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“The derivation of the
Schwarzschild metric”
Muhammad Hassaan Saleem
(PHYMATHS)
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)
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
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.
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!!!
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)
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
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)).
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.
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.
The Ricci tensor formulas are as follows
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
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!!!
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)
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)
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)
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
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.
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.
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
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
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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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