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From: Sue... on 9 Mar 2010 09:41
On Mar 9, 9:26 am, PD <thedraperfam...(a)gmail.com> wrote:
> On Mar 9, 7:49 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
>
>
> > On 8 Mar, 22:55, PD <thedraperfam...(a)gmail.com> wrote:
>
> > > On Mar 6, 5:32 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > What confuses me is that, if the clocks run slow by 2% for all the
> > > > time that they are moving, how does one reconcile this with the fact
> > > > that, if one uses the frame of one of the moving clocks, say clock B,
> > > > then it seems to be to be your argument that there is no slowdown at
> > > > all for B, and it is the other clocks, A and C, that slow down (i.e..
> > > > *disregarding* both acceleration and propagation delays).
>
> > > Be careful. The acceleration profiles are common between B and C, but
> > > they are not common to A. So while there is no difference between B
> > > and C due to the acceleration, you CANNOT say that the acceleration
> > > has no effect whatsoever. In fact, it is the indisputable fact that B
> > > and C accelerate and A does NOT accelerate that makes the situation
> > > nonsymmetric for A. This is what makes the worldline for A straight,
> > > and the worldline for B and C kinked.
>
> > Yes, but we're supposed to have isolated the effect of acceleration,
> > and disregarded it.
>
> No, we did not. We said that it cannot account for the DIFFERENCE
> between B and C, but this does not discount or remove acceleration
> from further consideration, particularly with regard to how clock A's
> rate is seen by B.
>
>
>
> > And in any event, the more important question is
> > the discrepancy between B and C.
>
> > > Two places where I will try to intercept misconceptions.
> > > 1. The first temptation is to say, well, if the kink is what's
> > > responsible for the time dilation, then all the dilating must happen
> > > during the acceleration. That is not the case. Note the time dilation
> > > is different for B and C, even though they have the same kink (the
> > > same acceleration profile). The fact that there IS a kink is what
> > > makes the elapsed time less on B and C than it is on A (where there is
> > > no kink), but how much less depends on the steepness and length of the
> > > straight parts of the worldline on either side of the kink.
>
> > As I say, I've stipulated that we are measuring on the outbound
> > journey, before any of the clocks have turned back. So we've had one
> > episode of acceleration and now B and C are travelling at the same
> > speed away from the origin point, but in opposite directions. What
> > amount of time dilation does C suffer relative to B? Nil? 2%? 4%?
>
> It's not that simple, and to get a number, you need to use the math.


Indeed! The maths!

<< 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...

[...]


> Sorry, but that's the facts. The 2% is the Lorentz dilation factor,
> and that is given by a function 1/sqrt(1-v^2/c^2), and you can see
> that doubling the speed will not double the Lorentz dilation factor.
> Furthermore, the speed of C relative to B is not twice the velocity of
> B with respect to A, because the relation for combining velocities is
> (v1+v2)/(1+v1*v2/c^2).
>
> To find out what the Lorentz dilation factor is for C relative to B,
> then you simply need to put in the numbers and crank.
>
>
>
> > > 2. The second temptation is to say, well, in B's frame it does not
> > > accelerate, and A does accelerate. This is not correct.
>
> > I know.
>
> > > Now, it IS fair to say that while the clock B is on its outward
> > > journey at constant speed, it can look back at clock A and discern
> > > that clock A is running slow relative to B. And the same goes on the
> > > inbound journey. And so it's a fair question to say, how does it
> > > happen then that by the time B lands back at A, it is BEHIND clock A?
> > > And here is where things do get interesting in terms of what B sees in
> > > terms of the clock reading on A in the turnaround.
>
> > No, the question is between B and C, not B and A. I'm using B and C
> > specifically in order to allow us to disregard any effects of
> > acceleration.
>
> > And we've already stipulated that, if B and C have the same
> > acceleration profiles and travel the same distance before returning to
> > A, then they will be in agreement with each other, but both will have
> > slowed relative to A. The question is, while they are still on the
> > outbound journey, what do B and C report about each other? Do they
> > both agree with each other still? Or not?
>
>

From: bert on 9 Mar 2010 09:48
On Mar 9, 9:41 am, "Sue..." <suzysewns...(a)yahoo.com.au> wrote:
> On Mar 9, 9:26 am, PD <thedraperfam...(a)gmail.com> wrote:
>
>
>
>
>
> > On Mar 9, 7:49 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > On 8 Mar, 22:55, PD <thedraperfam...(a)gmail.com> wrote:
>
> > > > On Mar 6, 5:32 am, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > > > > What confuses me is that, if the clocks run slow by 2% for all the
> > > > > time that they are moving, how does one reconcile this with the fact
> > > > > that, if one uses the frame of one of the moving clocks, say clock B,
> > > > > then it seems to be to be your argument that there is no slowdown at
> > > > > all for B, and it is the other clocks, A and C, that slow down (i..e.
> > > > > *disregarding* both acceleration and propagation delays).
>
> > > > Be careful. The acceleration profiles are common between B and C, but
> > > > they are not common to A. So while there is no difference between B
> > > > and C due to the acceleration, you CANNOT say that the acceleration
> > > > has no effect whatsoever. In fact, it is the indisputable fact that B
> > > > and C accelerate and A does NOT accelerate that makes the situation
> > > > nonsymmetric for A. This is what makes the worldline for A straight,
> > > > and the worldline for B and C kinked.
>
> > > Yes, but we're supposed to have isolated the effect of acceleration,
> > > and disregarded it.
>
> > No, we did not. We said that it cannot account for the DIFFERENCE
> > between B and C, but this does not discount or remove acceleration
> > from further consideration, particularly with regard to how clock A's
> > rate is seen by B.
>
> > > And in any event, the more important question is
> > > the discrepancy between B and C.
>
> > > > Two places where I will try to intercept misconceptions.
> > > > 1. The first temptation is to say, well, if the kink is what's
> > > > responsible for the time dilation, then all the dilating must happen
> > > > during the acceleration. That is not the case. Note the time dilation
> > > > is different for B and C, even though they have the same kink (the
> > > > same acceleration profile). The fact that there IS a kink is what
> > > > makes the elapsed time less on B and C than it is on A (where there is
> > > > no kink), but how much less depends on the steepness and length of the
> > > > straight parts of the worldline on either side of the kink.
>
> > > As I say, I've stipulated that we are measuring on the outbound
> > > journey, before any of the clocks have turned back. So we've had one
> > > episode of acceleration and now B and C are travelling at the same
> > > speed away from the origin point, but in opposite directions. What
> > > amount of time dilation does C suffer relative to B? Nil? 2%? 4%?
>
> > It's not that simple, and to get a number, you need to use the math.
>
> Indeed! The maths!
>
> << 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...
>
> [...]
>
>
>
> > Sorry, but that's the facts. The 2% is the Lorentz dilation factor,
> > and that is given by a function 1/sqrt(1-v^2/c^2), and you can see
> > that doubling the speed will not double the Lorentz dilation factor.
> > Furthermore, the speed of C relative to B is not twice the velocity of
> > B with respect to A, because the relation for combining velocities is
> > (v1+v2)/(1+v1*v2/c^2).
>
> > To find out what the Lorentz dilation factor is for C relative to B,
> > then you simply need to put in the numbers and crank.
>
> > > > 2. The second temptation is to say, well, in B's frame it does not
> > > > accelerate, and A does accelerate. This is not correct.
>
> > > I know.
>
> > > > Now, it IS fair to say that while the clock B is on its outward
> > > > journey at constant speed, it can look back at clock A and discern
> > > > that clock A is running slow relative to B. And the same goes on the
> > > > inbound journey. And so it's a fair question to say, how does it
> > > > happen then that by the time B lands back at A, it is BEHIND clock A?
> > > > And here is where things do get interesting in terms of what B sees in
> > > > terms of the clock reading on A in the turnaround.
>
> > > No, the question is between B and C, not B and A. I'm using B and C
> > > specifically in order to allow us to disregard any effects of
> > > acceleration.
>
> > > And we've already stipulated that, if B and C have the same
> > > acceleration profiles and travel the same distance before returning to
> > > A, then they will be in agreement with each other, but both will have
> > > slowed relative to A. The question is, while they are still on the
> > > outbound journey, what do B and C report about each other? Do they
> > > both agree with each other still? Or not?- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -

It would not be a photon if it did not go at c It would not be a
photon if it bounced. TreBert
From: Dono. on 9 Mar 2010 10:02
On Mar 9, 1:33 am, "Inertial" <relativ...(a)rest.com> wrote:
> "Dono." <sa...(a)comcast.net> wrote in message
>
> news:9a5eb50c-6ed1-4435-9493-0a0fef9039df(a)v20g2000prb.googlegroups.com...
>
> > On Mar 8, 6:06 pm, "Inertial" <relativ...(a)rest.com> wrote:
>
> >> In LET the speed of light is only isotropically c in the aether frame.
>
> > ...then all the modern tests that put severe constraints on light
> > speed anisotropy falsify LET.
>
> No .. because we measure the speed as isotropic due to the distorted rulers
> and malfunctioning clocks that movement thru the aether causes (according to
> LET).
>

You are contradicting yourself , in the earlier post you claimed
(correctly) that LET predicts light speed to be isotropic ONLY in the
preferrential frame of the "aether". In ALL other frame, light speed
is ANISOTROPIC.
So, you are now contradicting yourself and your new post is wrong.



From: Dono. on 9 Mar 2010 10:05
On Mar 9, 2:49 am, "Inertial" <relativ...(a)rest.com> wrote:
> "Jerry" <Cephalobus_alie...(a)comcast.net> wrote in message
>
>
> No reason they cannot. No reason why gravity waves and EM waves cannot both
> travel at c. Any test showing that the do both travel at c will simply
> refute your assertion that they must be different .. it won't refute aether
> theory itself. And if necessary, aether will be modified yet again to
> account for it.

Bad answer: EM waves are TRANSVERSE whereas gravitational wave are
LONGITUDINAL waves. The same medium cannot propagate both, so, you
need AT LEAST two different "aethers"

From: Dono. on 9 Mar 2010 10:06
On Mar 9, 4:16 am, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
wrote:
> If and when the speed of gravitation is ever measured, if the
> speed turns out to be identical to the speed of light, that
> would be an incredible finding that casts severe questions on
> the viability of aether theories in general. Aether theories
> explain wave propagation as due to the mechanical properties
> of the particular aether in question, and there is no reason
> whatsoever why two theories should share the same wave speed.
> Certainly the gravitational and luminiferous aethers shouldn't.
>
> Jerry
>
> ___________________________
> The confirmation of the existence of the strong and weak forces and QM
> generally is itself a strong argument against LET at a number of levels.
>
> The argument that Lorentz put forward is based upon in the absence of
> gravity there are physical objects, and electromagnetism. Maxwell's eqns
> transformed according to the Lorentz equations. As experiments had shown
> that physical objects and EM transform the same way, and everybody was
> pretty confident about Maxwell, the obvious (and correct) answer was that
> physical objects must also obey a Lorentz transform. The putative mechanism
> was that the movement through the ether compressed physical objects.
>
> Now, that's all well and good, but how do you reconcile this with the
> existence of other fields, such as the strong and weak forces? It has long
> been known that radioactive decay rates (from say cosmic rays) and other
> processes that are mediated by the strong and weak force follow exactly the
> same transformations eg time dilation.
>
> So now this ether is doing more than compress physical objects to make them
> have the same transformation rules as EM, it is also compressing weak fields
> and strong fields in exactly the same way to force them to align with EM.
>
> By the time you have run this kludge three times (for physical objects,
> strong force, weak force) to align them with Maxwell, its pretty obvious
> that its much, much simpler to assume space itself is changing.



Nice post .
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