Re: Relativists cannot count to three.
From: Todd (nope_at_nospam.com)
Date: 12/27/04
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Date: Mon, 27 Dec 2004 22:12:03 GMT
"Androcles" <dummy@dummy.net> wrote in message
news:fC_zd.63636$ef5.48925@fe1.news.blueyonder.co.uk...
>
> "Todd" <nope@nospam.com> wrote in message
> news:3YZzd.28519$k25.6716@attbi_s53...
>>
>> "Androcles" <dummy@dummy.net> wrote in message
>> news:OcVzd.60125$ef5.15664@fe1.news.blueyonder.co.uk...
>>> If I stand before a vertical surface and throw a ball 'o' with speed
>>> 'u', catching it as it returns, I need a minumum of two coordinates
>>> to describe the events of emission, reflection and reception.
>>> Ao----------(+u - >)------------* emission
>>> *--------------------------------oB reflection
>>> Ao----------( -u <- )------------* reception
>>>
>>> In agreement with experience it is clear that the time for the ball
>>> to reach the wall and back again is given by 2AB/(t'A-tA) = u,
>>> the velocity of the ball.
>>>
>>>
>>> If I do this whilst riding a moving platform travelling at velocity v,
>>> the wall also moving, I need a minimum of three coordinates
>>> for a stationary observer to evaluate the events.
>>>
>>> Ao----------(v+u)-----------* emission
>>> *---------------------------oB reflection
>>> Co---------(v-u)------------* reception
>>>
>>> In the frame of the observer, the ball travels from A to B'
>>> at v+u and returns from B' to C at v-u.
>>>
>>> Einstein only uses two cordinates, A = C = (0,0,0) and B = (x',0,0).
>>>
>>> Proof:
>>> "From the origin of system k let a ray be emitted at the time tau0
>>> along the X-axis to x', and at the time tau1 be reflected thence
>>> to the origin of the co-ordinates, arriving there at the time tau2"
>>> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
>>> Reference :
>>> http://www.fourmilab.ch/etexts/einstein/specrel/www/
>>>
>>> Hence Einstein and his relativist disciples cannot count to
>>> three, QED.
>>>
>>> "the velocity of light in our theory plays the part, physically,
>>> of an infinitely great velocity." -Einstein.
>>>
>>> Androcles.
>>>
>>
>>
>> To understand Einstein's formula you have to realize and keep in mind
>> that Einstein is defining tau as a function of (x',y,z,t) rather than
>> (x,y,z,t). x' is defined by the equation
>> x'=x-vt
>
> Of course. I works for a ball bouncing, too. There is no magic to it.
>
>
>> tau is the time of an event as measured in Einstein's 'moving' frame K.
>
> Yep.
>
>
>> (x,y,z,t) are the coordinates of the event as measured in the
>> 'stationary' frame k, and v is the speed of frame k relative to frame K.
>
> Yep. Simple, isn't it?
>
>>
>> Einstein introduced x' merely for mathematical convenience.
>
> Sure. Writing x-vt every time get old.
>
>
>> Note that x' does not represent the x-coordinate of the event in frame k
>> (unless t happens to be zero) nor does it represent the x-coordinate of
>> the event as measured in the 'moving' frame K.
>
> 'Moving' frame K? What's that then? K is the 'stationary' frame.
> You are getting mixed up.
>
You're right! Sorry for mixing up the notation.
>> The physical interpretation of x' is that it represents the x-distance
>> between the location of the event and the origin of the 'moving' frame K
>> as measured in the 'stationary' frame k.
>
> You've got your frames crossed. Read the paper again.
> K is the stationary frame, k is the 'moving frame'.
>
Yes, as noted above.
>
>> (Failure to understand the meaning of x' is a primary stumbling block in
>> following Einstein's derivation.)
>
> Not to me.. It might be harder for someone that confuses K-frame
> with 'moving' frame though.
>
>>
>> With this in mind and letting A, B, and C denote the events of emission,
>> reflection, and reception, respectively, of the light signal, then we can
>> write Einstein's formula as
>>
>> (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B)
>
> That's right. Well done.
>
>
>
>> This equation merely expresses Einstein's clock synchronization in frame
>> K.
> Well, not really. That half at the beginning is nonsense. It certainly
> wouldn't be true if I threw a ball against the wind and i never caught up
> with the wall.
>
But Einstein was considering a light signal where the equation above does
indeed express that the clocks in the moving frame are
Einstein-synchronized.
>>
>> Now x'_A represents the x-distance (as measured in frame k) between the
>> point of emission of the light signal and origin of frame K.
>
> Yep, that's right.
>
>
> This is
>> clearly zero since the signal was emitted from the origin of K.
>
> No no, you are still getting your frames crossed over.
Yes, I switched the notations k and K compared to Einstein throughout my
post. Again, sorry for the inconvenience. If you like, I can repost it
with the notation fixed.
> Quote:
> From the origin of system k let a ray be emitted...
> Unquote. Reference
> http://www.fourmilab.ch/etexts/einstein/specrel/www/
>
> It is zero in the stationary frame K, though, so you are almost right.
>
>>
>> Likewise, x'_C represents the x-distance (as measured in k) between the
>> point of reception of the light signal and the origin of frame K. Again
>> this is clearly zero since the signal was received at the origin of K.
>
> I can't really answer that since you have your frames crossed over.
> I did provide a diagram, though.
> Ao----------(v+u)-----------* emission
> *---------------------------oB reflection
> <==vt==>Co---------(v-u)------------* reception
>
> As you can see, C is displaced from A by a distance vt.
>
OK, but in looking at Einstein's equation we need to keep in mind that he
uses the symbol t to stand for the time in the stationary frame that the
light signal was emitted. So, to be more precise, it might be better to say
that C is displaced from A according to the stationary frame by the distance
x_C - x_A = v*(t_C - t_A).
> Putting that into your equation
> (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B)
> we have
> (1/2)[tau(x'_A,0,0,t_A)+tau(x'_A-vt,0,0, t_C)] = tau(x'_B,0,0,t_B)
>
No, by definition we have
x'_C = x_C - v*t_C and
x'_A = x_A - v*t_A
Using these relations in x_C - x_A = v*(t_C - t_A) merely yields
x'_A - x'_C = 0 or x'_A = x'_C.
But we already know this since, in fact, x'_A and x'_B are both zero. But
you mangled my equation by substituting x'_C = x'_A-vt which is clearly
incorrect.
> but obviously Einstein doesn't have your equation, does he?
>
Well, he naturally doesn't have your incorrect modification of my equation.
>
>> Thus, x'_A = x'_C = 0.
> No no, that is why I use a ball instead of light and said that relativists
> can't count to three. x'_C =0-vt , x'_A = 0.
>
Even with the light replaced by a ball, we would still have x'_A=x'_C=0.
Remember, Einstein uses x' to denote the x-distance (as measured in the
stationary frame) between an event and the origin of the moving frame.
Since A and C for the ball both occur at the origin of the moving frame, we
must have x'_A=x'_C=0.
Todd
> Snip remainder, you've blundered.
>
> Androcles.
>
>
>
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