# The gravitational Klein-Gordon Equation

1. Jan 31, 2010

### Altabeh

Hi

Following a request by FunkyDwarf (I don't know if dwarfs have odor!!) in https://www.physicsforums.com/showpost.php?p=2556673&postcount=11" thread, regarding how one can get the Klein-Gordon equation (KGE) for free particles in a gravitational field, i.e. the equation

$$-g^{\mu \nu} \partial_{\mu}\partial_{\nu} \psi + g^{\mu\nu}\Gamma^{\rho}_{\mu \nu}\partial_{\rho} \psi + \frac{m^2c^2}{\hbar^2}\psi =0,$$ (1)

I'm going to explain the details of calculation here. In almost every textbooks on GR, if this stuff is brought up, it just is a reference to the eq. (1) itself with a little hint at the way it can be deduced. Among all, the following books throw some words about KGE:

1- General Relativity: An Introduction for Physicists by Hobson, Cambridge Press, 2006. pp 431-2.
2- Gravitation and Spacetime by Ohanian, W. W Norton & Company, 1963. p 434.
3- Relativity, Astrophysics and Cosmology by Israel, D. Reidel Publisliing Company, 1973. pp 269-70
4- Exact solutions and Scalar fields in Gravity by Marcias, Kluwer Academic Publishers, 2002. p 178.
etc.

So let's get to work: Take $$\psi$$ to be a massive scalar field with mass $$m$$. The Lagrangian of this field $$L$$ in the absence of the gravitation reads

$$L=\frac{1}{2}( {\eta}^{\mu \nu} \partial_{\mu}\psi \partial_{\nu} \psi -\frac{m^2c^2}{\hbar^2}\psi ).$$ (2)

The question is how does one make sure that (2) has the right form? Applying (2) in the Euler-Lagrange equations,

$${\partial}_{x^{\nu}}\frac{\partial L}{\partial {\psi}_{,\nu}}-\frac{\partial L}{\partial {\psi}}=0,$$

where "," represents the ordinary derivative, yields

$$({\square}+m^2c^2/{\hbar}^{2})\psi=0,$$ (3)

with $${\square}=\frac{1}{c^2}\frac{{\partial}^2}{\partial t^2}-{\nabla}^2\equiv {\eta}^{\mu \nu}{\partial}_{\mu}{\partial}_{\nu}$$ being the D'Alembert operator. Eq. (3) is clearly the KG equation by which we can confirm that the choice of Lagrangian (2) is suitable.

Now for GR, or in the presence of gravitation, you have to replace the Minkowski metric $${\eta}^{\mu \nu}$$ by a general metric tensor $${g}^{\mu \nu}$$ and must note that in GR the situation needs much of your attention to be gone through successfully:
Since in SR the quantity $$\sqrt{-\eta}$$ or the square root of the determinant of metric tensor equals one, then the Lagrangian $$L$$ and the Lagrangian density $$\mathfrak{L}$$ are equivalent:

$$\mathfrak{L}=\sqrt{-\eta}L\equiv L$$.

This is the reason why I didn't make it hard for me to introduce the Lagrangian density for Minkowski spacetime and I think you know that all quantized field equations are guaranteed to exist through a good choice of Lagrangian density and this is what makes the new theories distinctive from classical physics of Newton dealing with (scalar) fields in the background of Euclidean geometry along with having a hand from the use of an absolute time. Thus the appropriate scalar Lagrangian density for our case in GR is

$$L= \frac{1}{2}\sqrt{-g}\left( {g}^{\mu \nu} \partial_{\mu}\psi \partial_{\nu} \psi -\frac{m^2c^2}{\hbar^2}\psi \right) .$$ (2')

Which is sometimes written as

$$\mathfrak{L}= \frac{1}{2}\left( {\mathfrak{g}^{\mu \nu} \partial_{\mu}\psi \partial_{\nu} \psi -\sqrt{-g}\frac{m^2c^2}{\hbar^2}\psi \right) ,$$ (2'')

where $$\mathfrak{g}^{\mu \nu}$$ is called the "dual metric density" or "reciprocal metric density". Working this scalar Lagrangian density into the Euler-Lagrange equation and dividing all terms by $$\sqrt{-g}$$ gives

$$\frac{1}{\sqrt{-g}}{\partial}_{{\mu }}(\mathfrak{g}^{\mu\nu}{\partial}_{\nu}\psi) -\frac{m^2c^2}{\hbar^2}\psi =0.$$

Using the chain rule, this can be written in a more expanded way as

$${g}^{\mu\nu}{\partial}_{\mu}{\partial}_{\nu}\psi + \frac{1}{\sqrt{-g}}{\partial}_{\mu}{\mathfrak{g}^{\mu\nu}}{\partial}_{\nu}\psi -\frac{m^2c^2}{\hbar^2}\psi =0.$$ (4)

Now we want to find an equivalent expression for $${\partial}_{\mu}{\mathfrak{g}^{\mu\nu}}$$. We know that the covariant derivative of the metric tensor is always zero. Using this fact and that

$$\sqrt{-g}_{,\kappa}=\sqrt{-g}\Gamma^{\alpha}_{\alpha\kappa},$$

one can easily prove (exercise!)

$$\partial}_{\mu}{\mathfrak{g}^{\mu\nu}} =-\mathfrak{g}^{\alpha\kappa}\Gamma^{\nu}_{\alpha\kappa}.$$ (5)

Ultimately introducing (5) into (4) results in (1) and we are done.

AB

Last edited by a moderator: Apr 24, 2017
2. Jan 31, 2010

### FunkyDwarf

Coolbeans, cheers!