A constant speed of light in all reference frames? Surely you can't be serious.
in [Relativity]
|
From: Peter Webb on 15 Feb 2010 01:08 "mpc755" <mpc755(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/thread/2bb6ac5e1cbf17ed# _____________________________________________ 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?
From: Peter Webb on 15 Feb 2010 01:18 "mpc755" <mpc755(a)gmail.com> wrote in message news:cc37a395-3b16-4471-9964-d9db63246254(a)v20g2000prb.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 the speed of light and how do you determine it is different from 'c'? Are you using a mirror or synchronized clocks? _______________________________ Use whatever measuring apparatus you like. What you are incapable of understanding is everything is under the effects of the aether. As I said in one of my original posts which it would help you understand the point I am making. The atomic clocks the Observers on the train are using are offset because of their state with respect to the aether. So, I will ask you again. How is the light to be measured? ________________________________ You must already have some means of measuring light speed, or you couldn't claim the speed was constant relative to the ether. Use that.
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 06:52 On 14 Feb, 20:06, YBM <ybm...(a)nooos.fr.invalid> wrote: > Ste a écrit : > > > Of course I realise that. The point I'm making is that where you're > > dealing with two different frames that are moving relative, the same > > object (i.e. the photon, or whatever) cannot physically have the same > > measured velocity in both. > > I still can't make sense of what you intend by "physically", but what > is remarquable here is that your "physically" is in contradiction with > "according to experiments". Whatever you mean by "physical" it is > contradictory with "reality". That's because you, rather foolishly in my view, seem to think that observation is not open to interpretation. You can't imagine how wearing it is to keep being told "the sun goes round the Earth, because that is what we observe", as though it is impossible that the observation could possibly have more than one explanation. Indeed, it's even more serious than that. It's like talking to children, who can't distinguish between the physical movement of the Sun, and their own movement when they spin themselves in a circle, and who are constantly self-referential in their answers. So when the child spins on the spot, they cry "the sky is spinning". And when the sky is *really* spinning (due to a previously unheard-of astronomical phenomenon), they cry "the sky is spinning". And when you ask, "which is physically spinning", then child looks at you and says "what do you mean 'physical'?"
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. |