How long does it take to reach the singularity in a free fall at r=0?

[LAHLH]

Let's say you start outside the event horizon r>2m, how long does it take according to your own wristwatch to hit the singularity at r=0 (ignoring the fact the tidal forces rip you apart...), assuming you just free fall and plunge in radially?

I'm thinking you need to consider the timelike radial geodesics since you are in free fall. These can be simplified by noting that K^{\mu}=(1,0,0,0) is a Killing Vector, so we have a constant of motion E=-K_{\mu}\tfrac{dx^{\mu}}{d\tau} =(1-\tfrac{2M}{r})\tfrac{dt}{d\tau}, then expanding g_{\mu\nu}U^{\mu}U^{\nu}=-1 and using the energy constant. I get: \left(\frac{dr}{d\tau}\right)^2=E^2-1+\tfrac{2m}{r}

Can I now just invert this to get \tfrac{d\tau}{dr} and integrating from the starting value of r to r=0? and get total proper time? \Delta\tau=\int^{0}_{R} \frac{d\tau}{dr}dr

(I calculate the above in ingoing Eddington Finkelstein coords ds^2=-(1-\frac{2m}{r})dv^2+2dvdr+angles)

 

[K^2]

You do need to use the Killing Vector for this, that's correct, but I'm not sure where you are getting this.

E = -K_{\mu}\frac{dx^{\mu}}{d\tau}

The momentum vector along Killing Vector, K_μp^μ, should be conserved. What you have up there is a proper velocity, and p^μ=mu^μ. The problem here is that m is not constant for object falling into a black hole. You can easily verify that by using the above constraint with the fact that u_μu^μ=c².

Take correction for variable m, and you should have it.

 

[LAHLH]

You do need to use the Killing Vector for this, that's correct, but I'm not sure where you are getting this.

E = -K_{\mu}\frac{dx^{\mu}}{d\tau}

The momentum vector along Killing Vector, K_μp^μ, should be conserved. What you have up there is a proper velocity, and p^μ=mu^μ. The problem here is that m is not constant for object falling into a black hole. You can easily verify that by using the above constraint with the fact that u_μu^μ=c².

Take correction for variable m, and you should have it.

thanks for the reply. hmm, I got the equation from Carroll chapter 5 (5.61). I believe he also states that E is energy per unit mass, so maybe that fixes things. In general he seems to use
K_{\mu}\frac{dx^{\mu}}{d\lambda}=constant however e.g. (5.54)...(text is free online if you don't have a copy). Although the derivation in chp3 does seem to indicate it's for p not u.

 

[K^2]

What's lambda?

The equation you stated is valid for constant mass, except the constant isn't going to be energy. It's going to be energy per unit of mass. It would be sufficient to find time to event horizon, but mass is not constant, and that's the whole point.

Use K_μp^μ = const, and u_μu^μ=c². You should be able to derive it from just these two things and the metric.

 

[LAHLH]

What's lambda?

The equation you stated is valid for constant mass, except the constant isn't going to be energy. It's going to be energy per unit of mass. It would be sufficient to find time to event horizon, but mass is not constant, and that's the whole point.

Use K_μp^μ = const, and u_μu^μ=c². You should be able to derive it from just these two things and the metric.

Well lambda is the affine parameter parametrising the geodesic. If we call the tangent vector to the geodesic u^a and consider \tfrac{d}{d\lambda} (K_a u^a)=u^c \nabla_c (K_a u^a)=u^a u^c \nabla_c K_a+K_a u^c \nabla_c u^a.

The second term here is zero by the geodesic equation (usually anyway in my proof perhaps this is where things are different in this case). Then the uu is symmetric so picks out the symmetric part in first term:

=u^a u^c \nabla_{(c} K_{a)} =0 if K is a KV.So anyway, I suspect you're going to say this proof falls because of the u^c \nabla_c u^a=0 by the geodesic equation part? because it should be p^c \nabla_c p^a =0 if m is not a constant along the motion. I can't argue with this, but am somewhat confused at why my notes and Carrol use the above. How is Carroll legitimate in using the form in my original post? Also the fact that \epsilon=-g_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda} is constant along the path seems to rely on a similar proof that doesn't quite work.
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EDIT: isn't the m in p^a=m_0 u^a actually the rest mass? so is a constant and everything works except the energy E I had is just the energy per unit mass (which Carroll states later on in the chapter.) 0=p^c \nabla_c p^a =m_0 u^c \nabla_c (m_0 u^a) which implies u^c \nabla_c u^a=0

 

[George Jones]

I often use R for Eddington-Finkelstein coordinates instead of the more usual r. Even though R = r is one of the coordinate transformation equations from Schwarzschild to Eddington-Finkelstein, \partial / \partial R \ne \partial / \partial r, and, unlike r, R is everywhere a lightlike coordinate! Also, v has the same causal properties as t, since \partial / \partial v = \partial / \partial t, even though v \ne t!

Let's say you start outside the event horizon r>2m, how long does it take according to your own wristwatch to hit the singularity at r=0 (ignoring the fact the tidal forces rip you apart...), assuming you just free fall and plunge in radially?

I'm thinking you need to consider the timelike radial geodesics since you are in free fall. These can be simplified by noting that K^{\mu}=(1,0,0,0) is a Killing Vector

Since \partial / \partial v = \partial / \partial t, this coordinate expression gives the same Killing vector in both coordinat systems.

so we have a constant of motion E=-K_{\mu}\tfrac{dx^{\mu}}{d\tau} =(1-\tfrac{2M}{r})\tfrac{dt}{d\tau}

But in the Eddington-Finkelstein coordinate system that you have used, there is no "t". What is (the same) E in the Eddington-Finkelstein coordinate system?

then expanding g_{\mu\nu}U^{\mu}U^{\nu}=-1 and using the energy constant. I get: \left(\frac{dr}{d\tau}\right)^2=E^2-1+\tfrac{2m}{r}

Can I now just invert this to get \tfrac{d\tau}{dr} and integrating from the starting value of r to r=0? and get total proper time? \Delta\tau=\int^{0}_{R} \frac{d\tau}{dr}dr

Yes. The constant E is determined by initial conditions. A typical iinitial condition is that the observer is initially at rest with respect to a hovering observer.

(I calculate the above in ingoing Eddington Finkelstein coords ds^2=-(1-\frac{2m}{r})dv^2+2dvdr+angles)

 

[Parkur]

Let's say you start outside the event horizon r>2m, how long does it take according to your own wristwatch to hit the singularity at r=0 (ignoring the fact the tidal forces rip you apart...), assuming you just free fall and plunge in radially?

For a particle released from rest at a Schwarzschild distance r=R from the origin, the proper time to fall to the origin r=0 is (pi R/2)sqrt(R/2m). You can find the derivation in any good book on relativity. See for example http://www.mathpages.com/rr/s6-04/6-04.htm