Geodesic in Weak Field Limit: Introducing Einstein's Relativity

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• GR191511
In summary: Yes, that is correct. The term ##d^2x^\alpha/dt^2## is ##O(\varepsilon)## because it is multiplied by ##\varepsilon/c^2## when the equation is divided by ##\varepsilon##. This leads to the final equation having a factor of ##1+O(\varepsilon)## instead of ##1+O(\varepsilon^2)##.
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##))"?

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)##.

PeroK, topsquark and vanhees71
strangerep said:
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)##"

GR191511 said:
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##.

PeterDonis said:
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))##

GR191511 said:
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).

topsquark and vanhees71
strangerep said:
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##

Dale
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.)

Dale and vanhees71
strangerep said:
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!

GR191511 said:
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.

GR191511

1. What is the geodesic in the weak field limit?

The geodesic in the weak field limit is a curved trajectory that follows the shortest distance between two points in spacetime. It is a fundamental concept in Einstein's theory of general relativity, which describes how gravity affects the curvature of spacetime.

2. How does the weak field limit differ from the strong field limit?

The weak field limit refers to situations where the gravitational effects are relatively small, such as in our everyday lives. In contrast, the strong field limit refers to situations where the gravitational effects are significant, such as near massive objects like black holes.

3. How does Einstein's theory of relativity explain the geodesic in the weak field limit?

Einstein's theory of general relativity explains the geodesic in the weak field limit by describing how matter and energy affect the curvature of spacetime. In the weak field limit, this curvature is relatively small, and the geodesic is essentially a straight line. However, in the strong field limit, the curvature is significant, and the geodesic can be highly curved.

4. What are some real-world applications of understanding the geodesic in the weak field limit?

Understanding the geodesic in the weak field limit is crucial for many real-world applications, such as GPS systems, which rely on precise measurements of time and space to function correctly. It also helps us understand the behavior of objects in our solar system, such as the orbits of planets around the sun.

5. Can the geodesic in the weak field limit be observed or measured?

Yes, the geodesic in the weak field limit can be observed and measured through various experiments and observations. For example, the bending of starlight near massive objects, such as the sun, is a direct observation of the geodesic in the weak field limit. Additionally, scientists can use precise measurements of the orbits of planets and other celestial bodies to confirm the predictions of general relativity.

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