- #1

fab13

- 318

- 6

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

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