Fermi Normal coordinates for an infalling observer

[pervect]

I thought I'd present some plots for the Fermi-normal coordinates (only in the r-t plane) for someone falling into a black hole "from infinity".

Fermi-normal coordinates radiate a set of space-like geodesics from some point on the worldine of an object - in this case, the worldline of an observer falling into a black hole. The set of points lying on space-like geodesics that pass through the observer's worldline at a proper time of tau represent the set of points with a time coordinate of tau. The distance coordinate is given directly by the affine parameter of the geodesic, which is equivalent to measuring the distance to the point along the arc-length of the geodesic curve connecting the observer to the observee.

In our diagram, the blue dashed line represents our infalling observer. The red lines are the space-like geodesics - they are "lines of simultaneity" for the fermi-normal coordinate system. Boxes are placed at unit intervals along the space-like geodesics to display the affine parameter or the "distance away from the observer".

We assume for simplicity that m=1/2, i.e the line element is

ds^2 = -(1-1/r)dt^2 + dr^2 / (1-1/r)Then, for the infalling observer the defining equations are:

$$\frac{dt}{d\tau} = \frac{1}{ \left( 1 - 1/r \right) }}$$

$$\frac{dr}{d\tau} = \frac{1}{\sqrt{r}}$$

for the space-like geodesics, passing through some time t0, r0 the defining equations are:

$$\frac{dt}{ds} = \frac{1}{\sqrt{r0} \, \left(1 - 1/r \right) }$$

$$\frac{dr}{ds} = \sqrt{1 + \frac{1}{r0} - \frac{1}{r} }$$

These equations were derived first by noting that (dt/ds) / (1-1/r) must be constant, just as (dt/dtau) / (1-1/r) is constant.

dr/ds is found by solving g_00 (dt/ds)^2 + g_11 (dr/ds)^2 = 1, similarly to the way dr/dtau is found by setting g_00 (dt/dtau)^2 + g_11 (dr/dtau)^2 = -1.

On a highly technical note, $u^{a} \nabla_a u^{b}$ was computed and confirmed to be zero to confirm that the vector fields above are the tangent fields of geodesics.

 

[yuiop]

In our diagram, the blue dashed line represents our infalling observer. The red lines are the space-like geodesics - they are "lines of simultaneity" for the fermi-normal coordinate system. Boxes are placed at unit intervals along the space-like geodesics to display the affine parameter or the "distance away from the observer".

We assume for simplicity that m=1/2, i.e the line element is

ds^2 = -(1-1/r)dt^2 + dr^2 / (1-1/r)Then, for the infalling observer the defining equations are:

$$\frac{dt}{d\tau} = \frac{1}{ \left( 1 - 1/r \right) }}$$

$$\frac{dr}{d\tau} = \frac{1}{\sqrt{r}}$$

OK, I can see that the dashed blue curve of the infalling observer can be obtained like this:

$$\frac{dt}{d\tau} \frac{d\tau}{dr} = \frac{dt}{dr} = \frac{\sqrt{r}}{ \left( 1 - 1/r \right) }}$$

Integrate the above with respect to r:

$$t = \int_r^0 \frac{\sqrt{r}}{ \left( 1 - 1/r \right) }} dr = log\left(\frac{\sqrt{r}+1}{\sqrt{r}-1} \right) - \frac{2}{3}\sqrt{r}(r+3)$$

and simply plot t versus r.

I think I understand that much, but I have not played Fermi normal coordinates before so I am not sure what equations you used to plot the red normal curves or to locate the boxes at unit intervals. I had a go but the end equation seemed very large and messy so I figured I was on the wrong track. Also, shouldn't the parameter t0 appear in at least one of the ds equations? Did you use parametric equations to plot the red curves?

P.S. You have had nearly 30 views of your diagram in a couple of days, when most people barely get half a dozen views of their diagrams in a couple of months. You must have celebrity status on this forum

 

[JesseM]

I thought I'd present some plots for the Fermi-normal coordinates (only in the r-t plane) for someone falling into a black hole "from infinity".

Fermi-normal coordinates radiate a set of space-like geodesics from some point on the worldine of an object - in this case, the worldline of an observer falling into a black hole. The set of points lying on space-like geodesics that pass through the observer's worldline at a proper time of tau represent the set of points with a time coordinate of tau. The distance coordinate is given directly by the affine parameter of the geodesic, which is equivalent to measuring the distance to the point along the arc-length of the geodesic curve connecting the observer to the observee.

A few conceptual questions about this:

1. If we have a slower-than-light particle in freefall, then if we know its instantaneous velocity at a single point (tangent vector at that point), in GR there should be a unique geodesic worldline for the particle as long as it remains in freefall. Would the same be true for a hypothetical tachyon whose instantaneous velocity at a point is FTL (spacelike tangent vector)?

2. If so, then could we think of the "space-like geodesics" representing surfaces of simultaneity in Fermi Normal coordinates as the worldlines of tachyons radiated from the falling object, such that in the locally inertial rest frame of the falling object at each point on it's worldline, the tachyon emitted at that point is moving parallel to the local inertial frame's line of simultaneity? In other words it has an instantaneous speed of infinity in the freefalling object's locally inertial rest frame at the moment it's emitted, and then by 1) we can figure out the tachyon's entire worldline based on its instantaneous velocity (tangent vector) at the moment it departs from the falling object--does that work?

3. Finally, if we pick two events which lie on the same space-like geodesic, would the coordinate distance between them be equal to the "proper distance" along that geodesic (i.e. the integral of the metric line element ds along it)?

 

[Ich]

You must have celebrity status on this forum

Sure. I bet I'm not the only one who is glad that he's around again.

 

[pervect]

I think I understand that much, but I have not played Fermi normal coordinates before so I am not sure what equations you used to plot the red normal curves or to locate the boxes at unit intervals. I had a go but the end equation seemed very large and messy so I figured I was on the wrong track. Also, shouldn't the parameter t0 appear in at least one of the ds equations? Did you use parametric equations to plot the red curves?

Yes. I couldn't integrate those equations either, so I used the parametric equations I gave in the original post to generate the plots.

Specifically:

$$\frac{dt}{ds} = \frac{1}{\sqrt{r0} \, \left(1 - 1/r \right) }$$

$$\frac{dr}{ds} = \sqrt{1 + \frac{1}{r0} - \frac{1}{r} }$$

are the equations that generate the red curves, which are spatial geodesics. The derivation I gave for those equations wasn't very detailed, though I did give a few notes at the end, I could expand on that more if there is interest.

I didn't do this by hand, of course, I used Maple's "dsolve" routine in the numeric mode to solve for the curves and to plot them. Getting Maple to plot the points was the hardest part , but that was a Maple specific issue (and there is probably a cleverer way to do it than the way I eventually found).

Conceptually, however the point is that when s=0, you are on the worldline of the infalling observer, the blue curve, and that the value of s measures your "distance" away from the infalling observer in in the Fermi-normal coordinates. (Though as I point out, you can also think of s as the arc-length of the red curve between the blue line of the observer's world-line and the black box at the point being measured).

The value of t0 does not enter into the equations for the geodesics because the Schwarzschild geometry is static, it is the same at time t0 as at some later time t1. Thus the equations for a space-like geodesic don't change with time.

Graphically, the boxed points are when s has an integer value - somewhere in the range of -7 to +7.

 

[pervect]

A few conceptual questions about this:

1. If we have a slower-than-light particle in freefall, then if we know its instantaneous velocity at a single point (tangent vector at that point), in GR there should be a unique geodesic worldline for the particle as long as it remains in freefall. Would the same be true for a hypothetical tachyon whose instantaneous velocity at a point is FTL (spacelike tangent vector)?

Yes. A point and the tangent vector defines the geodesic curve.

2. If so, then could we think of the "space-like geodesics" representing surfaces of simultaneity in Fermi Normal coordinates as the worldlines of tachyons radiated from the falling object, such that in the locally inertial rest frame of the falling object at each point on it's worldline, the tachyon emitted at that point is moving parallel to the local inertial frame's line of simultaneity?

Yes. Though most textbooks will consider the mater in terms of geodesics and not infinite velocity tachyons :-). But the important point is that given a point in space-time, and a velocity (or tangent vector) one uniquely defines the curve.

In other words it has an instantaneous speed of infinity in the freefalling object's locally inertial rest frame at the moment it's emitted, and then by 1) we can figure out the tachyon's entire worldline based on its instantaneous velocity (tangent vector) at the moment it departs from the falling object--does that work?

Yep - at least as far as I can tell, it should work just fine.

3. Finally, if we pick two events which lie on the same space-like geodesic, would the coordinate distance between them be equal to the "proper distance" along that geodesic (i.e. the integral of the metric line element ds along it)?

Yes - as long as you agree that the points on the space-like geodesic are simultaneous. Sometimes this doesn't happen - cosmologists, for instance, insist that simultaneity be defined by "time elapsed from the big bang in the CMB frame". However, a space-like geodesic curve in a FRW universe doesn't lie in the surface of simultaneity defined in this manner :-(. So the cosmological notion of "proper distance" isn't the same as the notion of Fermi-Walker distance for a co-moving observer.

If I recall correctly, one reason cosmologists do things in this way is so that Hubble's law works, so in the end it's driven by convenience and experiment.

The general process of defining distance seems to be that one defines one's notion of simultaneity first. After you've split up space-time into space+time, you define distances as lying entirely in the space-like hypersurface of simultaneity. Technically, you use your original space-time metric to calculate the "induced metric" on your space-like hypersurface, (which has one less dimension than your original space-time, so there will be one less coordinate in the metric expression - and it will have a Euclidean signature.). You still use the idea of a geodesic curves to define distances in this hypersurface, but they are geodesics on the hypersurface, using the induced metric, not necessarily geodesics of the original space-time.

Without any obvious external definition of simultaneity, though, Fermi-Walker coordinates provide a way of generating a notion of simultaneity, and of distance, given the worldline of an observer and a metric.

 

[George Jones]

If I recall correctly, one reason cosmologists do things in this way is so that Hubble's law works, so in the end it's driven by convenience and experiment.

And the Copernican principle. Cosmologists' hypersurfaces of simultaneity (in FRW universes) are maximally symmetric spaces that are spatially homogeneous and isotropic. Each cosmological hypersurface of simultaneity has six Killing vectors appropriate for the symmetries.

 

[yuiop]

Yes. I couldn't integrate those equations either, so I used the parametric equations I gave in the original post to generate the plots.

Specifically:

$$\frac{dt}{ds} = \frac{1}{\sqrt{r0} \, \left(1 - 1/r \right) }$$

$$\frac{dr}{ds} = \sqrt{1 + \frac{1}{r0} - \frac{1}{r} }$$

are the equations that generate the red curves, which are spatial geodesics. The derivation I gave for those equations wasn't very detailed, though I did give a few notes at the end, I could expand on that more if there is interest.

I for one am interested. I am sure there are others that would be.

I understand how "dr/dtau is found by setting g_00 (dt/dtau)^2 + g_11 (dr/dtau)^2 = -1".

I partly understand how "dr/ds is found by solving g_00 (dt/ds)^2 + g_11 (dr/ds)^2 = 1" as follows:

$$-(1-1/r)(dt/ds)^2 + (1-1/r)^{-1} (dr/ds)^2 = 1$$

$$dr/ds = \sqrt{(1+(1-1/r)(dt/ds)^2)(1-1/r) }$$

Now substitute the expression you gave for dt/ds into the above equation:

$$\frac{dr}{ds} = \sqrt{\left(1+\frac{1}{r0(1-1/r)}\right)(1-1/r) }$$

$$\frac{dr}{ds} = \sqrt{1+\frac{1}{r0} - \frac{1}{r}}$$

==================================================

Now am not sure how the original expression for dt/ds that I used in the substitution is obtained in the first place, namely:

$$\frac{dt}{ds} = \frac{1}{\sqrt{r0} \, \left(1 - 1/r \right) }$$

Could you elaborate on how r0 is inserted into/determined for the above equation?

 

[pervect]

Now am not sure how the original expression for dt/ds that I used in the substitution is obtained in the first place, namely:

$$\frac{dt}{ds} = \frac{1}{\sqrt{r0} \, \left(1 - 1/r \right) }$$

Could you elaborate on how r0 is inserted into/determined for the above equation?

As a result of the geodesic equations, we know that
$$\frac{d}{ds} \left( \frac{dt/ds}{1-1/r} \right)= 0$$

which implies that anywhere on a geodesic

(dt/ds) / (1-1/r) = constant

An abbreviated and not-too-terribly rigorous derivation of the geodesic equations goes as follows. Let rdot = dr/ds, and tdot = dt/ds. Then consider the quantity

$$L(r,t,rdot, tdot) = \int \sqrt{-\left( 1- 1/r \right) tdot^2 + rdot^2 / \left(1-1/r\right) }$$

L , which is just the formula for the distance along the geodesic, will be extremized when it satisfies the Euler-Lagrange equations

$$\frac{d}{ds} \left( \frac{ \partial L }{\partial tdot} \right) = \frac{ \partial L}{\partial t}$$

$$\frac{d}{ds} \left( \frac{ \partial L }{\partial rdot} \right) = \frac{ \partial L}{\partial r}$$

The first equation is the one of interest. Because there is no time dependency of L on t, the right hand side is zero.

Evaluating this , we get:

$$-{\frac { {\it tdot} \left( 1-1/r \right) }{\sqrt {- \left( 1-1/r \right) {{\it tdot}}^{2}+{{\it rdot}}^{2} / \left( 1-1/r \right) }}}$$

Now we have the freedom to parameterize the curve however we want - when we parameterize it so that it's tangent vector d/ds is of unit length everywhere, i.e. in terms of the affine parameterization of length, the denominator is constant and we get the much simpler expression

(d/ds) (1-1/r) tdot = 0

which implies that tdot = constant / (1-1/r) for any geodesic. Note that this quantity is constant along any given geodesic. Specifically, it's constant for our time-like geodesics (the blue curve) with one value anywhere on the blue curve, and it's also constant (but with a different value than on the blue curve) anywhere on our red curve, our spacelike geodesics.

For the time-like curve, we can consider this constant to be an energy. Noether's theorem tells us that any sort of symmetry gives us a conserved quantity, and additionally it's commonly known that when the symmetry is a time-translation symmetry, the corresponding conserved quantity is an energy.

So - we know the expression above is constant. How did we figure out its value? Well, to figure out the constant, we need a point on the curve, and the value of (dt/dtau) or (dt/ds) at that point. For our time like geodesics, that was r=infinity and dt/dtau = 1, giving us a value of 1 on the time-like geodesic. But the space-like geodesic will have some different constant.

For the space-like curve, what we require is that, at some point, specifically the point (t0,r0), where the blue and red curves cross, that our space-like geodesic be "normal" to our timelike geodesic. (r0,t0) is the point where we create the red curves, and shoot them out. We have to decide "what direction" to shoot them out. Because these are "fermi normal" coordinates, the tangent to the space-like geodesic must be normal to the tangent to the time-like geodesic where the two cross.

We can thus write

g_00 (dt/ds) (dt/dtau) + g_11 (dr/ds)(dr/dtau) = 0 at (t0,r0)

here dt/dtau and dr/dtau represent the tangent to our blue timelike worldline, and dt/ds and dr/ds represent the tangent to our red, spacelike worldline.

If you are familiar with four vectors, it should be obvious why this is the right choice. But let's step back a bit and see why this expression works.

The first point of this expression is that it's written to be independent of coordinates, or motion, because of the fact that it's a tensor.

The second point is that if we consider some stationary observer , they would have for their local time axis, by definition

dr/dtau = 0

and for their local space axis, they would have, by definition

dt/ds = 0

So if we plug the above into the dot product, we get

g_00 (dt/dtau) (dt/ds = 0) + g_11 (dr/dtau = 0) (dr/ds), which implies that this expression is zero for a stationary observer. Because it's a tensor expression, if it works for a stationary observer, it works for _all_ observers. So that's our motivation for requiring that space and time be normal, or orthogonal, to each other.

If you work out the details, you'll find that for the timelike geodesic the tangent vector is:

(dt/dtau) = 1 (dr/dtau) = 1/sqrt(r)

The vector that's orthogonal to this and of unit length is

(dt/ds) = (1/sqrt(r0)) (dr/ds) = 1

And we know it's orthogonal because

g_00 (dt/ds) (dt/dtau) + g_11 (dr/ds)(dr/dtau) = 0

and it's unit length because

g_00 (dt/ds)^2 + g_11 (dr/ds)^2 = 1

This then gives us the needed constant to continue with the derivation, which I think you've already followed.

 

[pervect]

Well, one correction - and some more results. It looks like I got the sign of the time displacement wrong in my plots. If I have time, I'll go back and fix this.

The new results:

While we can't find a tractable closed form solution for the geodesic equation, we can do a reasonable series expansion.

This gives a nonlinear series expansion for the r-coordinate which is a distance s away from the falling observer at R0 along a space-like geodesic:

R' = R0 + s + (1/4) s^2/R0^2 - (1/6) s^3 / R0^3We can now actually write down a map between Fermi - normal coordinates (tau,s) to Schwarzschild coordinates (T,R) that's valid for "small s".

We define R0(tau), a function that gives the height of our observer "falling from infinity" at proper time tau as:

$$R0(\tau) = \left(-\frac{3}{2} \tau \right) ^ {\frac{2}{3}}$$

where tau = $\tau$ < 0, and I've capitalized R as the lower case r looks too much like tau.

Then we can write

$$R(\tau,s) = R0(\tau) + s + \frac{1}{4} \left( \frac{s}{R0(\tau)} \right)^2 - \frac{1}{6} \left( \frac{s}{R0(\tau)} \right) ^3$$

The solution for T(tau,s) is a lot messier. It's also valid to the third order in s:

$$-\frac{2}{3}R0(\tau)^{\frac{3}{2}} - 2 \sqrt{R0(\tau)} + ln\left(\frac{\sqrt{R0(\tau)}+1}{\sqrt{R0(\tau)}-1}\right) - \frac{s}{\sqrt{R0(\tau)}\left(1-R0(\tau)\right)} + \frac{s^2}{2\sqrt{R0(\tau)}\left(R0(\tau)-1\right)^3}+ \frac{s^3 \left( 4 R0(\tau)^2 - R0(\tau) + 1 \right)}{12 R0(\tau)^{\frac{5}{2}}\left(R0(\tau) - 1 \right)^3}$$

Plugging this into the transform equations gives rise to a local metric in (tau,s) that's flat to the third order in s, except for g_00, which has one term of order s^2 due to the tidal forces

$$g_{00} = -1 + \frac{s^2 }{ R0(\tau)^3} + O(s^3)$$
$$g_{01} = O(s^3)$$
$$g_{11} = 1 + O(s^3)$$

 

[yuiop]

Yes. I couldn't integrate those equations either, so I used the parametric equations I gave in the original post to generate the plots.

I have finally managed to get a closed form exact symbolic integration of the normal curve using your earlier equations for dt/ds and dr/ds but using the symbol R for r0:

$$\frac{dt}{ds} \frac{ds}{dr} = \frac{dt}{dr} = \frac{1}{\sqrt{R} \, (1 - 1/r ) \sqrt{1 + 1/R - 1/r }}$$

Integrate the above with respect to r:

$$t = \int \frac{dr}{\sqrt{R} \, (1 - 1/r ) \sqrt{1 + 1/R - 1/r }}$$

then the symbolic indefinite integration of the normal curve is:

t = (sqrt(1+R-R/r)*r/(R+1) + log((r-1)/(r*(R+2+2*sqrt(1+R-R/r))-R)) + (3*R/2+1)/(R+1)^(3/2)*log(2*r*(R+sqrt(1+R-R/r)*sqrt(R+1)+1)-R)) + k [Eq1]

We already know that the wordline of the observer is given by:

t = -2/3*sqrt(r)*(r+3)-log((sqrt(r)-1)/(sqrt(r)+1)) [Eq2]

R is the intersection of the of the two curves and is the same as your r0 and k is the constant of integration. As you can see, not too pretty and that is after some manual simplification!

The constant of integration can be found by setting r = R and using k = [Eq2] - [Eq1] to obtain:

The constant of integration is:

k = -(sqrt(R)*(6 + 11*R + 2*R^2))/(3*(1 + R)) - log((-1 + sqrt(R))^2/(R*(1 + sqrt(R))^2)) - ((2 + 3*R)*log(R*(1 + 2*R + 2*sqrt(R)*sqrt(1 + R))))/(2*(1 + R)^(3/2))

The complete equation for the normal curve is then:

t = (3*r*sqrt(1+R-R/r) - sqrt(R)*(6 + 11*R + 2*R^2))/(3*(1 + R)) +log(R*(r-1)*(sqrt(R)+1)^2/((sqrt(R)-1)^2*(r*(R+2*sqrt(1+R-R/r)+2)-R))) + (3*R+2)/(2*(R+1)^(3/2))*log((2*r*(1+R+sqrt(R+1)*sqrt(1+R-R/r))-R)/(R*(1+2*R+2*sqrt(R+1)*sqrt(R))))

Sorry about the lack of Tex, but the complexity of the equation means that it won't mean much "visually" anyway and the above form has the advantage that it can be copied and pasted directly into plotting software. I used the free Compass and Ruler (C.a.R.) geometrical software to plot the curves and the above equation exactly duplicates your red "line of simultaneity" normal curves and correctly intersects the wordline of the falling observer at (r0,t0).

 

[yuiop]

Is it possible to obtain an equation that will allow us to plot the paths of falling clocks that maintain constant distance from the "primary free falling observer"? These other clocks will of course be experiencing proper acceleration.

I was going to produce a corrected plot of the Fermi normal lines with unit spatial and temporal markers (the boxes on your chart) similar to the one you did in the OP, but although I can plot the normal lines, I can only obtain the unit markers individually by an iterative process and this would be very time consuming for more than a handful of markers. Does your software allow you automate this process and produce a plot in less manual fashion?

To update my previous post to allow for the correct sign of the slope:

The symbolic indefinite integration of the normal curve is:

-(sqrt(1+R-R/r)*r/(R+1) + log((r-1)/(r*(R+2+2*sqrt(1+R-R/r))-R)) + (3*R/2+1)/(R+1)^(3/2)*log(2*r*(R+sqrt(1+R-R/r)*sqrt(R+1)+1)-R)) [1]

The wordline of the free falling observer is given by:

t = -2/3*sqrt(r)*(r+3)-log((sqrt(r)-1)/(sqrt(r)+1)) [2]

The constant of integration C is found from setting all occurrences of r equal to R in the above expressions and calculating C = [E2](r=R) - [E1](r=R)

The full equation for the Fermi normal curve that intersects the primary free falling observers curve at R is then t = [E2] + C.

P.S. Is the worlds largest physically meaningful equation that contains a single variable? Is nature that complicated?

P.P.S. I notice you claim to have produced a map between Fermi normal coordinates and Schwarzschild coordinates. Obviously from the point of view of the primary free falling Fermi observer the other falling observers are stationary and therefore are simplky represented by vertical lines, but is it possible to plot how "stationary" observers (and the EH) in the Schwarzschild metric "move" from the point of view the primary Fermi observer?

 

[pervect]

I missed this somehow the first go-around. Good work on finding an analytical expression for the curves, even though you did have to sacrifice the affine parameterization of the distances to do so.

I don't have a fast way of generating the unit markers, my plots were generated using numerical integration built into Maple.

I do have a map for fermi normal coordinates to Schwarzschild coordinates, but it's in the form of a series expansion rather than a closed form map. The series is an expansion in terms of the parameter s, which is assumed to be small.

I can add that the event horizon will appear more or less like an incoming light ray to the infalling observer. This isn't terribly surprising since the event horizon is not the worldline of any actual observer, rather it's a null worldline, i.e. the worldline of a light-ray rather than the worldline of an object.

I can give you the approximate metric (as a series) for the fermi-normal observer as a function of tau and s. This might help you figure out how things appear to them.

The metric is g_00 dtau^2 + g_11 ds^2 where

g_00 is approximately -1 + s^2 / r0(tau)^3 - s^3 / r0(tau)^4 +O(s^4)
g_11 = 1 (at least to order s^5, and I believe it's exact by construction)

There appears to be no off-diagonal terms (at least for the orders I calculated).

Note that it was assumed that M=1/2 for the above expressions. This will also give you the approximate acceleration of an observer as a function of s.

r0(tau) is the schwarzschild R coordinate of the observer at time tau and is equal to
r0(tau) = ((-3/2)*tau)^(2/3);