Deduce Geodesics equation from Euler equations

[fab13]

I am using from the following Euler equations :

$$\dfrac{\partial f}{\partial u^{i}}-\dfrac{\text{d}}{\text{d}s}\bigg(\dfrac{\partial f}{\partial u'^{i}}\bigg) =0$$

with function $f$ is equal to :

$$f=g_{ij}\dfrac{\text{d}u^{i}}{\text{d}s}\dfrac{\text{d}u^{j}}{\text{d}s}$$

and we have $$f=g_{ij} u'^{i}u'^{j}=1$$ (with $u'^{i}=\dfrac{\text{d}u^{i}}{\text{d}s}$)

My issue is this :

starting from the expression of function $f$ and Euler equations, I would like to get :

$$\dfrac{\text{d}}{\text{d}s}(g_{ij}u'^{j})-\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}=g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+(\partial_{k}g_{ij}-\dfrac{1}{2}\partial_{i}g_{jk})u'^{j}u'^{k}=0$$

but I can't obtain it.

Finally, I should get the general form of geodesics equation, i.e :

$$g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+\Gamma_{ijk}u'^{j}u'^{k}$$

For the moment, concerning this problem (which is about factor 1/2), IF I SUBTRACT INTENTIONALLY THE TERM $\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}$ to the expression $\dfrac{\text{d}}{\text{d}s}(g_{ij}u'^{j})$, so I can write :

$$\dfrac{\text{d}}{\text{d}s}(g_{ij}u'^{j})-\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}=$$

$$\dfrac{\partial g_{ij}}{\partial u^{k}}u'^{j}\dfrac{\text{d}u^{k}}{\text{d}s}+g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}-\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}=$$

$$\partial_{k} g_{ij} u'^{k}u'^{j}+g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}-\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}=$$

$$g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+(\partial_{k}g_{ij}-\dfrac{1}{2}\partial_{i}g_{jk})u'^{j}u'^{k}=0$$

from which I deduce :

$$g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+\Gamma_{ijk}u'^{j}u'^{k}$$

BUT I HAVE SUBTRACTED INTENTIONALLY THE TERM $\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}$ ,

Now I would like to know how this term( $\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}$ ) can be obtained from Euler equations , especially the factor $\dfrac{1}{2}$), I think it comes from the following term of Euler equations :

$$\dfrac{\partial f}{\partial u^{i}}$$

From my point of view, $$\dfrac{\partial f}{\partial u^{i}}=\partial_{i}g_{jk}u'^{j}u'^{k}$$

and not $$\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}$$.

but I don't know to make appear $\dfrac{1}{2}$ factor.

If someone could help me to solve this little issue, Regards

 

[Orodruin]

I do not understand the problem, this

I would like to get :

$$\dfrac{\text{d}}{\text{d}s}(g_{ij}u'^{j})-\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}=g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+(\partial_{k}g_{ij}-\dfrac{1}{2}\partial_{i}g_{jk})u'^{j}u'^{k}=0$$

is what you should get directly from applying the EL equations. Can you write out what you get when you write down the EL equations?

 

[fab13]

@Orodruin : For the moment, I try to deduce Geodesics equations :

$$g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+\Gamma_{ijk}u'^{j}u'^{k}$$

from the Euler equations $\dfrac{\partial f}{\partial u^{i}}-\dfrac{\text{d}}{\text{d}s}\bigg(\dfrac{\partial f}{\partial u'^{i}}\bigg) =0$

taking :

$$f=g_{ij}\dfrac{\text{d}u^{i}}{\text{d}s}\dfrac{\text{d}u^{j}}{\text{d}s}$$

with $s$ an affine parameter.

This demonstration is extracted from a book but it is not detailed : author starts from

$$\dfrac{\partial f}{\partial u^{i}}-\dfrac{\text{d}}{\text{d}s}\bigg(\dfrac{\partial f}{\partial u'^{i}}\bigg) =0\,\,\,\,\,\,\,\,\,\,\,\,(eq1)$$

and, just after this definition, writes :

$$\dfrac{\text{d}}{\text{d}s}(g_{ij}u'^{j})-\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}=g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}+(\partial_{k}g_{ij}-\dfrac{1}{2}\partial_{i}g_{jk})u'^{j}u'^{k}=0\,\,\,\,\,\,\,\,\,\,\,\,(eq2)$$

without putting the steps between $(eq1)$ and $(eq2)$.

So, I tried to find the different steps of this demo. First, I can write for the second term of Euler equations :

$$\dfrac{\text{d}}{\text{d}s}\bigg(\dfrac{\partial f}{\partial u'^{i}}\bigg)=\dfrac{\text{d}}{\text{d}s}(g_{ij}u'^{j})$$

$$=\dfrac{\partial g_{ij}}{\partial u^{k}}u'^{j}\dfrac{\text{d}u^{k}}{\text{d}s}+g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}$$

$$=\partial_{k} g_{ij} u'^{k}u'^{j}+g_{ij}\dfrac{\text{d}u'^{j}}{\text{d}s}$$

And for the first term of Euler equations :

$$\dfrac{\partial f}{\partial u^{i}}=\dfrac{\partial g_{jk}u'^{j}u'^{k}}{\partial u^{i}}=\partial_{i}g_{jk}u'^{j}u'^{k}\,\,\,\,\,\,\,\,\,\,\,\,(eq3)$$

But $eq(3)$ is not equal to the expected second term on left hand side of equation $(eq2)$ (the one of the book) :

$$\dfrac{1}{2}\partial_{i}g_{jk}u'^{j}u'^{k}$$

indeed, there is a factor $\dfrac{1}{2}$ which appears in $eq(2)$ that I can't get and I don't know how to find it.

If you could help me ...

Hoping you understand better.

 

[Orodruin]

I see what you are doing wrong. In the function $f = g_{ij} \dot u^i \dot u^j$ (I suggest using dots instead of primes for readability), the indices are both dummy indices (aka summation indices). It does not matter what you call these indices and you could just as well write $f = g_{ab} \dot u^a \dot u^b$ or with any other index names. In the derivative $df/d\dot u^i$, the $i$ is not a dummy index or summation index. When you introduce your function $f$, you should not introduce any dummy indices that are already used in the expression, hence
$$
\frac{\partial f}{\partial\dot u^i} = \frac{\partial}{\partial \dot u^i}(g_{ab} \dot u^a \dot u^b)
$$
is a correct way of writing it while
$$
\frac{\partial f}{\partial\dot u^i} = \frac{\partial}{\partial\dot u^i}(g_{ij} \dot u^i \dot u^j)
$$
is not. Also note that
$$
\frac{\partial\dot u^a}{\partial\dot u^i} = \delta^a_i,
$$
since it is equal to one if $a = i$ and zero otherwise.

As an illustration, consider the case of a two-dimensional manifold where $f = g_{11} \dot u^1 \dot u^1 + 2g_{12} \dot u^1 \dot u^2 + g_{22} \dot u^2 \dot u^2$. It is here clear that
$$
\frac{\partial f}{\partial \dot u^1} = 2g_{11} \dot u^1 + 2 g_{12} \dot u^2 = 2 g_{1i} \dot u^i.
$$
A similar relation holds for $\partial f/\partial\dot u^2$.

 

[fab13]

Great ! I think that I can write :

$$\dfrac{\partial f}{\partial u'^{i}}=\dfrac{\partial g_{jk}u'^{j}u'^{k}}{\partial u'^{i}}$$

$$=g_{jk}\bigg(\dfrac{\partial u'^{j}}{\partial u'^{i}}u'^{k}+u'^{j}\dfrac{\partial u'^{k}}{\partial u'^{i}}\bigg)$$

$$=g_{jk}(\delta_{ij}u'^{k}+\delta_{ik}u'^{j})$$

$$=2\,g_{jk}\,u'^{k}$$ since we count double by taking into account the inverting between $j$ and $k$ index.

Is it valid ? Thanks

 

[Orodruin]

Yes, apart from the fact that there is no such thing as $\delta_{ik}$ unless you are dealing with strictly Cartesian coordinates. In general coordinates, the Kronecker delta is a type (1,1) tensor and therefore has one covariant and one contravariant index, i.e., $\delta^i_k$. The end result is correct though.

Edit: Well, that is a truth with modification. There is no tensor for which $\delta_{ik}$ are the components in all coordinate systems.