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Topic: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Replies: 38   Last Post: Apr 13, 2013 11:57 PM

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 Hetware Posts: 148 Registered: 4/13/13
Re: Misner, Thorne and Wheeler, Exercise 8.5 (c)
Posted: Apr 9, 2013 12:23 AM

On 4/8/2013 10:03 PM, Lord Androcles, Zeroth Earl of Medway wrote:
> "Hetware" wrote in message
> news:CeSdneS1Wf0B-_7MnZ2dnUVZ_vGdnZ2d@megapath.net...
>
> On 4/7/2013 11:34 PM, Lord Androcles, Zeroth Earl of Medway wrote:

>> "Hetware" wrote in message
>> news:AuGdnVu7TbVslv_MnZ2dnUVZ_g6dnZ2d@megapath.net...
>>
>> This is the geodesic equation under discussion:
>>
>> d^2(r)/dt^2 = r(dp/dt)^2
>>
>> d^2(p)/dt^2 = -(2/r)(dp/dt)(dr/dt).
>>
>> r is radius in polar coordinates, p is the angle, and t is a path
>> parameter.
>>
>> The authors ask me to "[S]olve the geodesic equation for r(t) and p(t),
>> and show that the solution is a uniformly parametrized straight
>> line(x===r cos(p) = at+p for some a and b; y===r sin(p) = jt+k for some
>> j and k).
>>
>> I tried the following:
>>
>> (d^2(p)/dt^2)/(dp/dt) = -(2/r)(dr/dt)
>>
>> f=dp/dt
>>
>> (df/dt)/f = -(2/r)(dr/dt)
>>
>> -1/2 ln(f) + k = ln(r)
>>
>> a(f^(1/2)) = r
>>
>> a(dp/dt)^(1/2) = r
>>
>> And substitute for r in:
>>
>> d^2(r)/dt^2 = r(dp/dt)^2
>>
>> to get
>>
>> d^2(r)/dt^2 = a(dp/dt)^(3/2)
>>
>> But there I'm stuck.
>>
>> How should the problem be handled?
>> =============================================
>> What you have is a second order differential equation.
>> Unlike the solution to the general polynomial equation,
>> ax +bx^2 + cx^3 + ... + kx^n = 0, where you seek a value
>> for x given values for a,b,c etc., the solution for a
>> differential equation is a FUNCTION.
>> In other words you cannot obtain a numerical or algebraic
>> value (you don't have enough information and that is not
>> the idea anyway) but you can find functions r(t) and p(t) .
>> The authors have already told you the solution is a straight
>> line, which is of course a function.
>> http://search.snap.do/?q=solving+differential+equations&category=Web
>> HTH, because we don't do homework for you.

>
> That was the kick in the head that I needed.
>
> I was getting close. I remembered what Wheeler told me: "Anytime you
> are struggling to understand or explain something, draw a picture."
>
> http://www.speakeasy.org/~hattons/chapter08.nb
>
> I worked the problem in pencil today while away from my computer. Just
> use p=arctan(t/r_0) where r_0 is a radial vector perpendicular to the
> "curve". r=(r_0^2+t^2)=r_0/cos(p). The rest is just a matter of
> plugging these into the geodesic equation.
> ==========================================
> Good. Note that what you are doing is strictly NOT anything
> to do with relativity or gravitation or even physics, it is just
> math.
> Pity Einstein didn't work the examples or he'd have made fewer
> blunders like this one:
> http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF
>
> -- This message is brought to you from the keyboard of
> Lord Androcles, Zeroth Earl of Medway.
> When the fools chicken farmer Wilson and Van de faggot present an
> argument I cannot laugh at I'll retire from usenet.
>
>
>
>
>
>

I don't have time to follow up on this, but this is from a guy who
doesn't spend his time trying to prove how smart he is. He just shows
you what he things in the clearest manner available to him. The mark of
a true genius.

http://people.oregonstate.edu/~drayt/Courses/MTH437/2007/hw/geodesic.pdf