I Geodesic in Weak Field Limit: Introducing Einstein's Relativity

[GR191511]

I'm reading《Introducing Einstein's Relativity_ A Deeper Understanding Ed 2》on page 180,it says:

since we are interested in the Newtonian limit,we restrict our attention to the spatial part of the geodesic equation,i.e.when a=$\alpha$$\quad$,and we obtain,by using $\frac{d\tau}{dt}=1+O(\varepsilon^2)$:
$\frac {\partial^2 x^\alpha} {\partial t^2}+\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}(1+O(\varepsilon))=0$.$\quad$I wonder how the auther gets "(1+O($\varepsilon$))"?

 

[strangerep]

The authors (D'Inverno & Vickers) explained that on p179, just before eq(10.43). Do you not understand how eq(10.43) was obtained? Their preceding (unnumbered) equation is $$c^2 d\tau^2 ~=~ c^2 dt^2 (1 - \varepsilon^2) ~.$$ Take the square root of both sides. You might have to expand $\sqrt{1 - \varepsilon^2}$ as a Taylor series in $\varepsilon$ to see why it ends up as $\,1 - O(\varepsilon^2)$.

 

[GR191511]

You might have to expand $\sqrt{1 - \varepsilon^2}$ as a Taylor series in $\varepsilon$ to see why it ends up as $\,1 - O(\varepsilon^2)$.

Thanks！But it doesn't end up as"$1+O(\varepsilon^2)$",it is"$1+O(\varepsilon)$"

 

[PeterDonis]

it doesn't end up as"$1+O(\varepsilon^2)$",it is"$1+O(\varepsilon)$"

Check your work. Doing a Taylor series expansion of $\sqrt{1 + \varepsilon^2}$ does not mean you take the square root of $\varepsilon^2$.

 

[GR191511]

Check your work. Doing a Taylor series expansion of $\sqrt{1 + \varepsilon^2}$ does not mean you take the square root of $\varepsilon^2$.

I mean the geodesic "$\frac {\partial^2 x^\alpha} {\partial t^2}+\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}(1+O(\varepsilon))=0$ "it ends up as $(1+O(\varepsilon))$

 

[strangerep]

I mean the geodesic "$\frac {\partial^2 x^\alpha} {\partial t^2}+\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}(1+O(\varepsilon))=0$ "it ends up as $(1+O(\varepsilon))$

You make it much harder to help you if you don't show your work. I'm guessing you haven't correctly followed the author's instructions after eq(10.46).

 

[GR191511]

You make it much harder to help you if you don't show your work. I'm guessing you haven't correctly followed the author's instructions after eq(10.46).

$\frac {d^2 x^\alpha} {d \tau^2}+\Gamma^\alpha{}_{bc}\frac{dx^b}{d\tau}\frac{dx^c}{d\tau}=0\Rightarrow\frac{d}{d\tau}(\frac{dx^\alpha}{dt}\frac{dt}{d\tau})+\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}\frac{dt}{d\tau}\frac{dt}{d\tau}=0\Rightarrow$
$\left[\frac{d}{dt}(\frac{dx^\alpha}{dt}\frac{dt}{d\tau})\right]\frac{dt}{d\tau}+\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}\frac{dt}{d\tau}\frac{dt}{d\tau}=0\Rightarrow\frac{d}{dt}(\frac{dx^\alpha}{dt}\frac{dt}{d\tau})+$$\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}\frac{dt}{d\tau}=0\Rightarrow\frac{d^2x^\alpha}{dt^2}\frac{dt}{d\tau}+\frac{dx^\alpha}{dt}\frac{d}{dt}(\frac{dt}{d\tau})$$+\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}\frac{dt}{d\tau}=0$$\Rightarrow\frac{d^2x^\alpha}{dt^2}\frac{dt}{d\tau}+0+\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}\frac{dt}{d\tau}=0\Rightarrow\frac{d^2x^\alpha}{dt^2}+\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}$$=0$$\Rightarrow\frac{1}{c^2}\frac{d^2x^\alpha}{dt^2}+\frac{1}{c^2}\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}=0$$\quad$So where is the $1+O(\varepsilon)$?

 

[strangerep]

OK, it's actually a lot simpler than what you've written in post #7.

From (10.43) we have $d\tau = dt + O(\varepsilon^2)$. That means you can replace all the $d\tau$'s in the derivatives by $dt$. (Think of those derivatives as ratios of differentials.)

The trick they don't seem to explain is that $d^2 x^\alpha/dt^2$ is $O(\varepsilon)$, as is $\Gamma^a_{~bc}$, which is shown in (10.45) as an $\varepsilon$ term plus $O(\varepsilon^2)$. In the geodesic equation after (10.46), they divided throughout by 1 factor of $\varepsilon$.

(Is that explanation sufficient? I'll agree they probably could have been clearer.)

 

[GR191511]

OK, it's actually a lot simpler than what you've written in post #7.

From (10.43) we have $d\tau = dt + O(\varepsilon^2)$. That means you can replace all the $d\tau$'s in the derivatives by $dt$. (Think of those derivatives as ratios of differentials.)

The trick they don't seem to explain is that $d^2 x^\alpha/dt^2$ is $O(\varepsilon)$, as is $\Gamma^a_{~bc}$, which is shown in (10.45) as an $\varepsilon$ term plus $O(\varepsilon^2)$. In the geodesic equation after (10.46), they divided throughout by 1 factor of $\varepsilon$

(Is that explanation sufficient? I'll agree they probably could have been clearer.)

Why is "$d^2 x^\alpha/dt^2$ is $O(\varepsilon)$"related with this question?
Do you mean it is $\frac{\varepsilon}{c^2}\frac{d^2x^\alpha}{dt^2}+\frac{\varepsilon}{c^2}\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}(1+O(\varepsilon))=0$before" they divided throughout by 1 factor of $\varepsilon$"?Thank you!

 

[strangerep]

Why is "$d^2 x^\alpha/dt^2$ is $O(\varepsilon)$"related with this question?
Do you mean it is $\frac{\varepsilon}{c^2}\frac{d^2x^\alpha}{dt^2}+\frac{\varepsilon}{c^2}\Gamma^\alpha{}_{bc}\frac{dx^b}{dt}\frac{dx^c}{dt}(1+O(\varepsilon))=0$before" they divided throughout by 1 factor of $\varepsilon$"?Thank you!

That's the general idea. But... (sigh)... now you've got me wanting to rewrite that stuff properly without skipping steps. Probably not today though.