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### Schwarzchild solution derivation

• 1. “The derivation of the Schwarzschild metric” Muhammad Hassaan Saleem (PHYMATHS)
• 2. Some prominent solutions of Einstein equation 𝑅 𝜇𝜈 − 1 2 𝑅𝑔 𝜇𝜈 = 8𝜋𝐺𝑇𝜇𝜈 Solutions Uncharged Charged Non Spinning Schwarzschild (1916) Reissner Nordstrom (~1918) Spinning Kerr (1963) Kerr Newman (1965)
• 3. The problem Find the metric around a non spinning, uncharged mass. The 𝑇𝜇𝜈 = 0 around the mass. 𝑅 𝜇𝜈 − 1 2 𝑅𝑔 𝜇𝜈 = 0 ⇒ 𝑅 𝜇𝜈 = 0 Proof: 𝑅 𝜇𝜈 − 1 2 𝑅𝑔 𝜇𝜈 = 0 𝑔 𝜇𝜈 𝑅 𝜇𝜈 − 1 2 𝑅𝑔 𝜇𝜈 = 0 ⇒ 𝑅 − 1 2 𝑅 4 = 0 ⇒ 𝑅 = 0 ⇒ 𝑅 𝜇𝜈 = 0
• 4. Things to know The metric should be static (i.e. time independent). So, it should be invariant in the coordinate change 𝑡 → −𝑡  The metric is spherical symmetric. So, the metric should be invariant in the transformations 𝜃 → −𝜃 and 𝜙 → −𝜙 because of spherical symmetry.
• 5. The form of the metric The spherical symmetry and the symmetric property of the metric 𝑔 𝛼𝛽 = 𝑔 𝛽𝛼 gives the following metric 𝑔 𝛼𝛽 = 𝑔00 𝑔01 𝑔01 𝑔11 𝑔02 𝑔03 𝑔12 𝑔13 0 0 0 0 𝑟2 0 0 𝑟2 sin2 𝜃 The metric can be proved to be diagonal too!!!
• 6. The form of the metric For a coordinate transformation from 𝜉0, 𝜉1, 𝜉2, 𝜉3 to (𝜉0 ′ , 𝜉1 ′ , 𝜉2 ′ , 𝜉3 ′ ), the metric transforms as 𝑔 𝜇𝜈 ′ = 𝜕𝜉 𝛼 𝜕𝜉′𝜇 𝜕𝜉 𝛽 𝜕𝜉′𝜈 𝑔 𝛼𝛽 Now, if we just change the 𝜙 coordinate as 𝜙 → −𝜙 then 𝜉3 = 𝜙 and 𝜉3 ′ = −𝜙. Then, we can conclude that (verify) 𝑔′03 = −𝑔03 But the metric should be invariant in this transformation. So, 𝑔03 ′ should be 𝑔03. So, we conclude that 𝑔03 = 0. Similarly, we can show that 𝑔13 = 𝑔23 = 0. (Check it)
• 7. The form of the metric Exercise: Use the transformations 𝑡 → −𝑡 to conclude that 𝑔01 = 𝑔02 = 𝑔03 = 0 Use the transformation 𝜃 → −𝜃 to conclude that 𝑔02 = 𝑔12 = 𝑔23 = 0 So, the metric becomes 𝑔 𝜇𝜈 = 𝑔00 0 0 𝑔11 0 0 0 0 0 0 0 0 𝑟2 0 0 𝑟2 sin2 𝜃
• 8. The form of the metric In the terms of the line element, we have the metric 𝑑𝑠2 = −𝐴 𝑟 𝑑𝑡 2 + 𝐵 𝑟 𝑑𝑟 2 + 𝑟2 𝑑Ω2 Where 𝐴 𝑟 = −𝑔00 𝐵 𝑟 = 𝑔11 𝑑Ω2 = 𝑑𝜃 2 + sin2 𝜃 𝑑𝜙 2 So, the metric becomes diagonal with two unknown function only (i.e. 𝐴 𝑟 and 𝐵(𝑟)).
• 9. Bring in the Ricci tensor The Ricci tensor is defined as (locally) 𝑅 𝛼𝛽 = 𝜕𝜌Γ𝛼𝛽 𝜌 − 𝜕 𝛽Γ𝛼𝜌 𝜌 + Γ𝜌𝜆 𝜌 Γ𝛼𝛽 𝜆 − Γ𝛽𝜆 𝜌 Γ𝛼𝜌 𝜆 We will use a Ricci worksheet for diagonal metric How to use the sheet Assume that a diagonal metric is given as 𝑑𝑠2 = −𝐴 𝑑𝑥0 2 + 𝐵 𝑑𝑥1 2 + 𝐶 𝑑𝑥2 2 + 𝐷 𝑑𝑥3 2 Where 𝐴, 𝐵, 𝐶, 𝐷 can be the functions of 𝑥0, 𝑥1, 𝑥2 and 𝑥3. Then the Ricci tensor elements can be calculated off the sheet.
• 10. Relation to our problem In our problem, 𝑥0 = 𝑡 , 𝑥1 = 𝑟, 𝑥2 = 𝜃 , 𝑥3 = 𝜙 𝐶 = 𝑟2 , 𝐷 = 𝑟2 sin2 𝜃 Moreover, the metric depends on 𝑟 and 𝜃 only (The 𝜃 dependence is in 𝑟2 sin2 𝜃 only). Note that subscripts will mean derivatives e.g. 𝐵1 and 𝐵11 mean single and double derivatives with respect to 𝑥1 i.e. 𝐵1 = 𝜕𝐵 𝜕𝑥1 𝑎𝑛𝑑 𝐵11 = 𝜕2 𝐵 𝜕 𝑥1 2 So, we can use the Ricci tensor sheet in the next few pages.
• 11. The Ricci tensor formulas are as follows
• 12.
• 13.
• 14.
• 15.
• 16.
• 17.
• 18. A sample calculation Lets calculate 𝑅00 The following terms will be zero The terms with derivatives w.r.t 𝑡 (i.e.𝑥1) e.g. 𝐴1, 𝐵1 The terms with derivatives w.r.t 𝜙 (i.e. 𝑥3) e.g. 𝐴3, 𝐵3 The terms with derivatives w.r.t 𝜃(i.e. 𝑥2) except 𝐷2 as for our problem, 𝐷 = 𝑟2 sin2 𝜃 and thus, 𝐷3 ≠ 0. So, 𝑅00 has the following non zero terms (verify it) 𝑅00 = 1 2𝐵 𝐴11 − 1 4𝐴𝐵 𝐴1 𝐴1 − 1 4𝐵2 𝐴1 𝐵1 + 1 4𝐵𝐶 𝐴1 𝐶1 + 1 4𝐵𝐷 𝐴1 𝐷1
• 19. A sample calculation For our problem, 𝐶 = 𝑟2 and 𝐷 = 𝑟2 sin2 𝜃 Using this, we get 𝑅00 = 1 2𝐵 𝐴11 − 1 4𝐴𝐵 𝐴1 𝐴1 − 1 4𝐵2 𝐴1 𝐵1 + 1 𝐵𝑟 𝐴1 The eq. we want to solve is 𝑅 𝜇𝜈 = 0 So, 𝑅00 = 0. We get after some simplification, 2𝑟𝐴𝐵𝐴11 − 𝐵𝑟 𝐴1 2 − 𝐴𝑟𝐴1 𝐵1 + 4𝐴𝐵𝐴1 = 0 (𝑖) We can repeat this for all the ten components of Ricci tensor. DO WE HAVE TO??? NOOO!!!
• 20. The other components Exercise Verify that the off diagonal components of the Ricci tensor i.e. 𝑅01, 𝑅02, 𝑅03, 𝑅12, 𝑅13, 𝑅23 don’t have a single non zero term in our problem. So, the equation 𝑅 𝜇𝜈 = 0 for these equations just gives us 0 = 0. Exercise Verify that the equation 𝑅11 = 0 gives 4𝐴2 𝐵1 − 2𝑟𝐴𝐵𝐴11 + 𝑟𝐴𝐴1 𝐵1 + 𝑟𝐵 𝐴1 2 = 0 (𝑖𝑖) Verify that the equation 𝑅22 = 0 gives 𝑟𝐴𝐵1 + 2𝐴𝐵2 − 2𝐴𝐵 − 𝑟𝐵𝐴1 = 0 (𝑖𝑖𝑖)
• 21. The equations Exercise Verify that the equation 𝑅33 = 0 just gives us sin2 𝑅22 = 0. So, it implies that 𝑅22 = 0. So, 𝑅33 = 0 isn’t an independent equation here. The equations are 2𝑟𝐴𝐵𝐴11 − 𝐵𝑟 𝐴1 2 − 𝐴𝑟𝐴1 𝐵1 + 4𝐴𝐵𝐴1 = 0 (𝑖) 4𝐴2 𝐵1 − 2𝑟𝐴𝐵𝐴11 + 𝑟𝐴𝐴1 𝐵1 + 𝑟𝐵 𝐴1 2 = 0 (𝑖𝑖) 𝑟𝐴𝐵1 + 2𝐴𝐵2 − 2𝐴𝐵 − 𝑟𝐵𝐴1 = 0 (𝑖𝑖𝑖)
• 22. Solving the equations For a better notation, 𝐴1 = 𝐴′ , 𝐴11 = 𝐴′′ , 𝐵1 = 𝐵′ , 𝐵11 = 𝐵′′ Now, add (𝑖) and (𝑖𝑖) to get 4𝐴2 𝐵′ + 4𝐴𝐵𝐴′ = 0 ⇒ 4𝐴 𝐴𝐵′ + 𝐴′ 𝐵 = 0 ⇒ 𝑑 𝐴𝐵 𝑑𝑟 = 0 So, 𝐴𝐵 = 𝐿 where 𝐿 is a constant So, we have 𝐵 = 𝐿 𝐴 (𝑖𝑣)
• 23. Solving the equations 𝐵′ = 𝐿 𝐴 ′ = − 𝐿𝐴′ 𝐴2 𝐵′′ = −𝐿 𝐴′ 𝐴2 ′ = − 𝐿 𝐴𝐴′ − 2𝐴′2 𝐴3 Exercise Use the above formulas in (𝑖) to get (after a convenient cancellation ) 𝑟𝐴′′ + 2𝐴′ = 0
• 24. Solving the equations The equation we got was 𝑟𝐴′′ + 2𝐴′ = 0 Let 𝐴′ = 𝑥 to get 𝑑𝑥 𝑥 = −2 𝑑𝑟 𝑟 Which integrates to give 𝑥 = 𝑘 𝑟2 Where 𝑘 is a constant.
• 25. Solving the equations 𝑥 = 𝑘 𝑟2 ⇒ 𝑑𝐴 𝑑𝑟 = 𝑘 𝑟2 ⇒ 𝐴 = 𝐴0 − 𝑘 𝑟 𝑣 𝑎𝑛𝑑 𝐵 = 𝐿 𝐴0 − 𝑘 𝑟 (𝑣𝑖) Where 𝐴0 is a constant. Exercise Show that (𝑣) and (𝑣𝑖) satisfy (𝑖𝑖) identically.
• 26. Solving the equations Lets use (𝑣) and (𝑣𝑖) in 𝑖𝑖𝑖 . (𝑖𝑖𝑖) is 𝑟𝐴𝐵1 + 2𝐴𝐵2 − 2𝐴𝐵 − 𝑟𝐵𝐴1 = 0 ⇒ 𝑟 𝐴0 − 𝑘 𝑟 − 𝐿𝑘 𝑟2 1 𝐴0 − 𝑘 𝑟 2 + 2𝐿2 𝐴0 − 𝑘 𝑟 − 2𝐿 − 𝑟𝐿 𝐴0 − 𝑘 𝑟 𝑘 𝑟2 = 0 ⇒ 𝐴0 − 𝑘 𝑟 = 𝐿 − 𝑘 𝑟 ⇒ 𝐴0 = 𝐿 So, 𝐴 = 𝐿 − 𝑘 𝑟 𝐵 = 𝐿 𝐿 − 𝑘 𝑟 −1
• 27. Solving the equations We know that as 𝑟 → ∞, the 𝑔 𝜇𝜈 → 𝜂 𝜇𝜈. The line element for 𝜂 𝜇𝜈 is 𝑑𝑠2 = −𝑑𝑡2 + 𝑑𝑟2 + 𝑟2 𝑑Ω2 So, 𝐴 = 𝐿 − 𝑘 𝑟 → 1 as 𝑟 → ∞. So, 𝐿 = 1 So, we have 𝐴 𝑟 = 1 − 𝑘 𝑟 & 𝐵 𝑟 = 1 1 − 𝑘 𝑟
• 28. The solution The line element for the Schwarzschild metric is given as 𝑑𝑠2 = − 1 − 𝑘 𝑟 𝑑𝑡2 + 1 − 𝑘 𝑟 −1 𝑑𝑟2 + 𝑟2 𝑑Ω2 What about k? 𝑘 is fixed by comparing to a problem in weak gravity. (May be in another video)  𝑘 turns out to be rs = 2𝐺𝑀 where 𝑀 is the mass at the origin. 𝑟𝑠 is also referred to as the Schwarzschild radius. So, we have the following line element 𝑑𝑠2 = − 1 − 2𝐺𝑀 𝑟 𝑑𝑡2 + 1 − 2𝐺𝑀 𝑟 −1 𝑑𝑟2 + 𝑟2 𝑑Ω2
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