Dirk Van de moortel wrote:

snip

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)

If the travler was an electrical charge, and not a ship
would you still apply these gamma factors ?

I heard muons are like cousins to electrical charges.

Sue...

"Sue..." wrote in message ups.com...
Dirk Van de moortel wrote:

snip

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)

If the travler was an electrical charge, and not a ship
would you still apply these gamma factors ?

I heard muons are like cousins to electrical charges.

Sue...

Your potential is retarded and retired, Dennis.

Dirk Vdm

"Dirk Van de moortel" wrote
in message ...


| then the transformation equations are, in units where c = 1:
| t' = g ( t - v x )
| x' = g ( x - v t )

HAHAHAHAHAHA!!!
Too funny, Dork!
ROFLMAO!
http://www.fourmilab.ch/etexts/einst...ures/img53.gif
You ****in' raving lunatic arsehole! You don't know the difference between
x' and xi,
after all this time, you moron!
What a FUMBLE! Definitely one for the immortel fumbles. Greek to tough for
you?
Almost as good as Humpty Roberts:
This is PHYSICS, not math or logic, and "proof" is completely irrelevant.

What a pair of Dorks! LOL!
**** off, ****head, quit spreading your stupid lies.
Androcles.

Sorcerer wrote:
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.

Ah, yes, here we see that Androcles has learned nothing. For reference,
see
http://groups.google.com/group/sci.p...34988d8?hl=en&

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
| |
|

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

I can't find any problems with your math, and your methodology makes
sense according to my understanding of relativity theory. But I do have
a problem with what the mathematical conclusions are saying:

From the perspective of the space station, at t=0, the space ship is at
x=0. Again, it is at x=0 at t = T (1-v^2). It moves and reaches x=vT at
the time t=T. This contradicts our assumption that the space ship
travels at constant speed. If you have a watch aboard the ship, what
time would it say it is at the instant when the space ship is at x=0?

From the perspective of the space ship, at t=0, the space station is at
x=0. It moves to x=-vT sqrt(1-v^2) at t= T sqrt(1-v^2). But also, it is
at x=0 when t=T sqrt(1-v^2). How can the space station be two places at
once? This contradicts what everyone perceives to be reality.

How would a relativity theorist address these issues? And what happens
in between the events that you have calculated? And is there even a
such thing as "in between" in the land of Relativity?

Perhaps when used properly, relativity theory is able to predict many
things about our universe that Newtonian mechanics can't. So as a tool,
it may be good. But if you actually believe relativity theory, then it
seems to run into problems, since relativity theory also seems to make
predictions which are completely opposed to our understanding of
reality and our experience. Isn't there a better theory out there than
relativity?

Craig

wrote in message oups.com...

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

I can't find any problems with your math

Never mind the math, check the physics.
The math is just there to explain what is really happening.

, and your methodology makes
sense according to my understanding of relativity theory. But I do have
a problem with what the mathematical conclusions are saying:

From the perspective of the space station, at t=0, the space ship is at
x=0. Again, it is at x=0 at t = T (1-v^2).

No, the *station* is at x=0 at t = T (1-v^2). It is at x=0 at every value
of t, since the station is the origin of the coordinate system (x,y,z,t).
The event E3 happens on the station - not on the ship. This event is
simultaneous with event E2, according to the ship (the events
E2 and E3 have the same t'). But E2 happens on the ship (x' = 0),
and E3 happens on the station (x=0).

It moves and reaches x=vT at
the time t=T.

Yes, that is event E2 (on the ship where x'=0).

This contradicts our assumption that the space ship
travels at constant speed. If you have a watch aboard the ship, what
time would it say it is at the instant when the space ship is at x=0?

That is event E1 with (x,t) = (x',t') = (0,0). So the watch
on the ship says t' = 0 and the watch in the station says t = 0.


From the perspective of the space ship, at t=0, the space station is at
x=0.

If you talk "from the perspective of the ship", you use x' and t'.
So, at t'=0 the station is at x'=0, which is event E1, which indeed
correpsonds to the station being at x=0 (which it *always* is) at
it's own t=0, since that is the event where station and ship coincided
and set their watche to 0.

It moves to x=-vT sqrt(1-v^2) at t= T sqrt(1-v^2).

No.
Same he It moves to x' = -v T sqrt(1-v^2) at t'= T sqrt(1-v^2),
which is event E3, which has x=0 at t = T (1-v^2) )

But also, it is
at x=0 when t=T sqrt(1-v^2). How can the space station be two places at
once? This contradicts what everyone perceives to be reality.

Actually, it shows that you haven't fully understood the meaning of
the variables x, t, x', and t'. You see, it's not the math that counts. It's
the physics of the meaning of the variables.

Try going over everything again, but make sure you fully understand
what the variables mean. You can reply to my previous message again.

Dirk Vdm



How would a relativity theorist address these issues? And what happens
in between the events that you have calculated? And is there even a
such thing as "in between" in the land of Relativity?

Perhaps when used properly, relativity theory is able to predict many
things about our universe that Newtonian mechanics can't. So as a tool,
it may be good. But if you actually believe relativity theory, then it
seems to run into problems, since relativity theory also seems to make
predictions which are completely opposed to our understanding of
reality and our experience. Isn't there a better theory out there than
relativity?

Craig

On 7/7/06 22:32, in article
, "
wrote:

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

I can't find any problems with your math, and your methodology makes
sense according to my understanding of relativity theory. But I do have
a problem with what the mathematical conclusions are saying:

From the perspective of the space station, at t=0, the space ship is at
x=0. Again, it is at x=0 at t = T (1-v^2). It moves and reaches x=vT at
the time t=T. This contradicts our assumption that the space ship
travels at constant speed. If you have a watch aboard the ship, what
time would it say it is at the instant when the space ship is at x=0?

My guess here is you're talking about E1 and E3. E3 is an event that happens
about the station hence x will always be zero in the coordinates as defined
by the station.

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


so you can imagine some event E3, that takes place on the
station, with coordinates


From the perspective of the space ship, at t=0, the space station is at
x=0. It moves to x=-vT sqrt(1-v^2) at t= T sqrt(1-v^2). But also, it is
at x=0 when t=T sqrt(1-v^2). How can the space station be two places at
once? This contradicts what everyone perceives to be reality.


You don't seem to have your head around frames.

How would a relativity theorist address these issues? And what happens
in between the events that you have calculated? And is there even a
such thing as "in between" in the land of Relativity?

Perhaps when used properly, relativity theory is able to predict many
things about our universe that Newtonian mechanics can't. So as a tool,
it may be good. But if you actually believe relativity theory, then it
seems to run into problems, since relativity theory also seems to make
predictions which are completely opposed to our understanding of
reality and our experience. Isn't there a better theory out there than
relativity?

It doesn't run into problems. The problems are your interpretation of them.

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wrote in message
ups.com...
|
| 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.

Think about this, too:
http://www.androcles01.pwp.blueyonde...ork/trojan.htm

Androcles.

"Todd" wrote in message
m...
|
| 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'.

It appears to me that you are total ****head.
Androcles.

"Dirk Van de moortel" wrote
in message ...
|
| "Sue..." wrote in message
ups.com...
| Dirk Van de moortel wrote:
|
| snip

Of course he did, but he really means [anip]
Androcles