A constant speed of light in all reference frames? Surely you can't be serious. [Physics]

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From: Jerry on 8 Mar 2010 07:28

On Mar 7, 11:11 pm, "Inertial" <relativ...(a)rest.com> wrote:

> We only need one aether to do that. That is enough to make all aethers (if
> any) undetectable (if they have the same propagation speed).

As a supporter of the "one aether hypothesis", please provide me
with a mechanistic description of a single aether that can
support both the inverse-squared (gravity and light) and diverse
non-inverse-squared laws (strong and weak forces).

Also, you will have to prove that W and Z particles travel at
the speed of light. :-)

Jerry

From: Jerry on 8 Mar 2010 07:49

On Mar 8, 6:21 am, "Inertial" <relativ...(a)rest.com> wrote:
> "Jerry" <Cephalobus_alie...(a)comcast.net> wrote in message
> news:7ec422c5-c1be-4bb7-a26a-e694cc7a4e34(a)t41g2000yqt.googlegroups.com...

> > Classically,
> > supporters of the aether tied themselves into knots trying to
> > come up with an aether with the necessary mechanical properties
> > to support all the known characteristics of light.
>
> It would have to have properties unlike any other media .. contracting all
> matter and slowing all processes in particular :)
>
> Basically .. in LET you give the aether whatever properties you need to to
> explain whatever you want. Its quite ad-hoc in that respect.

No. Present-day advocates of aether theory may invest the aether
with godlike magical properties, but Lorentz did not do that.
Lorentz's aether had a highly specific set of properties.

> However, despite all that, you've still not come up with a way to falsify
> aether.

Not within LET itself, but within the meta-theory of which LET
must represent merely one of multiple media of highly diverse
properties.

Unless you wish to contend that LET is also a theory of gravity,
for example, and explains gravity waves? Who claims that?

Jerry

From: Peter Webb on 8 Mar 2010 08:31

Unless you wish to contend that LET is also a theory of gravity,
for example, and explains gravity waves? Who claims that?

___________________________
That would be a somewhat stronger argument if we had ever actually detected
a gravity wave. As it is, their absence may prove a MM moment for GR sooner
or later.

From: Jerry on 8 Mar 2010 08:56

On Mar 8, 7:31 am, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
wrote:
>> Unless you wish to contend that LET is also a theory of gravity,
>> for example, and explains gravity waves? Who claims that?
>
> ___________________________
> That would be a somewhat stronger argument if we had ever actually detected
> a gravity wave. As it is, their absence may prove a MM moment for GR sooner
> or later.

Of course, Hulse Taylor makes the existence of gravitational
waves all but certain.

My contention is that aether theories impose no necessary
constraint on the speed of gravitational waves versus light waves.
There is no existing aether theory that explains them as
manifestations of the same phenomenon. Aether theories as
presently understood therefore predict different speeds for
gravitational waves versus light waves.

Note: PREdict, not POSTdict. The speed of gravitation has not
been measured.

If the speed of sound were measured to be the same in air, water
and steel, that observation would disprove all current theories
of sound.

If the speed of gravitational waves is ever shown identical to
the speed of light waves, that observation would disprove all
current aether theories...

Jerry

From: Ste on 8 Mar 2010 09:35

On 7 Mar, 02:51, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
wrote:
> "Ste" <ste_ro...(a)hotmail.com> wrote in message
>
> news:651a713d-7ae4-4048-bafb-f1b3219ee4fc(a)v20g2000yqv.googlegroups.com...
>
>
>
>
>
> > On 6 Mar, 12:47, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au>
> > wrote:
> >> > This should make perfect sense to you. If a clock is running 2%
> >> > slower, then it is running 2% slower regardless of distance. But if,
> >> > as a result of running 2% slower, it falls behind 6 minutes after
> >> > running a certain amount of time, then it will fall behind 12 minutes
> >> > after running for twice as long.
>
> >> Agreed.
>
> >> The question now is, if we agree that both clocks suffer time dilation
> >> in this way, then when they return to the start point, how do they
> >> each reconcile the fact that (after accounting for the effects of
> >> acceleration) it ought to be the other clock which is slow, when in
> >> fact one clock (the one that went furthest from the start point) will
> >> be slower than the other? And a third clock, left at the start point,
> >> will be running ahead of both?
>
> >> _________________________
> >> They know that the operations were not symmetric. Only one clock remained
> >> in
> >> the same inertial reference frame throughout. The other two clocks spent
> >> different amounts of time in at least 3 different inertial reference
> >> frames.
> >> Everybody can see this is true, and so nobody expects that the clocks
> >> will
> >> remain synchronised.
>
> > Yes, but the important question here is whether they agree *after* the
> > effects of acceleration are taken into account. I mean, if we said
> > that each travelling clock slows by 2% when moving away from the start
> > point at a certain speed, then by rights both travelling clocks should
> > slow equally. Yes?
>
> As I understand your thought experiment, no.
>
> In SR, time dilation is a function of relative speed and the time for which
> they are moving at the speed. It is not a function of accleration.
>
> A doesn't move. B moves at speed v for time t, and its clock will read x
> behind A. C moves at speed v for time 2t, and its clock will read 2x behind
> A.

The question is this. We'll deal with only the outbound trip (in other
words, the clocks are on the move, but time 't' has not yet elapsed,
so there has been no further acceleration). I agree with your answer
above, as it concerns A's frame.

The question is, from the frame of B, what will the slowdown be on
clock C, *after* having accounted for the increased distances between
them (i.e. having accounted for the increased propagation delays). It
seems to me that the natural answer is to say "4%".