Help with a derivation from S. Carroll's GR book

[sfoulkes2@gmail.com]

I am a recent PhD in Exp. Particle Physics and am trying to learn GR
by working my way through Sean Carroll's Spacetime and Geometry book.
On page 101 he states --

It is straightforward to show that the Christoffel connection
satisfies

Lambda^{mu}_{mu lambda} = 1/sqrt{|g|} \partial_{mu} sqrt{|g|}
(Equation 3.33)

I don't see how you can derive this from the definition of the
Christoffel connection
(Eq. 3.27 on page 99 with the upper and the first lower index the
same).

I don't need all the details, but a good hint would be nice.

s. foulkes

[lalbatros]

Could you write all the equations in clear.
I don't have this book, and I think there is a mistake there:

Lambda^{mu}_{mu lambda} = 1/sqrt{|g|} \partial_{mu} sqrt{|g|}

[Dirk Van de moortel]

<sfoulkes2@gmail.com> wrote in message news:1171414889.808029.59900@m58g2000cwm.googlegro ups.com...
>I am a recent PhD in Exp. Particle Physics and am trying to learn GR
> by working my way through Sean Carroll's Spacetime and Geometry book.
> On page 101 he states --
>
> It is straightforward to show that the Christoffel connection
> satisfies
>
> Lambda^{mu}_{mu lambda} = 1/sqrt{|g|} \partial_{mu} sqrt{|g|}
> (Equation 3.33)

I don't have the book, only the course notes, but that's probably
Gamma^{mu}_{mu lambda} = 1/sqrt{|g|} \partial_{lambda} sqrt{|g|}


>
> I don't see how you can derive this from the definition of the
> Christoffel connection
> (Eq. 3.27 on page 99 with the upper and the first lower index the
> same).
>
> I don't need all the details, but a good hint would be nice.
>
> s. foulkes
>

Writing it with (Gamma -> G, mu -> m, lambda -> n) as
G^m_mn = 1/sqrt{|g|} @_m} sqrt{|g|}

Starting from
G^s_mn = 1/2 g^sr ( @_m g_nr + @_n g_rm - @_r g_mn )
you get
G^m_mn = 1/2 g^mr ( @_m g_nr + @_n g_rm - @_r g_mn )
The first and third terms vanish because
g^mr ( @_m g_nr - @_r g_mn )
= g^mr @_m g_nr - g^mr @_r g_mn
= g^mr @_m g_nr - g^rm @_m g_rn (swap r and m in 2nd)
= g^mr @_m g_nr - g^mr @_m g_nr (symmetry of g_mn)
= 0
so
G^m_mn = 1/2 g^mr @_n g_rm
= 1/2 g^mr @g_rm / @x^n

This is the easy part.

>From my private notes (from WAY-back!)

You probably know the definition
|g| = g_mr C^mr (som over r only!)
where C^mr are the cofactors in the metric matrix
and that g^mr is defined such that
g^mr = C^mr / |g|
so
@|g|/@g_mr = C^mr = |g| g^mr
so
1/|g| @|g|/@g_mr = g^mr
so
@ln(|g|)/@g_mr = g^mr
so
@ln(|g|)/@x^n = @ln(|g|)/@g_mr @g_mr/@x^n
= g^mr @g_mr/@x^n

So the expressin for G^m_mn becomes
G^m_mn = 1/2 g^mr @g_rm / @x^n
= 1/2 @ln(|g|) / @x^n
= @ln(sqrt(|g|)) / @x^n
= 1/sqrt(|g|) @sqrt(|g|) / @x^n

Hope this helps (and that I didn't make too many typos)

Dirk Vdm

[Dirk Van de moortel]

"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com.telenet-ops.be> wrote in message
news:shCAh.8619$9y4.154689@phobos.telenet-ops.be...
>
> <sfoulkes2@gmail.com> wrote in message news:1171414889.808029.59900@m58g2000cwm.googlegro ups.com...
>>I am a recent PhD in Exp. Particle Physics and am trying to learn GR
>> by working my way through Sean Carroll's Spacetime and Geometry book.
>> On page 101 he states --
>>
>> It is straightforward to show that the Christoffel connection
>> satisfies
>>
>> Lambda^{mu}_{mu lambda} = 1/sqrt{|g|} \partial_{mu} sqrt{|g|}
>> (Equation 3.33)
>
> I don't have the book, only the course notes, but that's probably
> Gamma^{mu}_{mu lambda} = 1/sqrt{|g|} \partial_{lambda} sqrt{|g|}
>
>
>>
>> I don't see how you can derive this from the definition of the
>> Christoffel connection
>> (Eq. 3.27 on page 99 with the upper and the first lower index the
>> same).
>>
>> I don't need all the details, but a good hint would be nice.
>>
>> s. foulkes
>>
>
> Writing it with (Gamma -> G, mu -> m, lambda -> n) as
> G^m_mn = 1/sqrt{|g|} @_m} sqrt{|g|}
>
> Starting from
> G^s_mn = 1/2 g^sr ( @_m g_nr + @_n g_rm - @_r g_mn )
> you get
> G^m_mn = 1/2 g^mr ( @_m g_nr + @_n g_rm - @_r g_mn )
> The first and third terms vanish because
> g^mr ( @_m g_nr - @_r g_mn )
> = g^mr @_m g_nr - g^mr @_r g_mn
> = g^mr @_m g_nr - g^rm @_m g_rn (swap r and m in 2nd)
> = g^mr @_m g_nr - g^mr @_m g_nr (symmetry of g_mn)
> = 0
> so
> G^m_mn = 1/2 g^mr @_n g_rm
> = 1/2 g^mr @g_rm / @x^n
>
> This is the easy part.
>
>>From my private notes (from WAY-back!)
>
> You probably know the definition
> |g| = g_mr C^mr (som over r only!)
> where C^mr are the cofactors in the metric matrix
> and that g^mr is defined such that
> g^mr = C^mr / |g| so
> @|g|/@g_mr = C^mr = |g| g^mr so
> 1/|g| @|g|/@g_mr = g^mr
> so
> @ln(|g|)/@g_mr = g^mr

Actually, this should be
@ln(||g||)/@g_mr = g^mr
where ||g|| is notation for the absolute value of the determinant |g|.
I had been using the notation |g| for the determinant here, and I
forgot to take the absolute value when pulling |g| into the log.

So a better way, writing g for the determinant, would be:
The determinant is defined as
g = g_mr C^mr (sum over r only!)
where C^mr are the cofactors in the metric matrix
and g^mr is defined such that
g^mr = C^mr / g
so
@g/@g_mr = C^mr = g g^mr
so
1/g @g/@g_mr = g^mr
so
@ln(|g|)/@g_mr = g^mr

Notationwise, the remainder is okay...

> so
> @ln(|g|)/@x^n = @ln(|g|)/@g_mr @g_mr/@x^n
> = g^mr @g_mr/@x^n
>
> So the expressin for G^m_mn becomes
> G^m_mn = 1/2 g^mr @g_rm / @x^n
> = 1/2 @ln(|g|) / @x^n
> = @ln(sqrt(|g|)) / @x^n
> = 1/sqrt(|g|) @sqrt(|g|) / @x^n
>
> Hope this helps (and that I didn't make too many typos)
>
> Dirk Vdm
>