SR is an incomplete aether theory.
in [Relativity]
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From: Sam Wormley on 9 Jul 2010 11:31 On 7/9/10 8:21 AM, kenseto wrote: > The experimental basis would be the same as that for SR except that > IRT includes the possibility that an observed clock can run faster > than the observer's clock. Absent differential gravitational wells, that is never observed in nature. Take, for example two astronaut with clocks in intergalactic space, where we can ignore gravitational effects. We will also ignore the effects of Doppler shift here. A and B are observers with identical clocks. That is A and B's clocks ticked synchronously when they were together. ∆t represent a time interval between tick of the clocks. Special relativity predicts that observer A will measure that ∆t_B' = γ ∆t_B where ∆t represent a time interval, v is the relative velocity between A and B, and γ = 1/√(1-v^2/c^2) . Furthermore, special relativity predicts that observer B will measure that ∆t_A' = γ ∆t_A Neither measure the other's clock running fast. But each measures the other's clock running slow. Here's the part your brain can't seem to handle. You can be observer A or you can be observer B. You cannot be both simultaneously. There is no contradiction. Special Relativity correctly predicts the observations. Physics FAQ: What is the experimental basis of special relativity? http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
From: Sam Wormley on 9 Jul 2010 11:40 On 7/9/10 7:30 AM, kenseto wrote: > Hey idiot when you assert that all the clocks moving wrt you are > running slow you are assuming that you are at rest in the aether > frame. > The correct assumption is that a clock moving wrt you can run slow or > fast compared too your clock. No, I don't have to be a rest with respect to anything. If there is a clock moving with respect to me with velocity v, disregarding gravitational affects, accelerations, and Doppler effects, I measure that clock's time interval, ∆t_clock' = γ ∆t_clock where γ = 1/√(1-v^2/c^2) . The measurement agrees with the predictions of special relativity. Physics FAQ: What is the experimental basis of special relativity? http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
From: Michael Moroney on 9 Jul 2010 12:34 kenseto <kenseto(a)erinet.com> writes: >On Jul 7, 12:23 pm, PD <thedraperfam...(a)gmail.com> wrote: >> The principle of equivalence explicitly says that there is no >> preferred frame. .... >Yes they are the exclusive properties of the aether frame....every LET >observer accept that the aether frame exists and claims these >exclusive properties....that's why LET and SR have the same math. The phrase "the exclusive properties of the aether frame" DIRECTLY conflicts with the phrase "there is no preferred frame". Since the latter is part of the principle of equivalence in SR, your claim that "SR uses the exclusive properties of the aether frame" is completely false. Why can't you see that? Also, I've asked you repeatedly for proof of your claim "SR uses the exclusive properties of the aether frame", where SR states that and you cannot do so. Yet you still repeat it as truth. Put up or shut up.
From: PD on 9 Jul 2010 12:52 On Jul 9, 8:27 am, kenseto <kens...(a)erinet.com> wrote: > On Jul 8, 9:57 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Jul 8, 7:09 am, kenseto <kens...(a)erinet.com> wrote: > > > > On Jul 7, 12:23 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Jul 7, 8:41 am, kenseto <kens...(a)erinet.com> wrote: > > > > > > SR is an incomplete aether theory. > > > > > You must be talking about a theory other than SR, because: > > > > > > Here's why: > > > > > 1. The principle of relativity (PoR) says that all inertial frames > > > > > including the unique absolute rest frame of the aether are > > > > > equaivalent. > > > > > The principle of equivalence explicitly says that there is no > > > > preferred frame. > > > > No it says that all frames are equivalent > > > Yes. > > > > and the only way for that to > > > happen is that they all use the same aether frame to derive the math. > > > No, that's what YOU say. That's not what SR says. SR says you don't > > need to use the aether frame, you can use any frame, because all > > frames are equivalent. That should be obvious. > > Hey idiot when an SR observer asserts that all the clocks in the > universe are running slow he is assuming that he is at rest in the > aether frame. No, he's not asserting that at all. YOU are asserting that. But that is completely counter to what SR asserts. You get very confused when you mix together what SR says and what YOU say. > > > It's when you mix what SR says with what YOU say that nonsense > > results. > > > > > Your statement is equivalent to saying, "No man is better than any > > > > other man, including the man that is better than other men." > > > > > > 2. This allows every SR observer to choose any frame to do physics and > > > > > the rest frame of the aether is choosed because it is the simplest > > > > > frame to do physics. > > > > > No, SR does NOT choose that frame. It uses any frame equally. > > > > You call the aether frame as an inertial frame and that's why you can > > > use any frame equally. > > > No, I don't. The aether frame is a preferred frame. I've already told > > you that the preferred frame is the one that is NOT equivalent to the > > equivalent inertial frames. > > Then why does every SR observer uses the exclusive properties of the > preferred frame to do physics????? Only YOU say those properties are exclusive to the preferred frame. Physicists do not. It's when you mix up what SR say with what YOU say that nonsense results. > > > YOU say that inertial frames are really the aether frame. SR does not > > say that. > > > It's when you mix what SR says with what YOU say that nonsense > > results. > > > > > > 3. Choosing the aether frame to do physics allows every SR observer to > > > > > claim the exclusive properties of the aether frame which are: All the > > > > > clocks moving wrt to an SR observer are running slow and all the ruler > > > > > moving wrt him are contracted. > > > > > That is not the exclusive property of the aether frame. That statement > > > > is one you made up. > > > > Yes they are the exclusive properties of the aether frame. > > > Thats what YOU say. That's not what physicists mean by the aether > > frame or the preferred frame. > > So the physicists are wrong. I'm sorry, Ken, but nonsense results when you use YOUR meanings of physics terms when they are used in other people's theories, like special relativity. You cannot possibly understand what SR says unless you learn and use terms the way they are used in physics. When you see words and make up your own meanings for the words used, you end up with contradictory nonsense. That is exactly what has happened to you for 15 years. > You cannot assume the properties of the > preferred frame and at the same time deny the existence of the > preferred frame. > > Ken Seto >
From: kenseto on 9 Jul 2010 14:07
On Jul 9, 11:31 am, Sam Wormley <sworml...(a)gmail.com> wrote: > On 7/9/10 8:21 AM, kenseto wrote: > > > The experimental basis would be the same as that for SR except that > > IRT includes the possibility that an observed clock can run faster > > than the observer's clock. > >   Absent differential gravitational wells, that is never observed in >   nature. Take, for example two astronaut with clocks in intergalactic >   space, where we can ignore gravitational effects. We will also ignore >   the effects of Doppler shift here. > >   A and B are observers with identical clocks. That is A and B's >   clocks ticked synchronously when they were together. > >   ât represent a time interval between tick of the clocks. > >   Special relativity predicts that observer A will measure that >    ât_B' = γ ât_B Wormy you are the stupidest runt of the Srians. I told you many times: ât_B' = γ ât_A Ken Seto > >   where ât represent a time interval, v is the relative velocity >   between A and B, and γ = 1/â(1-v^2/c^2) . > >   Furthermore, special relativity predicts that observer B will >   measure that >    ât_A' = γ ât_A > >   Neither measure the other's clock running fast. But each measures >   the other's clock running slow. > >   Here's the part your brain can't seem to handle. You can be observer >   A or you can be observer B. You cannot be both simultaneously. There >   is no contradiction. Special Relativity correctly predicts the >   observations. > >   Physics FAQ: What is the experimental basis of special relativity? >    http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html |