This is a follow up to a post "Where is the flaw?" I had before about
relativity. I want to make my question as simple as possible; I felt
the question of that post before was not simple enough:

Suppose that a space station is located at point
P0=(x0,y0,z0,t0)=(0,0,0,0). There is a space ship moving from point
P1=(x1,y1,z1,t1)=(0,1,0,0) to point P2=(x2,y2,z2,t2)=(1,1,0,1), with
respect to the inertial frame of reference of the space station. So the
velocity of the space ship with respect to the space station is
v=(1,0,0). The units of measurement are assumed to be kilometers and
years.

So the time T that it takes for the space ship to get from point P1 to
point P2 from the point of view of the space station is T=1 year. Let
us calculate the time T' that it takes for the space ship to get from
point P1 to point P2 from the point of view of the space ship using the
time dilation formula. We get:

T'=T*sqrt(1-v^2/c^2)=sqrt(1-1/c^2) years, where c is the speed of light
in kilometers per year.

Now, let us think of the space station as moving and the space ship at
rest. (Since there is no prefered frame of reference according to
relativity theory, this is certainly a legitimate way of understanding
this scenario.) So the time that point P1 leaves the space ship to the
time that P2 reaches the space ship from the point of view of the space
ship is T'. Now, let us apply the time dilation formula to calculate
the time T" that the space station measures for point P1 to leave the
space ship to the time that P2 reaches the space ship. Since the
velocity of the space station with respect to the space ship is
v=(-1,0,0), we get:

T''=T' *sqrt(1-v^2/c^2)=1-1/c^2 years, where c is the speed of light in
kilometers per year.

But T'', which is defined to be the time that the space station
measures for point P1 to leave the space ship to the time that P2
reaches the space ship is really the same thing as T, the time that the
space ship goes from point P1 to point P2 from the perspective of the
space station.

So we have 1-1/c^2 years = T'' = T = 1 year. So 1-1/c^2=1, which
implies that -1/c^2=0, which implies that the speed of light c is
infinity. What is going on here? I would really appreciate if someone
could tell me where the flaw is in my argument. As I said before, I am
not out to claim that Einstein is wrong; I am giving Einstein the
benefit of doubt and assuming that I am wrong in my understanding of
relativity or that I made a simple mistake in my calculations
somewhere.

Thank you,
Craig

wrote in message
oups.com...
| This is a follow up to a post "Where is the flaw?" I had before about
| relativity. I want to make my question as simple as possible; I felt
| the question of that post before was not simple enough:
|
| Suppose that a space station is located at point
| P0=(x0,y0,z0,t0)=(0,0,0,0). There is a space ship moving from point
| P1=(x1,y1,z1,t1)=(0,1,0,0) to point P2=(x2,y2,z2,t2)=(1,1,0,1), with
| respect to the inertial frame of reference of the space station. So the
| velocity of the space ship with respect to the space station is
| v=(1,0,0). The units of measurement are assumed to be kilometers and
| years.
|

One kilometer a year?
2.74 meters a day, 11 cm/hour.
Snails are MUCH faster.
Androcles




| So the time T that it takes for the space ship to get from point P1 to
| point P2 from the point of view of the space station is T=1 year. Let
| us calculate the time T' that it takes for the space ship to get from
| point P1 to point P2 from the point of view of the space ship using the
| time dilation formula. We get:
|
| T'=T*sqrt(1-v^2/c^2)=sqrt(1-1/c^2) years, where c is the speed of light
| in kilometers per year.
|
| Now, let us think of the space station as moving and the space ship at
| rest. (Since there is no prefered frame of reference according to
| relativity theory, this is certainly a legitimate way of understanding
| this scenario.) So the time that point P1 leaves the space ship to the
| time that P2 reaches the space ship from the point of view of the space
| ship is T'. Now, let us apply the time dilation formula to calculate
| the time T" that the space station measures for point P1 to leave the
| space ship to the time that P2 reaches the space ship. Since the
| velocity of the space station with respect to the space ship is
| v=(-1,0,0), we get:
|
| T''=T' *sqrt(1-v^2/c^2)=1-1/c^2 years, where c is the speed of light in
| kilometers per year.
|
| But T'', which is defined to be the time that the space station
| measures for point P1 to leave the space ship to the time that P2
| reaches the space ship is really the same thing as T, the time that the
| space ship goes from point P1 to point P2 from the perspective of the
| space station.
|
| So we have 1-1/c^2 years = T'' = T = 1 year. So 1-1/c^2=1, which
| implies that -1/c^2=0, which implies that the speed of light c is
| infinity. What is going on here? I would really appreciate if someone
| could tell me where the flaw is in my argument. As I said before, I am
| not out to claim that Einstein is wrong; I am giving Einstein the
| benefit of doubt and assuming that I am wrong in my understanding of
| relativity or that I made a simple mistake in my calculations
| somewhere.
|
| Thank you,
| Craig
|

Sorcerer wrote:
wrote in message
oups.com...
| This is a follow up to a post "Where is the flaw?" I had before about
| relativity. I want to make my question as simple as possible; I felt
| the question of that post before was not simple enough:
|
| Suppose that a space station is located at point
| P0=(x0,y0,z0,t0)=(0,0,0,0). There is a space ship moving from point
| P1=(x1,y1,z1,t1)=(0,1,0,0) to point P2=(x2,y2,z2,t2)=(1,1,0,1), with
| respect to the inertial frame of reference of the space station. So the
| velocity of the space ship with respect to the space station is
| v=(1,0,0). The units of measurement are assumed to be kilometers and
| years.
|

One kilometer a year?
2.74 meters a day, 11 cm/hour.
Snails are MUCH faster.
Androcles

Yes, that is true. Still, the argument still works (or perhaps, does
not work) in any units. There are no approximations in the argument.





| So the time T that it takes for the space ship to get from point P1 to
| point P2 from the point of view of the space station is T=1 year. Let
| us calculate the time T' that it takes for the space ship to get from
| point P1 to point P2 from the point of view of the space ship using the
| time dilation formula. We get:
|
| T'=T*sqrt(1-v^2/c^2)=sqrt(1-1/c^2) years, where c is the speed of light
| in kilometers per year.
|
| Now, let us think of the space station as moving and the space ship at
| rest. (Since there is no prefered frame of reference according to
| relativity theory, this is certainly a legitimate way of understanding
| this scenario.) So the time that point P1 leaves the space ship to the
| time that P2 reaches the space ship from the point of view of the space
| ship is T'. Now, let us apply the time dilation formula to calculate
| the time T" that the space station measures for point P1 to leave the
| space ship to the time that P2 reaches the space ship. Since the
| velocity of the space station with respect to the space ship is
| v=(-1,0,0), we get:
|
| T''=T' *sqrt(1-v^2/c^2)=1-1/c^2 years, where c is the speed of light in
| kilometers per year.
|
| But T'', which is defined to be the time that the space station
| measures for point P1 to leave the space ship to the time that P2
| reaches the space ship is really the same thing as T, the time that the
| space ship goes from point P1 to point P2 from the perspective of the
| space station.
|
| So we have 1-1/c^2 years = T'' = T = 1 year. So 1-1/c^2=1, which
| implies that -1/c^2=0, which implies that the speed of light c is
| infinity. What is going on here? I would really appreciate if someone
| could tell me where the flaw is in my argument. As I said before, I am
| not out to claim that Einstein is wrong; I am giving Einstein the
| benefit of doubt and assuming that I am wrong in my understanding of
| relativity or that I made a simple mistake in my calculations
| somewhere.
|
| Thank you,
| Craig
|

wrote:
This is a follow up to a post "Where is the flaw?" I had before about
relativity. I want to make my question as simple as possible; I felt
the question of that post before was not simple enough:

Suppose that a space station is located at point
P0=(x0,y0,z0,t0)=(0,0,0,0). There is a space ship moving from point
P1=(x1,y1,z1,t1)=(0,1,0,0) to point P2=(x2,y2,z2,t2)=(1,1,0,1), with
respect to the inertial frame of reference of the space station. So the
velocity of the space ship with respect to the space station is
v=(1,0,0). The units of measurement are assumed to be kilometers and
years.

So the time T that it takes for the space ship to get from point P1 to
point P2 from the point of view of the space station is T=1 year. Let
us calculate the time T' that it takes for the space ship to get from
point P1 to point P2 from the point of view of the space ship using the
time dilation formula. We get:

T'=T*sqrt(1-v^2/c^2)=sqrt(1-1/c^2) years, where c is the speed of light
in kilometers per year.

Now, let us think of the space station as moving and the space ship at
rest. (Since there is no prefered frame of reference according to
relativity theory, this is certainly a legitimate way of understanding
this scenario.) So the time that point P1 leaves the space ship to the
time that P2 reaches the space ship from the point of view of the space
ship is T'. Now, let us apply the time dilation formula to calculate
the time T" that the space station measures for point P1 to leave the
space ship to the time that P2 reaches the space ship. Since the
velocity of the space station with respect to the space ship is
v=(-1,0,0), we get:

T''=T' *sqrt(1-v^2/c^2)=1-1/c^2 years, where c is the speed of light in
kilometers per year.

But T'', which is defined to be the time that the space station
measures for point P1 to leave the space ship to the time that P2
reaches the space ship is really the same thing as T, the time that the
space ship goes from point P1 to point P2 from the perspective of the
space station.

So we have 1-1/c^2 years = T'' = T = 1 year. So 1-1/c^2=1, which
implies that -1/c^2=0, which implies that the speed of light c is
infinity. What is going on here? I would really appreciate if someone
could tell me where the flaw is in my argument. As I said before, I am
not out to claim that Einstein is wrong; I am giving Einstein the
benefit of doubt and assuming that I am wrong in my understanding of
relativity or that I made a simple mistake in my calculations
somewhere.

Thank you,
Craig

Einstein is correct that the conflict with Maxwell's equation and
the principle of invariance is only apparent. The way he
resolves the conflict is wrong and in fact mathematiclly absurd.

The correct application of the Lorentz transform to resolve the
conflict is discussed he

"Retarded potential"
http://farside.ph.utexas.edu/teachin...es/node50.html
"Incident Wave Impedance"
http://www.conformity.com/0102reflectionsfig3.gif
http://www.conformity.com/0102reflections.html

The many absurd 'resolutions' and their flaws are discussed
he
C. S. Unnikrishnan
http://www.iisc.ernet.in/currsci/dec252005/2009.pdf


Just understanding the nature of the problem requires
more than primary school motion calculations and
understanding the solution requires some fairly advanced
electromagnetism. It should be accessable to third
year undergraduate in a science or engineering curriculum.

So if you are not up to comprehending material on this
level:
"Visualizations"
http://web.mit.edu/8.02t/www/802TEAL3D/teal_tour.htm
.... then you will be chasing your tail with an absurdity
for ever.

The short answer is:
Clocks and birthday cakes are not affected by motion and
Einstein is correct that the postulates are resolvable with
a rigourous application of an imaginary axis.

Sue...

wrote in message
oups.com...
|
| Sorcerer wrote:
| wrote in message
| oups.com...
| | This is a follow up to a post "Where is the flaw?" I had before about
| | relativity. I want to make my question as simple as possible; I felt
| | the question of that post before was not simple enough:
| |
| | Suppose that a space station is located at point
| | P0=(x0,y0,z0,t0)=(0,0,0,0). There is a space ship moving from point
| | P1=(x1,y1,z1,t1)=(0,1,0,0) to point P2=(x2,y2,z2,t2)=(1,1,0,1), with
| | respect to the inertial frame of reference of the space station. So
the
| | velocity of the space ship with respect to the space station is
| | v=(1,0,0). The units of measurement are assumed to be kilometers and
| | years.
| |
|
| One kilometer a year?
| 2.74 meters a day, 11 cm/hour.
| Snails are MUCH faster.
| Androcles
|
| Yes, that is true. Still, the argument still works (or perhaps, does
| not work) in any units. There are no approximations in the argument.


Son, if you want to understand magic, you have to study it. It is all
about deception. Smoke and mirrors. You say you are not out to show
Einstein was wrong. Well, Houdini wasn't 'wrong' either, they were both
successful stage magicians. It takes a real sorcerer to know how the trick
is done, that's all. It's no good coming out of the theatre after the show
is over and asking the audience, most of them think the woman really
was sawn in half, especially the ones that think they can do it but dare
not try.
The cuckoo transformations
xi = (x-vt)/sqrt(1-v^2/c^2)
tau = t.sqrt(1-v^2/c^2)
were derived from a mirror and a smokescreen, 2AB/(t'A-tA) = c.
See, the light leaves A and bounces at a mirror B, returning to A.
On the way along the ruler is passes 1,2,3,...,27,28,29, and arrives at
30. Then it goes back again, 29,28,27,...3,2,1,0 to where it started.
So the distance traveled is (at least in a coordinate system) is zero.
The light hasn't gone anywhere, it is back where it started.
Hence c = 0/0.
Division by zero isn't legitimate in mathematics, even when
disguised as algebra. For example,
v = c
Multiply by c
cv = c^2
Subtract v^2
cv-v^2 = c^2-v^2
Factorize:
v(c-v) = (c+v)(c-v)
Divide by c-v
v = c+v
but v = c, so substitute
c = c+c
c = 2c
1 = 2.
That's how Einstein did his magic.
Proof:
http://www.androcles01.pwp.blueyonde...mart/Smart.htm

It's up to you to decide whether Einstein was an idiot or a charlatan,
that's
a matter of opinion, but the math is undeniable. Do not be fooled into
thinking c = 300,000 km/sec, it is 0/0. The speed of light is 300,000
km/sec,
relative to the source, but c is not the speed of light, that is the lie.
He wasn't a real magician, he was a muggle. A con artist. A ****bag. A LIAR.
Being Mr. Nice Guy is the stock-in-trade of the huckster.
"Please believe me" when I say "In agreement with experience WE further
ASSUME the quantity 2AB/(t'A-tA) = c.
That is very different to Newton, who is in your face. "If you deny it, that
is
against the supposition." Newton was not Mr. Nice Guy, he didn't need to me.
Newton was a mathematician who devised the calculus, Einstein was a beggar.
Einstein's disciples are incompetent fools trying to appear smart. They are
as
scruffy as he was.

Androcles Dumbledore, Headmaster, hogwarts.physics school for
zauberlehrlings.







|
|
|
|
|
| | So the time T that it takes for the space ship to get from point P1 to
| | point P2 from the point of view of the space station is T=1 year. Let
| | us calculate the time T' that it takes for the space ship to get from
| | point P1 to point P2 from the point of view of the space ship using
the
| | time dilation formula. We get:
| |
| | T'=T*sqrt(1-v^2/c^2)=sqrt(1-1/c^2) years, where c is the speed of
light
| | in kilometers per year.
| |
| | Now, let us think of the space station as moving and the space ship at
| | rest. (Since there is no prefered frame of reference according to
| | relativity theory, this is certainly a legitimate way of understanding
| | this scenario.) So the time that point P1 leaves the space ship to the
| | time that P2 reaches the space ship from the point of view of the
space
| | ship is T'. Now, let us apply the time dilation formula to calculate
| | the time T" that the space station measures for point P1 to leave the
| | space ship to the time that P2 reaches the space ship. Since the
| | velocity of the space station with respect to the space ship is
| | v=(-1,0,0), we get:
| |
| | T''=T' *sqrt(1-v^2/c^2)=1-1/c^2 years, where c is the speed of light
in
| | kilometers per year.
| |
| | But T'', which is defined to be the time that the space station
| | measures for point P1 to leave the space ship to the time that P2
| | reaches the space ship is really the same thing as T, the time that
the
| | space ship goes from point P1 to point P2 from the perspective of the
| | space station.
| |
| | So we have 1-1/c^2 years = T'' = T = 1 year. So 1-1/c^2=1, which
| | implies that -1/c^2=0, which implies that the speed of light c is
| | infinity. What is going on here? I would really appreciate if someone
| | could tell me where the flaw is in my argument. As I said before, I am
| | not out to claim that Einstein is wrong; I am giving Einstein the
| | benefit of doubt and assuming that I am wrong in my understanding of
| | relativity or that I made a simple mistake in my calculations
| | somewhere.
| |
| | Thank you,
| | Craig
| |
|

Sue... wrote:

Clocks and birthday cakes are not affected by motion and
Einstein is correct that the postulates are resolvable with
a rigourous application of an imaginary axis.

Sue...
ahhh
axis
spin

r u married sue?

malibu wrote:
Sue... wrote:

Clocks and birthday cakes are not affected by motion and
Einstein is correct that the postulates are resolvable with
a rigourous application of an imaginary axis.

Sue...
ahhh
axis
(sqrt -1)
spin
None in Maxwell's

r u married sue?
Been the Done that. :-(

Sue...

wrote in message oups.com...
This is a follow up to a post "Where is the flaw?" I had before about
relativity. I want to make my question as simple as possible; I felt
the question of that post before was not simple enough:

Suppose that a space station is located at point
P0=(x0,y0,z0,t0)=(0,0,0,0).

That is not a point. It is an event.
The point is where the space station is and it has coordinates
(x,y,z) = (x0,y0,z0) = (0,0,0) .
You can look at this as the 'world line' of the station.
An arbitrary event anywhere anywhen has coordinates
(x,y,z,t) .
Every tick on a clock on the station is an event with coordinates
(x,y,z,t) = (0,0,0,t) .

But see below.

There is a space ship moving from point
P1=(x1,y1,z1,t1)=(0,1,0,0) to point P2=(x2,y2,z2,t2)=(1,1,0,1), with
respect to the inertial frame of reference of the space station.

Those are not points but events, taking place on the spaceship.
The world line of the ship is given by the station coordinates
(x,y,z) = ( v t, 1, 0 )
or by the ship coordinates
(x',y',z') = (0,0,0) .
An arbitrary event anywhere anywhen has coordinates
(x',y',z',t') .
Every tick on a clock on the ship is an event with coordinates
(x',y',z',t') = (0,0,0,t') .

But see below.

So the
velocity of the space ship with respect to the space station is
v=(1,0,0). The units of measurement are assumed to be kilometers and
years.

So the time T that it takes for the space ship to get from point P1 to
point P2 from the point of view of the space station is T=1 year. Let
us calculate the time T' that it takes for the space ship to get from
point P1 to point P2 from the point of view of the space ship using the
time dilation formula. We get:

T'=T*sqrt(1-v^2/c^2)=sqrt(1-1/c^2) years, where c is the speed of light
in kilometers per year.

Yes, but see below.


Now, let us think of the space station as moving and the space ship at
rest. (Since there is no prefered frame of reference according to
relativity theory, this is certainly a legitimate way of understanding
this scenario.) So the time that point P1 leaves the space ship to the
time that P2 reaches the space ship from the point of view of the space
ship is T'. Now, let us apply the time dilation formula to calculate
the time T" that the space station measures for point P1 to leave the
space ship to the time that P2 reaches the space ship. Since the
velocity of the space station with respect to the space ship is
v=(-1,0,0), we get:

T''=T' *sqrt(1-v^2/c^2)=1-1/c^2 years, where c is the speed of light in
kilometers per year.

But T'', which is defined to be the time that the space station
measures for point P1 to leave the space ship to the time that P2
reaches the space ship is really the same thing as T, the time that the
space ship goes from point P1 to point P2 from the perspective of the
space station.

So we have 1-1/c^2 years = T'' = T = 1 year. So 1-1/c^2=1, which
implies that -1/c^2=0, which implies that the speed of light c is
infinity. What is going on here? I would really appreciate if someone
could tell me where the flaw is in my argument. As I said before, I am
not out to claim that Einstein is wrong; I am giving Einstein the
benefit of doubt and assuming that I am wrong in my understanding of
relativity or that I made a simple mistake in my calculations
somewhere.

Aw.

With the setup you have given, the roles of y and z are irrelevant,
and the choice of units is a bit awkward, so we will work with
variables instead of unit values.

The station's world line is
x = 0
The ship's world line is
x = v t

If station and ship are together at event
(x,t) = (x',t') = (0,0)
then the transformation equations are, in units where c = 1:
t' = g ( t - v x )
x' = g ( x - v t )
and the inverse
t = g ( t' + v x' )
x = g ( x' + v t' )
with
g = 1/sqrt(1-v^2)

You can verify that combining the world lines with these
equations, that, in the ship's coordinates, the station's world line
is given by
x' = - v t'
and the ship's world line by
x' = 0

The first event E1 (the ship leaves place P1) is an event with
station coordinates
E1: (x,t) = (0,0)
The second event E2 (the ship arrives at place P2) is an event with
station coordinates
E2: (x,t) = (L,T)
where, according to the station, L is the distance covered by the ship
and T the time it takes.
Since the velocity is v, we know that L = v T, so the second event
can be written as
E2: (x,t) = ( v T, T )

If you insert these values into the transformations, you find
E1: (x',t') = ( 0, 0 )
and
E2: (x',t') = ( 0, T sqrt(1-v^2) )
so, indeed, according to the transformation, the time it takes for
the ship is
T' = T sqrt(1-v^2)

When you look at the station's coordinates from the point of
view of the ship, you see that they always satisfy the equation
x' = - v t',
so when event E2 happens (on the ship!), the station is at
location
x' = - v T sqrt(1-v^2)
so the station has covered a distance
v T sqrt(1-v^2) ,
so you can imagine some event E3, that takes place on the
station, with coordinates
E3: (x',t') = ( -v T sqrt(1-v^2), T sqrt(1-v^2) )
Inserting this into the transformation gives
E3: (x,t) = ( 0, T (1-v^2) ).

Now compare the relationship between the coordinates of events
E1 and E2
with those of the events
E1 and E3 .
You see:
E1: (x,t) = (x',t') = ( 0, 0 )
and
E2: (x,t) = ( v T, T )
E2: (x',t') = ( 0, T sqrt(1-v^2) )
and
E3: (x',t') = ( -v T sqrt(1-v^2), T sqrt(1-v^2) )
E3: (x,t) = ( 0, T (1-v^2) ).

What do you notice?

Dirk Vdm

Dirk Van de moortel wrote:
wrote in message oups.com...
This is a follow up to a post "Where is the flaw?" I had before about
relativity. I want to make my question as simple as possible; I felt
the question of that post before was not simple enough:

Suppose that a space station is located at point
P0=(x0,y0,z0,t0)=(0,0,0,0).

That is not a point. It is an event.
The point is where the space station is and it has coordinates
(x,y,z) = (x0,y0,z0) = (0,0,0) .
You can look at this as the 'world line' of the station.
An arbitrary event anywhere anywhen has coordinates
(x,y,z,t) .
Every tick on a clock on the station is an event with coordinates
(x,y,z,t) = (0,0,0,t) .

But see below.

There is a space ship moving from point
P1=(x1,y1,z1,t1)=(0,1,0,0) to point P2=(x2,y2,z2,t2)=(1,1,0,1), with
respect to the inertial frame of reference of the space station.

Those are not points but events, taking place on the spaceship.
The world line of the ship is given by the station coordinates
(x,y,z) = ( v t, 1, 0 )
or by the ship coordinates
(x',y',z') = (0,0,0) .
An arbitrary event anywhere anywhen has coordinates
(x',y',z',t') .
Every tick on a clock on the ship is an event with coordinates
(x',y',z',t') = (0,0,0,t') .

But see below.

So the
velocity of the space ship with respect to the space station is
v=(1,0,0). The units of measurement are assumed to be kilometers and
years.

So the time T that it takes for the space ship to get from point P1 to
point P2 from the point of view of the space station is T=1 year. Let
us calculate the time T' that it takes for the space ship to get from
point P1 to point P2 from the point of view of the space ship using the
time dilation formula. We get:

T'=T*sqrt(1-v^2/c^2)=sqrt(1-1/c^2) years, where c is the speed of light
in kilometers per year.

Yes, but see below.


Now, let us think of the space station as moving and the space ship at
rest. (Since there is no prefered frame of reference according to
relativity theory, this is certainly a legitimate way of understanding
this scenario.) So the time that point P1 leaves the space ship to the
time that P2 reaches the space ship from the point of view of the space
ship is T'. Now, let us apply the time dilation formula to calculate
the time T" that the space station measures for point P1 to leave the
space ship to the time that P2 reaches the space ship. Since the
velocity of the space station with respect to the space ship is
v=(-1,0,0), we get:

T''=T' *sqrt(1-v^2/c^2)=1-1/c^2 years, where c is the speed of light in
kilometers per year.

But T'', which is defined to be the time that the space station
measures for point P1 to leave the space ship to the time that P2
reaches the space ship is really the same thing as T, the time that the
space ship goes from point P1 to point P2 from the perspective of the
space station.

So we have 1-1/c^2 years = T'' = T = 1 year. So 1-1/c^2=1, which
implies that -1/c^2=0, which implies that the speed of light c is
infinity. What is going on here? I would really appreciate if someone
could tell me where the flaw is in my argument. As I said before, I am
not out to claim that Einstein is wrong; I am giving Einstein the
benefit of doubt and assuming that I am wrong in my understanding of
relativity or that I made a simple mistake in my calculations
somewhere.

Aw.

With the setup you have given, the roles of y and z are irrelevant,
and the choice of units is a bit awkward, so we will work with
variables instead of unit values.

The station's world line is
x = 0
The ship's world line is
x = v t

If station and ship are together at event
(x,t) = (x',t') = (0,0)
then the transformation equations are, in units where c = 1:
t' = g ( t - v x )
x' = g ( x - v t )
and the inverse
t = g ( t' + v x' )
x = g ( x' + v t' )
with
g = 1/sqrt(1-v^2)

You can verify that combining the world lines with these
equations, that, in the ship's coordinates, the station's world line
is given by
x' = - v t'
and the ship's world line by
x' = 0

The first event E1 (the ship leaves place P1) is an event with
station coordinates
E1: (x,t) = (0,0)
The second event E2 (the ship arrives at place P2) is an event with
station coordinates
E2: (x,t) = (L,T)
where, according to the station, L is the distance covered by the ship
and T the time it takes.
Since the velocity is v, we know that L = v T, so the second event
can be written as
E2: (x,t) = ( v T, T )

If you insert these values into the transformations, you find
E1: (x',t') = ( 0, 0 )
and
E2: (x',t') = ( 0, T sqrt(1-v^2) )
so, indeed, according to the transformation, the time it takes for
the ship is
T' = T sqrt(1-v^2)

When you look at the station's coordinates from the point of
view of the ship, you see that they always satisfy the equation
x' = - v t',
so when event E2 happens (on the ship!), the station is at
location
x' = - v T sqrt(1-v^2)
so the station has covered a distance
v T sqrt(1-v^2) ,
so you can imagine some event E3, that takes place on the
station, with coordinates
E3: (x',t') = ( -v T sqrt(1-v^2), T sqrt(1-v^2) )
Inserting this into the transformation gives
E3: (x,t) = ( 0, T (1-v^2) ).

Now compare the relationship between the coordinates of events
E1 and E2
with those of the events
E1 and E3 .
You see:
E1: (x,t) = (x',t') = ( 0, 0 )
and
E2: (x,t) = ( v T, T )
E2: (x',t') = ( 0, T sqrt(1-v^2) )
and
E3: (x',t') = ( -v T sqrt(1-v^2), T sqrt(1-v^2) )
E3: (x,t) = ( 0, T (1-v^2) ).

What do you notice?

Dirk Vdm

Thank you, Dirk. I'll definitely think about this.

Craig

wrote in message
oups.com...
This is a follow up to a post "Where is the flaw?" I had before about
relativity. I want to make my question as simple as possible; I felt
the question of that post before was not simple enough:

Suppose that a space station is located at point
P0=(x0,y0,z0,t0)=(0,0,0,0). There is a space ship moving from point
P1=(x1,y1,z1,t1)=(0,1,0,0) to point P2=(x2,y2,z2,t2)=(1,1,0,1), with
respect to the inertial frame of reference of the space station. So the
velocity of the space ship with respect to the space station is
v=(1,0,0). The units of measurement are assumed to be kilometers and
years.

So the time T that it takes for the space ship to get from point P1 to
point P2 from the point of view of the space station is T=1 year. Let
us calculate the time T' that it takes for the space ship to get from
point P1 to point P2 from the point of view of the space ship using the
time dilation formula. We get:

T'=T*sqrt(1-v^2/c^2)=sqrt(1-1/c^2) years, where c is the speed of light
in kilometers per year.

Now, let us think of the space station as moving and the space ship at
rest. (Since there is no prefered frame of reference according to
relativity theory, this is certainly a legitimate way of understanding
this scenario.) So the time that point P1 leaves the space ship to the
time that P2 reaches the space ship from the point of view of the space
ship is T'. Now, let us apply the time dilation formula to calculate
the time T" that the space station measures for point P1 to leave the
space ship to the time that P2 reaches the space ship. Since the
velocity of the space station with respect to the space ship is
v=(-1,0,0), we get:

T''=T' *sqrt(1-v^2/c^2)=1-1/c^2 years, where c is the speed of light in
kilometers per year.

But T'', which is defined to be the time that the space station
measures for point P1 to leave the space ship to the time that P2
reaches the space ship is really the same thing as T, the time that the
space ship goes from point P1 to point P2 from the perspective of the
space station.

So we have 1-1/c^2 years = T'' = T = 1 year. So 1-1/c^2=1, which
implies that -1/c^2=0, which implies that the speed of light c is
infinity. What is going on here? I would really appreciate if someone
could tell me where the flaw is in my argument. As I said before, I am
not out to claim that Einstein is wrong; I am giving Einstein the
benefit of doubt and assuming that I am wrong in my understanding of
relativity or that I made a simple mistake in my calculations
somewhere.

Thank you,
Craig


You might be misinterpreting the symbols in the time dilation formula

T' = T * Sqrt(1-v^2/c^2).

It appears to me that you are interpreting T as the time as measured in the
reference frame that is being considered to be 'at rest', and T' as the time
as measured in the reference frame that is considered to be 'in motion'.

But, that's not the correct interpretation. T' is the time between two
events as measured in the reference frame in which the two events occur at
the same x-coordinate in that frame. T is the time between those events as
measured in the reference frame in which the two events occur at different
values of x.

In your example, the two events (namely, the event where the space ship is
coincident with P1 and the event where the spaceship is coincident with P2)
both occur at the same place relative to the space ship reference frame and
the two events occur at different places relative to the space station. And
this is true no matter which frame of reference you wish to consider as
being 'at rest'. So, T' must be the time as measured by the space ship and
T is the time as measured by the space station.

So, you didn't set up the equation involving T' and T'' correctly. You
should have written

T' = T'' * Sqrt(1 - v^2/c^2)

And of course this is consistent with your earlier equation

T' = T * Sqrt(1 - v^2/c^2)

if T'' = T, which must be true since T and T'' both represent the same
thing - viz., the time between the events as measured in the space station
frame.

Todd