Radial motion in the Schwarzshild spacetime

[Theactualbman]

Hello,

The following has been confusing my friends and I, I want to make sure I have this clear as it is fairly elementary. (note set c = 1)

Ed is falling radially into a black hole, the Schwarzschild metric is:

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

his proper time is dτ² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

If Simon is stationary and is coincident with Ed at some radius then he measures the proper distances and times given by:

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

I think that's all right. But then is Simon's metric dt_s² - dr_s² ≈ ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr² or are these really meant to be different bases?

Many thanks.

 

[xox]

Hello,

The following has been confusing my friends and I, I want to make sure I have this clear as it is fairly elementary. (note set c = 1)

Ed is falling radially into a black hole, the Schwarzschild metric is:

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

his proper time is dτ² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

correct (there are no rotational terms)

If Simon is stationary and is coincident with Ed at some radius then he measures the proper distances and times given by:
dr_s = (1 - 2μ/r)^-1/2 * dr. I think that's all right.

The above doesn't seem right. For Simon, dr=0.

 

[Theactualbman]

Yes that's what I thought, but if one solves the geodesic equations you find dr/dτ = -√2μ/r. If instead of perameterizing the metric in terms of τ, we instead use the proper time and proper distance of Simon then we find the corresponding derivative dr_s/dt_s = -√2μ/r, I found this in a bunch of textbooks (e.g. General Relativity (Hobson et al)). The real dilemma for me is whether the dr_s and dt_s correspond to different parameterizations of the metric, or if they're part of the same one I wrote above.

 

[George Jones]

If Simon is stationary and is coincident with Ed at some radius then he measures the proper distances and times given by:

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

I think that's all right.

These relations apply to different situations.

Suppose that Simon is standing on a spherical shell that has r constant on its shell. and that two pennies plop down at Simon's. The two plops happen at the same place, dr = 0, but at different times. The second equation relates the time difference between plops according to Simon's watch to the difference in Schwarzschild coordinate time.

Now suppose that there is a second spherical shell just bellow Simon's spherical shell. Simon drills a hole in his shell, sticks a metre stick through the hole, and measures the spatial distance between the shells. The two positions of the shells against the metre stick are taken simultaneously, dt =0. The first equation relates the metre stick distance to the difference in Schwarzschild r.

 

[xox]

The real dilemma for me is whether the dr_s and dt_s correspond to different parameterizations of the metric.

They do. You cannot combine them into one metric as you tried.

or if they're part of the same one I wrote above

You cannot combine them into one metric as you tried.

 

[Theactualbman]

Okay, so then in Simon's small patch of the spacetime, he sees Ed go past at dr_s/dt_s.

How does Simon perameterize the invariant ds². Is it even the same 'invariant' as Ed's?

And I still don't quite understand why you can't write

dt_s² - dr_s² ≈ ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr² , to my mind it follows by direct substitution and describes the flat tangent space to the point where Simon is.

(also, forgive me if I seem argumentative, I'm sure you're right, but I want this to be clear)

 

[WannabeNewton]

$d\tau^2 - d\rho^2$ is not Simon's metric (I am denoting proper time and proper distance by $\tau,\rho$). Why would you think this to be the case? $d\tau^2$ and $d\rho^2$ are obtained from two entirely different situations along Simon's world-line.

 

[PAllen]

If Simon set up local coordinates following the conventions used setting up standard inertial coordinates in SR, he would would end up with a local metric that would approach the Rindler metric very near him (actually, the Rindler metric translated to have an origin away from the horizon). If Ed did the same, his metric would very locally be the standard SR Minkowski metric. These statements are true independent of the global geometry (SC or otherwise). They are direct consequences of the definition of pseudo-Riemannian manifold, specifically that the tangent space is everywhere Minkowski space.

 

[PeterDonis]

if one solves the geodesic equations you find dr/dτ = -√2μ/r.

Simon is not moving on a geodesic trajectory (though Ed is). He is not in free fall; he feels acceleration (he has to in order to stay at a constant altitude).

 

[xox]

Okay, so then in Simon's small patch of the spacetime, he sees Ed go past at dr_s/dt_s.

That would mean that, according to your calculations, Simon measures the radial speed v=\frac{1}{1-2 \mu/r} \frac{dr}{dt} for Ed. This is not only incorrect, it is also useless, since it is dependent on \frac{dr}{dt}.
The correct derivation is considerably more complicated, you can find it on page 27 of "Black Holes, an Introduction" by Raines and Thomas. I highly recommend this little book, it is excellent.

PS: the correct answer is :v=\sqrt{\frac{\mu/r}{1-2 \mu/r}}. I could derive it for you, if you are really interested but I think that you could benefit a lot more from buying the book.

 

[Theactualbman]

So Simon cannot construct a Minkowski metric around him since he isn't in freefall, that makes a lot more sense.

 

[PeterDonis]

So can Simon not construct a Minkowski metric around him since he isn't in freefall?

He can, but it will only be valid in a small patch of spacetime around a particular event on his worldine. And his worldline will not look like a straight line in the Minkowski metric in that small patch; it will look like a hyperbola because he is accelerated (Ed's worldline will look like a straight line).

And is that why he can't have a metric (I should have said line element, but the intended metric follows pretty simply) of the form I said?

If you mean a metric in which Simon's worldline is a straight line, then yes, it can't be of the form you said. A metric in which Simon's worldline is a straight line will be, as PAllen pointed out, a Rindler metric.

 

[George Jones]

PS: the correct answer is :v=\sqrt{\frac{\mu/r}{1-2 \mu/r}}.

This isn't correct. If Ed falls from rest from a large distance, then Ed's passing speed, as meausred by Simon, is
$$v= \sqrt{\frac{2 \mu}{r}}.$$

 

[xox]

This isn't correct. If Ed falls from rest from a large distance, then Ed's passing speed, as meausred by Simon, is
$$v= \sqrt{\frac{2 \mu}{r}}.$$

The book I quoted seems to disagree. As far as I know, \sqrt{\frac{2 \mu}{r}} is Ed's proper speed, not the coordinate speed as measured by Simon.

 

[PAllen]

The book I quoted seems to disagree. As far as I know, \sqrt{\frac{2 \mu}{r}} is Ed's proper speed, not the coordinate speed as measured by Simon.

If Simon sets up standard local coordinates, e.g. Fermi-Normal (which would take the form of a translated Rindler metric in this case), then the coordinate speed using these locally physically meaningful coordinates would match the proper speed. More importantly, this is the speed any reasonable experimental procedure would measure.

 

[Theactualbman]

v=\frac{1}{1-2 \mu/r} \frac{dr}{dt} then gives that result doesn't it (applying the geodesic equations to Ed who follows a geodesic). Is it the case that in the neighbourhood of Simon the metric I suggest holds, but only extremely locally?

 

[George Jones]

The book I quoted seems to disagree. As far as I know, \sqrt{\frac{2 \mu}{r}} is Ed's proper speed, not the coordinate speed as measured by Simon.

Okay, so then in Simon's small patch of the spacetime, he sees Ed go past at dr_s/dt_s.

This means actualbman is considering what you (xox) call "proper speed". Note also that I wrote

"Ed's passing speed, as measured by Simon", whereas you did not qualify your speed. Unqualified, I took v to be physical speed. What else would would Simon measure? Coordinate speed? Why and how would Simon measure this?

I gave what Raine and Thomas call $v_{\mathrm{loc}}$, "the velocity of the object relative to the shell" on page 39 of their second edition. My first edition is at home, so I don't know the page in it.

 

[Theactualbman]

Also xox isn't the result you quoted for an observer at infinity?

 

[xox]

Also xox isn't the result you quoted for an observer at infinity?

No, it isn't.

 

[xox]

This means actualbman is considering what you (xox) call "proper speed". Note also that I wrote

"Ed's passing speed, as measured by Simon", whereas you did not qualify your speed.

I qualified it by referring the reader to page 27 , the 2006 edition.

I gave what Raine and Thomas call $v_{\mathrm{loc}}$, "the velocity of the object relative to the shell" on page 39 of their second edition.

I cannot locate anything like the above in my 2006 book.

 

[WannabeNewton]

The book I quoted seems to disagree. As far as I know, \sqrt{\frac{2 \mu}{r}} is Ed's proper speed, not the coordinate speed as measured by Simon.

An observer cannot measure coordinate velocity; an observer can measure only relative velocity, a measurement that is necessarily done through his local lattice of rulers and clocks yielding always physical observables (mathematically they project Lorentz indices of vector and tensor fields onto the local Lorentz frame of the observer).

Incidentally, there is no difference in value between coordinate velocity and relative velocity when one measures the velocity of a radially freely falling observer who starts from rest at spatial infinity relative to a stationary observer at some $r$. This is of course just a coincidence in this case as the local ruler distance and local time of the stationary observer are such that their being affected by the space-time geometry happen to cancel one another when calculating said relative velocity.

More explicitly, the local Lorentz frame of this stationary observer is simply $e_t = (1 - 2M/r)^{-1/2}\partial_t, e_r = (1 - 2M/r)^{1/2}\partial_r$ where I have ignored the irrelevant local polar and azimuthal axes. The 4-velocity of a radially freely falling observer, who starts at rest at spatial infinity, written relative to the unphysical coordinate frame $\{\partial_t, \partial_r\}$ is simply $u = (1 - 2M/r)^{-1}\partial_t - (2M/r)^{1/2}\partial_r$ which, expressed in terms of the stationary observer's local frame, becomes $u = (1 - 2M/r)^{-1/2}e_t -(2M/r)^{1/2}(1 - 2M/r )^{-1/2}e_r$ from which it is clear that $v = |\frac{\langle u, e_r\rangle}{\langle u, e_t\rangle} | = (2M/r)^{1/2}$ which is coincidentally also the coordinate speed $|\frac{\langle u, \partial_r\rangle}{\langle u, \partial_t\rangle}|$ of the freely falling observer.

 

[xox]

The 4-velocity of a radially freely falling observer, who starts at rest at spatial infinity, written relative to the unphysical coordinate frame $\{\partial_t, \partial_r\}$ is simply $u = (1 - 2M/r)^{-1}\partial_t - (2M/r)^{1/2}\partial_r$ which, expressed in terms of the stationary observer's local frame, becomes $u = (1 - 2M/r)^{-1/2}e_t -(2M/r)^{1/2}(1 - 2M/r )^{-1/2}e_r$ from which it is clear that $v = |\frac{\langle u, e_r\rangle}{\langle u, e_t\rangle} | = (2M/r)^{1/2}$ which is coincidentally also the coordinate speed $|\frac{\langle u, \partial_r\rangle}{\langle u, \partial_t\rangle}|$ of the freely falling observer.

Sorry, I was looking at the wrong motion (orbital, rather than radial). My bad.

 

[pervect]

THe thread has gotten long already, but I thought I'd revisit the very first question.

Hello,

The following has been confusing my friends and I, I want to make sure I have this clear as it is fairly elementary. (note set c = 1)

Ed is falling radially into a black hole, the Schwarzschild metric is:

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

his proper time is dτ² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

Using the Schwarzschild metric, anyone's proper time (not just Ed's) is given by that formula. Note that you have toi integrate $d\tau$ along some particular curve to get a proper time. For Ed, you integrate $d\tau$ along Ed's worldline, to get a proper time interval.

If Simon is stationary and is coincident with Ed at some radius then he measures the proper distances and times given by:

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

I'm not sure what you mean by dr_s, you wrote this formula down without fully explaining what it was you wanted to compute, but I would agree that you integrate $\sqrt {\ \left| ds^2 \right| }$ to get the length of a curve, and that if your curve is timelike this length is a proper time, and if your curve is spacelike this length is a proper length. Depending on sign conventions, ds^2 is positive for space and negative for time, or vica-versa, hence the absolute value. The metric you specified has ds^2 positive for timelike intervals and negative for spacelike intervals.

I think that's all right. But then is Simon's metric dt_s² - dr_s² ≈ ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr² or are these really meant to be different bases?

I don't understand this question, sorry. What do you mean by "Simon's metric"? Simon can use any metric he wants, it appears that you have been using the Schwarazschild metric all along, so I assumed that's what Simon would be using. Are you thinking the metric is physically determined someohw? The metric is not physically determined by anything, it's determined by one's choice of coordinates which is a human choice, and hence nothing physical.

I suspect you MAY be trying to ask "how do you get a locally inertial frame" for Ed and Simon, but you didn't actually ask that, so I may have misunderstood the intent of your question.

If you WERE asking about locally inertial frames, the various posts that suggest that you use Fermi Normal coordinates is a good answer.

I'll add that Ed's Fermi-Normal coordinates are not the same as Simons, and that the metric depends (as always) on the coordinate choice, so Ed's Fermi Normal coordinates, Simon's Fermi Normal coordinates, and Schwarzschild coordinates all have different metrics.

We could try and go into Fermi Normal coordinates in more depth if that is actually the question - I'm not sure if the phrase means anything to you , or is just "word soup". But if you're not interested, a long exposition would be a waste at this point.

 

[George Jones]

I am at home now, so I have the first edition, but I can't look closely, as I am heading out with my family.

I qualified it by referring the reader to page 27 , the 2006 edition.

I gave what Raine and Thomas call $v_{\mathrm{loc}}$, "the velocity of the object relative to the shell" on page 39 of their second edition.

I cannot locate anything like the above in my 2006 book.

On page 35 of the first edition, look at the paragraph containing equation (2.33).

The equation that you posted from page 27 of the first edition is prefaced by "What is the orbital speed of a particle as measured by a hovering observer?", i.e., the equation is a physical speed as measured by Simon, but it is not (in general) the speed of a radially freely falling Ed, it is Ed's speed (as measured by Simon) if Ed is in a freely falling circular orbit that just "skims" Simon's shell.

Note v_esc in problem 17 just below this. By symmetry, this Ed's speed, as measured by Simon, if Ed falls freely and radially from rest at a large distance (infinity).

 

[xox]

I am at home now, so I have the first edition, but I can't look closely, as I am heading out with my family.
On page 35 of the first edition, look at the paragraph containing equation (2.33).

The equation that you posted from page 27 of the first edition is prefaced by "What is the orbital speed of a particle as measured by a hovering observer?", i.e., the equation is a physical speed as measured by Simon, but it is not (in general) the speed of a radially freely falling Ed, it is Ed's speed (as measured by Simon) if Ed is in a freely falling circular orbit that just "skims" Simon's shell.

Note v_esc in problem 17 just below this. By symmetry, this Ed's speed, as measured by Simon, if Ed falls freely and radially from rest at a large distance (infinity).

Yes, I was looking at the orbital (not radial) speed, my bad.
Coordinate speed for radial motion for a distant observer is \frac{dr}{dt}=-\sqrt{\frac{2m}{r}} (1-\frac{2m}{r})
Proper speed for radial motion (for a distant observer) is \frac{dr}{d \tau}=\frac{dr}{dt} \frac{dt} {d \tau}=-\sqrt{\frac{2m}{r}} (1-\frac{2m}{r}) \frac{1}{1-2m/r}=-\sqrt{\frac{2m}{r}}
Proper speed for radial motion for the shell observer (Simon) is the same as the proper speed for the distant observer, -\sqrt{\frac{2m}{r}}

 

[Theactualbman]

That's okay, I don't know anything about Fermi normal coordinates, and I've only used Rindler coordinates once.

I was confused since I thought I could just construct Minkowski coordinates for Simon (corresponding to the proper distances and proper times which they don't) but he's not following a geodesic. The really important thing is getting the radial speed for Ed that Simon would measure, but if when Simon and Ed are cooincident we know that Simon measures proper distances r_s and proper times t_s then it seems to stand to reason that he'll measure the velocity of Ed in freefall as dr_s/dt_s.

 

[Theactualbman]

Which gives the same result as WannabeNewton and the others have.

 

[pervect]

That's okay, I don't know anything about Fermi normal coordinates, and I've only used Rindler coordinates once.

I was confused since I thought I could just construct Minkowski coordinates for Simon (corresponding to the proper distances and proper times which they don't) but he's not following a geodesic.

You can construct locally Minkowskii coordinates for the stationary observer (Simon) just by scaling the Schwarzschild coordinates by a constant factor. They won't be globally Minkowskian, however, that's impossible.

You can also you the basis vector formalism, by constructing an orthonormal set of basis vectors. This is what most textbooks do, the wrinkle here is the basis vectors really span the tangent space. If you are satisfied with approximations that cover a small area, you can ignore the distinction between the tangent space and the manifold, just as you can navigate on the curved manifold of the Earth's surface by constructing a flat plane tangent to it, and navigating as if you were on the flat plane.

An typical example of a basis vector in the basis vector formailsm would be $\hat{r}$, a unit-length vector in the R direction. You'll see them sometimes written with different notation, that may be less intuitive, as well, for instance $\partial / \partial r$.
[/quote]

The really important thing is getting the radial speed for Ed that Simon would measure, but if when Simon and Ed are cooincident we know that Simon measures proper distances r_s and proper times t_s then it seems to stand to reason that he'll measure the velocity of Ed in freefall as dr_s/dt_s.

There are several posts on how to get the radial speed, I think all of the science advisors got the same (right) answer, some non-SA's got different answers and may still be arguing about it.

 

[xox]

The really important thing is getting the radial speed for Ed that Simon would measure, but if when Simon and Ed are cooincident we know that Simon measures proper distances r_s and proper times t_s then it seems to stand to reason that he'll measure the velocity of Ed in freefall as dr_s/dt_s.

No, it doesn't. It is a pure coincidence that dr_s/dt_s gives the right answer, this was pointed out to you repeatedly by different posters. The correct derivation is considerably more involved than dividing dr_s by dt_s. The first page of this thread is occupied by posts that show your approach to be unfounded.

 

[WannabeNewton]

Which gives the same result as WannabeNewton and the others have.

See if this helps: https://www.physicsforums.com/showpost.php?p=4586813&postcount=55

 

[George Jones]

No, it doesn't. It is a pure coincidence that dr_s/dt_s gives the right answer, this was pointed out to you repeatedly by different posters. The correct derivation is considerably more involved than dividing dr_s by dt_s. The first page of this thread is occupied by posts that show your approach to be unfounded.

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

I have posted that there is more going on than meets the eye, but it is more than a coincidence the division $dr_\mathrm{S}/dt_\mathrm{S}$ gives the physical velocity. I"m not sure, but WannabeNewton might be gnashing his teeth over what seems to be a particularly egregious example of physicists' handwaving mathematics (often useful!). Actually, the division is mathematically rigourous(!), and does give the answer in a somewhat intuitive way.

the 4-velocity of a radially freely falling observer, who starts at rest at spatial infinity, written relative to the unphysical coordinate frame $\{\partial_t, \partial_r\}$ is simply $u = (1 - 2M/r)^{-1}\partial_t - (2M/r)^{1/2}\partial_r$

The trick is take what Theactualbman defined seriously as differential forms, and to evaluate these differential forms using Ed's 4-velocity. Then, the division makes mathematical sense. There is a slight abuse of notation in using "d" on the right of Theactualbman's definitions, and in suppressing the notation for the evaluation using Ed's 4-velocity $u$.

$$\frac{dr_\mathrm{S} \left(u\right)}{dt_\mathrm{S} \left(u\right)} = -\sqrt{ \frac{2M}{r}}$$

 

[grav-universe]

Hello,

The following has been confusing my friends and I, I want to make sure I have this clear as it is fairly elementary. (note set c = 1)

Ed is falling radially into a black hole, the Schwarzschild metric is:

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

his proper time is dτ² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

If Simon is stationary and is coincident with Ed at some radius then he measures the proper distances and times given by:

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

I think that's all right. But then is Simon's metric dt_s² - dr_s² ≈ ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr² or are these really meant to be different bases?

Many thanks.

SR is valid locally, so right, Simon measures the metric as just

ds² = dt_s² - dr_s²

Using Schwarzschild coordinates, the relation between the local stationary observer and that of a distant observer at infinity is

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

as you wrote, so the metric according to what the distant observer infers becomes

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

as you have also. The Schwarzschild metric is formulated according to what the distant observer infers. And yes,

dr_s / dt_s

is the locally measured velocity.

 

[xox]

Simon measures the metric as just

ds² = dt_s² - dr_s²

Several of us spent a full page (the first page) showing that this is not true. There is no justification for claiming the above. While this is trivially true:

Using Schwarzschild coordinates, the relation between the local stationary observer and that of a distant observer at infinity is

dr_s = (1 - 2μ/r)^-1/2 * dr and dt_s = (1 - 2μ/r)^1/2 dt

...this isn't:

so the metric according to what the distant observer infers becomes

ds² = (1-2μ/r) dt² - (1- 2μ/r)^-1 dr²

as you have also.

 

[grav-universe]

Several of us spent a full page (the first page) showing that this is not true. There is no justification for claiming the above.

While this is trivially true:

...this isn't:

It is correct. Why wouldn't it be?

 

[xox]

It is correct. Why wouldn't it be?

Simple, because you can't make a valid metric out of two differentials calculated for two DIFFERENT conditions. This is explained by several different posters on page 1.

 

[grav-universe]

Simple, because you can't make a valid metric out of two differentials calculated for two DIFFERENT conditions. This is explained by several different posters on page 1.

It is the same condition, or rather the same invariant, as calculated by three different observers: dτ² as measured by the freefaller, dt_s² - dr_s² as measured by a stationary observer that coincides with the freefaller, and (1-2μ/r) dt² - (1- 2μ/r)^-1 dr² as inferred by a distant observer using Schwarzschild coordinates.

 

[xox]

It is the same condition,

No, it isn't:

dr_s is calculated from the metric by setting the condition dt=0 in ds^2=\frac{dr^2}{1-2 \mu/r}-(1-2 \mu/r)dt^2
dt_s is calculated from the metric by setting a DIFFERENT condition, dr=0, into ds^2=(1-2 \mu/r)dt^2-\frac{dr^2}{1-2 \mu/r}

Three of us pointed out this kind of error on the first page, please read it, no point in repeating the same mistake over and over.

 

[grav-universe]

No, it isn't:

dr_s is calculated from the metric by setting the condition dt=0 in ds^2=\frac{dr^2}{1-2 \mu/r}-(1-2 \mu/r)dt^2
dt_s is calculated from the metric by setting a DIFFERENT condition, dr=0, into ds^2=(1-2 \mu/r)dt^2-\frac{dr^2}{1-2 \mu/r}

Three of us pointed out this kind of error on the first page, please read it, no point in repeating the same mistake over and over.

Well right, time and distance are two different measurements, but when combined, they give the same metric for the overall spacetime.

 

[xox]

Well right, time and distance are two different measurements, but when combined, they give the same metric for the overall spacetime.

You CAN'T "combine" bits of two different solutions obtained for two different conditions, this is the mistake that you keep repeating.