Prev: ETHERISTS LESS INSANE THAN EINSTEINIANS
Next: (only $20-$38,free shipping) for newest Nike air max --Paypal
From: nobody1357 on 18 May 2010 01:13
Hi, please consider this setup

A B
______________

______ is the lab frame, A and B are identical cars carrying same
amount of fuel.

They start their engines at the same time and accelerate in the same
direction until they run out the fuel.

Now, according to the lab frame, the distance between A and B has not
changed, it's same as when they started, since there is no reason for
identical cars would have moved differently.

Now, If A and B were connected, they would have lorentz contracted as
a system, but since they are not connected, according to them, the
distance must have increased (as opposed to the contracted state which
their distance would seem the same to them).

So far so good... However, suppose A sends a light pulse to B, which
gets reflected back to A, and A measures the duration of this. Since
their clocks are slowed down, A must now find that this duration has
decreased, compared to the beginning of the experiment, and therefore
concludes their distance have decreased.

So there is the paradox, was the distance increased or decreased
according to A?
Thanks for your help in advance.
From: Paul B. Andersen on 18 May 2010 08:53
On 18.05.2010 07:13, nobody1357 wrote:
> Hi, please consider this setup
>
> A B
> ______________
>
> ______ is the lab frame, A and B are identical cars carrying same
> amount of fuel.
>
> They start their engines at the same time and accelerate in the same
> direction until they run out the fuel.
>
> Now, according to the lab frame, the distance between A and B has not
> changed, it's same as when they started, since there is no reason for
> identical cars would have moved differently.
>
> Now, If A and B were connected, they would have lorentz contracted as
> a system, but since they are not connected, according to them, the
> distance must have increased (as opposed to the contracted state which
> their distance would seem the same to them).
>
> So far so good... However, suppose A sends a light pulse to B, which
> gets reflected back to A, and A measures the duration of this. Since
> their clocks are slowed down, A must now find that this duration has
> decreased, compared to the beginning of the experiment, and therefore
> concludes their distance have decreased.
>
> So there is the paradox, was the distance increased or decreased
> according to A?
> Thanks for your help in advance.

I will assume that the pulse is emitted when the cars have ran out
of fuel, and are moving with a constant speed v.

-> v
A---------B------>x'
|---------|------>x
0 L


Let the distance between the cars be L as measured
in the unprimed (x,t) frame.
Let the pulse be emitted at t_0 = 0 when A is at x_a0 = 0
and B is at x_b0 = L in the the unprimed frame.
Let A's clock t_a0' = 0 at this event.

Let g = 1/sqrt(1-v^2/c^2)

Your first question:
--------------------
What is the distance between the cars in the primed
(x',t') frame where the cars are stationary?

x_b0' = g(x_b0 - v*t_0) = L/sqrt(1-v^2/c^2)

The distance between the cars is increased in the car-frame.
(See Bell's spaceship paradox)

Why should this be a paradox?
Let's calculate what A's clock shows when the light hits it.

Calculated in the unprimed frame:
----------------------------------
When the light hits B at the time t_1, car B will have
moved a distance v*t_1.
So we have:
c*t_1 = L + v*t_1
t_1 = L/(c-v)
B's position is x_b1 = c*t_1 = L/(1-v/c)

When the light hits A at the time t_2, A will be
at the position x_a2 = v*t_2
The light has moved from the position x_b1 during
the time (t_2-t_1)
So we have:
(x_b1 - x_a2) = c*(t_2-t_1)
L/(1-v/c)-v*t_2 = c*(t_2-L/(c-v))
t_2 = (2L/c)/(1-v^2/c^2)

To find what A's clock show at this event, we
must transform its time coordinate to the primed frame:

t_a2' = g(t_2 - v*x_a2/c^2) = g*t_2(1 - v^2/c^2)
t_a2' = t_2*sqrt(1-v^2/c^2) (so clock 'runs slow')
t_a2' = (2L/c)/sqrt(1-v^2/c^2)

Calculated in the primed frame:
-------------------------------
The cars are stationary, and the distance is L/sqrt(1-v^2/c^2)
So:
t_a2' = 2(L/sqrt(1-v^2/c^2))/c = (2L/c)/sqrt(1-v^2/c^2)

Same result, no paradox.

--
Paul

http://home.c2i.net/pb_andersen/
From: Sue... on 18 May 2010 09:39
On May 18, 1:13 am, nobody1357 <nobody1...(a)operamail.com> wrote:
> Hi, please consider this setup
>
>       A        B
> ______________
>
> ______ is the lab frame, A and B are identical cars carrying same
> amount of fuel.

>
====================

> They start their engines at the same time and accelerate in the same
> direction until they run out the fuel.

<< Application of Noether's theorem allows physicists to
gain powerful insights into any general theory in physics,
by just analyzing the various transformations that would
make the form of the laws involved invariant. For example:

* the invariance of physical systems with respect
to spatial translation (in other words, that the laws
of physics do not vary with locations in space) gives
the law of conservation of linear momentum;
* invariance with respect to rotation gives the law
of conservation of angular momentum;
* invariance with respect to time translation gives
the well-known law of conservation of energy >>


http://en.wikipedia.org/wiki/Noether%27s_theorem#Applications


>
> Now, according to the lab frame, the distance between A and B has not
> changed, it's same as when they started, since there is no reason for
> identical cars would have moved differently.
>
> Now, If A and B were connected, they would have lorentz contracted as
> a system, but since they are not connected, according to them, the
> distance must have increased (as opposed to the contracted state which
> their distance would seem the same to them).
>
> So far so good... However, suppose A sends a light pulse to B, which
> gets reflected back to A, and A measures the duration of this. Since
> their clocks are slowed down, A must now find that this duration has
> decreased, compared to the beginning of the experiment, and therefore
> concludes their distance have decreased.
>
> So there is the paradox, was the distance increased or decreased
> according to A?
> Thanks for your help in advance.

That paradox might be a good reason to avoid Lorentz ether
theory.

http://en.wikipedia.org/wiki/Lorentz_ether_theory

====================

< Einstein's relativity principle states that:

All inertial frames are totally equivalent
for the performance of all physical experiments.

In other words, it is impossible to perform a physical
experiment which differentiates in any fundamental sense
between different inertial frames. By definition, Newton's
laws of motion take the same form in all inertial frames.
Einstein generalized[1] this result in his special theory of
relativity by asserting that all laws of physics take the
same form in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html

[1]<< the four-dimensional space-time continuum of the
theory of relativity, in its most essential formal
properties, shows a pronounced relationship to the
three-dimensional continuum of Euclidean geometrical space.
In order to give due prominence to this relationship,
however, we must replace the usual time co-ordinate t by
an imaginary magnitude

sqrt(-1)

ct proportional to it. Under these conditions, the
natural laws satisfying the demands of the (special)
theory of relativity assume mathematical forms, in which
the time co-ordinate plays exactly the same rôle as
the three space co-ordinates. >>
http://www.bartleby.com/173/17.html

<< where epsilon_0 and mu_0 are physical constants which
can be evaluated by performing two simple experiments
which involve measuring the force of attraction between
two fixed charges and two fixed parallel current carrying
wires. According to the relativity principle, these experiments
must yield the same values for epsilon_0 and mu_0 in all
inertial frames. Thus, the speed of light must be the
same in all inertial frames. >>
http://farside.ph.utexas.edu/teaching/em/lectures/node108.html

Sue...


From: Sue... on 18 May 2010 09:52
On May 18, 8:53 am, "Paul B. Andersen" <paul.b.ander...(a)somewhere.no>
wrote:
> On 18.05.2010 07:13, nobody1357 wrote:
>
>
>
> > Hi, please consider this setup
>
> >        A        B
> > ______________
>
> > ______ is the lab frame, A and B are identical cars carrying same
> > amount of fuel.
>
> > They start their engines at the same time and accelerate in the same
> > direction until they run out the fuel.
>
> > Now, according to the lab frame, the distance between A and B has not
> > changed, it's same as when they started, since there is no reason for
> > identical cars would have moved differently.
>
> > Now, If A and B were connected, they would have lorentz contracted as
> > a system, but since they are not connected, according to them, the
> > distance must have increased (as opposed to the contracted state which
> > their distance would seem the same to them).
>
> > So far so good... However, suppose A sends a light pulse to B, which
> > gets reflected back to A, and A measures the duration of this. Since
> > their clocks are slowed down, A must now find that this duration has
> > decreased, compared to the beginning of the experiment, and therefore
> > concludes their distance have decreased.
>
> > So there is the paradox, was the distance increased or decreased
> > according to A?
> > Thanks for your help in advance.
>
> I will assume that the pulse is emitted when the cars have ran out
> of fuel, and are moving with a constant speed v.
>
>      -> v
>   A---------B------>x'
>   |---------|------>x
>   0         L
>
> Let the distance between the cars be L as measured
> in the unprimed (x,t) frame.
> Let the pulse be emitted at t_0 = 0 when A is at x_a0 = 0
> and B is at x_b0 = L in the the unprimed frame.
> Let A's clock t_a0' = 0 at this event.
>
> Let g = 1/sqrt(1-v^2/c^2)
>
> Your first question:
> --------------------
> What is the distance between the cars in the primed
> (x',t') frame where the cars are stationary?
>
> x_b0' = g(x_b0 - v*t_0) = L/sqrt(1-v^2/c^2)
>
> The distance between the cars is increased in the car-frame.
> (See Bell's spaceship paradox)

<<Generally speaking, the aim of pseudoscience is to
rationalize strongly held beliefs, rather than to
investigate or to test alternative possibilities.
Pseudoscience specializes in jumping to "congenial
conclusions," grinding ideological axes, appealing to
preconceived ideas and to widespread misunderstandings.

Pseudoscience is indifferent to criteria of valid
evidence. The emphasis is not on meaningful, controlled,
repeatable scientific experiments. Instead it is on
unverifiable eyewitness testimony, stories and tall
tales, hearsay, rumor, and dubious anecdotes. Genuine
scientific literature is either ignored or misinterpreted.>>
http://www.quackwatch.org/01QuackeryRelatedTopics/pseudo.html

4-velocity and 4-acceleration
http://farside.ph.utexas.edu/teaching/em/lectures/node115.html


Sue...

[...]




From: nobody1357 on 18 May 2010 16:17
On May 18, 3:53 pm, "Paul B. Andersen" <paul.b.ander...(a)somewhere.no>
wrote:
> On 18.05.2010 07:13, nobody1357 wrote:
>
>
>
> > Hi, please consider this setup
>
> > A B
> > ______________
>
> > ______ is the lab frame, A and B are identical cars carrying same
> > amount of fuel.
>
> > They start their engines at the same time and accelerate in the same
> > direction until they run out the fuel.
>
> > Now, according to the lab frame, the distance between A and B has not
> > changed, it's same as when they started, since there is no reason for
> > identical cars would have moved differently.
>
> > Now, If A and B were connected, they would have lorentz contracted as
> > a system, but since they are not connected, according to them, the
> > distance must have increased (as opposed to the contracted state which
> > their distance would seem the same to them).
>
> > So far so good... However, suppose A sends a light pulse to B, which
> > gets reflected back to A, and A measures the duration of this. Since
> > their clocks are slowed down, A must now find that this duration has
> > decreased, compared to the beginning of the experiment, and therefore
> > concludes their distance have decreased.
>
> > So there is the paradox, was the distance increased or decreased
> > according to A?
> > Thanks for your help in advance.
>
> I will assume that the pulse is emitted when the cars have ran out
> of fuel, and are moving with a constant speed v.
>
> -> v
> A---------B------>x'
> |---------|------>x
> 0 L
>
> Let the distance between the cars be L as measured
> in the unprimed (x,t) frame.
> Let the pulse be emitted at t_0 = 0 when A is at x_a0 = 0
> and B is at x_b0 = L in the the unprimed frame.
> Let A's clock t_a0' = 0 at this event.
>
> Let g = 1/sqrt(1-v^2/c^2)
>
> Your first question:
> --------------------
> What is the distance between the cars in the primed
> (x',t') frame where the cars are stationary?
>
> x_b0' = g(x_b0 - v*t_0) = L/sqrt(1-v^2/c^2)
>
> The distance between the cars is increased in the car-frame.
> (See Bell's spaceship paradox)
>
> Why should this be a paradox?
> Let's calculate what A's clock shows when the light hits it.
>
> Calculated in the unprimed frame:
> ----------------------------------
> When the light hits B at the time t_1, car B will have
> moved a distance v*t_1.
> So we have:
> c*t_1 = L + v*t_1
> t_1 = L/(c-v)
> B's position is x_b1 = c*t_1 = L/(1-v/c)
>
> When the light hits A at the time t_2, A will be
> at the position x_a2 = v*t_2
> The light has moved from the position x_b1 during
> the time (t_2-t_1)
> So we have:
> (x_b1 - x_a2) = c*(t_2-t_1)
> L/(1-v/c)-v*t_2 = c*(t_2-L/(c-v))
> t_2 = (2L/c)/(1-v^2/c^2)
>
> To find what A's clock show at this event, we
> must transform its time coordinate to the primed frame:
>
> t_a2' = g(t_2 - v*x_a2/c^2) = g*t_2(1 - v^2/c^2)
> t_a2' = t_2*sqrt(1-v^2/c^2) (so clock 'runs slow')
> t_a2' = (2L/c)/sqrt(1-v^2/c^2)
>
> Calculated in the primed frame:
> -------------------------------
> The cars are stationary, and the distance is L/sqrt(1-v^2/c^2)
> So:
> t_a2' = 2(L/sqrt(1-v^2/c^2))/c = (2L/c)/sqrt(1-v^2/c^2)
>
> Same result, no paradox.
>
> --
> Paul
>
> http://home.c2i.net/pb_andersen/

I made a wrong assumption.. I thought the duration wouldn't change if
cars are stationary or moving uniformly. but 2L/c isn't equal to L/(c
+v) + L/(c-v).. thanks
 |  Next  |  Last
Pages: 1 2
Prev: ETHERISTS LESS INSANE THAN EINSTEINIANS
Next: (only $20-$38,free shipping) for newest Nike air max --Paypal