ANSWERED on Sun 16 Nov 2008 - 2:57 pm UTC by mathtalk
Question: Derivation involving calculus (Part 1)
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Asked by kline on Mon 10 Nov 2008 - 1:04 pm UTC:
I need to know how this parametric equation for the radial motion of a free falling particle in proper time is derived, r = R/2(cos(θ)+1) , tau = R/2 sqrt(R/2m)(sin(θ)+θ) where tau is the proper time of the free falling particle and R is the apogee height. Please see http://www.mathpages.com/rr/s6-04/6-04.htm for background information. It appears to be derived from equation (5) on that page but it is not clear to me how the solutions are obtained. My calculus is very rusty so please provide links to tutorials or FAQs to explain any advanced technical jargon you use. I do not need to know anything about the tensors or Christoffel symbols, just the algebra and calculus involved. I also need to know if the solution is correct and if it is a unique solution. (For example square roots and quadratics normally have more than one solution) A hint provided to me on a free forum suggested substituting θ for dr/dt and d^2r dθ dr dθ dθ ------ = ---- = -------- = θ ---- dt^2 dt dt dr dr to obtain the parametric solution but that does not seem to help much.
Request for clarification by Researcher mathtalk on Wed 12 Nov 2008 - 7:53 pm UTC:
The writeup to which you linked uses the letter t in the text (but not in the equations) to refer both to "coordinate time t" and "proper time tau", so there is a little confusion arising from that. However if you simply want a calculus based confirmation that the parametric solution does solve the equation (5) there (a second order nonlinear differential equation in r and tau), that is tractable. Some discussion about uniqueness and the presence of square roots would be in order, but of course there are many different ways of parameterizing a solution. So uniqueness has to be understood in terms of the r vs. tau relationship, omitting an influence of how that relationship is parameterized. regards, mathtalk
Question clarification by kline on Thu 13 Nov 2008 - 5:28 am UTC:
I would have preferred a direct derivation showing all solutions but since I had no other offers I will accept a detailed confirmation that the parametric solution is a solution of the second order differential equation (equation 5) in the mathpages writeup. As I understand it parametric curves have an unambiguous direction assoitiated with them unlike explicit equations. Can you offer any insight into the direction of the solution for values of r less than 2m? If you offer additional insights into the uniqueness of the solution including lost constants of integration and multiple roots that would be much appreciated. For example the arccos function can have positive and negative roots, cubes roots have three roots and the Lambert W function can have an infinite number of solutions. Basically I am looking for a counter proof to the mathpages solution but I do not wish to bias your answer and only wish for a unbiased rigorous opinion. Thanks.
Comment by kline on Thu 13 Nov 2008 - 5:36 am UTC:
P.S. I had another look at the mathpages text and can not find the places where they have used t ambiguously to mean coodinate time t or proper time tau.
Comment by kline on Thu 13 Nov 2008 - 5:48 am UTC:
I have Maple software so if it makes your work any easier you can present your answers in the form of a Maple document or as series of inputs that can be used in a Maple document.
Comment by kline on Thu 13 Nov 2008 - 1:46 pm UTC:
Mathtalk, I am impressed by your previous work for others so just do what you can and I'll happily pay the $25.
Comment by Researcher mathtalk on Thu 13 Nov 2008 - 4:52 pm UTC:
Your confidence in my work is generous, but let me say a few words about the uniqueness of the solution. The author of the mathpages text is only claiming that the parametic "cycloid" solution is unique for the intial conditions: r = R at τ = 0 dr/dτ = 0 at τ = 0 where 0 < R < 2m has the physical interpretation of the radius at apogee. Because this (nonlinear) ODE (5) is second order, two initial conditions are needed to impose uniqueness. However the equation is "autonomous", meaning that τ does not appear explicitly, so that a one-dimensional family is generated by translating τ by any constant amount (forward or backward). In physical terms it doesn't matter when we set the clock on board our orbiting platform to zero, but the given parametric solution is for setting τ = 0 when the platform is at apogee. [Autonomous system (mathematics) -- Wikipedia] http://en.wikipedia.org/wiki/Autonomous_system_(mathematics) Time translation and the free parameter R (radius at apogee) account for our two degrees of freedom in the solution space for (5), but this only covers solutions where 0 < r < 2m for all proper time τ. One can ask about mathematical solutions where r > 2m, but that would invoke a separate "patch" of the solution space not covered by the "cycloid" parametric solutions. regards, mt
Comment by Researcher mathtalk on Thu 13 Nov 2008 - 5:04 pm UTC:
Another thing perhaps worth clarifying: In your question, you say the equation is "for the radial motion of a free falling particle in proper time." Actually the mathpages article assumes a purely radial motion, i.e. no angular motion allowed. This is narrower than the solving for the radial component of a more general motion. In the physical interpretation an "apple" accelerates toward the center of the gravitational field at r = 0. Conventionally radius is nonnegative distance, so when it reaches the origin and passes through, it's okay that our parametric solution for r remains nonnegative. regards, mt
Question clarification by kline on Fri 14 Nov 2008 - 5:27 pm UTC:
>> The author of the mathpages text is only claiming that the parametic >> "cycloid" solution is unique for the intial conditions: r = R at τ = 0 >> and dr/dτ = 0 at τ = 0. That seems reasonable to me. >> but this only covers solutions where 0 < r < 2m for all proper time τ. >> One can ask about mathematical solutions where r > 2m, but that would >> invoke a separate "patch" of the solution space not covered by the >> the "cycloid" parametric solutions. How is this reconciled with the statement in mathpages that "we can use equation (5) to numerically integrate the geodesic path from any given initial trajectory, and it confirms that the radial coordinate passes smoothly through r = 2m as a function of the proper time t." ? This is one the aspects I am very interested in and one of the motivations for the original question. The plot of the parametric equation on mathpages is from r/2 = 10 to r = 0 covering both sides of r = 2m. Is this a rigourously valid plot? mathpages also states that "dr/dt is invariably forced to 1 - 2m/R precisely at r = 2m, so the quantity in the square brackets goes to zero, canceling the zero in the denominator." Does this not imply that the solution for r = 2m is undetermined (0/0) or is this resolved by applying L'Hopitals rule? >> Another thing perhaps worth clarifying: In your question, you say the >> equation is "for the radial motion of a free falling particle in proper >> time." Actually the mathpages article assumes a purely radial motion, >> i.e. no angular motion allowed. I am aware of this and have no problem with addressing the purely radial aspects of the motion especially if makes the analysis simpler. >> Conventionally radius is nonnegative distance, so when it reaches the >> origin and passes through, it's okay that our parametric solution for >> r remains nonnegative. I think it acceptable to assume (r >= 0 , R >= 0, m>0) and hopefully this makes the problem more tractable. Finally, if you have a solution for the mathpages equation (5) in terms of r and proper time tau (and possibly a parametric variable) that you think is valid, even it does not agree with the mathpages solution, then I would be interested in that. Best regards..
Question clarification by kline on Fri 14 Nov 2008 - 5:30 pm UTC:
Errata: from r/2 = 10 to r = 0 covering both sides of r = 2m. That should read "from r/2 = 10m to r = 0 covering both sides of r = 2m"
Question clarification by kline on Sat 15 Nov 2008 - 7:19 pm UTC:
It looks like the mathpages page http://www.mathpages.com/rr/s6-04/6-04.htm has been updated (after I asked this question) to include a derivation of the parametric equation. Does it look reasonable to you?
Answer by Researcher mathtalk on Sun 16 Nov 2008 - 2:57 pm UTC:
Let's begin by showing that for 0 < R < 2m, the mathpages
"cycloid" parametric solution:
r = (R/2) (1 + cos(α))
τ = (R/2) sqrt(R/2m) (α + sin(α))
satisfies the 2nd order ordinary differential equation:
d²r m
̶̶̶ = ̶̶̶̶̶̶̶̶̶ [(dr/dτ)² - (1 - 2m/R)] (*)
dτ² r(r - 2m)
and the initial conditions at τ = 0:
r = R
dr/dτ = 0
Note r,τ are analytic functions (convergent power series) of
the parameter α. As we show below, despite isolated points
at which dτ/dα = 0, τ is a strictly increasing function of
α that takes on all real values. In particular τ = 0 only
when α = 0.
Clearly when τ = 0, we have r = (R/2) (1 + cos(0)) = R.
To express derivative of r with respect to τ, use the
chain rule:
dr/dα = (dr/dτ)(dτ/dα)
so that:
dr/dα
dr/dτ = ̶̶̶̶̶̶̶̶̶
dτ/dα
Differentiating with respect to parameter α:
dr/dα = -(R/2) sin(α)
dτ/dα = (R/2) sqrt(R/2m) (1 + cos(α))
we now see that dτ/dα = 0 precisely when α is an odd
multiple of π, and that these "flat places" occur with
intervening intervals of increase, dτ/dα > 0. Thus τ
is a strictly increasing function of parameter α, as
we previously claimed.
Generally we have, except for α an odd multiple of π:
dr/dτ = -sin(α)/[sqrt(R/2m) (1 + cos(α))]
Since dτ/dα > 0 in particular at α = 0, we don't have a
problem with division by zero in evaluating the initial
condition. I.e. when τ = 0:
dr/dτ = -sin(0)/[sqrt(R/2m) (1 + cos(0))] = 0
as desired.
Thus we've verified that both initial conditions are
satisfied.
To show the differential equation (*) is satisfied, we
again use the chain rule to express a second derivative:
d(dr/dτ)
̶̶̶̶̶̶̶̶
d²r dα
̶̶̶ = ̶̶̶̶̶̶̶̶̶̶
dτ² dτ/dα
The derivative in the numerator turns out to simplify
considerably (see end of post for more details about
this differentiation, omitted here for the sake of
continuity in presentation):
d(dr/dτ)
̶̶̶̶̶̶̶̶ = -1/[sqrt(R/2m) (1 + cos(α))]
dα
where the derivative in the denominator is one we found
earlier.
Thus:
d²r -1/[sqrt(R/2m) (1 + cos(α))]
̶̶̶ = ̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶
dτ² (R/2) sqrt(R/2m) (1 + cos(α))
= -4m/[R² (1 + cos(α))²]
except when α is an odd multiple of π and the division
indicated thus would be by zero.
Let's compare this to the terms in the right-hand side
of (*) once we substitute expressions for r and dr/dτ:
m
̶̶̶̶̶̶̶̶̶ [(dr/dτ)² - (1 - 2m/R)]
r(r - 2m)
m
=
̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶
(R/2)(1 + cos(α))[(R/2)(1 + cos(α)) - 2m]
sin²(α)
* [ ̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶ - (1 - 2m/R) ]
(R/2m)(1 + cos(α))²
4m
=
̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶
R (1 + cos(α)) [ R (1 + cos(α)) - 4m ]
sin²(α) + (1 + cos(α))²
* [ ̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶ - 1 ]
(R/2m)(1 + cos(α))²
4m
=
̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶
R (1 + cos(α)) [ R (1 + cos(α)) - 4m ]
2(1 + cos(α))
* [ ̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶ - 1 ]
(R/2m)(1 + cos(α))²
4m
=
̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶
R (1 + cos(α)) [ R (1 + cos(α)) - 4m ]
4m
* [ ̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶ - 1 ]
R (1 + cos(α))
4m
=
̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶
R (1 + cos(α)) [ R (1 + cos(α)) - 4m ]
4m - R (1 + cos(α))
* ̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶
R (1 + cos(α))
= -4m/[R² (1 + cos(α))²]
This agrees with our expression for d²r/dτ², so the
parametric solution satisfies the equation (*).
Finally, a few comments as promised on simplifying
the expression for:
d(dr/dτ)
̶̶̶̶̶̶̶̶ = -1/[sqrt(R/2m) (1 + cos(α))]
dα
Recall that unless α is an odd multiple of π:
dr/dτ = -sin(α)/[sqrt(R/2m) (1 + cos(α))]
Let's take out the constant factor -1/sqrt(R/2m)
and focus on differentiating:
sin(α)/(1 + cos(α))
Although there are ways to get around using the
quotient rule for derivatives, it seems to be the
most direct approach here:
cos(α)(1 + cos(α)) - sin(α)(-sin(α))
̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶̶
(1 + cos(α))²
= (1 + cos(α))/(1 + cos(α))²
= 1/(1 + cos(α))
Restoring the omitted constant factor -1/sqrt(R/2m)
gives us the answer as before.
regards, mathtalk
Comment by kline on Mon 17 Nov 2008 - 9:33 am UTC:
Nice work, mathtalk. Thanks.
Accepted and rated by kline on Mon 17 Nov 2008 - 10:30 am UTC:
Mathtalk gave a nice detailed explantion to a non trivial problem that has defeated experts on other forums and took the extra time to address most of the extra concerns and issues I raised. I am happy to pay the fee and a 100% bonus. Mathtalk certainly knows his stuff!
Accepted and rated by kline on Mon 17 Nov 2008 - 10:32 am UTC:
Mathtalk gave a nice detailed explantion to a non trivial problem that has defeated experts on other forums and took the extra time to address most of the extra concerns and issues I raised. I am happy to pay the fee and a 100% bonus. Mathtalk certainly knows his stuff!
Accepted and rated by kline on Mon 17 Nov 2008 - 10:33 am UTC:
Comment by Researcher Roger B (eiffel) on Mon 17 Nov 2008 - 9:24 pm UTC:
I cannot imagine how long it took mathtalk to get all those formulae formatted correctly in Uclue's primitive text input box. Truly impressive!
Comment by Researcher mathtalk on Tue 18 Nov 2008 - 2:47 pm UTC:
I used Unicode characters to compose my formulas, but while the Greek letters seem to be correctly displayed in all the browsers I've tried (Firefox, IE7, and Chrome on Windows XP, Firefox on Linux), the viniculum (horizontal stroke for division) is passably rendered only by Firefox on Linux. Even there some of the occurrences toward the end of the post are incorrectly left-justified in the display. What I tried was using multiple copies of Unicode U+0336 (Combining Long Stroke Overlay). All the Windows based browsers appear to collapse each occurrence down to a single character (with similar left-justification problems toward the end of the post as in Linux). experimentally, mathtalk
Answer clarification by Researcher mathtalk on Tue 18 Nov 2008 - 11:30 pm UTC:
Let me tie up a couple of loose threads in this Question. >> One can ask about mathematical solutions where r > 2m, but that would >> invoke a separate "patch" of the solution space not covered by the >> the "cycloid" parametric solutions. > How is this reconciled with the statement in mathpages that "we can use > equation (5) to numerically integrate the geodesic path from any given > initial trajectory, and it confirms that the radial coordinate passes > smoothly through r = 2m as a function of the proper time t." ? In general the boundary r = 2m separates two mathematical domains, and solutions of (5) are defined only up to this boundary, since (5) is not defined there. However in the special application, physically motivated, the constant R that appears in the differential equation (5) is also the radius at apogee R that appears in the intial conditions. For these special solutions, the symbolic solution (cycloid parameterization) does cross smoothly over r = 2m (provided R > 2m). The author of the mathpages article is discussing this when he/she turns to "numerically integrat[ing] the geodesic path" as a way of confirming that the radial coordinate is continuous "through r = 2m as a function of the proper time [tau]." From a physical interpretation, r = 2m is the event horizon of a black hole, and it is often stated that an observer crossing this would not observe anything special (other than those pesky tidal forces ripping one limb from limb!) as time continues to flow normally. An "outside" observer however sees time slowing to a crawl for the falling body. More here: [The Schwarzchild Metric and Event Horizons by Stefan Waner] http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec15.html * * * * * * * * * * * * * * * * * * * The other item I wanted to respond to was the query about the mathpages article being updated recently(?) and whether a derivation there of the parametric equation appears valid. I'm not sure what addition you mean, as the site seems the same to me as when I first looked at it. The Page Info reported by Firefox suggests it hasn't been modified since Nov. 4, 2007, but that may not be reliable. regards, mathtalk
Comment by User johnfrommelbourne on Wed 19 Nov 2008 - 1:31 pm UTC:
So what sort of IQ does one have to have to replicate the work of Mathtalk. I am guessing Mathtalk has an IQ measured at about 150+. Suspect Roger and a few other members of Roger & Staff Pty Ltd are up there also.
Comment by kline on Wed 19 Nov 2008 - 5:00 pm UTC:
>> From a physical interpretation, r = 2m is the event horizon of a black >> hole, and it is often stated that an observer crossing this would not >> observe anything special .... I am aware this is the conventional interpretation, but I have many good reasons to believe it not true and that is the motivation for my enquiries. Obviously I am going to have to dig deeper even if that means learning tensors (gulp). >> [The Schwarzchild Metric and Event Horizons by Stefan Waner] >> http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec15.html Thanks for the link. It looks interesting. >> The other item I wanted to respond to was the query about the mathpages >> article being updated recently(?) and whether a derivation there of the >> parametric equation appears valid. I'm not sure what addition you >> mean, as the site seems the same to me as when I first looked at it. >> The Page Info reported by Firefox suggests it hasn't been modified >> since Nov. 4, 2007, but that may not be reliable. It may be that your computer has cached an old version of the page. You might have to delete the copy of the page in your internet history to see the new version. There are a couple of mathematical typos in the new version of the mathpages page, but none look too serious. In the paragraph starting "In terms of this new variable, the integral is", Kevin Brown (The author of mathpages) states: τ = R/2*√(R/2/m)*int{√[(1+cos(α))/(1+cos(α))]*sin(alpha),α,0,α'} where int{x,α,0,α'} means the integral of x with respect to α from α=0 to α=α'. That should read: τ = R/2*√(R/2/m)*int{√[(1+cos(α))/(1-cos(α))]*-sin(alpha),α,0,α'} The result of that integration is: τ = R/2*√(R/2/m)*(α' + sin(α')*tan(α'/2)/√(tan²(α'/2)) Kevin Brown has simplified that to τ = R/2*√(R/2/m)*(α' + sin(α') but the simplification is only true for positive values of α. Later he states "A plot of this r versus t corresponds to the position of a point on the rim of a rolling wheel of radius R, where α is the angle of the wheel." but I am fairly sure he should have said a wheel of radius R/2 or a wheel of diameter R, but that is a minor point. What I would like to know (although you are under no obligation to answer) is the reasoning behind the beginning of Kevin Brown's derivation (updated) where he states "To solve equation (5) analytically, note that the derivative (with respect to t) of the quantity in the square brackets is.." Why does he take that step?
Comment by Researcher mathtalk on Wed 19 Nov 2008 - 8:59 pm UTC:
I think you are right that I was seeing a cached image, because after I hit the refresh button once I navigated back to the mathpages site, then the additional content appeared. I'll be happy to take a look. Yes, he should have said "diameter R" for the interpretation of r = (R/2)(1 + cos(α)), and this is one of the places in his text where Greek τ in the formula above has turned into Latin t in the text. regards, mt
Answer clarification by Researcher mathtalk on Thu 20 Nov 2008 - 4:10 pm UTC:
> What I would like to know (although you are under no obligation
> to answer) is the reasoning behind the beginning of Kevin Brown's
> derivation (updated) where he states "To solve equation (5)
> analytically, note that the derivative (with respect to t) of
> the quantity in the square brackets is.."
Notice that he has recast equation (5) in the following form:
d²r (1-2m/R) - (dr/dτ)²
--- = -(m/r²) * -------------------
dτ² (1-2m/r)
The argument given is a very interesting one. Basically he shows
that if (5) holds, then the second factor on the right-hand side
is a constant. Using the initial conditions (at apogee), the value
of the constant must be 1. Thus as a function of proper time τ,
r also satisfies the classic Newton equation for gravitational
acceleration:
d²r
--- = -(m/r²)
dτ²
Furthermore the constancy of the expression being 1 leads to the
parameteric "cycloid" solution for r and τ, as well as a parametric
expression for coordinate time t.
BTW, do you still wish an answer on part 2 regarding the parametric
expression for coordinate time t?
regards, mathtalk
Comment by kline on Fri 21 Nov 2008 - 3:51 am UTC:
>> Basically he shows that if (5) holds, then the second factor on the >> right-hand side is a constant. Using the initial conditions (at >> apogee), the value of the constant must be 1. O.K. I understand now that if the derivative of an expression is zero then the expression must be a constant and the value of that constant is 1 anywhere, then by definition it is 1 everywhere. I get that much. I am not sure about the next bit where hefinds the derivative with respect to tau of the quantity in equation 5 in square brackets. (You can answer this your reply to my part 2 question that is still open as it relates to that too.) I tried finding the derivative of the quantity in square brackets [call it x] using maple and web mathematica using the following method. Let a = dr/dτ dx/dτ = dx/dτ*a/a = dx/dτ*dτ/dr*a = dx/dr*a so the derivative with respect to τ of the quantity in square brackets is the same as derivative with respect to r multiplied by a=dr/dτ (if I have reasoned correctly but I am no means sure about this) The result I get is d[((1-2m/R)-a²)/(1-2m/r)]/dτ = -2(a)/(r-2m)^2*m*(1-2m/R)-a²) which is similar to the mathpages result, but missing the last term. ------------------------------------------------------------------- Another thing I would like your opinion on, is would you agree there is some ambiguity to the paramatric equation? For example is the parametric equation is treated as a simultaneous equation and solved for tau while eliminating alpha, there are four solutions. This comes about, I believe, because the sign of a square root and the inverse-cosine of a sine can both be either plus or minus. The four solutions are: τ = (+mR*√(2R/m)*acos((2r-R)/R)+2*√(2mRr(R-r)))/(4m) τ = (+mR*√(2R/m)*acos((2r-R)/R)-2*√(2mRr(R-r)))/(4m) τ = (-mR*√(2R/m)*acos((2r-R)/R)+2*√(2mRr(R-r)))/(4m) τ = (-mR*√(2R/m)*acos((2r-R)/R)-2*√(2mRr(R-r)))/(4m) ----------------------------------------------------------------------- >> Yes, he should have said "diameter R" If you save a copy of the cached mathpage you have currently as an archive .mht file or as a complete .html file that includes the latex gif images and forward it to me at kevspage2001atyahoo.co.uk, I would be grateful. After that, if you refresh the page you will see that the mathpages page has been updated in the last 24 hours to state "a wheel of radius R/2". Spooky, eh! The update also now includes a derivation of the radial equation in Schwarzschild coordinate time I asked for in the open part 2 question. You might notice that the absolute function in the last term of the equation that I said "looked suspicious" has now gone. I am starting to get the strong impression that Kevin Brown must be a reader of uclue.com By the way, if Kevin Brown is the sole author of mathpages then it is very impressive. Is it possible for a single person to know that much about geometry, maths and physics, including the history and philosophy behind those subjects? Anyway, if you could address the questions I put here, as part of your answer to the open part 2 question and maybe add some insights (or a different approach) to the mathpages derivation of the radial motion in Schwarzchild coordinate time, then I am still happy to pay the original fee. You have already gone beyond the call of duty ;) Thanks
Answer clarification by Researcher mathtalk on Fri 21 Nov 2008 - 5:04 pm UTC:
> Let a = dr/dτ > dx/dτ = dx/dτ*a/a = dx/dτ*dτ/dr*a = dx/dr*a > so the derivative with respect to τ of the quantity > in square brackets is the same as derivative with > respect to r multiplied by a=dr/dτ (if I have reasoned > correctly but I am no means sure about this) > The result I get is > d[((1-2m/R)-a²)/(1-2m/r)]/dτ > = -2(a)/(r-2m)^2*m*(1-2m/R)-a²) > which is similar to the mathpages result, but missing > the last term. Your procedure doesn't seem to take into account that a is a function of τ, which is where we would get the missing term involving the second derivative of r wrt τ. I'll put a differentiation walk-thru into the answer of the other question. > Another thing I would like your opinion on, is would > you agree there is some ambiguity to the paramatric > equation? For example is the parametric equation is > treated as a simultaneous equation and solved for tau > while eliminating alpha, there are four solutions. > This comes about, I believe, because the sign of a > square root and the inverse-cosine of a sine can both > be either plus or minus. The only "ambiguity" of the parametric solution is that there can be many equivalent parameterizations of the same relation between r and τ. Because τ is a monotone increasing function of parameter α, the relation between r and τ constitutes a function r of "independent variable" τ, but not the other way round. So when you try to "solve" for τ in terms of r, you will run into infinitely many non-unique solutions (because of the periodicity of cosine and the fact that r is an even function of α). By uniqueness of r as a function of τ, I mean that there is only one solution for the ODE (5) together with the initial conditions r = R and dr/dτ = 0 at τ = 0. Any good undergraduate text on ODEs such as Boyce & DiPrima will have the basic results for existence and uniqueness of solutions. Existence can be a bit tricky. The proof assumes smoothness (Hölder continuity) of the nonlinear function, which fails at the singularities of equation (5), and moreover only gives a "local" existence of the solution (via a fixed point theorem). But existence is not at issue here because we've got an explicit solution (well, parametric solution, but that is explicit enough for the purpose of concluding that a solution exists up to the singularity & is analytic across the r = 2m boundary). However the contraction mapping that underlies the proof of existence gives uniqueness of the solution quite easily. In effect the distance between two solutions of (5) can be bounded by the distance between their initial conditions at τ = 0, so two solutions with equal initial conditions must be equal everywhere. > If you save a copy of the cached mathpage you have > currently as an archive .mht file or as a complete > .html file that includes the latex gif images and > forward it to me... Sorry. On my Windows machine I still have a cached copy of the site as I saw it when you first posted this question, while on my Linux machine I have the fully updated one that says "a rolling wheel of radius R/2." Yes, I agree that the author of the mathpages site is impressive. Perhaps that particular page was getting a lot of activity on his server logs, and he looked at the referring site and saw your notes. If so, credit is due to him for such quick response. regards, mathtalk
Comment by kline on Sat 22 Nov 2008 - 3:18 pm UTC:
>> Your procedure doesn't seem to take into account that a is a >> function of τ, Your right. I suspected that was the case, but I was not sure how to solve it. I think I have it figured out now. Let x be the original equation in square brackets that required differentiating with respect to τ : x = [((1-2m/R)-a²)/(1-2m/r)] x can be split into 2 terms (call them y and z) for convenience: where y = (1-2m/R)/(1-2m/r) and z = -a²/(1-2m/r) We know that dx/dτ = dy/dτ + dz/dτ y can be rewritten as (1-2m/R)(1-2m*r^(-1))^(-1) which can be differentiated with respect to τ using implicit differentiation with r as a function of τ and using the chain rule to give: dy/dτ = -2m(1-2m/R)(dr/dτ)/(r-2m/r)² = -2ma(1-2m/R)/(r-2m/r)² using the definition of dr/dτ = a, used in an earlier post. z can be differentiated with respect to τ using implicit differentiation with the understanding that both r and a are functions of τ Using the quotient rule dz/dτ = [d(-a²)/dτ (1-2m/r)-(-a²)d(1-2m/r)/dτ]/(1-2m/r)² dz/dτ = [-2a(da/dτ)(1-2m/r)+a²(2m(dr/dτ)/r^2)]/(1-2m/r)² dz/dτ = [-2a(da/dτ)(1-2m/r)r^2+a²(2m(dr/dτ))]/(r-2m)² Adding dy/dt and dz/dt together gives dx/dτ: dx/dτ = [-2ma(1-2m/R) -2a(da/dτ)(1-2m/r)r² +2ma²(dr/dτ)]/(r-2m)² Simplifying: dx/dτ = (-2a[m((1-2m/R)-a²)+(da/dτ)(1-2m/r)r²])/(r-2m)² which is basically the mathpages result when dr/dτ is substituted back in for a. Hopefully, I have saved you some work in explaining that bit to me.
Comment by kline on Sat 22 Nov 2008 - 10:32 pm UTC:
Just correcting a typo in my last post. The last line of this: > y can be rewritten as (1-2m/R)(1-2m*r^(-1))^(-1) > which can be differentiated with respect to τ using implicit > differentiation with r as a function of τ and using the chain rule > to give: > dy/dτ = -2m(1-2m/R)(dr/dτ)/(r-2m/r)² = -2ma(1-2m/R)/(r-2m/r)² should have read: dy/dτ = -2m(1-2m/R)(dr/dτ)/(r²(1-2m/r)²) = -2ma(1-2m/R)/(r-2m)²
Comment by User montecristo on Fri 6 Mar 2009 - 3:02 pm UTC:
Wow, what a fantastic read, even for someone like me who has no idea what it's all about.
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Comment by kline on Wed 12 Nov 2008 - 2:04 pm UTC: