Comparing Weak Field Approx, Sign Question, & Derivatives

[binbagsss]

Comparing two sources one has $\frac{d^{2}x^{i}}{dt^{2}}=-\frac{1}{2}\epsilon\bigtriangledown_{i}h_{00}$ and the other has $\frac{d^{2}d^{i}}{dt^{2}}=\frac{1}{2}\epsilon\bigtriangledown^{i}h^{00}$, And the one using the lower index has the Newton-Poisson equation as $\frac{d^{2}}{dx^{i}}=-\bigtriangledown_{i}\Phi$ and the source using the upper index for the partials has $\frac{d^{2}}{dx^{i}}=-\bigtriangledown^{i}\Phi$. I.e- there is no sign change in the Newton-Poison equation.

Questions:
- Thus they are led to the same identification of $g_{00}$ but differering by a sign- how is the physics unchanged?
- I'm really confused about what this raisin and lowering partial derivative means- I know that the contravaiant and covariant derivaitve differ in their definition for the sign of the connection term, however the partial derivative term has the same sign in both cases.
- Why is there no change in the Newton-Poisson equation, is it because the operators $\bigtriangledown$ are denoting differnt things in Newtonian mechanics and GR? So in Newtonian mechanics it's just partial derivatives. How is it differing in the GR sense?

Thanks in advance.
(The two sources are Tod and Hughston, intro to GR, and http://www.mth.uct.ac.za/omei/gr/chap7/node3.html )

 

[PeterDonis]

the one using the lower index has the Newton-Poisson equation as $\frac{d^{2}}{dx^{i}}=-\bigtriangledown_{i}\Phi$

I think you mean $\frac{d^2 x^i}{dt^2}=-\nabla_{i}\Phi$, correct? (I would think a similar comment would apply to the other source, but I don't have it so I can't check there.)

they are led to the same identification of $g_{00}$ but differering by a sign

There are two metric sign conventions in the literature; in one, for a diagonal metric, $g_{00}$ is negative and the other three coefficients are positive; in the other, $g_{00}$ is positive and the other three are negative. The source you gave a link to uses the first sign convention; if the other source uses the second sign convention, that would explain the difference.

how is the physics unchanged

Changing the metric sign convention changes a bunch of signs in the equations; the changes all happen in concert so that all actual physical observables are unchanged.

I'm really confused about what this raisin and lowering partial derivative means

In Newtonian mechanics, upper and lower indexes mean the same thing; which way an index is written is just typographical convenience. The Poisson equation is a Newtonian equation, so the placement of the index $i$ there doesn't mean anything physically. Both sources appear to treat the $i$ index on the equations involving $h_{00}$ similarly, i.e., where the index is placed does not mean anything physically.

A proper treatment in a relativistic context would involve carefullly distinguishing vectors and covectors, which requires some background in differential geometry. The textbooks on GR that I'm familiar with (MTW and Wald) both go into this in some detail. But I don't think either of the sources you're using are going to that level of rigor.

I know that the contravaiant and covariant derivaitve differ in their definition for the sign of the connection term

Your terminology here is a bit unusual. What you are referring to, I take it, is this:

$$
\nabla_{\mu} X^{\nu} = \partial_{\mu} X^{\nu} + \Gamma^{\nu}{}_{\mu \rho} X^{\rho}
$$

as compared to this:

$$
\nabla_{\mu} X_{\nu} = \partial_{\mu} X_{\nu} - \Gamma^{\rho}{}_{\mu \nu} X_{\rho}
$$

The sign on the connection coefficient term depends, not on where the index is on the derivative operator, but on where the index is on the object the derivative operator is being applied to. This sign difference has nothing to do with any sign conventions for the metric, or even with index conventions for the derivative operator itself.

is it because the operators $\bigtriangledown$ are denoting differnt things in Newtonian mechanics and GR?

As far as I can tell, both of your sources are just using $\nabla$ as an alternate notation for $\partial$, i.e., it is just the partial derivative operator in both cases (Newtonian and GR). This is not really standard notation in GR (although it is often used in Newtonian mechanics); in GR, a careful distinction is drawn between the partial derivative $\partial$ and the covariant derivative $\nabla$--the latter includes the connection coefficient term (or terms, if the covariant derivative of a multi-index tensor is being taken). But as far as I can tell, neither of your sources talks about or uses covariant derivatives at all; they appear to be assuming a coordinate chart in which the connection coefficients are zero, so partial and covariant derivatives are the same.

 

[binbagsss]

There are two metric sign conventions in the literature; in one, for a diagonal metric, $g_{00}$ is negative and the other three coefficients are positive; in the other, $g_{00}$ is positive and the other three are negative. The source you gave a link to uses the first sign convention; if the other source uses the second sign convention, that would explain the difference.

.

I did consider a differing sign convention of the metric as an explanation, however within a single source I thought there was a sign change associated with a $^{i}$ to a $_{i}$: http://www.mth.uct.ac.za/omei/gr/chap7/node3.html comparing equations 29 and 45.

(https://www.physicsforums.com/threa...tion-quick-sign-question.805452/#post-5057911, post 2)

Thanks.

 

[PeterDonis]

within a single source I thought there was a sign change associated with a ${}^{i}$ to a ${}_{i}$:

It does look like there's a sign change going from (29) to (45) there, but I don't think it can be from lowering an index, because lowering a spatial index should not change the sign (since the spatial metric coefficients are positive). It looks to me like this source is just being sloppy, or else leaving out steps (which is disappointing since it appears to be course notes). I suggest Carroll's lecture notes, chapter 4 (the part starting with equation 4.35) for a better treatment of the same material.

 

[binbagsss]

or else leaving out steps

Any idea on what sort of steps these are?

 

[PeterDonis]

Any idea on what sort of steps these are?

Not really, I just mentioned it as a general possibility. At some point, if one is having trouble seeing how a particular derivation is done, one has to cut one's losses and just find another derivation that is easier to follow.