Equations of motion por a free particle in curved spacetime

[MManuel Abad]

Hi there, Physics lovers!

I've got some questions for you!

Denoting by

(1) [tex]ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}=c^{2}d\tau^{2}[/tex]

the interval (and [tex]\tau[/tex] the proper time) and using the signature (+---), we have that the equations of motion for a free particle are:

(2) [tex]\frac{d^{2}x^{\mu}}{d\tau^{2}}+\Gamma^{\mu}_{\nu\lambda}\frac{dx^{\nu}}{d\tau}\frac{dx^{\lambda}}{d\tau}=0[/tex]

; the so called "geodesic (timelike) equations", (those capital Gamma are the Christoffel symbols).

Obviously we have the relation:

(3) [tex]g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}=c^{2}[/tex]

'Cuz the LHS of (3) gives us the square of the magnitude of the 4-velocity, and that's always PLUS c²

(FIRST QUESTION: using the -+++ signature we get that LHS of (3) is equal to -c², don't we?)

Everything is fine. But then I get in trouble. Using an action based on an auxiliary field I found in some place, I get that for a MASSLESS particle, the equations of motion are:

(4) [tex]g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}=0[/tex]

; which people call the "geodesic (lightlike) equations".

SECOND (set of) QUESTION(s): So, we have TWO geodesic equations? (2) for massive particles and (4) for massless particles? If that's so, why does (2) doesn't involve mass m??

THIRD (set of) QUESTION(s):

And why is LHS of (4) equal to zero? isn't that the square of the magnitude of the 4-velocity for a massless particle? shouldn't it be c²?? I thought EVERY PARTICLE (massive or massless) travels in spacetieme with 4-speed equal to that of light!

FOURTH QUESTION:

I've read in many places:

"Well, obviously we get from (4) that:

(5) [tex]ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}=0[/tex]

Sorry, but that's not obvious to me. Could someone derive it for me? Please? I know that (5) would imply that the massless particle travels at the speed of light c, even if LHS of (4) is equal to zero, so I THINK that would solve question 4. But I just don't know how to get (5) from (4).

Well, that's all, folks. I hope you help me. Thank you so much since now :)

 

[jfy4]

(FIRST QUESTION: using the -+++ signature we get that LHS of (3) is equal to -c², don't we?)

Yes

Everything is fine. But then I get in trouble. Using an action based on an auxiliary field I found in some place, I get that for a MASSLESS particle, the equations of motion are:

(4) [tex]g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}=0[/tex]

; which people call the "geodesic (lightlike) equations".

SECOND (set of) QUESTION(s): So, we have TWO geodesic equations? (2) for massive particles and (4) for massless particles? If that's so, why does (2) doesn't involve mass m??

What you called the second geodesic equation is actually the condition on 4-velocity for massless particles. The geodesic equation has an analogous form for massless particles

[tex]\frac{d^{2}x^{\mu}}{d\lambda^{2}}+\Gamma^{\mu}_{\nu\lambda}\frac{dx^{\nu}}{d\lambda}\frac{dx^{\lambda}}{d\lambda}=0[/tex]

Where the [itex]\lambda[/itex] is an affine parameter, since we can't use proper time with massless particles.

THIRD (set of) QUESTION(s):

And why is LHS of (4) equal to zero? isn't that the square of the magnitude of the 4-velocity for a massless particle? shouldn't it be c²?? I thought EVERY PARTICLE (massive or massless) travels in spacetime with 4-speed equal to that of light!

On a space-time diagram light travels in such a way that with a Lorentzian signature, the temporal part always equals the spatial part.

FOURTH QUESTION:

I've read in many places:

"Well, obviously we get from (4) that:

(5) [tex]ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}=0[/tex]

Sorry, but that's not obvious to me. Could someone derive it for me? Please? I know that (5) would imply that the massless particle travels at the speed of light c, even if LHS of (4) is equal to zero, so I THINK that would solve question 4. But I just don't know how to get (5) from (4).

With

[tex]g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}=0[/tex]

Multiply through by [itex]d\lambda^2[/itex] to get[tex]ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}=0[/tex]

Given what I've said here, I'm still just a practicing student, and it would be good to have a professional verify/correct/expand on this. I hope this helps though.

 

[MManuel Abad]

Wow! Thanks a lot! You killed two birds with one shot! I didn't know about that affine parameter! Yeah. (5) wasn't obvious for me because I thought: I CANNOT multiply by d[tex]\tau²[/tex], because i'ts zero for a massless particle!

So you use an affine parameter for getting that problem away, and that stuff is then fixed! Thank you!

Wow, so there's just ONE geodesic equation. For a massive particle we can use the proper time as an affine parameter, and for a massless particle we use that lambda, a different affine parameter. And that's because we want to avoid that proper time, which is zero for those massless particles.

So, that condition for the velocities of massless particles is just the same as that of the interval being zero, isn't it? They're equivalent. Thank you so much. You've been very kind, clear and straight :D