Ricci components in Schwarzschild solution

In summary, the conversation discusses the Ricci tensor components and their comparison to various texts and literature. The participants also mention various definitions and sign conventions, as well as provide references for further study. The main issue at hand is a disagreement on the proper sign convention for one of the expressions. The participants continue to discuss and provide input on the matter, with one participant sharing a link to a helpful resource.
  • #1
rickphysics
6
0
Attached is a word document where I am setting out the Ricci tensor components to solve for the terms in the line element equation and get to the Schwarzschild solution. I am overlooking something as I have a sign off in the Ricci components in comparison to the many texts and literature I have referenced on this. Hopefully someone, if they have Word, can point out my error. If you do not have Word but are interested, please post a response and format you like, and I will change it. Thanks in advance.
 

Attachments

  • Schwarzschild Ricci components.doc
    30.5 KB · Views: 316
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  • #2
I tossed your metric into GRTensorII, and it spits out

[tex]
R_{tt} =
-1/4\,{\frac { \left( {\frac {d}{dr}}A \left( r \right) \right)
\left( {\frac {d}{dr}}B \left( r \right) \right) A \left( r \right)
r-2\, \left( {\frac {d^{2}}{d{r}^{2}}}A \left( r \right) \right) B
\left( r \right) A \left( r \right) r+ \left( {\frac {d}{dr}}A
\left( r \right) \right) ^{2}B \left( r \right) r-4\, \left( {\frac
{d}{dr}}A \left( r \right) \right) B \left( r \right) A \left( r
\right) }{ \left( B \left( r \right) \right) ^{2}A \left( r \right)
r}}
[/tex]

for whatever it's worth (it doesn't seem to match either one of your expressions).

The Christoffel symbols all looked OK
 
  • #3
Why didn't you choose the more famous Ansatz

[tex]ds^{2}=e^{2\nu (r)}dt^{2}-e^{2\lambda (r)}dr^{2}-r^{2}d\theta^{2}-r^{2}\sin^{2}\theta d\phi^{2} [/tex]

It leads to nicer ODE-s.You can follow and check your calculations with Dirac's book [1].

Daniel.

[1]P.A.M.Dirac,"General Relativity",1975.
 
  • #4
Schwarzschild Ricci components

Thanks for the responses. I have gone through every kind of starting point
Ansatz, et al, but come to the same point of one term in the end being off by a sign. I attached another Word document with an abbreviated version to better identify my problem. Unfortuneately I am not around academia and am just reading GR on my own and do not have anyone to help with what seems to be a simple misunderstanding of the Ricci tensor signage. Once again thanks in advance.
 

Attachments

  • Schwarzschild Ricci components (2).doc
    26 KB · Views: 256
  • #5
Just quickly skimming over your red equations... is there a [relative] sign-error in distributing the minus signs? In the brackets, for example, shouldn't the u^2 and uv terms have different signs (since u and v do in the preceding equation)?

What are the definitions/sign-conventions/signature-conventions of the text you are comparing with?
 
  • #6
I think your expression for the Ricci is wrong.

MTW "Gravitation" pg 340 gives the Riemann

[tex]
R^u{}_{vab} = \partial_a \Gamma^u{}_{vb} - \partial_b\Gamma^u{}_{va}+ \Gamma^u{}_{pa}\Gamma^p{}_{vb} - \Gamma^u{}_{pb}\Gamma^p{}_{va}
[/tex]

where [tex]\partial_a = \frac{\partial}{\partial{x^a}} [/tex] in the original, and I've also replaced greek with latin.

The Ricci is just

[tex]
R_{uv} = R^x{}_{uxv}
[/tex]

I don't think this set of expressions is equivalent because from the first term (the only one with a single positive Christoffel symbol), I derive the mapping from my letters to yours

a->v
u->sigma
v->u
b->sigma

and substituting this into the left hand side, you are calculating [tex]R^\sigma{}_{uv\sigma}[/tex], which is not the Ricci, the Ricci should be [tex]R^\sigma{}_{u{\sigma}v}[/tex]
 
  • #7
pervect said:
I think your expression for the Ricci is wrong.

MTW "Gravitation" pg 340 gives the Riemann

[tex]
R^u{}_{vab} = \partial_a \Gamma^u{}_{vb} - \partial_b\Gamma^u{}_{va}+ \Gamma^u{}_{pa}\Gamma^p{}_{vb} - \Gamma^u{}_{pb}\Gamma^p{}_{va}
[/tex]

where [tex]\partial_a = \frac{\partial}{\partial{x^a}} [/tex] in the original, and I've also replaced greek with latin.

The Ricci is just

[tex]
R_{uv} = R^x{}_{uxv}
[/tex]

I don't think this set of expressions is equivalent because from the first term (the only one with a single positive Christoffel symbol), I derive the mapping from my letters to yours

a->v
u->sigma
v->u
b->sigma

and substituting this into the left hand side, you are calculating [tex]R^\sigma{}_{uv\sigma}[/tex], which is not the Ricci, the Ricci should be [tex]R^\sigma{}_{u{\sigma}v}[/tex]

Either just inside the front cover, or just inside the the back cover, of MTW, there is a table of sign conventions. Unfortunately, I don't have my copy of MTW handy :frown: - does MTW list different sign conventions for Ricci?

Regards,
George
 
  • #8
There are large arrows pointing to the plus signs on the lefthand sides of the following two expressions

[tex]
+R^u{}_{vab} = \partial_a \Gamma^u{}_{vb} - \partial_b\Gamma^u{}_{va}+ \Gamma^u{}_{pa}\Gamma^p{}_{vb} - \Gamma^u{}_{pb}\Gamma^p{}_{va}
[/tex]
[tex]
+R_{uv} = R^x{}_{uxv}
[/tex]

I interpret this as suggesting that some authors might put a minus sign where the plus sign is.
 
  • #9
Thanks,

pervect is correct here. I lost the proper notation somewhere going back and forth between all the literature. Thanks for the MTW reference George. I bought mine when I was 13 years old as I thought it was neat and I might one day understand it. 31 years later that is starting to happen.

Rick
 
  • #10
I'm working this out for myself (https://www.physicsforums.com/showthread.php?t=80565) and I ended up with all the same Christoffel symbols except for one:

[tex]\Gamma^1{}_{4 4} = \frac{-A'}{2B}[/tex] (Note: If you're looking at my thread, I defined my A and B in reverse. So your A is my B and vice versa)

Our answers for this symbol differ by a negative sign. I've looked back on it again and I think it should have a negative sign. Is this the "proper notation" you may have lost?


(By the way... I'm an undergrad in physics, and haven't heard of this MTW. What is it exactly? It seems to be an equations manual of some sort. It sounds really useful, so would you recommend I obtain one?)
 
Last edited:
  • #11
MTW is also called "the telephone book" and is

Ch.Misner,K.Thorne,J.A.Wheeler,"Gravitation",1972.

Daniel.
 
  • #12
I had a chance to recalulate my Ricci components in light of pervect's input, which I saw would change my sign, but ultimately I only got the negative of everything I originally had.

I am going to need a little more help. I will reference

"An exact solution" from Math Pages" at
http://www.mathpages.com/rr/s6-01/6-01.htm

of which I have attached an exert ( Word attachment) with my comments in red.

In here as everywhere else I am still not coming up with the correct sign convention on one of my expressions. If you are inclined, please look at the attached, where I have carefully substituted as shown by the author, but do not get the author's expression. It seems to be simple math, and this is not unique to this author, it is the same error I am making in comparing to all others, including Dirac, which dexercioby graciously pointed out.

Once again thanks for input.
 

Attachments

  • schwarzschild ricci components (3).doc
    47.5 KB · Views: 238
  • #13
rickphysics said:
In here as everywhere else I am still not coming up with the correct sign convention on one of my expressions. If you are inclined, please look at the attached, where I have carefully substituted as shown by the author, but do not get the author's expression. It seems to be simple math, and this is not unique to this author, it is the same error I am making in comparing to all others, including Dirac, which dexercioby graciously pointed out.

Once again thanks for input.

I think your problem is that you incorrectly differentiated [itex]\Gamma^r{}_{tt}[/tex]

Representing derivatives as subscripts, you want

d/dr (e^(2u-2v) u_r), where u and v are functions of r.

The chain rule gives us.

e^(2u-2v) u_rr + d/dr(e^(2u-2v)) u_r

Let x = 2u-2v, then d/dr (e^x) = d/dx (e^x) * dx/dr = e^x dx/dr

Thus we have for the final result

e^(2u-2v) * (u_rr + (2u_r - 2v_r)*u_r)

Or, using Maxima
Code:
(%i1) depends([u,v],r);
(%o1) 				 [u(r), v(r)]
(%i2) e1 :: exp(2*u-2*v)*diff(u,r);
				du   2 u - 2 v
(%o2) 			        -- %E
				dr
(%i3) diff(e1,r);
						2
		du   2 u - 2 v	  du	 dv    d u   2 u - 2 v
(%o3) 	        -- %E	       (2 -- - 2 --) + --- %E
		dr		  dr	 dr	 2
					       dr
(%i4)
 
  • #14
pervect is correct again. I failed to use the chain rule and only differentiated the first derivative again, whilst the other part of the expression is also a function of r. This gives me he correct answer for all of my various starting cases. I knew it was an obvious error which lied somewhere between arithimetic and tensor calculus.

Thanks again.
 

1. What are Ricci components in the Schwarzschild solution?

The Ricci components in the Schwarzschild solution refer to the mathematical quantities that describe the curvature of spacetime around a non-rotating, spherically symmetric mass. These components are part of the larger Ricci tensor, which is a key element in Einstein's theory of general relativity.

2. How do Ricci components relate to the Schwarzschild radius?

The Schwarzschild radius is a measure of the size of the event horizon of a black hole. It is closely related to the Ricci components in the Schwarzschild solution, as the values of these components determine the shape and size of the event horizon. In fact, the Schwarzschild radius can be calculated using the Ricci components and the mass of the black hole.

3. What is the significance of the Ricci components in the Schwarzschild solution?

The Ricci components play a crucial role in understanding the gravitational effects of massive objects, such as black holes, in the context of general relativity. They provide a mathematical description of the curvature of spacetime, which is directly related to the distribution of matter and energy in the universe.

4. Can the Ricci components be measured or observed?

No, the Ricci components are not directly measurable or observable. They are purely mathematical quantities used in theoretical models of gravity, such as the Schwarzschild solution. However, their effects on the curvature of spacetime can be observed through phenomena such as gravitational lensing.

5. How do the Ricci components change in the presence of matter or energy?

The Ricci components are affected by the distribution of matter and energy in the universe. In the Schwarzschild solution, the Ricci components become larger as the mass of the central object increases, which results in a stronger curvature of spacetime. Additionally, the Ricci components can change over time as matter and energy move and interact in the universe.

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