The David Twin paradox - for STRICH

[Spaceman]

On Sep 5, 3:16 pm, "Dirk Van de moortel"

On Sep 5, 2:27 pm, "Dirk Van de moortel"

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?
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-

Frame A: age A = 0, age B = -2L/c,

http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrapReconstruction.html
Loading Image...

When A has age t' = 0,
then B has, according to A age
t" = 2 L v / ( g (1+v^2) )

Moron!
A says B is the same age as him all the time.
B says that A is the same age.
At least one clock simply malfunctioned.
Only morons can not understand a clock malfunction
and the twins being the same revolutions of Earth
WRT the Sun old at all times.

Dirk is one of the clock illiterates that ignore the Entire
Universal clock and refuse to learn what screwed up the
traveling clock.
Dirk.
you should read this link and after you do such..
refresh it and read a different version that is also correct.
http://www.hyperdeath.co.uk/spaceman/message.html

(unless of course they changed the page because they hate
when I use it.)
:)

[s***@gmail.com]

On Sep 5, 4:39 pm, "Dirk Van de moortel"

On Sep 5, 3:16 pm, "Dirk Van de moortel"

On Sep 5, 2:27 pm, "Dirk Van de moortel"

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?
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-

Frame A: age A = 0, age B = -2L/c,

   http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrapReconstruc...
   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 ;-)
Dirk Vdm-

Dirk, you forget, at time t'=0 for A, relative velocity is 0 (zero),
so solving your t'' for B: t''=2Lv/(g(1+v^2)) = 2L * zero /
(g(1+zero^2)) = zero / g = zero.

According to Dirk's clumsy calculation, t'=0 for A and t''=0 for B.

WRONG.

Let me show you the elegant answer:

At time t'=0 for A, t''=2L/c for B.

Remember basic Minkowski diagrams, everything outside a frame is in
its past, due to the finite speed of the information transmitting
metric, light.