# 'Crossing' speed: Lemaitre vs. 'dropped' observer

1. Jan 21, 2014

### Jorrie

We are trying to find the relative velocity in a Lemaitre observer's frame, when said observer passes an observer dropped from rest at Painleve radius $r_0$. I have searched for related previous posts, found some, but none that clarified this particular situation.

Let the passing happens at $0 < r <= r_0$. We used Painleve coordinates, since this reaches the "interior" of the black hole. Since the relativistic velocity addition does not seem to work for "inside" velocities, exceeding c, we have tried proper velocities and rapidity. We would appreciate feedback as to the validity of the approach (or where we went of the rails).

The Lemaitre observer measures static (shell) observers passing him at proper velocity

$$w_{Lem(r)} = dr/d\tau =\sqrt{r_s/r}$$
where $r_s = 2GM$, the event horizon (Schwarzschild) radius.

His constant mechanical energy is $E/m = 1$. The observer dropped from $r_0$ has a constant $E_0/m = \sqrt{1-r_s/r_0}$ and she measures static (shell) observers passing her at proper velocity

$$w_{drop(r)} = \sqrt{E_0^2-(1-r_s/r)} =\sqrt {{\frac {r_s}{r}}-{\frac {r_s}{r_0}}}$$
Both $w$ are valid for any $r \le r_0$, except for $r \le 0$.

Now the question is: can we convert these two $w$ to rapidity $\eta(w) = \sinh^{-1}{(w)}$, subtract them for "relative rapidity" and then get a relative proper velocity by taking the $\sinh$ of that difference?

We have done this and there seems to be no "blow-ups" at or inside the horizon. This graph resulted:

https://www.physicsforums.com/attachment.php?attachmentid=65884&d=1390302266

The observers are dropped from R=2.5 and 1.1, with the bright red and green curves their respective (proper) speeds relative to the Lemaitre observer, at the instant his passing them. The light green and amber curves are their proper speeds on the Painleve chart.

Looks pretty, but is the method valid?

-J

Last edited: Jan 21, 2014
2. Jan 21, 2014

### Mentz114

It is not possible to have a stationary observer at $r\leq r_s$, so you can't carry this calculation past the horizon.

I haven't had time to consider if your approach works for $r_0>r_s$. The relative speed between two worldlines $\xi, \xi'$ when they coincide is given by

$\frac{||h_a^b \xi'_b||}{||k_a^b \xi'_b||}$

where $k$ and $h$ are the temporal and spatial projection tensors for $\xi$.

See Malament's notes, page 122. http://www.lps.uci.edu/lps bios/dmalamen [Broken].

Last edited by a moderator: May 6, 2017
3. Jan 21, 2014

### WannabeNewton

Yes but only if all the quantities are evaluated at the same event so as to employ standard SR formulas such as addition of rapidities. Therefore your method will only work outside of the event horizon because static observers only exist in that region.

EDIT: As a side note, did you intend to write $\tanh^{-1}(w)$? The rapidity $\eta$ corresponding to a relative velocity $w$ is $\eta(w) = \tanh^{-1}(w)$.

Last edited: Jan 21, 2014
4. Jan 21, 2014

### Jorrie

I was hoping that going to Painleve coordinates and proper velocities ($dr/d\tau$) would free us from the need for stationary observers. After all, we are only concerned with the relative velocity between two inertial frames here - the Lemaitre observer and the "dropped" observer. They should have a valid relative velocity even inside the horizon, except at the central singularity. How do we get that relative velocity?

-J

5. Jan 21, 2014

### Jorrie

The Lemaitre observer and the "dropped" inertial observer pass each other one place and time. Does that not qualify as the same event, even inside the horizon? I am interested in their SR relative velocity, either as coordinate $v$ or proper $w$, it does not matter.

AFAIK, this is for coordinate velocities; does proper velocities not require $\sinh$?

-J

6. Jan 21, 2014

### WannabeNewton

Your method makes explicit use of a static observer but static observers only exist outside of the event horizon so you can't apply your method inside the event horizon. Mentz already mentioned, in post #2, a method that is valid everywhere not only in Schwarzschild space-time but also arbitrary space-times.

I have personally never seen the term "proper velocity" used before. Regardless, the formula is for the relative velocity between two arbitrary observers at an event coincident on both their world lines. In the local Lorentz frame of one of the observers, this corresponds to making an infinitesimal measurement of length (using a ruler carried by said observer) in an infinitesimal increment of proper time (as read by an ideal clock carried by said observer) along the other observer's instantaneous line of motion. Looking back at your formula $\frac{dr}{d\tau} = \sqrt{\frac{2M}{r}}$, which is the velocity of a Painleve observer relative to a static observer whom the Painleve observers passes by, it seems that you are using the term "proper velocity" to mean relative velocity and this is exactly what the standard rapidity formula is meant for.

7. Jan 21, 2014

### Mentz114

I calculated the value of the relative velocity I cited above, for the case of the faller and stationary observer in Schwarzschild and it comes out as

$\sqrt {{\frac {r_s}{r}}-{\frac {r_s}{r_0}}}$

which is no surprise. Details available if required.

8. Jan 21, 2014

### Mentz114

I misunderstood, the problem ( as usual) so I'll recalculate the proper case for fun.

9. Jan 21, 2014

### WannabeNewton

There was no misunderstanding. What you calculated is $\xi^{\mu}\xi'_{\mu} = -\gamma$ which is exactly the relative Lorentz factor between the two observers at a coincident event on their world lines. From this we get the magnitude of the relative velocity of one observer with respect to the other at the coincident event. Nowhere in this calculation does coordinate velocity show up. Relative velocity is the exact same thing as the OP's "proper velocity" based on the OP's expressions in post #1. What you did is analogous to what I did here: https://www.physicsforums.com/showthread.php?t=724848&page=4

Last edited: Jan 21, 2014
10. Jan 21, 2014

### Mentz114

The OP has 2 falling observers. My calculation was for one static and one falling.

With two falling observers, one released at infinity, and one at $r_0$ there is an interesting result. With
$\xi = \frac{r}{2\,M-r}\partial_t - \frac{\sqrt{2}\,\sqrt{M}}{\sqrt{r}}\partial_r$
and
$\xi' = \frac{\sqrt{r}\,\sqrt{2\,\left( 1-K\,r\right) \,M-2\,M+r}}{2\,M-r}\partial_t -\frac{\sqrt{2}\,\sqrt{1-K\,r}\,\sqrt{M}}{\sqrt{r}}\partial_r$
where $K=1/r_0$ is a constant. I get ( I can't get the latex to display. The bad radical extends over the parentheses).

$\frac{||h_{ab}\xi'^b||}{||k_{ab}\xi'^b||}=\sqrt{\frac{4\,\sqrt{r}\,\sqrt{(1-K\,r)}\,M\,\sqrt{r-2\,Kc\,r\,M}+4\,K\,r\,{M}^{2}+\left( 2\,K\,{r}^{2}-4\,r\right) \,M}{4\,\sqrt{r}\,\sqrt{1-Kc\,r}\,M\,\sqrt{r-2\,K\,r\,M}+\left( 4\,K\,r-4\right) \,{M}^{2}+2\,K\,{r}^{2}\,M-{r}^{2}}}$

If $K=0$ this is zero, as expected because $\xi^a=\xi'^a$. If $r=2M$, the result is 1 !

As a check, I also solved $\frac{1}{1-\beta^2}=\left(\xi^a \xi'_a \right)^2$ for $\beta$ and got the same answer.

I guess I made a mistake but I'll stick with it until someone else calculates a different result.

11. Jan 21, 2014

### WannabeNewton

Yes but what the OP was doing was calculating the relative velocity between a static observer and a Painleve observer, the relative velocity between the same static observer and a radially infalling observer starting from some finite radius, and using addition of rapidities at a coincident event for all three observers in order to get the relative velocity between the Painleve observer and the "dropped" observer. Your method sidesteps the need to use a static observer (which is only valid outside the event horizon) by directly calculating the relative velocity between the Painleve and "dropped" observers, so I'm referring to your method and not your explicit calculation(s); I should have made that clearer before.

12. Jan 21, 2014

### Jorrie

OK thanks. I'll check out Mentz's method, especially his later explicit solution.

On proper velocity, I was under the impression that the Painleve velocity $dr/d\tau= \sqrt{r_s/r}$ is not a relative velocity, because it exceeds c inside the horizon, which is a property of "proper velocity". I notice the term here and there in the literature, e.g. http://www.lecture-notes.co.uk/susskind/special-relativity/lecture-5/proper-velocity-and-momentum/.

I can now see that it may not be applicable to the problem that I stated.

-J

13. Jan 21, 2014

### WannabeNewton

Ah ok, I see the distinction, thanks! In the context of your specific calculations, the distinction only arises when considering relative velocities that involve static observers so if you stick to talking about only the Painleve (Lemaitre) observer and the "dropped" observer then you won't have to worry about the distinction between "proper velocity" and relative velocity that you elucidated.

MTW has a number of exercises that have you perform these kinds of calculations (going to the local Lorentz frame of a specific observer and using local rods and clocks to make measurements of local observables such as relative velocities) so let me see if I can find some that will be of use to you.

14. Jan 22, 2014

### Jorrie

I have stuck some values into your result:

It gives me the same numbers all the way into the horizon than what a naive use of the SR velocity addition yields, so it seems correct, with (c=GM=1):

$v_1 = \sqrt\frac{2}{r}$
$v_2 = \frac{dr}{d\tau} = \frac{\sqrt{E^2-(1-2/r)}}{E}$
$v_{rel} = \frac{v_1-v_2}{1-v_1v_2}$

I understand that the SR velocity addition suffers the same issue of having to refer to static observers, but I think one can view it as not "static observers inside $r_s$", but rather as a fixed Painleve chart $r$, which is regular for any $r > 0$.

According to Taylor & Wheeler's "Exploring Black Holes", Chapter 7, Fig. 1, the Lemaitre observer can in principle determine his instantaneous $r$ coordinate locally. Knowing his local time, he can in principle determine his Painleve coordinate velocity $\Delta r/\Delta \tau$. Whether this is also possible for a "dropped" observer is not clear to me.

-J

15. Jan 22, 2014

### pervect

Staff Emeritus
I used ingoing eddington-finklestein coordinates to find the doppler shift from a light source "at infinity" in https://www.physicsforums.com/showthread.php?p=4637332#post4637332

Basically, I wrote the geodesic equations as a function of proper time $\lambda$, then reparameterized them in terms of r, using the chain rule.

You should be able to find the relative gamma factor by dividing two expressions - i.e. if one observer has a 2:1 doppler shift from a light beam (in this case, a light beam from radial infinity), and a collocated observer has a 4:1 doppler shift from the exact same light beam, the relative gamma should be 2. This is easy to see just by imagining one observer retransmitting the recieved signal to the other, but can also be derived formally by the relativistic velocity addition formula and the relativistic doppler shift formula.

If you have observers with a maximum height of R1 and R2, R1 > R2, the relative gamma factor seems complicated in general. At the event horizon it's a special case that's a lot easier, and I get:

$\gamma = \sqrt \frac{1-2m/R1}{1-2m/R2}$

which doesn't depend on r as in this special case it factors out. (But r doesn't seem to factor out in general).

This was an extension of some work I previously did which expands it to cover the more general case. I checked it as best as I could, but it sort of grew into a fairly big project.

16. Jan 22, 2014

### Mentz114

I hope you spotted the spurious 'c''s in my latex. Sorry that was a typo because K was called %c originally. Corrected above.

I'm glad to hear you get sensible numbers. I wasn't pleased that the result was not 'pretty'.

That was in Schwarzschild coordinates. I'm having some difficulty with a 'dropped' faller in Painleve coords, where $ds^2=-(1-r_s/r)dt^2-2\sqrt{r_s/r}dr dt+dr^2 + r^2d\Omega^2$
The faller from infinity has 4-velocity $\partial_t-\sqrt{r_s/r}\partial_r$ but the dropped geodesic is not so simple and does not give the above in the limit dropped height goes to infinity.

I'll try to sort this out but pushed for time.

Last edited: Jan 22, 2014