r.t himself.
CONGRATULATIONS!
AGAIN!
> this time whereas A undergoes time
> dilation proportional to V.
Thus A ages slower than B as measured by B.
> Thus a contradiction.
B ages slower than A as measured by A.
A ages slower than B as measured by B.
When they get together, they have the same age.
Thus no contradiction.
You look smaller to your spaceman twin from a distance.
Your spaceman twin looks smaller to you from a distance.
When you get together, you have the same height.
Thus no contradiction.
Infinite imbecile.
>
> Clarification: Linear algebra did not fail. Relativity failed.
You didn't specify their initial ages.
So, quoting myself again:
Feel free to provide the complete analysis, and compare
with mine. Don't forget to show every detail of your calculation,
so I can help you when you get stuck.
Don't forget that we are talking about arbitrary velocities and we
are working in special relativity, so make sure you take
V = 2 v/(1+v^2)
- you probably recognise this quantity in the expressions:
( 2 L v + T/g ) / ( g (1+v^2) )
In order not to get stuck, I advise you to use the same variables
As I have, namely (x',t') for A and (x",t") for B.
This way you can verify what happens if you manage to
eliminate variables x and t from [LTA] and [LTB] and compare
with you find.
You just can't do it ;-)
Dirk Vdm
iqgo...@gmail.com
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> iqgoo...@gmail.com <iqgoo...@gmail.com> wrote in message
>
> f83b689b-158f-495a-ab5f-6b89cf81f...@l42g2000hsc.googlegroups.com
You were doing well until the last line. What happened? Did you
blank out for a second?
Let me do this slowly for you:
B ages slower than A, as measured by A.
When B arrives at A, B is obviously seen younger by A.
A ages slower than B, as measured by B.
When A arrives at B, A is obviously seen younger by B.
I'll let you provide the conclusion. Don't blank out this time.
Dirk Van de moortel
That depends on what they reckon their initial ages are
at certain times and how they look at simultaneity.
You did not specify their initial ages.
I did in my calculation.
>
> A ages slower than B, as measured by B.
> When A arrives at B, A is obviously seen younger by B.
That depends on what they reckon their initial ages are
at certain times and how they look at simultaneity.
You did not specify their initial ages.
I did in my calculation.
>
> I'll let you provide the conclusion. Don't blank out this time.
So, quoting myself again:
Feel free to provide the complete analysis, and compare
with mine. Don't forget to show every detail of your calculation,
so I can help you when you get stuck.
Don't forget that we are talking about arbitrary velocities and we
are working in special relativity, so make sure you take
V = 2 v/(1+v^2)
- you probably recognise this quantity in the expressions:
( 2 L v + T/g ) / ( g (1+v^2) )
In order not to get stuck, I advise you to use the same variables
As I have, namely (x',t') for A and (x",t") for B.
This way you can verify what happens if you manage to
eliminate variables x and t from [LTA] and [LTB] and compare
with you find.
You just can't do it ;-)
Dirk Vdm
iqgo...@gmail.com
<dirkvandemoor...@nospAm.hotmail.com> wrote:
>
> >> B ages slower than A as measured by A.
> >> A ages slower than B as measured by B.
> >> When they get together, they have the same age.
>
> > You were doing well until the last line. What happened? Did you
> > blank out for a second?
>
> > Let me do this slowly for you:
>
> > B ages slower than A, as measured by A.
> > When B arrives at A, B is obviously seen younger by A.
>
> That depends on what they reckon their initial ages are
> at certain times and how they look at simultaneity.
> You did not specify their initial ages.
> I did in my calculation.
>
>
>
> > A ages slower than B, as measured by B.
> > When A arrives at B, A is obviously seen younger by B.
>
> That depends on what they reckon their initial ages are
> at certain times and how they look at simultaneity.
> You did not specify their initial ages.
> I did in my calculation.
>
>
>
> > I'll let you provide the conclusion. Don't blank out this time.
>
> So, quoting myself again:
>
>
> Dirk Vdm-
I'm glad you did not jump to the same wrong conclusion this time.
That was a good start...
But, you dig yourself a deeper hole.
Perhaps you lost the link to my answer in the past, since you only
keep the bad links. See below.
http://www.physicsbanter.com/theory-relativity/111256-dirk-answers-david-twin-paradox.html
They are twins, so they are the same age. But, technically speaking,
in frame A, twin B starts the journey in time -2L/c, so B ages by (-2L/
c + T/g).
Similarly, in frame B, A starts the journey at time -2L/c, so A ages
by (-L/c + T/g). In both cases, the correction factor reduces the
starting age of each twin as judged by the other, further reducing the
age of each apporaching twin as seen from the frame of the other. That
is:
With respect to frame A, B ages -2L/c + T/g, which is less than T for
B.
With respect to frame B, A ages -2L/c + T/g, which is less than T for
A.
I'll let you finish the last line. You may look in my previous answer
in the link for help. Try to get it right this time.
Dirk Van de moortel
That depends on the reference frame, and you don't have the guts
to specify it - and you know why ;-)
stric...@gmail.com
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> > On Sep 5, 2:27 pm, "Dirk Van de moortel"
> > <dirkvandemoor...@nospAm.hotmail.com> wrote:
>
> >>>> B ages slower than A as measured by A.
> >>>> A ages slower than B as measured by B.
> >>>> When they get together, they have the same age.
>
> >>> You were doing well until the last line. What happened? Did you
> >>> blank out for a second?
>
> >>> Let me do this slowly for you:
>
> >>> B ages slower than A, as measured by A.
> >>> When B arrives at A, B is obviously seen younger by A.
> >>> When A arrives at B, A is obviously seen younger by B.
>
> >> That depends on what they reckon their initial ages are
> >> at certain times and how they look at simultaneity.
> >> You did not specify their initial ages.
>
> to specify it - and you know why ;-)
>
Do I have to spell everything out for you:
Frame A: age A = 0, age B = -2L/c, aging A = T, aging B = T/g
If you do not want to start at time 0 for A, you can add an arbitrary
age E (age at start of travel), but this is uniform, as they are
twins, and is added to both ages.
Frame A: age A = E+0, age B = E-2L/c, aging A = T, aging B = T/g
Frame A: age A at meet: E+0+T; age B at meet E-2L/c+T/g
Comparison E+0+T greater than E-2L/c+T/g
I'll skip the frame B to avoid confusing you, but if you want to know,
substitute A for B and vice verse (that would be B for A).
Since you get stuck in the conclusion, i'll finish it this time.
Frame A: A older than B
Frame B: B older than A
So either linear algebra is wrong, or relativity is wrong. I would
bet on the latter. Fanatics are welcome to rewrite linear algerba.
Dirk Van de moortel
> On Sep 5, 3:16 pm, "Dirk Van de moortel"
> <dirkvandemoor...@nospAm.hotmail.com> wrote:
>>> On Sep 5, 2:27 pm, "Dirk Van de moortel"
>>> <dirkvandemoor...@nospAm.hotmail.com> wrote:
>>
>>>>>> B ages slower than A as measured by A.
>>>>>> A ages slower than B as measured by B.
>>>>>> When they get together, they have the same age.
>>
>>>>> You were doing well until the last line. What happened? Did you
>>>>> blank out for a second?
>>
>>>>> Let me do this slowly for you:
>>
>>>>> B ages slower than A, as measured by A.
>>>>> When B arrives at A, B is obviously seen younger by A.
>>>>> A ages slower than B, as measured by B.
>>>>> When A arrives at B, A is obviously seen younger by B.
>>
>>>> That depends on what they reckon their initial ages are
>>>> at certain times and how they look at simultaneity.
>>>> You did not specify their initial ages.
>>
>> That depends on the reference frame, and you don't have the guts
>> to specify it - and you know why ;-)
>>
>> Dirk Vdm-
>
> Do I have to spell everything out for you:
>
> Frame A: age A = 0, age B = -2L/c,
Wrong, imbecile:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrapReconstruction.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DavidTwinParadox.gsp.png
When A has age
t' = 0,
then B has, according to A age
t" = 2 L v / ( g (1+v^2) )
If you can't do it properly, then you better don't do it ;-)
Dirk Vdm
stric...@gmail.com
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> strich....@gmail.com <strich....@gmail.com> wrote in message
>
>
>
>
>
>
> > On Sep 5, 3:16 pm, "Dirk Van de moortel"
> > <dirkvandemoor...@nospAm.hotmail.com> wrote:
> >>> On Sep 5, 2:27 pm, "Dirk Van de moortel"
> >>> <dirkvandemoor...@nospAm.hotmail.com> wrote:
>
> >>>>>> B ages slower than A as measured by A.
> >>>>>> A ages slower than B as measured by B.
> >>>>>> When they get together, they have the same age.
>
> >>>>> You were doing well until the last line. What happened? Did you
> >>>>> blank out for a second?
>
> >>>>> Let me do this slowly for you:
>
> >>>>> B ages slower than A, as measured by A.
> >>>>> When B arrives at A, B is obviously seen younger by A.
> >>>>> A ages slower than B, as measured by B.
> >>>>> When A arrives at B, A is obviously seen younger by B.
>
> >>>> That depends on what they reckon their initial ages are
> >>>> at certain times and how they look at simultaneity.
> >>>> You did not specify their initial ages.
>
> >> That depends on the reference frame, and you don't have the guts
> >> to specify it - and you know why ;-)
>
> >> Dirk Vdm-
>
> > Do I have to spell everything out for you:
>
> > Frame A: age A = 0, age B = -2L/c,
>
> Wrong, imbecile:
> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DavidTwinParad...
> When A has age
> t' = 0,
> then B has, according to A age
> t" = 2 L v / ( g (1+v^2) )
>
> If you can't do it properly, then you better don't do it ;-)
>
Clumsy...
At time t'=0 for A, relative velocity = 0 (zero) for B, so t'' for B =
2 L v / ( g (1+v^2) ) = 2 L * zero / ( g (1+zero^2) ) = 0/g=0.
WRONG.
Do it correctly:
At time t'=0 for A, t''=-2L/c for B.
Remember Minkowski diagrams, everything outside a frame is in its
past. See that minus sign for t''?
Dirk Van de moortel
>> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DavidTwinParadox.gsp.png
>> When A has age
>> t' = 0,
>> then B has, according to A age
>> t" = 2 L v / ( g (1+v^2) )
>>
>> If you can't do it properly, then you better don't do it ;-)
>>
>> Dirk Vdm-
>
> Clumsy...
>
> At time t'=0 for A, relative velocity = 0 (zero) for B, so t'' for B =
> 2 L v / ( g (1+v^2) ) = 2 L * zero / ( g (1+zero^2) ) = 0/g=0.
Deadly imbecile :-)
You are arguing with 4th grade analytic geometry and linear algebra:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrapReconstruction.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/DavidTwinParadox.gsp.png
Dirk Vdm
stric...@gmail.com
<dirkvandemoor...@nospAm.hotmail.com> wrote:
> You are arguing with 4th grade analytic geometry and linear algebra:
>
4th grade... and I had to teach you.