Re: How do you Interpret the Lorentz transform
- From: The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 30 Oct 2005 05:00:06 GMT
In sci.physics.relativity, King Coffee
<king.coffee@xxxxxxx>
wrote
on Sun, 30 Oct 2005 03:56:06 GMT
<qvX8f.2820$zb5.2429@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
> Hi,
>
> I think you are saying:
>
> O see the first pulse at O(x, t) = O(x0, t0) = O(0, 0)
> then A see this same pulse at A(x', t') = A(0, 0)
>
> The second pulse occurs at O(t1, 0)
Uh, no. O(0, t1). The light in O's coordinate space is not moving,
relative to O. Then again, it depends on the problem setup; are
you assuming two blinking endpoints of a rod, perhaps?
> (you said A is fix, but you use the O observation coordinate as fix)
> So, A see it at A(x'_1a, t'_1a) = A(-v*t_1*g, t_1)
More or less. If A had a series of detectors in his coordinate
space (and properly compensates for the distance between said
detectors and himself), he might see the event at the predicted
point.
In any event, A is moving relative to O, and A's entire (x,y,z)
coordinate system is moving along with A. (Actually, it's
more distorted than moving, if one looks at it from the (x,t)
or (x',t') viewpoint.)
>
> --------------------------------------------------------------
> But I'm using the observation of a pulse at
>
> O(x, t) = O(-x_1, 0) and O(x, t) = (x_1, 0)
> Observer O sees a pulse equal distance apart on the x axis at t = 0.
>
> These points map to observation in A at
> A(x', t') = A(-g*x_1, g*x_1*v/c^2) and A(g*x_1, -g*x_1*v/c^2)
They might, if A is capable of making that observation, by
using a network of photoscanners spread out over an area
and moving with him.
>
> sorry, I noting below I forget to put the gamma term on the time transform.
>
> My concern is:
>>> One of the points have a negative time. That would imply a future
>>> interval.
>>> That makes no sense, because the stationary observer k, already received
>>> the
>>> photon information at the origin and k' is moving away of the origin, so
>>> he
>>> should have already received the information too.
>
> Em I using the Transforms correctly.
The transforms appear to be correct, though the interpretation may
need some work in light of A being fixed at his origin. However,
that's mostly a point of clarifying how A observes the event.
>
> Thanks for replying,
> King
>
> "The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
> news:8r9d33-lmc.ln1@xxxxxxxxxxxxxxxxxxxxxxxxxx
>> In sci.physics.relativity, King Coffee
>> <king.coffee@xxxxxxx>
>> wrote
>> on Sat, 29 Oct 2005 21:52:33 GMT
>> <BaS8f.1264$zb5.177@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
>>> Hello,
>>>
>>> I want to determine -- what would observer k observe in the k' reference
>>> frame moving with velocity v, parallel to the x axis and the two
>>> coordinate
>>> system origins coincide at t = 0, a strait forward transform.
>>>
>>> I assumed that lighting strikes were observed at -x and x simultaneously
>>> in
>>> the k coordinate system, at t = 0.
>>>
>>> So, I have the two observation points (-x, 0) and (x, 0).
>>>
>>> let: A = 1 / sqrt( 1- (v*v)/(c*c)), per Lorentz definition.
>>>
>>> That maps to the coordinate point pairs in the k' system as ( -Ax,
>>> x*v/(c*c) ) and ( Ax, -x*v/c*c) ).
>>>
>>> One of the points have a negative time. That would imply a future
>>> interval.
>>> That makes no sense, because the stationary observer k, already received
>>> the
>>> photon information at the origin and k' is moving away of the origin, so
>>> he
>>> should have already received the information too.
>>>
>>> However, if I assuming -- at t =0, the lighting just struck at -x and x,
>>> the
>>> math will probably work out. But than observer k would not have yet
>>> observe
>>> the events at t = 0.
>>>
>>> Do coordinate point pair ( x, t) corresponded to the observation of an
>>> event
>>> in the k system? I thought propagation delay time was implicit to the
>>> Lorentz Transforms.
>>>
>>> Can you tell me how to use (interpret) the Lorentz Transforms.
>>>
>>> King
>>>
>>
>> At the risk of A****c**s calling me something insulting, allow me. :-)
>> This is more or less the standard solution of how to derive
>> the relationships
>>
>> l/l_0 = sqrt(1-v/c) / sqrt(1+v/c)
>> f/f_0 = sqrt(1+v/c) / sqrt(1-v/c)
>>
>> from the Lorentz
>>
>> x' = (x-vt)/sqrt(1-v^2/c^2)
>> t' = (t-vx/c^2)/sqrt(1-v^2/c^2)
>>
>> So let O have the coordinate system (x,t), and A (x',t'), with
>> the origins coincident at the instant x = x' = t = t' = 0.
>> O thereafter emits two instantaneous lightpulses (or, if one
>> prefers, the crests of two waves), one at time t = 0, one at time
>> t = t_1 = 1/f_0, as measured (or designed) by O, with t_1 > 0.
>>
>> How does A see these pulses? The first pulse is easy; t'_0 = 0.
>> With the second pulse, one gets
>>
>> x'_1a = (0 - v * t_1) * g
>> t'_1a = (t_1 - v * 0/c^2) * g
>>
>> where I've written g = 1/sqrt(1-v^2/c^2) for simplicity.
>>
>> The problem is that A is locked to his origin, so he does not see
>> the pulse as occurring at t'_1a; it occurs some time later. How
>> much later? Well, lightspeed is c in all reference frames in SR,
>> so one has to work out abs(x'_1a) / c, which is simple since t_1
>> is positive and therefore x'_1a is negative. The time A sees the
>> second pulse is therefore
>>
>> t'_1 = t'_1a + abs(x'_1a) / c
>> = (t_1 - v * 0/c^2) * g + (v * t_1 / c) * g
>> = (t_1 + v * t_1 / c) / sqrt(1-v^2/c^2)
>> = (1 + v/c) * (t_1) / sqrt(1-v^2/c^2)
>> = (1 + v/c) * (t_1) / (sqrt(1+v/c) * sqrt(1-v/c))
>> = sqrt(1 + v/c) * t_1 / sqrt(1 - v/c)
>>
>> (we are assuming 0 < v < c, so that 1 - v/c is positive and
>> therefore this is quite legitimate).
>>
>> Since f = 1/t, f/f_0 = sqrt(1 - v/c) / sqrt(1 + v/c).
>> Since l = ct, l/l_0 = sqrt(1 + v/c) / sqrt(1 - v/c).
>>
>> Had we specified a negative t_2 (i.e., O emits two pulses at time
>> t_2 and 0, instead of time 0 and t_1), t'_2 would have
>> been
>>
>> x'_2a = (0 - v * t_2) * g
>> t'_2a = (t_2 - v * 0/c^2) * g
>>
>> and since x'_2a is *positive*, one gets
>>
>> t'_2 = t'_2a + abs(x'_2a) / c
>> = (t_2 - v * 0/c^2) * g - (v * t_2 / c) * g
>> = (t_2 - v * t_2 / c) / sqrt(1-v^2/c^2)
>> = (1 - v/c) * (t_2) / sqrt(1-v^2/c^2)
>> = (1 - v/c) * (t_2) / (sqrt(1+v/c) * sqrt(1-v/c))
>> = sqrt(1 - v/c) * t_2 / sqrt(1 + v/c)
>>
>> There is a very pleasing aesthetic duality to this expression, IMO. :-)
>>
>> As an exercise, assume O sends out a light pulse with equation
>> (x^2 - c^2t^2) from his origin at time zero; show that
>> A sees (x'^2 - c^2t'^2) as well. This shows that one can derive
>> lightspeed constancy from the Lorentz. For an example of
>> deriving the Lorentz from lightspeed constancy, I will refer you
>> to a translation of Einstein's original paper on the topic:
>>
>> http://www.fourmilab.ch/etexts/einstein/specrel/www/ .
>>
>> HTH
>>
>> --
>> #191, ewill3@xxxxxxxxxxxxx
>> It's still legal to go .sigless.
>
>
--
#191, ewill3@xxxxxxxxxxxxx
It's still legal to go .sigless.
.
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