sci.physics.relativity

On Mon, 24 Sep 2007 05:53:18 -0700, George Dishman
wrote:

>On 23 Sep, 22:09, HW@....(Henri Wilson) wrote:
>> On Sun, 23 Sep 2007 12:33:14 +0100, "George Dishman" wrote:
>> >"Henri Wilson" wrote in message>news:kigbf317ahrqpdeknk5lrk0rsv0t7oc9f3@ ...
>> >> On Fri, 21 Sep 2007 01:08:23 +0100, "George Dishman"
>> >> wrote:
>> >>>"Henri Wilson" wrote in message
>
>> to run away>
>
>>>> On Wed, 19 Sep 2007 21:36:10 +0100, "George Dishman"
>>>> wrote:
>>>>>"Henri Wilson" wrote in message
>>>>>news:qj2ue3dda74t2eg93ev1c66sdv5hfa5144@ ...
>>>>>> On Mon, 17 Sep 2007 19:23:54 +0100, "George Dishman" wrote:
>
>> >>>>>>>Multiple images start at what I called the critical
>> >>>>>>>distance given by d = c^2/a where a is the peak radial
>> >>>>>>>acceleration. Variation of mags means a ratio of
>> >>>>>>>4:1 in luminosity which means the speed unification
>> >>>>>>>distance is about 75% of the critical distance. Other
>> >>>>>>>values are
>> >>>>>>>
>> >>>>>>> Percentage of
>> >>>>>>> Mag 1:n critical
>> >>>>>>> 1 %
>> >>>>>>> %
>> >>>>>>> 2 %
>> >>>>>>> 3 %
>> >>>>>>> 4 %
>> >>>>>>> 5 %
>> >>>>>>> 6 %
>> >>>>>>> 7 %
>> >>>>>>> 8 %
>> >>>>>>> 9 %
>> >>>>>>>
>> >>>>>>>Isn't it magical how we see variations of up to 9 mags
>> >>>>>>>in some stars yet NEVER see multiple images.
>> >>>>>>>
>> >>>>>> It is true some stars are reported to vary by 7-9 mags. This cannot be
>> >>>>>> explained solely by c+v effects.
>> >>>>>
>> >>>>>Of course it can Henry, that's the point. Suppose
>> >>>>>we take an arbitrary figure of 10 light years
>> >>>>>for the speed equalisation distance in the space
>> >>>>>surrounding some star. To get mag variation
>> >>>>>you need a critical distance of light years.
>> >>>>>To get 9 mag that needs to be light years,
>> >>>>>just a 25% increase in the peak orbital
>> >>>>>acceleration.
>> >>>>
>> >>>> What the hell are you talking about?
>> >>>
>> >>>Check the numbers, see for yourself. For a given
>> >>>value of peak acceleration, draw a graph of peak
>> >>>variation as a function of distance.
>>
>> >> What do you think my program does?
>>
>> >If you think it can tell you the above results,
>> >use it. Tell me what distance as a percentage
>> >of (c^2/a) you get a variation peak variation
>> >of 8 magnitudes (peak-to-peak will be about
>> > magnitudes more).
>>
>> >Tell me whether you would get multiple images if
>> >the distance were 1% greater.
>>
>> >If that is at an inclination of 45 degrees, would
>> >we see multiple images if it were 1 degree closer
>> >to edge-on?
>>
>> It never gets to that point.
>
>Stop ducking the issue Henry, you claimed my numbers
>were wrong so prove it, put up or shut up.

George, use a computer program ....and stop wasting your time with equations.

>> >>>> Your numbers are completely wrong...
>>
>> >Then use your program and post your alternatives.
>>
>> You can use it. ...
>
>Why, don't you know how? I think you are so hopeless
>that even this trivial arithmetic is beyond you, but
>you claimed my numbers were wrong so it is up to you
>to back that up, I'm not going to do it for you. Oh
>and if your program gives something other than those
>values, you will still need to prove it is your code
>that is right and not my numbers, IMO your code is
>highly suspect. Bear in mind the numbers are peak,
>not peak-to-peak which I think is what your program
>usually produces as the summary.

George, I'm way ahead of you on this...
You don't even understand the basics...

>George

Henri Wilson. ASTC,BSc,DSc(T)

/hewn/