🚀 go-pugleaf

"Henri Wilson" <HW@....> wrote in message news:l4gvr2tiilng82746s3lq3l4nvhja93ksn@4ax.com...
> On Tue, 30 Jan 2007 05:15:05 GMT, "Dumbledore_" <Headmaster@hogwarts.physics_q>
> wrote:
> 
>>
>>"Henri Wilson" <HW@....> wrote in message news:ho0rr2pg16khti8c2ahfoubmhlob1u3h9f@4ax.com...
>>> On Mon, 22 Jan 2007 16:51:03 GMT, "MathMagician" <Math@Magic.co.uk> wrote:
>>> 
>>>>
>>>><rotchm@gmail.com> wrote in message news:1169478732.985047.109090@q2g2000cwa.googlegroups.com...
>>>>>> However, the basic reasoning exemplified in the above quoted text
>>>>>> implies instead of contraction of moving objects the exact contrary:
>>>>>> contraction of the rigid ether with respect to moving objects. In
>>>>>> LET (inasfar as it is not "a clever restatement" of SR), contraction
>>>>>> results from the motion through the ether [3], so contraction of the
>>>>>> ether with respect to moving objects is impossible.
>>>>> 
>>>>> Correct. Your quoted example is wrong. It should be the other way.
>>>>> 
>>>>>>Since he is moving at speed v, the length of the object is v*dt, where
>>>>>>dt is the time it takes him to traverse the length of the object - but
>>>>>>which  "dt" will he use?  Naturally if he bases his length estimate on
>>>>>> the  measure of the time interval recorded on a ground clock, he will
>>>>>>have dt = L/v, so he will judge the object to be v*(L/v) = L
>>>>> 
>>>>> That is totally wrong as you noted. The author of that claim has either
>>>>> screwed up or
>>>>> is....  <cough, cough>.
>>>>> 
>>>>> LET and SR perdict that a measurement of the lenght of the object will
>>>>> be v*dt where dt is the measured time (clock) and v the measured speed.
>>>>> The v*dt will yield L/gamma in both theories.
>>>>
>>>>Who cares?,  they are both ridiculous.
>>>>  http://www.androcles01.pwp.blueyonder.co.uk/mmx4dummies.htm
>>> 
>>> That is not a good demo.
>>
>>It's better than any you've made.
> 
> How about: http://www.users.bigpond.com/hewn/movingframe.exe
> 
> You are really demonstrating the light clock principle, not the MMX.
> 
> The MMX requires that you explain the difference in travel times between the
> parallel and transverse rays for different speeds through the 'aether'.
> 
>>> In the MMX, the observer is always at rest wrt the apparatus.
>>
>>Yeah, so ? 
>>
>>"To the "stationary" observer on the left the yellow path of the "tip of the ray" as Einstein calls it (which he/she never observes) "
>>
>>Can't you fucking read, dumbfuck jealous ozzie?
> 
> There is only one observer who is always stationary wrt the apparatus.

Bullshit, Albert Michelson had his buddy Edward Morley take a look.
You think he was jumping up and down?



> 
> Why don't you demonstrate a light clock?

I did, I even built one. It's still working. 
  http://www.androcles01.pwp.blueyonder.co.uk/lightclock.gif
LEDs and NAND gates are pretty cheap, the mirrors were 
more expensive.  I took the capacitor out, it wasn't needed.
 http://www.allaboutcircuits.com/vol_4/chpt_3/6.html


That reminds me, I've got to get the step ladder and dust it.
I put it close to the ceiling so as not to interrupt the beam,
it goes right around the walls with mirrors.
Oh well, that can wait until Spring.