Re: 1c+1c Closing Velocity...,answer to Henri Wilson

From: Paul B. Andersen (paul.b.andersen_at_deletethishia.no)
Date: 02/25/05 

Date: Fri, 25 Feb 2005 16:38:31 +0100

Henri Wilson wrote:
> On Wed, 23 Feb 2005 23:07:08 +0100, "Paul B. Andersen"
> <paul.b.andersen@deletethishia.no> wrote:
>>
>>Here are my calculations are again.
>>Please point out exactly what is wrong.
>>Please be specific.
>>"This is wrong", "absolute nonsense"
>>or "I hope Androcles reads this"
>>won't do.
>>
>>The kinetic energy for each degree of freedom is kT/2.
>>That means that along the line of sight, we have:
>>mv^2/2 = kT/2
>>v^2 = kT/m
>>where v^2 is the average square of the velocity
>>component along the line of sight.
>>If we assume that T = 6000K and that the atom is H,
>>we get v = 7 km/s.
>>The speed is distributed according to the Maxwellian
>>distribution:
>>http://my.unidata.ucar.edu/content/staff/blynds/maxw.html
>>Note that this is the speed in any direction, and we
>>are interested only in the radial speed component
>>because that is what matters for the photons that
>>will reach us. We have accounted for this above.
>>The distribution of the speed towards us will be
>>qualitatively the same, and the average is ca. 7 km/s.
>
>
> Oh jesus, this is funny!!! :)
>
> Paul, the average molecular speed is zero in any direction.

It obviously is.
If you look below, you will see that I never said otherwise.

> Have you ever seen a box of hot air flying off at 7km/sec without any rocket?
>
> You have invented the perfect Maxwell's demon. Perpetual motion... to hell with
> the second law of thermodynamics!!! Paul Andersen's hot air engine has it all.
> All the molecules have a speed in one direction.
>
> Hahahahahahaa!!!!!!

The distribution drawn below is wrong, but the average
speed IS zero

>>So the distribution of the radial velocity component
>>will be something like this:
>>
>>
>> * *
>> * * * *
>> * * * *
>> * * * *
>> * *
>> * *
>>---------|---|------|------|----|---------->
>> -10 -7 0 7 10
>>
>>(Pointing out the that the average is zero
>>is of course a ridiculous remark! :-) )
>
>
> Paul, you are going to be so embarrassed by this you will have to at least
> change your name. You are really making an absolute fool of yourself.

Yes, this ditribution is wrong.
I was wrong!
But as opposed to you, I always admit having done
a blunder as soon as I realize having done so.
And I will not have to change my name.
As opposed to you with your phoney name, I always
post under my real name.

The distribution of the velocity component in one
particular direction is like this. And this is drawn
to scale as far as it is possible in this medium:

T = 6000K
Vertical scale is relative populaton

    |
1.0| *
    | * *
0.8| * *
    | * *
0.6| * *
    | * *
0.4| * *
    | * *
    | * *
0.1| * *
    -----|----|----|----|----|----|----|---- v
        -15 -10 -5 0 5 10 15 km/s

>>Now the the light from the atoms with zero
>>velocity will reach as after a time d/c, while
>>the light from an atom with the speed v will
>>reach us after a time d/(c+v).
>>The difference is ca. dv/c^2.
>>If d = 1 LY, the light from an atom moving at 10km/s
>>towards us will reach us 0.6 hours before the light
>
>>from an atom moving at 10km/s away from us.
>
>>This time is proportional to the distance.
>>
>>So the light emitted from the star at a specific
>>instant will be received with a temporal distribution
>>something like this:
>>
>>
>> * *
>> * * * *
>> * * * *
>> * * * *
>> * *
>> * *
>>--------------------|--------------------->
>> |<------------------->|
>> 0.6 hour per light year
>>
>>
>>Of course the time will vary with the temperature of
>>the star, but not very much, it will be of this order
>>of magnitude for all stars.
>>
>>The rather obvious consequence of this is that according
>>to the ballistic theory, any periodicity shorter than
>>0.6 hours per light year distance should be be pretty well
>>masked out.
>
>
> Paul, the light will reach any point like this:
>
>> *
>> * *
>> * *
>> * *
>> * *
>> * *
>> * *
>>-----------------|------------------->
>> c+v-x (c+v) c+v+x

Yes, something like this.
And the consequence will be exactly as explained above,
the light emitted from the star at a specific instant will
be received with a temporal distribution like this:

    |
1.0| *
    | * *
0.8| * *
    | * *
0.6| * *
    | * *
0.4| * *
    | * *
    | * *
0.1| * *
    -----|----|----|----|----|----|----|---- v
              |< 0.6 hour per LY >|
                   |<0.4h/LY>|

The conclusion remains qualitatively the same:
According to the ballistic theory, any periodicity shorter than
0.4 hours per light year distance should be be pretty well
masked out.

Paul

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