# Homework Help: The equivalent lagrangian and the derivative

1. Jan 19, 2010

hi everyone!
I have just posted a thread which was about the equivalent lagrangian.

and I think I have another problem with it too!

again in section 11.6 d'inverno, it is said that if you use equation 11.37 then you can achive

(∂L ̅)/(∂g_(,c)^ab )=Γ_ab^c-1/2 δ_a^c Γ_bd^d-1/2 δ_b^c Γ_ad^d

[I couldn't type it well.this is not L. this is L^- as you can see in the book.]

I couldn't achieve this and here is what I did.I combined 11.39 and 11.36 and I got:

2L ̅=-g_(,c)^ab Γ_ab^c+g_(,b)^ab Γ_ac^c
so
L ̅=1/2{-g_(,c)^ab Γ_ab^c+g_(,b)^ab Γ_ac^c}

(∂L ̅)/(∂g_(,c)^ab )=1/2(∂̅)/(∂g_(,c)^ab ){-g_(,c)^ab Γ_ab^c+g_(,b)^ab Γ_ac^c}

=1/2[-Γ_ab^c-g_(,c)^ab ∂/(∂g_(,c)^ab ) Γ_ab^c+δ_c^b Γ_ac^c+g_(,b)^ab ∂/(∂g_(,c)^ab ) Γ_ac^c]

and here is the problem:
what should I do with the term ∂/(∂g_(,c)^ab ) Γ_ab^c ?

once I tried to use equation 11.41 but it really didn't work!

2. Jan 19, 2010

### Altabeh

Ughhh! That is a very long calculation that only must be done through starting from the equation (11.38). You don't have any other alternative and your calculation, I think, is based on the wrong derivative $$\frac{\partial} {\partial g^{ab}_{,c}}$$ whereas (11.43) says that is $$\frac{\partial} {\partial \mbox{g}^{ab}_{,c}}$$.

AB

Last edited: Jan 19, 2010
3. Jan 19, 2010

Why working with (11.38)?
what I mean is that I don't know either (∂̅)/(∂g_(,c)^ab )of Γ_ab^c or (∂̅)/(∂g_(,c)^ab ) of g^ab !!
Why my way is wrong?
I'm just using the equations and by playing with them I can derive L bar.
can you tell me why it should only be done through starting from (11.38)?

4. Jan 19, 2010

### Altabeh

What are you taking derivative of g^ab or Γ_ab^c with respect to? The point is that I don't know what exactly you mean by (∂)/(∂g_(,c)^ab)! If it is the derivative wrt to g^{ab}_{,c}, then we have

(∂g^ab)/(∂g_(,c)^ab)= 0 and

http://img136.imageshack.us/img136/9596/cascyy.jpg [Broken]

But if it is the derivative wrt to $$\mbox{g}^{ab}_{,c}$$ where $$\mbox{g}^{ab}$$ being a contravariant density, so rest assured that both of the aforementioned derivatives vanish.

Please clarify for me what are you looking for.

Last edited by a moderator: May 4, 2017
5. Jan 20, 2010

Last edited by a moderator: May 4, 2017
6. Jan 20, 2010

### Altabeh

No! You didn't notice that here we are given a product of $$\mathfrak{g}^{ab}$$ and Christoffel symbols. So your treatment doesn't seem to be true as you first need to multiply $$\mathfrak{g}^{ab}$$ by each term in parenthesis and decompose the tensor density and expand Christoffel symbols to acquire terms including $$\mathfrak{g}^{ab}_{,c}$$.

Please give it a go again and if you were unable to get the result, I would start with the details of this long calculation.

Last edited by a moderator: May 4, 2017
7. Jan 20, 2010

### Altabeh

1- In an early post here, I explained how one can take the derivative of Christoffel symbols wrt the $$g^{ab}_{,c}$$. Here we will need the derivatives of $$\mathfrak{g}^{ab}_{,f}$$ wrt $$\mathfrak{g}^{ab}_{,c}$$ which can be computed in a similar way. Assume that the indices a and b of one of the tensor densities above differ from another, say, we want to calculate

$$\frac{\partial \mathfrak{g}^{ab}_{,f}}{\partial \mathfrak{g}^{hk}_{,c}}$$.

Since h and k are upper indices in the denominator of the differential fraction, so they would appear as covariant indices on the other side and the inverse is true for the lower index c. So the result of differentiation clearly is $$\delta^{a}_{h}\delta^{b}_{k}\delta^{f}_{c}$$. Now what about when one gets back to h=a and b=k? Here the situation is slightly different and must be treated acutely: if I were not smart enough to get over this thing, I would arrive at a seriously erroneous result of $$\delta^{a}_{a}\delta^{b}_{b}\delta^{f}_{c}=4\times4\delta^{f}_{c}$$ where the sum has been done over the dummy indices running from 0 to 3. (I've even seen many don't sum over the repeated indices and think they got something based upon the definition of Kronecker delta so they immediately turn to putting $$\delta^{a}_{a}=1$$.) The blind spot of this illogic calculation is that you are not allowed to substitue Kronecker deltas for one-to-one dummy indices during differentiation. Another important remark to be made here is that if the metric is supposed to be symmetric (as is in the treatment of GR by looking at the symmetry of its affine connections, or the so-called Christoffel symbols), then an exchange of a and b of one of the $$g$$'s does make no difference, which means

$$\frac{\partial \mathfrak{g}^{ab}_{,f}}{\partial \mathfrak{g}^{ab}_{,c}} = \delta^{c}_{f}=\frac{\partial \mathfrak{g}^{ab}_{,f}}{\partial \mathfrak{g}^{ba}_{,c}}$$.

This readily states that the symmetry of $$\mathfrak{g}^{ab}_{,c}$$ in a and b counts as one in any case after differentiation.

2- The aim of calculation is to recover terms including $$\mathfrak{g}^{ab}_{,c}$$ in the equation (11.38) of D'inverno's book without leaving any term which would cancel 'em out. Taking the derivative of any other term identified as being different or indicative of other nature would make them disappear after all.

Still continues...

Last edited: Jan 21, 2010
8. Jan 21, 2010

### Altabeh

Now we want to proceed to derive the equation

$$\bar{\mathfrank{L}}^{c}_{ab}\equiv\frac{\partial \bar{\mathfrank{L}}}{\partial \mathfrak{g}^{ab}_{,c}} =\Gamma^{c}_{ab}-1/2(\delta^{c}_b\Gamma^{d}_{ad}+\delta^{c}_a\Gamma^{d}_{bd})$$. (1)

To do so, first we drag some stuff from D'inverno's book right in here to pervent writing too much and spending more time on giving proofs that I bet it could be annoying and knocking off attaining your attention toward the rest of the subject. The following is the equation (11.41) in his book which actually describes that the contravariant density $$\mathfrak{g}^{ab}$$ is covariantly constant:

$$\mathfrak{g}^{ab}_{,c}=\Gamma^{d}_{cd}\mathfrak{g}^{ab}-\Gamma^{a}_{cd}\mathfrak{g}^{bd}-\Gamma^{b}_{cd}\mathfrak{g}^{ad}$$. (2)

Here putting b=c reduces this equation to the useful relation

$$\mathfrak{g}^{ab}_{,b}=-\Gamma^{a}_{bd}\mathfrak{g}^{bd}$$, (3)

that will be used in our later calculations.

In D'inverno's book the function $$\bar{\mathfrank{L}}$$ is defined as

$$\bar{\mathfrank{L}}=\mathfrak{g}^{ab}(-\Gamma^{c}_{ab}\Gamma^{d}_{cd}+\Gamma^{d}_{ac}\Gamma^{c}_{bd})$$. (4)

Well, I don't know how he does get to this equation with a wrong sign, but to reach (1) it is only possible to start with a different sign other than that given to us by D'inverno. (Perhaps it is just a typo or something though that wouldn't sound rational as when you drop a sign or type incorrectly, that wouldn't happen two times in one short line and shockingly in a single file). So I'm gently going to use this last equation with a (-1) multiplied by just the right-hand side of it.

Re-write (4) as

$$\bar{\mathfrank{L}}=(\mathfrak{g}^{ab}\Gamma^{c}_{ab})\Gamma^{d}_{cd}-\Gamma^{d}_{ac}(\Gamma^{c}_{bd}\mathfrak{g}^{ab}).$$

Now introduce into this equation the corresponding expressions of $$\mathfrak{g}^{ab}\Gamma^{c}_{ab}$$ and $$\Gamma^{c}_{bd}\mathfrak{g}^{ab}$$ from (3) and (2), respectively:

$$\bar{\mathfrank{L}}=-\mathfrak{g}^{kf}_{,f}\Gamma^{d}_{kd}-\Gamma^{d}_{kf}(\Gamma^{h}_{hd}\mathfrak{g}^{kf}-\Gamma^{k}_{hd}\mathfrak{g}^{hf}-\mathfrak{g}^{kf}_{,d})=$$
$$\bar{\mathfrank{L}}=-\mathfrak{g}^{kf}_{,f}\Gamma^{d}_{kd}+\Gamma^{d}_{kf}\mathfrak{g}^{kf}_{,d}+ other\ terms.$$ (5)

Here "other terms" do not contain any derivatives of the dual metric density so they won't contribute to $$\bar{\mathfrank{L}}^{c}_{ab}$$. Thus by making use of the first remark we talked about in an early post, we have from (5)

$\bar{\mathfrank{L}}^{c}_{ab}=-\delta^{k}_{a}\delta^{f}_{b}\delta^{c}_{f}\Gamma^{d}_{kd}+\delta^{k}_{a}\delta^{f}_{b}\delta^{c}_{d}\Gamma^{d}_{kf}=-\delta^{c}_{b}\Gamma^{d}_{ad}+\Gamma^{c}_{ab}$. (6)

As we anticipated winding up with (1), we would have been done through getting (6) after differentiation. But wait a minute. Let's be more patient and check if we are missing something. The missing point would be found in the second half of the remark 1 where one hits the problem of symmetry. If the equation (6) goes right through the test of symmetry it flunks because of the first term on the right side. We do know nothing as to whether $$\delta^{c}_{b}\Gamma^{d}_{ad}$$ is symmetric in a,b or not: we just know that the metric is symmetric so are the Christoffel symbols and so is the metric density and $$\bar{\mathfrank{L}}^{c}_{ab}$$. So what should we do? As the second term is symmetric in a,b, adding a term of the form $$\delta^{c}_{a}\Gamma^{d}_{bd}$$ to the right-hand side would be enough to attain the symmetry but this is illicit; nothing can pop out of nowhere in an equation. What about testing $$\frac{\partial \mathfrak{g}^{ab}_{,f}}{\partial \mathfrak{g}^{ab}_{,c}} = \delta^{c}_{f}=\frac{\partial \mathfrak{g}^{ab}_{,f}}{\partial \mathfrak{g}^{ba}_{,c}}$$? Let's do it:

$$\bar{\mathfrank{L}}^{c}_{ba}=-\delta^{k}_{b}\delta^{f}_{a}\delta^{c}_{f}\Gamma^{d}_{kd}+\delta^{k}_{b}\delta^{f}_{a}\delta^{c}_{d}\Gamma^{d}_{kf}=-\delta^{c}_{a}\Gamma^{d}_{bd}+\Gamma^{c}_{ba}$$.

Adding the sides of this equation with those of (6) correspondingly yields

$$\bar{\mathfrank{L}}^{c}_{ab}=-1/2(\delta^{c}_{a}\Gamma^{d}_{bd}+\delta^{c}_{b}\Gamma^{d}_{ad})+\Gamma^{c}_{ab}$$

which is (1) and of course has symmetry with respect to indices a and b. Q.E.D.

AB

Last edited: Jan 22, 2010
9. Jan 22, 2010

Hi my dear friend!
thanks for being so patient and helping me.

I have a serious problem
I wrote numbers beside them

1)how did you change with
and what is the minus sign for?!

2) why do you rule out this term?! I do not understand it.

10. Jan 22, 2010

hi my friend!
I have a quaestion
what do mean by 'covariantly constant'?

and thanks for the rest
I understood all the rest!

11. Jan 22, 2010

### Altabeh

Hi

Please be a little bit more stringent on the use of metric tensor $$g_{ab}$$ and metric density $$\mathfrak{g}_{ab}$$. All of g's in the my picture are of the nature of metric tensor, not metric density as your pictures show. There I have made use of the fact that the derivative of Kronecker delta vanishes:

$$g^{cf}g_{af,b} = (g^{cf}g_{af})_{,b} - g^{cf}_{,b}g_{af} = (\delta^{c}_{f})_{,b} - g^{cf}_{,b}g_{af} = -g^{cf}_{,b}g_{af}.$$ (1)

Because it cannot be written in pure terms of $$g^{cf}_{,f}$$ in the sense that if I took the path of (1) for the last term, then it would not appear to have a vanishing term, basically including a delta as in (1). So the whole term $$g^{cf}g_{ab,f}$$ won't contribute to the derivative. (Just try to write down the calculation (1) for $$g^{cf}g_{ab,f}$$ and you will observe that an extra term appears so it cancels out $$g^{cf}_{,f}g_{ab}$$.)

AB

Last edited: Jan 22, 2010
12. Jan 22, 2010

### Altabeh

If a tensor\vector\scalar is covariantly constant, then it vanishes under the operation of covariant differentiation. E.g. $$A^{ab}$$ being covariantly constant, so $$A^{ab}_{;c}=0.$$

AB

13. Jan 23, 2010

thanks alot...

14. Aug 2, 2010

### TerryW

I am working on D'Inverno too and Alatbeh has been helping me with my problems too. But I think he has got this wrong:
All D'Inverno does to get

$$\bar{\mathfrank{L}}$$ is to replace

$$+\mathfrak{g}^{ab}(\Gamma^{c}_{ab}\Gamma^{d}_{cd}-\Gamma^{d}_{ac}\Gamma^{c}_{bd})$$

with

$$-\bar{\mathfrank{L}}$$

so his sign is OK.

You might also be interested in my alternative solution to this problem:

$$\mathfrak{g}^{ab}_{,c}=\Gamma^{d}_{cd}\mathfrak{g}^{ab}-\Gamma^{a}_{cd}\mathfrak{g}^{bd}-\Gamma^{b}_{cd}\mathfrak{g}^{ad}$$.

Now add in some delta functions with appropriate changes to indices elsewhere:

$$\mathfrak{g}^{ab}_{,c}=\Gamma^{d}_{cd}\mathfrak{g}^{ab}-\Gamma^{a}_{cd}\delta^{d}_a\mathfrak{g}^{ba}-\Gamma^{b}_{cd}\delta^{d}_b\mathfrak{g}^{ab}$$.

Giving:

$$\frac{\partial \mathfrak{g}^{ab}_{,c}}{\partial \mathfrak{g}^{ab}}=\Gamma^{d}_{cd}-\delta^{d}_a\Gamma^{a}_{cd}-\delta^{d}_b\Gamma^{b}_{cd} = \Gamma^{d}_{cd}-\Gamma^{d}_{cd}-\Gamma^{d}_{cd} = -\Gamma^{d}_{cd}$$.

We can then use this result in:

$$\frac{\partial\bar\mathfrank{L}}{\partial\mathfrak{g}^{ab}}=\frac{\partial\bar\mathfrank{L}}{\partial\mathfrak{g}^{ab}_c}\frac{\partial\mathfrak{g}^{ab}_c}{\partial\mathfrak{g}^{ab}}$$

to give:

$$-\Gamma^{d}_{cd} \frac{\partial\bar\mathfrank{L}}{\partial\mathfrak{g}^{ab}_c} = \Gamma^{d}_{ac}\Gamma^{c}_{bd}-\Gamma^{d}_{cd}\Gamma^{c}_{ab}$$ (1)

Now we do a bit of manipulation on $$\Gamma^{c}_{bd}$$

$$\Gamma^{c}_{bd} =\delta^{c}_a \Gamma^{a}_{bd} = \delta^{c}_d\delta^{d}_a\Gamma^{a}_{bd} = \delta^{c}_d\Gamma^{d}_{bd}$$ (2)

also

$$\Gamma^{c}_{bd} =\delta^{c}_a \Gamma^{a}_{bd} = \delta^{c}_d\delta^{d}_a\Gamma^{a}_{bd} = \delta^{c}_d\Gamma^{d}_{bd} = \delta^{c}_d\delta^{c}_b\Gamma^{d}_{dc}$$ (3)

Now we rewrite (1) as

$$-\Gamma^{d}_{cd} \frac{\partial\bar\mathfrank{L}}{\mathfrak{g}^{ab}_c} = 1/2\Gamma^{d}_{ac}\Gamma^{c}_{bd} + 1/2\Gamma^{d}_{ac}\Gamma^{c}_{bd}-\Gamma^{d}_{cd}\Gamma^{c}_{ab}$$

Then we use our two results (2),(3) from above get:

$$-\Gamma^{d}_{cd} \frac{\partial\bar\mathfrank{L}}{\mathfrak{g}^{ab}_c} = 1/2\delta^{c}_a\Gamma^{d}_{dc}\Gamma^{d}_{bd} + 1/2\delta^{c}_b\Gamma^{b}_{bd}\Gamma^{d}_{ad}-\Gamma^{d}_{cd}\Gamma^{c}_{ab}$$

then divide throughout by $$-\Gamma^{d}_{cd}$$

To get the desired result.

Last edited: Aug 2, 2010