Re: Relativists cannot count to three.
From: Todd (nope_at_nospam.com)
Date: 12/28/04
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Date: Tue, 28 Dec 2004 16:43:32 GMT
----- Original Message -----
From: "Androcles" <dummy@dummy.net>
Newsgroups: sci.physics.relativity,sci.physics
Sent: Tuesday, December 28, 2004 3:07 AM
Subject: Re: Relativists cannot count to three.
>
> "Todd" <nope@nospam.com> wrote in message
> news:TM%zd.574417$wV.242935@attbi_s54...
>> "Androcles" <dummy@dummy.net> wrote in message
>> news:fC_zd.63636$ef5.48925@fe1.news.blueyonder.co.uk...
>>
>>>
>>> "Todd" <nope@nospam.com> wrote in message
>>> news:3YZzd.28519$k25.6716@attbi_s53...
>>>>
[snip some of the earlier parts]
>>>> 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.
>
> I don't care what he was considering, he doesn't get to say the journey
> time out equals the journey time back when two different speeds are used.
> If I fly to New York and get a rowboat back to London I don't estimate
> 7 hours + 193 hours divided by 2 = 100 hours each way. The half is pulled
> out of a hat and no self respecting mathematician or physicist would fall
> for such a silly story. I'm certainly not going to. He can keep his
> idiotic method
> of sychronization.
>
That's not what Einstein does. I've already shown how the (c-v) and (c+v)
quantities occur in Einstein's expressions. These quantities do no
represent the speed of light relative to either frame. They simply
represent the rate at which the distance between the light signal and the
origin of the moving frames is changing as observed in the stationary frame.
You have to understand the meaning of the x-primes.
>>>>
>>>> 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).
>
> That's (0,0,0) - (0,0,0) = v * (anything at all)
> According to those that cannot count to three, C = A. Hence v = 0
> and trivial.
>
No. x_C does not equal x_A. But the corresponding primed quantities are
equal (each is equal to zero). Again, you must distinguish the meaning of
the primed and the unprimed x's.
>>> 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.
>
> That's right, I fully agree.
> The origin of the k-frame doesn't move from the origin
> of the K-frame while the tip of the ray is in flight, so either
> a) v = 0
> or
> b) the speed of light is infinite.
> or
> c) both of the above.
> Relativists cannot count to three, as you have now demonstrated.
>
The origin of the k-frame certainly does move relative to K while the ray is
in flight.
It seems to me that you are not understanding the meaning of the primed x's.
>
>>
>> But we already know this since, in fact, x'_A and x'_B are both zero.
Whoops - another unfortunate typo on my part. I meant x'_C rather than
x'_B. The value of x'_B is definitely NOT zero as it measures the distance
of separation between the origin of the moving frame and the point of
reflection of the light as measured in the stationary frame.
>
> Not wise. The idiot wants to differentiate his equation, so he says
> "Hence, if x' be chosen infinitesimally small"
>
> f(x+h) - f(x)
> f'(x) = ----------------------
> h
> doesn't work if h is already zero.
>
>
To say x' is infinitesimal is not the same as saying it is zero.
>> But you mangled my equation by substituting x'_C = x'_A-vt which is
>> clearly incorrect.
>
> Only if you cannot count to three.
>
>>
>>> but obviously Einstein doesn't have your equation, does he?
>>>
>>
>> Well, he naturally doesn't have your incorrect modification of my
>> equation.
>
> That's why I said relativists cannot count to three. The picture clearly
> shows
> a separation between A and C which you've said above is zero.
> Quote:
> x'_A - x'_C = 0 or x'_A = x'_C.
> Unquote.
Again, you are not understanding the meaning of the primed x's. The
relation x'_A = x'_C does not mean that there is no spatial separation
between the events A and C in the stationary frame. x'_A = x'_C does not
imply x_A = x_C.
>>
>>>
>>>> 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.
>
> Fine.
> tau = (t - vx/(ball speed)^2) / sqrt ( 1 - v^2 / (ball speed)^2)
Sorry, that went right past me.
> I don't like to be deliberately obtuse, but I'm not the crackpot, I can
> count to three.
>
>> 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.
>
> Really?
> We are trying to find tau in the moving frame as seem from the stationary
> frame, so the parameters of the function tau() are all coordinates in K.
Einstein clearly states that tau is a function of x' rather than x. You
have to understand the meaning of x'. x' is not the x-coordinate in K.
> We don't get to use moving frame parameters in the calculation, which is
> what you are doing.
> |__A ____________x'__| k-frame tau = 0
> |__C ____________x __| K-frame t = 0
>
>
>
> |__A____________x'__| k-frame tau = 1
> <-vt1->|__C ____________x __| K-frame t = 1
>
>
> |__A____________x'__| k-frame tau = 2
> <-----vt2------>|__C ____________x __| K-frame t = 2
>
> Placing A = C.
> |__A____________x'__| k-frame tau = 2
> |__C____________x __| K-frame t = 2
>
> v = 0.
>
Nobody is saying that A and C are spatially coincident in K. x_A does not
equal x_C even though x'_A = x'_C. It is important to understand the
meaning of x'=x-vt.
> You've also said x'_B = 0 as well, and called it a fact.
> Quote:
> "But we already know this since, in fact, x'_A and x'_B are both zero.
As I noted above, I meant to say x'_A and x'_C are both zero. x'_B is not
zero. Sorry for the confusion.
Todd
> Unquote:
> I gotta be honest with ya. You are one of those that cannot count to
> three.
> Androcles.
>
>
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