Which term of the Einstein equation is larger?

[exponent137]

TL;DR Summary

How different terms of the Einstein equation are important?

https://en.wikipedia.org/wiki/Einstein_field_equations

If we study when the sun bent the light of the stars behind it, which above terms have the largest influence, and which can be neglected? (I think that the third term can be neglected for such phenomena.)

And what changes when we are close to a small black hole?

Probably the first and the second terms have different visualizations and meanings?

Probably these answers can be related to Baez's paper https://arxiv.org/pdf/gr-qc/0103044?

Probably they can be related also to Carroll's paper https://arxiv.org/pdf/gr-qc/9712019?

 

[Vanadium 50]

Kinetic energy is $m(v_x^2 + v_y^2 + v_z^2)/2$. Which term is largest?

 

[Nugatory]

I don't think your question as an answer of the sort that you're looking for.
You are right that the $\Lambda$ term is negligible in the situations you are considering. Also the right-hand side of the equation is zero in these cases, which pretty much requires that the two $R$ terms are always the same size. But...

1) That's not one equation. $\mu$ and $\nu$ take on values from zero to three, so we're looking at a compact representation of 64 different equations (although because of various symmetries it's really just 2010).
2) The various terms in any one of those ten equations don't add to produce the light-bending effect. The overall shape of the spacetime is described by the metric tensor $g_{\mu\nu}$ which you see in the equation so these equations determine the value of the various components of that tensor.
3) These are differential equations - both $R$ and $R_{\mu\nu}$ are complicated functions of the derivatives of the various $g_{\mu\nu}$ values.

You've quoted two good sources, but you might be better off starting with https://preposterousuniverse.com/wp-content/uploads/2015/08/grtinypdf.pdf

[Edited to correct "20" to "10"]

 

[anuttarasammyak]

As was explained, cosmological term of $\Lambda$ be zero here. Once you have gotten metric tensor $g_{\mu \nu}$, pass of light or geodesic is derived by its definition. So you should investigate how to get Schwartzshild metric $g_{\mu \nu}$ which tells us gravity around the celestrial body, which is BH or star does not matter. You may find in textbooks that Einstein equation in vaccum space where lights travel, $R_{\mu \nu}=0$, and Newton's gravitation law as approximation are used for its derivation.

 

[Vanadium 50]

Λ is zero...unless it's not.

Further, even if non-zero, it is negligible in most cases....but not all.

This is really unanswerable in the sane way my earlier message is unasnwerable.

 

[Ibix]

we're looking at a compact representation of 64 different equations

16, not 64. Given your existing correction I guess you were thinking of the Riemann tensor, which does have 64 components of which 20 are independent.

If we study when the sun bent the light of the stars behind it, which above terms have the largest influence, and which can be neglected?

In that case you'd be thinking of vacuum solutions, and the whole thing reduces to $R_{\mu\nu}=0$, which you'd solve for $g_{\mu\nu}$.

But if you're interested in gravitational lensing you'd probably already have the solution to this in mind, the Schwarzschild or Kerr metrics depending on how important the angular momentum of your lensing object is (negligible for the Sun, but maybe not for a black hole). You'd be interested in the geodesic equations, which describe the paths of free-falling objects in spacetime.

 

[PeterDonis]

If we study when the sun bent the light of the stars behind it, which above terms have the largest influence, and which can be neglected?

In the case you describe, all of the terms are zero.

 

[PeterDonis]

the right-hand side of the equation is zero in these cases, which pretty much requires that the two terms are always the same size.

No, it requires that (if we assume the $\Lambda$ term is negligible) both Ricci terms are zero. The simplest way to see this is to transform to the trace reversed form of the equation, which shows that we must have $R_{\mu \nu} = 0$ if $T_{\mu \nu} = 0$.

 

[Nugatory]

No, it requires that (if we assume the $\Lambda$ term is negligible) both Ricci terms are zero. The simplest way to see this is to transform to the trace reversed form of the equation, which shows that we must have $R_{\mu \nu} = 0$ if $T_{\mu \nu} = 0$.

Yes, that stronger statement is also true.