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

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From: Peter Webb on 15 Feb 2010 01:20

"mpc755" <mpc755(a)gmail.com> wrote in message
news:16bd20be-baaa-459a-90d2-f763cba4f366(a)b36g2000pri.googlegroups.com...
On Feb 15, 12:27 am, Sam Wormley <sworml...(a)gmail.com> wrote:
> On 2/14/10 11:23 PM, mpc755 wrote:
>
> > How do you measure your speed relative to the ether?
>
> What ether?

The aether which is the reason for the observed behaviors in every
double slit experiment ever performed.

A C-60 molecule is in the slit(s). Detectors are placed at the exits
to the slits while the C-60 molecule is in the slit(s). Every time the
C-60 molecule exits the slit(s) it is detected exiting a single slit.

When the detectors are placed and removed from the exits to the slits
the C-60 molecule is able to create an interference pattern.

How is this possible without the C-60 molecule having an associated
aether displacement wave?

______________________________________
Or, more to the point, how is this possible without the C-60 molecule having
an associated pan-galactic gargleblaster pressure wave? Well?

From: Ste on 15 Feb 2010 06:54

On 14 Feb, 23:46, mpalenik <markpale...(a)gmail.com> wrote:
> On Feb 14, 2:03 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>
> > I'm afraid you're easily satisifed Tom. As I say, I'm not really
> > interested in learning geometry, or talking about completely
> > hypothetical "grooves in spacetime".
>
> And as many people have repeatedly tried to explain to you, the answer
> simply is geometry. When you accellerate, you rotate in spacetime.
> Why? Because that's what accelleration means. That's what it means
> to be travelling with a certain velocity with respect to something
> else. It means that you're both "facing different directions". Every
> effect predicted by relativity can be explained simply by the fact
> that two different observers at different speeds are "facing different
> directions" in spacetime--because that's what it means to be moving
> with respect to something else. It means that you have a different t
> and x axis.

Mark, if you consider this an answer, then you simply haven't
understood the question.

From: Inertial on 15 Feb 2010 07:21

"Ste" <ste_rose0(a)hotmail.com> wrote in message
news:ed73b725-fdd3-4f16-bfab-b15738b748a9(a)e1g2000yqh.googlegroups.com...
> On 14 Feb, 23:46, mpalenik <markpale...(a)gmail.com> wrote:
>> On Feb 14, 2:03 pm, Ste <ste_ro...(a)hotmail.com> wrote:
>>
>> > I'm afraid you're easily satisifed Tom. As I say, I'm not really
>> > interested in learning geometry, or talking about completely
>> > hypothetical "grooves in spacetime".
>>
>> And as many people have repeatedly tried to explain to you, the answer
>> simply is geometry. When you accellerate, you rotate in spacetime.
>> Why? Because that's what accelleration means. That's what it means
>> to be travelling with a certain velocity with respect to something
>> else. It means that you're both "facing different directions". Every
>> effect predicted by relativity can be explained simply by the fact
>> that two different observers at different speeds are "facing different
>> directions" in spacetime--because that's what it means to be moving
>> with respect to something else. It means that you have a different t
>> and x axis.
>
> Mark, if you consider this an answer, then you simply haven't
> understood the question.

What was your question?

From: mpc755 on 15 Feb 2010 08:35

On Feb 15, 1:06 am, "Peter Webb"
<webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote:
> "mpc755" <mpc...(a)gmail.com> wrote in message
>
> news:21c1d72e-9898-436a-ba4e-05a849fc4efc(a)g8g2000pri.googlegroups.com...
> On Feb 15, 12:35 am, "Peter Webb"
>
>
>
> <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote:
> > "mpc755" <mpc...(a)gmail.com> wrote in message
>
> >news:e03b248e-5f49-4e80-9c4c-d542dd7e269e(a)k5g2000pra.googlegroups.com...
> > On Feb 15, 12:18 am, "Peter Webb"
>
> > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote:
> > > As I have said at least three times now,
> > > you cannot determine the speed of the aether.
> > > ____________________________________
>
> > > You said light moves at a constant velocity relative to the ether. So
> > > why
> > > can't you measure the speed of light, see how much it differs from c,
> > > and
> > > the difference is your speed relative to the ether? Why doesn't that
> > > procedure determine the speed of the ether?
>
> > How do you measure your speed relative to the ether?
>
> > As I have said at least four times now, you can't measure the speed of
> > the aether. If you can't measure the speed of the aether you can't
> > measure your speed relative to the aether.
>
> > Do you want to ask this same question again so I can answer it for a
> > fifth time?
>
> > ______________________________________
> > I just described how you *can* measure your speed relative to the ether..
> > You
> > measure the speed of light, see how much it differs from c, and the
> > difference is your speed relative to the ether.
>
> How do you measure the speed of light so it is not 'c'?
>
> _________________________________
> Anyway you like. Aren't you claiming that the speed of light is a constant
> relative to the speed of the ether, and not constant relative to the
> observer? So you can measure the speed of light in some way, to make this
> claim at all, right? So why not measure it, see how much it departs from c,
> and then the difference is the speed of the ether.
>
> Why won't that work?

I am asking you to state how it is you want to measure the speed of
light? Are you using mirrors? If you are using mirrors and the light
propagates from M' to A' and B' the light will take longer with
respect to the aether to get from M' to B' because the light waves are
propagating against the 'flow' of the aether relative to the train but
the Observer at B's clock is off set an behind the other Observers
clock so when the light gets to B' the clock at B' will read the same
time as the clock at A' even though the light gets to A' after the
light gets to B' with respect to the aether the train is moving
through. When the light is reflected the opposite occurs so the light
gets back to M' simultaneously.

You really need to read my previous posts to understand this.

From: mpc755 on 15 Feb 2010 08:36

On Feb 15, 1:08 am, "Peter Webb"
<webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote:
> "mpc755" <mpc...(a)gmail.com> wrote in message
>
> news:285f58e2-a468-4257-8051-fa7249dc0e72(a)m35g2000prh.googlegroups.com...
> On Feb 15, 12:35 am, "Peter Webb"
>
>
>
> <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote:
> > "mpc755" <mpc...(a)gmail.com> wrote in message
>
> >news:e03b248e-5f49-4e80-9c4c-d542dd7e269e(a)k5g2000pra.googlegroups.com...
> > On Feb 15, 12:18 am, "Peter Webb"
>
> > <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote:
> > > As I have said at least three times now,
> > > you cannot determine the speed of the aether.
> > > ____________________________________
>
> > > You said light moves at a constant velocity relative to the ether. So
> > > why
> > > can't you measure the speed of light, see how much it differs from c,
> > > and
> > > the difference is your speed relative to the ether? Why doesn't that
> > > procedure determine the speed of the ether?
>
> > How do you measure your speed relative to the ether?
>
> > As I have said at least four times now, you can't measure the speed of
> > the aether. If you can't measure the speed of the aether you can't
> > measure your speed relative to the aether.
>
> > Do you want to ask this same question again so I can answer it for a
> > fifth time?
>
> > ______________________________________
> > I just described how you *can* measure your speed relative to the ether..
> > You
> > measure the speed of light, see how much it differs from c, and the
> > difference is your speed relative to the ether. That is because according
> > to
> > you, light moves at a constant speed relative to the ether. So if you
> > measure the speed of light, and subtract if from c, that must give you
> > your
> > speed relative to the ether.
>
> > So say you measure that light is moving at 2 x 10^8 m/s relative to you..
> > We
> > know it is moving at 3 x 10^8 m/s relative to the ether, therefore you are
> > moving at 3 x 10^8 m/s - 2 x 10^8 m/s = 1 x 10^8 m/s relative to the
> > ether.
>
> > Why doesn't that procedure determine your speed relative to the ether?
>
> Not sure this link will work, but this is a link to the two posts I
> made having to do with the train and the embankment and the time on
> the clocks and the lightning strikes.
>
> I realize you are not going to understand what I have written, but
> this is why the light is not detected at other than 'c' for either the
> Observers on the embankment or the Observers on the train:
>
> http://groups.google.com/group/sci.physics.relativity/browse_frm/thre...
>
> _____________________________________________
> I didn't ask about trains, or embankments, or anything like that. I asked
> you why you can't measure the relative speed of the ether by the simple
> process I described above. Why can't you? Or can you?

In order to answer the question I used Einstein's train gedanken with
water/aether at rest with respect to the embankment.