# Coordinate velocity and free-falling past an event horizon

Ws can re-arrange the standard Schwarzschild metric as $$\left(\frac{d\tau}{dt}\right)^2=\frac{r-1}{r}\left[1-\left(\frac{r}{r-1}\frac{dr}{dt}\right)^2\right]$$ Where $\frac{r-1}{r}$ is the local speed of light at $r$ (http://www.physik.uni-augsburg.de/annalen/history/einstein-papers/1911_35_898-908.pdf). Does this equation not suggest that the rate at which a clock of a free falling observer ticks $\left(\frac{d\tau}{dt}\right)$ at $r$ is slower relative to the clock an observer at rest at $r$? Furthermore, since the difference in the square brackets must be positive in order for $\frac{d\tau}{dt}$ to be real, this should imply that $$\left|\frac{dr}{dt}\right|\le\frac{r-1}{r}$$ at a given $r$ (the coordinate velocity of an observer at $r$ must be less than or equal to the coordinate belocity of light at $r$. Therefore, we see from the first equation that the maximum $\frac{dr}{dt}$ of a falling observer must decrease to zero at the horizon, meaning that, even in the free-falling observer's frame, they will never cross the horizon because their coordinate velocity will be forced toward zero as they approach the horizon. Since and observer in motion at $r$ can never have a coordinate velocity greater than the speed of light at $r$, and the coordinate speed of light at $r$ is zero, shouldn't the curvature of the spacetime coordinates force the observer to a 'stop' as it approaches the horizon such that it can never cross (even in the free falling observer's frame). The equation seems to suggest that there is not a coordinate singularity at $r=1$, but rather the rate at which a free falling observer's clock ticks goes to zero at the horizon (i.e. the observer would need to follow a light-like interval to cross the horizon). It seems to me that this suggests that nothing can actually cross the horizon in any reference frame and possibly that black holes couldn't form at all (because the collapsing material could never collapse beyond the horizon). I've written a paper that builds on this idea, but have been unable to get it published, which tells me I'm missing something basic. I'm hoping someone here can help me understand where I've gone wrong.

So my question is: what is the flaw with interpreting the above arguments as evidence that a black hole horizon cannot be crossed, even in the free faller's reference frame?

## 1 Answer

A choice of coordinate system doesn't change the physics.

We can consider a time like curve heading towards the event horizon and notice that it has a finite amount of proper time along the section between the outside and the horizon.

So when you claim never for the frame of the infalling particle, no one will be impressed by some choice of coordinate system.

If you truly think other people that have already done this calculation made a mistake and your coordinate system shows their mistake and how to do it correctly, then show the mistake they made.

Now there is some room for irony. You might turn around and ask me to find your mistake. And that's why I asked if truly think other people made a mistake. And it sounds like you think you missed something basic.

And that's why I pointed out that a choice of coordinates doesn't change the physics.

Your coordinate system only cover $r\gt r_s.$ And the region outside that is covers is fine. Out there you can have a curve where $r$ decreases and you just have to make sure your $t$ changes too as you approach the surface $r=r_s.$ Keep in mind you'd have to so that even if you curve was spacelike in the coordinate system you chose.

Therefore, we see from the first equation that the maximum $\frac{dr}{dt}$ of a falling observer must decrease to zero at the horizon,

We see no such thing. Becasue your coordinatevsyatem does have an "at the horizon." If you select a cirve from outside the horizon that goes towards the horizon. Then there is an infinite amount of coordinate time elapsed between the outside event and the corresponding later event on the horizon along the finite proper time curve between them.

Sure. That happens but that's just a coordinate artefact of the particular way that your coordinate system doesn't extend there. That's the same thing I've been saying the entire time. There is a curve with a finite proper time where you travel an infinite amount of denominator coordinate. That just means your coordinate system is bad. In a Kruskal-Szekeres coordinate system the lines of finite Schwarzschild time are radial lines between the event horizons of the white and black holes. When going towards the black hole event horizon you pass through every single Schwarzschild coordinate time later than your starting Schwarzschild time before you hit the horizon.

It means nothing else. It means nothing physically. Here's an example. If you understand this example then all your questions should be answered.

Take flat empty regular Minkowski space with regular boring coordinates $t,$ $x,$ $y,$ and $z.$ Then choose new coordinates $T=\arctan (t/t_0),$ $X=x,$ $Y=y,$ and $Z=z.$ Now the space is just as boring as always. And now your coordinate system for $T,$ $X,$ $Y,$ and $Z$ is a coordinate system that only covers part of the spacetime. Just like for your example. And now you could argue that $dx/dt$ goes to zero as you approach the surface $t=t_0\pi/2$ and that you approach this surface after elapsing a finite amount of proper time.

But this is literally just a particle sitting at rest in Minkowski space for at most a Minkowksi time duration of $\pi t_0\lt\infty$ with nothing going on except you choose a coordinate system that required it go through an infinite amount of denominator coordinate before getting there. But there was nothing going on physically.

All that happens in your example is your surface of $T=\infty$ happens to be called the surface $r=r_s$ and you acted like it was a place.

But there is no content to your objection and there is nothing going on except that an infinite amount of coordinate gets traveled in a finite amount of proper time and that means nothing.

even in the free-falling observer's frame, they will never cross the horizon because their coordinate velocity will be forced toward zero as they approach the horizon.

That's just having an infinite amount of coordinate $T$ to travel in a finite amount of other coordinate it means nothing about anything.

Since and observer in motion at $r$ can never have a coordinate velocity greater than the speed of light at $r$,

Why does would anyone care about coordinate velocity? Especially when it goes to zero for a boring reason like having an infinite amount of denominator coordinate be cramed into a curve of finite proper time.

the coordinate speed of light at $r$ is zero, shouldn't the curvature of the spacetime coordinates

But that happened even in our Minkowski spacetime example where there was no curvature, just an infinite amount of denominator coordinate crammed into a curve of finite proper time.

The equation seems to suggest that there is not a coordinate singularity at $r=1$, but rather the rate at which a free falling observer's clock ticks goes to zero at the horizon (i.e. the observer would need to follow a light-like interval to cross the horizon).

It suggests no such thing. The equation suggests you can compute the proper time of a time like curve. And it clearly assigns no finite $t$ to the surface $r=r_s.$

If you still have trouble I have an example even closer to the Schwarzschild coordinate system.

Consider Minkowski spacetime again, this time 2d (just $x$ and $t$ to make it easier).

We decide to make a coordinate system that only covers events spacelike separated from the origin, which is fine, a single coordinate system doesn't have to cover all of spacetime.

So we are covering the points $(x,t)$ with $|ct|\lt |x|$.

In the original coordinates $c^2\tau^2=c^2dt^2-dx^2$ and clearly a particle can sit at rest and hit the the surface $|ct|=|x|$ after waiting a finite amount of proper time (or $t$ time) provided it started out with $|ct|\lt |x|$.

Now we are going to choose new coordinates. We choose the lines $x=vt$ for $|v|\gt c$ to be our curves of constant $T$ and choose specifically that $$T(x,t)=\arctan\left(\frac{\pi ct}{2|x|}\right).$$ And each of these lines is spacelike, so are perfectly fine for a line of constant $T.$

Next we choose hyperbolas of constant $x^2-(ct)^2$ to be curves of constant $X$ and choose specifically that $$X(x,t)=\frac{x}{|x|}\left(1+\sqrt{x^2-(ct)^2}\right).$$ And each of these hyperbolas has a tangent that is timelike so the are perfectly fine for surfaces of constant coordinate for a spacelike coordinate.

Our new $X,T$ coordinate system is perfectly fine for the events where $|ct|\lt |x|$. And now if you write down the metric in the new coordinates you'll see that a test particle of constant $x$ that hits the horizon at $X=1$ after a finite amount of proper time has $|dX/dT|$ approach zero as $X$ approaches the horizon $X=1.$

But all this means is that we set it up to have a finite coordinate amount of $X$ gets traversed as an infinite coordinate amount of $T$ gets traversed as you go from $(x,t)=(9m,0s)$ (or $(X,T)=(4,0)$) to $(x,t)=(9m,9m/c)$ (or $(X,T)=(1,+\infty)$).

Hopefully you can see that $X=1$ is $T=+\infty$ and that as your coordinate $X$ goes from a. Finite number bigger than 1 to 1 and the $T$ coordinate goes from a finite number to $T=+\infty$ you must have $|dX/dT|$ go to zero.

But it doesn't physically mean the particle doesn't reach the horizon. Its just a particle at in Minkowski space sitting there for time $9m/c$ and in its frame it definitely gets there.

Your main problem is thinking an infinite amount of coordinate means absolutely anything when a curve has a finite proper time. It doesn't mean anything. Unless maybe you want to relate that coordinate to something physical, such as what someone else sees. In which case an infinite coordinate would mean that the someone else doesn't see it. But it doesn't mean anything about someone that leaves that coordinate system in finite proper time.

• Thank you for the comment. I agree that it is absurd to think that people studying black holes would have missed anything basic, which is why I wrote a detailed question so that I could understand where I went wrong. As far as the dividing by zero, the point here is that if r/(r-1) goes to infinity while dr/dt goes to zero, then you do not have a coordinate sigularity at r=1 because zero times infinity can be a finite number in the limit (this is essentially how derivatives are defined). Since, in order for the equation to remain real away from the horizon, the allowable dr/dt must be – Chris L. Oct 17 '15 at 20:28
• less than or equal to (r-1)/r, we can see that in the limit, (r-1)/r*dr/dt goes to 1 because dr/dt goes to (r-1)/r when r=1. So I assume there's something flawed in that argument since the concensus is that there is a coordinate singularity at r=1, but I'm trying to figure out what the flaw is. Also, these coordinates are just the plain old Schwarzschild coordinates. – Chris L. Oct 17 '15 at 20:31
• @PetTaxi Use better coordinates, such as Kruskal-Szekeres if you want to see the horizon clearly. I've updated my answer. – Timaeus Oct 17 '15 at 22:16
• ok, so what you're saying is that the mistake I made at the beginning was factoring out (r-1)/r, which is not allowed. If that is forbidden then the rest of the argument falls apart. – Chris L. Oct 18 '15 at 15:21
• Also, when I was saying 'at r', I didn't mean 'at rest at r', I meant 'at an instant at r', so it should have been stated as 'at a given r and t' I think. – Chris L. Oct 18 '15 at 15:25