MManuel Abad
May4-11, 07:17 PM
Hi there, physics lovers!!
Sorry I'm bothering you again. The following may sound as stupid questions, but this is a very hard topic for me, so don't judge me that bad! :)
You see, when we're talking about light in curved spacetime, the geodesic equations (or equations of motion) use an affine parameter \lambda to parametrize its world line (actually I know this thanks to this forum).
Indeed, searching in Section 87 "Motion of a particle in a gravitational field" of the 2nd Volume "The Classical Theory of Fields", from the Course of Theoretical Physics by Landau and Lifshitz, I find that these geodesic equations (which are for light, and are for null geodesics, defined as those in which ds=0) look like:
\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\n u\sigma}\frac{dx^{\nu}}{d\lambda}\frac{dx^{\sigma} }{d\lambda}=0
with \Gamma^{\mu}_{\nu\sigma} being the Christoffel symbols. I also saw that, because the wave vector of light is always tangent to its path, we can write:
k^{\mu}=\frac{dx^{\mu}}{d\lambda}
and then, the geodesic equations look like:
\frac{dk^{\mu}}{d\lambda}+\Gamma^{\mu}_{\nu\sigma} k^{\nu}k^{\sigma}=0
In section 48, Landau defines, talking IN MINKOWSKI (FLAT) SPACE-TIME:
k^{\mu}=\left\{\frac{\omega}{c},\frac{\omega}{c}\t extbf{n}\right\}
where n is a unit vector along the direction of propagation of the wave. Then it is obvious that:
k^{\mu}k_{\mu}=0
Then, returning to section 87 and the geodesic equations, Landau says that this very same equation holds, even in a gravitational field (CURVED space-time).
QUESTION 1:
Is this 'cuz k^{\mu}k_{\mu} is a four-scalar and then it's invariant in changes of coordinates, even if they're for non-flat space-time? Or why?
QUESTION 2:
k^{\mu}=\left\{\frac{\omega}{c},\frac{\omega}{c}\t extbf{n}\right\} is the form of the wave vector ONLY IN FLAT SPACE-TIME, ISN'T IT? I mean, its form changes in curved space time, actually, the wave vector is what we want to find from the equations:
\frac{dk^{\mu}}{d\lambda}+\Gamma^{\mu}_{\nu\sigma} k^{\nu}k^{\sigma}=0
, isn't it?
Well, there it is. I hope you help me. I thank you very much your attention.
Sorry I'm bothering you again. The following may sound as stupid questions, but this is a very hard topic for me, so don't judge me that bad! :)
You see, when we're talking about light in curved spacetime, the geodesic equations (or equations of motion) use an affine parameter \lambda to parametrize its world line (actually I know this thanks to this forum).
Indeed, searching in Section 87 "Motion of a particle in a gravitational field" of the 2nd Volume "The Classical Theory of Fields", from the Course of Theoretical Physics by Landau and Lifshitz, I find that these geodesic equations (which are for light, and are for null geodesics, defined as those in which ds=0) look like:
\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\n u\sigma}\frac{dx^{\nu}}{d\lambda}\frac{dx^{\sigma} }{d\lambda}=0
with \Gamma^{\mu}_{\nu\sigma} being the Christoffel symbols. I also saw that, because the wave vector of light is always tangent to its path, we can write:
k^{\mu}=\frac{dx^{\mu}}{d\lambda}
and then, the geodesic equations look like:
\frac{dk^{\mu}}{d\lambda}+\Gamma^{\mu}_{\nu\sigma} k^{\nu}k^{\sigma}=0
In section 48, Landau defines, talking IN MINKOWSKI (FLAT) SPACE-TIME:
k^{\mu}=\left\{\frac{\omega}{c},\frac{\omega}{c}\t extbf{n}\right\}
where n is a unit vector along the direction of propagation of the wave. Then it is obvious that:
k^{\mu}k_{\mu}=0
Then, returning to section 87 and the geodesic equations, Landau says that this very same equation holds, even in a gravitational field (CURVED space-time).
QUESTION 1:
Is this 'cuz k^{\mu}k_{\mu} is a four-scalar and then it's invariant in changes of coordinates, even if they're for non-flat space-time? Or why?
QUESTION 2:
k^{\mu}=\left\{\frac{\omega}{c},\frac{\omega}{c}\t extbf{n}\right\} is the form of the wave vector ONLY IN FLAT SPACE-TIME, ISN'T IT? I mean, its form changes in curved space time, actually, the wave vector is what we want to find from the equations:
\frac{dk^{\mu}}{d\lambda}+\Gamma^{\mu}_{\nu\sigma} k^{\nu}k^{\sigma}=0
, isn't it?
Well, there it is. I hope you help me. I thank you very much your attention.