Experiment to test mutual time dilation
in [Relativity]
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From: kenseto on 12 Aug 2010 09:06 On Aug 11, 10:30 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) wrote: > kenseto <kens...(a)erinet.com> writes: > >On Aug 10, 12:09 pm, moro...(a)world.std.spaamtrap.com (Michael Moroney) > >wrote: > >> kenseto <kens...(a)erinet.com> writes: > >> >Why is B1 and B2 moving past A...why not A moving past B1 and B2 ??? > > >> Geez, you've been kooking in here for 15 years now but still don't > >> understand the basic fundamentals of relativity? All inertial frames > >> in SR have the same math, so of course an observer in A's frame can > >> do measurements with B1 and B2 moving past A by assuming A is stationary. > >No idiot he can't do any measurements.... he predicts. > > What do you mean "he can't do any measurements"? The gedanken is very > simple, and it explicitly describes the measurements. (Of course he can > predict as well) In real life no such measurement is possible....only predictions and you runts of the SRians keep on perpetuating the myth that predictions are measurements to give your theory more credence. > > >> (and yes, an observer in B will see the A clocks as moving, but that > >> was already discussed as the first gedanken, so it's not relevant here) > >It is relevant....B will use the same SR equation to predict that A > >runs slow and thus the bogus concept of nutual time dilation. > > But it's the first half of the gedanken. B observer sees A clocks moving > is first part. A observer seeing B clocks moving is second part. Here's the flaws in you logic: 1. B sees A clock is moving and thus A clock is running slow. 2. A sees B clock is moving and thus B clock is running slow. These conclusions include the bogus assumption that every observer's clock is running faster than every observed clock. > > Of course it makes perfect sense A and B measure the same thing if you > consider this: Clocks A1, A2 B1 B2 are identical, and observers A and B > are also identical. For anything else you can think of, let anything in > the two frames be identical. Therefore the two frames are identical so of > course they will both predict and measure the same thing. This makes no sense at all. It is based on the bogus assumption that every observer assumes that his clock is running at a faster rate than every observed clock. No observer can claim that all clocks moving wrt him are running slow. Therefore he must include all possibilities: that an observed clock can run slow by a factor of 1/gamma or run fast by a factor of gamma compared to the observer's clock. Ken Seto Ken Seto If not, there > would be a difference between them but I just said make everything in the > two frames be identical. > > >> You got to keep the A and B frames separate. This is where you always > >> get mixed up. > >I didn't mix up anything. A and B predicts each other's clock runs > >slow. > > And measure them as running slow. > > >There is no logic....A and B uses the same equation to predict each > >other's clock runs slow. > > And measure them as running slow. > > >> >This is wrong....B is the observer and B1 and B2 are in sych in the B > >> >frame. > > >> But not the A frame. And that is the point. > >We are not talking about the A frame....we are talking about what the > >B frame predicts. > > B observer sees B clocks as synchronized - but not the A clocks. > A observer sees A clocks as synchronized - but not the B clocks.
From: Sam Wormley on 12 Aug 2010 09:46 On 8/12/10 8:06 AM, kenseto wrote: > In real life no such measurement is possible....only predictions and > you runts of the SRians keep on perpetuating the myth that predictions > are measurements to give your theory more credence. Special relativity would not have survived for this more than a century now, had it not been confirmed by observation and experiment. Physics FAQ: What is the experimental basis of special relativity? http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html One very striking example of special relativity is the observed kinematics of accelerating particles in particle accelerators. See "The impact of Einstein's theory of special relativity on particle accelerators", http://iopscience.iop.org/0953-4075/38/9/020
From: Inertial on 12 Aug 2010 11:17 "Sam Wormley" wrote in message news:MK-dnb-BPPS0Zv7RnZ2dnUVZ_uqdnZ2d(a)mchsi.com... > Special relativity would not have survived for this more than a > century now, had it not been confirmed by observation and experiment. Why bother replying to him? You're just feeding the troll
From: Sam Wormley on 12 Aug 2010 12:48 On 8/12/10 10:17 AM, Inertial wrote: > "Sam Wormley" wrote in message > news:MK-dnb-BPPS0Zv7RnZ2dnUVZ_uqdnZ2d(a)mchsi.com... >> Special relativity would not have survived for this more than a >> century now, had it not been confirmed by observation and experiment. > > Why bother replying to him? You're just feeding the troll Perhaps someone will benefit that reads these newsgroups!
From: Tony M on 12 Aug 2010 23:01
On Aug 4, 1:48 pm, kenseto <kens...(a)erinet.com> wrote: > On Aug 4, 11:33 am, Tony M <marc...(a)gmail.com> wrote: > > > On Aug 4, 9:57 am, kenseto <kens...(a)erinet.com> wrote: > > > > So every observer > > > does not know if the observed clock is running slow or fast compared > > > to his clock. This means that he must include both possibilities when > > > predicting the rate of an observed clock as follows: > > > Observed clcok runs slow: > > > Delta(t')=gamma*Delta(t) > > > Observed clock runs fast: > > > Delta(t')=Delta(t)/gamma > > > Ken, why not Delta(t)/gamma <= Delta(t') <= gamma*Delta(t)? Think > > about it! > > No....Delta(t')=gamma*Delta(t) means that the passage of > Delta(t') on the t' clock is equal to the passage of > gamma*Delta(t) on the t clock....that means that the t' clock is > running slower than the t clock. > Similarly...Delta(t')=Delta(t)/gamma means that the passage of > Delta(t') on the t' clock is equal to the passage of > Delta(t)/gamma on the t clock....that means that the t' clock is > running faster than the t clock. Ken, I believe that's backwards, but let's try the following exercise. Let's use 3 observers, A, B and C, with their time intervals tA, tB and tC, all moving with the same relative velocity v, so we have the same gamma between each pair of observers. (Yes, that's possible.) From your theory the below observations should all be true: Observer A would measure: (1) tB=tA*gamma or (2) tB=tA/gamma (3) tC=tA*gamma or (4) tC=tA/gamma Observer B would measure: (5) tC=tB*gamma or (6) tC=tB/gamma Now, if (1) and (3) are true that means tB=tC, which contradicts both (5) and (6); observers A and B would disagree on their observations. If we take (1) and (4) as true then tB=tC*gamma^2, which contradicts both (5) and (6); observers A and B would disagree on their observations. The same goes for the other combinations. If we apply your theory to 3 observers none will agree with (all) each-other's observations. At least in SR they agree to disagree. |