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5 stars ANSWERED on Sat 29 Nov 2008 - 5:26 pm UTC by mathtalk

Question: Derivation involving calculus (Part 2)

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Asked by kline on Mon 10 Nov 2008 - 1:15 pm UTC:

I need to know how this equation for radial motion in Schwarzschild
coordinate time is derived:

t = (R/2+2m)Q(θ)+(R/2)Q sin(θ) + 2m ln(abs[(Q+tan(θ/2))/(Q-tan(θ/2))])

where Q = sqrt(R/2m-1).

Please see http://www.mathpages.com/rr/s6-04/6-04.htm and
http://www.mathpages.com/rr/s6-07/6-07.htm for background information.

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 unique.

(The absolute || function in the last term of the solution looks suspicious
to me.)


Question clarification by kline on Mon 10 Nov 2008 - 1:25 pm UTC:

The equation I wanted deriving got split over two lines so I'll try again:

t=(R/2+2m)Q(θ)+(R/2)Q sin(θ)+2m ln(abs[(Q+tan(θ/2))/(Q-tan(θ/2))])

where Q = sqrt(R/2m-1).

and it is at the bottom of this webpage
http://www.mathpages.com/rr/s6-04/6-04.htm

Uclue Researcher Request for clarification by Researcher mathtalk on Fri 21 Nov 2008 - 2:40 am UTC:

Given that we already have parametric representations for r and τ, we can
verify (or derive) the corresponding representation for t from the "metric"
given by equation (1) on that page.  Rewritten it says:

    (dt)² = (r/(r-2m))(dτ)² + (r/(r-2m))²(dr)²

More concretely one can obtain an expression for dt/dα, where α is the
parameter previously used to represent r and τ.


regards, mathtalk

Question clarification by kline on Fri 21 Nov 2008 - 3:58 am UTC:

Hi mathtalk,

I would like you continue with this question if possible. I have added a
comment to part 1 that is very relevant to this question, so please read
that before continuing.

Cheers,

K

Question clarification by kline on Mon 24 Nov 2008 - 2:19 am UTC:

Hi mathtalk,

I think I noticed a couple more errors in the mathpages page in question
http://www.mathpages.com/rr/s6-04/6-04.htm 

The last term of the equation just before the sentence "Now, making use of
the trigonometric identities" can be written as:

-4m*i*arctan[(cos(α)-1)/(sin(α)*sqrt(1-R/2/m)]

Substituting Q = sqrt(R/2/m-1) into the term gives:

-4mi arctan[(cos(α)-1)/(sin(α)*iQ)] 

where i = sqrt(-1)

Substituting the half angle identity tan(x/2) = (1-cos(x))/sin(x) as
suggested by K.Brown gives:

-4m*i*arctan[-tan(α/2)/(i*Q)]

Substituting the identity, arctan(x) = -i/2*ln[(1+ix)/(1-ix)] where in this
case x = -tan(α/2)/(i*Q) gives:

-4m*i*-i/2*ln[(1-tan(α/2)/Q)/(1+tan(α/2)/Q)]

which simplifies to:

-2m*ln[(Q-tan(α/2))/(Q+tan(α/2))]

which is similar to the last term given in the final mathpages equation for
the radial motion of a particle in Schwarzchild coordinate time, but the
signs are all wrong.

Tracing backwards reveals that the identity given for the arctangent is not
true when x is a negative imaginary number and it turns out that the
quantity x = -tan(α/2)/(i*Q) = -tan(α/2)/sqrt(1-R/2/m) does take on
negative imaginary values in the domain we are talking about (-pi/2 <α<
pi/2), for example when m = 1, R = 3 and α = -1 so this problem can not be
ignored.

I think the reason for the confusion is that the so called "fundamental
propeties" of natural logarithms do not always hold for complex values. For
example:

ln(a*b) = ln(a) + ln(b)  is not true if a = -1 and b = sqrt(-1) and 

ln(a/b) = ln(a) - ln(b)  is not true if a =-1 and b = -sqrt(-1)

However, playing around with signs and inserting the correct arctangent
identity does not fix the problem and the problem is deeper than that,
somewhere in the integration step I suspect.

Uclue Researcher Request for clarification by Researcher mathtalk on Mon 24 Nov 2008 - 4:18 pm UTC:

This is really just a comment, albeit perhaps one about my approach to your
question.  You wrote:

> I think the reason for the confusion is that the
> so called "fundamental propeties" of natural
> logarithms do not always hold for complex values.
> For example:

> ln(a*b) = ln(a) + ln(b)
>  is not true if a = -1 and b = sqrt(-1) and 
> ln(a/b) = ln(a) - ln(b)
>  is not true if a =-1 and b = -sqrt(-1)

> However, playing around with signs and inserting
> the correct arctangent identity does not fix the
> problem and the problem is deeper than that,
> somewhere in the integration step I suspect.

I think we should stick to real values for the
sake of clarity, and establish the validity of
the solution for r > 2m of the parametric value
of coordinate time t.

The extension of natural logarithm to complex
arguments produces a multi-valued "function",
an artifact of which is the absolute values
inserted at one point inside the logarithm.
Since the exponential function is periodic
in the complex plane (of period 2πi), taking
the inverse (natural logarithm) produces a
multi-layered surface.

regards, mathtalk

Question clarification by kline on Tue 25 Nov 2008 - 8:19 pm UTC:

Hi mathtalk,

I have satisfied myself that the mathpages integral is OK by by copying and
pasting this expression R/2*sqrt(R/2/m-1)*(1+cos(a))/(1-4*m/R/(1+cos(a)))
into the online web-mathematica solver here
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced
and solving for indefinite integral of a (which represents alpha here). As
far as I can tell, when a is zero the indefinite integral is zero so there
is no need to solve for the definite integral, which seems to give the
online solver a headache.

The result churned out by the solver is basically the mathpages integration
with the half angle formula already substituted. 

When the correct identity arctan(x) = i/2*ln[(1-ix)/(1+x)] which is correct
for all complex x, is used in the derivation, rather than the
not-strictly-correct identity: arctan(x) = -i/2*ln[ (1+ix)/(1-x) ] used by
mathpages, then it all works out fine and the final result for radial
motion in Schwarzschild coordinates is as claimed by K.Brown in the final
equation of the mathpage’s derivation.

(I hope he reads this and fixes the erroneous arctangent identity in the
derivation.) 

----------------------------------------------------------------------------

An alternative derivation starting from the mathpages integral with the
half angle formula and Q = sqrt(R/(2m)-1) already substituted is:

t = (R/2+2m)Qα+(R/2)Q*sin(α)-4m*i*arctan[-i*tan(α/2)/Q]

using the hyperbolic arctangent identity,  arctanh(z) = -i*arctan(iz) ref(
http://www.cs.umbc.edu/~squire/reference/identities.txt )

t = (R/2+2m)Qα+(R/2)Q*sin(α)+4m*arctanh[tan(α/2)/Q]

using the logarithmic representation of the hyperbolic arctangent,
arctanh(z) = ½*ln[(1+z)/(1-z)]  ref(
http://en.wikipedia.org/wiki/Inverse_hyperbolic_functions#Logarithmic_representation
)

t = (R/2+2m)Qα+(R/2)Q*sin(α)+4m*ln[(1+tan(α/2)/Q)/(1-tan(α/2)/Q)]

Simplifying to

t = (R/2+2m)Qα+(R/2)Q*sin(α)+4m*ln[(Q+tan(α/2))/(Q-tan(α/2))]

which is the same as the final result obtained by Ken Brown.

P.S. I am still interested in any thoughts you have on the mathpage’s
derivation (or my derivation) but the question is about to expire, so if
you do not have time do not worry about it too much. (No pressure)

Cheers.

Comment by kline on Tue 25 Nov 2008 - 8:26 pm UTC:

P.P.S Any thoughts on new paragraph recently added to the end of
http://www.mathpages.com/rr/s6-04/6-04.htm concerning the physical
interpretation of the equations?

Uclue Researcher Comment by Researcher mathtalk on Wed 26 Nov 2008 - 5:49 am UTC:

It's quite interesting and addresses a number of topics that occurred to
me.  For example the "metric" involves squares of infinitesimals.  In pure
math this is a bit questionable, and presumably one "means" something
like:

    (dt/dα)² = (r/(r-2m))(dτ/dα)² + (r/(r-2m))²(dr/dα)²

rather than simply stating:

    (dt)² = (r/(r-2m))(dτ)² + (r/(r-2m))²(dr)²

In any case the equations do not determine whether proper time and
coordinate time are to flow in the same direction or opposite directions.

I took the point of view that in the limit of an apogee R tending to +oo,
these time measurements should agree.  From this one deduces that they flow
in the "same" direction.  [By symmetry whether one considers them both to
move forward or both backward is a "distinction without a difference".]

regards, chip

Question clarification by kline on Wed 26 Nov 2008 - 1:49 pm UTC:

> For example the "metric" involves squares of infinitesimals. In pure
> math this is a bit questionable...

I assume you are referring to the orphaned numerators of the infinitesimal
quotients. I think one of the difficulties with handling these sort of
problems is that sometimes infinitesimals can be treated like ordinary
variables and sometimes they can not. For example the derivation of the
velocity of light from those coordinates involves finding the square roots
of those squared infinitesimals like this:

Stating with purely radial Schwarzschild metric, 

(dτ)^2 = [1-2m/r](dt)^2 - [1/(1-2m/r)](dr)^2

For light the proper time is zero so:

0 = [1-2m/r](dt)^2 - [1/(1-2m/r)](dr)^2

[1/(1-2m/r)](dr)^2 = [1-2m/r](dt)^2 

(dr)^2/(dt)^2  = [1-2m/r][1/(1-2m/r)] = [1-2m/r]^2

dr/dt = [1-2m/r] for a photon.

Question clarification by kline on Wed 26 Nov 2008 - 1:52 pm UTC:

P.S. I have made no comments on your comments of time. This is because I do
not want to be accused of "leading the (expert) witness".

Question clarification by kline on Thu 27 Nov 2008 - 6:11 am UTC:

Hi Chip,

You may have noticed that the mathpage has been updated again overnight.
The good new is that Kev Borown has taken onboard this comment of mine:

> When the correct identity arctan(x) = i/2*ln[(1-ix)/(1+x)] which is
> correct for all complex x, is used in the derivation, rather than the
> not-strictly-correct identity: arctan(x) = -i/2*ln[ (1+ix)/(1-x) ] used >
by mathpages, then it all works out fine ..

and corrected the arctan identity. The bad news is that the absolute
function is back in the final term of the equation for radial motion in
Schwarzschild coordinate time. Remember in my first post I said the
absolute function looked suspicious there? It was removed in one of the
updates and I do not see any mathematical justification for its comeback.
Do you?

As far as I can tell the only justification is that the absolute function
allows the curve to be extended down below r=2m as shown in the mathpage
plot in order to comply with the classical physical interpretation. This
has now become the single most important issue that I am interested in now.

Uclue Researcher 5 stars Answer by Researcher mathtalk on Sat 29 Nov 2008 - 5:26 pm UTC:

>  As far as I can tell the only justification is that the
>  absolute function allows the curve to be extended down
>  below r=2m as shown in the mathpage plot in order to
>  comply with the classical physical interpretation. This
>  has now become the single most important issue that I am
>  interested in now.

While I don't see how extending of the solution for coordinate
time t below r=2m (for a bounded trajectory with apogee R > 2m)
complies with any "classical physical interpretation," I can
can offer some mathematical observations on this extension.

The absolute value inside the logarithm is crucial to extending
the trajectory below r=2m, but its presence is sensible.  As a
quick explanation, the conventional antiderivative for 1/x is:

     d ln|x|
     ------- =  1/x        (x =/= 0)
       dx

valid on the real line except zero.  Of course a constant could
be added arbitrarily, and indeed one might use different constants
for x > 0 and x < 0, since we already have a discontinuity there.
But the above is evidently the simplest expression.

In Kevin Brown's writeup the parametric equation for t:

                                           |Q + tan(α/2)|
    t = Q*((R/2 + 2m)α + (R/2)sin(α)) + ln(|------------|)
                                           |Q - tan(α/2)|

where constant Q is:
         __________
    Q = √(R/2m) - 1     [assume R > 2m]

is to be compared with the corresponding equations:

    r = (R/2) (1 + cos(α))
               ____
    τ = (R/2) √R/2m (α + sin(α))

In the latter case we have a smooth passage of r as a function
of proper time τ across r=2m.  By smooth here we mean analytic
continuation, which implies continuous derivatives of all orders
across this boundary.

Indeed even the behavior of r(τ) as τ approaches π is actually
very nice as singularities go.  Recall from Part 1 that:
                     ____
   dr/dτ = -sin(α)/[√R/2m (1 + cos(α))]

Although this becomes infinite at odd multiples of π, those are
"simple poles" in the terminology of analytic functions, i.e.
the reciprocal dτ/dr has simple roots (zeros) at these places.

Given these mathematical facts, a physical interpretation of the
solution r(τ) as a periodic orbit seems unobjectionable to me.

However the singularity at r=2m in function r(t) of coordinate
time is much worse, an essential singularity in analytic function
theory.  Accordingly my faith in any physical interpretation of
the purely mathematical extension of r(t) across the boundary is
weak.  If anything such an interpretation would be supported best
by consistency with the smooth extension in terms of proper time.

That is, whatever metric exists within the "event horizon" r < 2m,
one could relate coordinate time there with the proper time τ and
radius r associated to the entering object.  But relating some
coordinate inside the event horizon to that outside seems quite
problematic.

For one thing as you showed above, light effectively stops at the
event horizon, so no external observational framework supports a
notion of relating spacetime coordinates across this boundary.

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *

If we permit speculation, a number of authors of both serious and
fictive literature have presented ideas that black holes may be
gateways to time travel.  Perhaps the simplest form of such a
theory would be an idea that time inside the event horizon runs
opposite to time outside.

But my impression is that the direction of time is mysterious
enough without dragging black holes into the picture.  An author I
like on this subject is Huw Price:

[Huw Price, Univ. of Sydney]
http://www.usyd.edu.au/time/price/

What we can say is that the parameterizations r(α) and τ(α) give
rise to a solution r(τ) in proper time that is continuous and
has period:
                      ____
   τ(π) - τ(-π) = Rπ √R/2m,

while the parameterization t(α) "blows up" at α = ±2 arctan(Q).

[Since Q involves a square root, we can consider what would be
the result if the negative root were chosen.  From the definition
of t(α) one sees that if Q and α are replaced by -Q and -α, then
exactly the same value for t is obtained.  Since r(α) is an even
function, this amounts to a different parameterization for the
same solution t(α).  So in all we would conclude as before that
the direction of coordinate time can go either way while still
satisfying the Schwarzchild metric equation.]

Evaluating t(α) between the limits ±2 arctan(Q) can be done
without applying the absolute value in the logarithm's argument.
Extension beyond these limits gets us a fraction with negative
values, so that the absolute value has to be applied before the
(real) logarithm can be taken.

If α is slightly increased above 2 arctan(Q), then the numerator
Q + tan(α/2) is little changed and the denominator Q - tan(α/2)
switches from a small positive value to a small negative one.
After applying absolute values, the argument of the logarithm
is still a large (positive) value, and taking the logarithm will
give a large positive result for t(α).  However we have now
entered a region in which t(α) decreases with increasing α, a
perhaps different sense in which one could claim time runs
backward.

Similarly an argument could be made that moving backward along
the curve t(α) until α is decreased slightly below -2 arctan(Q)
will bring us to the other branch where r < 2m, but with t(α)
increasing (as the argument to the logarithm passes thru zero).

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *  *

Is this an artifact of applying the absolute values to the
argument of logarithm?  Well, yes it is.  But there are many
things that are artifacts of following ones mathematical nose,
which is all that is being done here.  Sometimes it means a
solution to a word problem involving negative ages for family
children (which clearly has no real world meaning) and other
times it gives us negative weights for the amount of sugar to
buy (which might seem meaningless at first, but could turn
out to mean an advantage from selling sugar!).

For me the plausibility of this approach to theoretical time
travel is countermanded by two facts, one mathematical and
one physical.  The math fact has already been mentioned, and
that is the essential singularity of the logarithm we must
"pass thru" at either limit of α = ±2 arctan(Q).

The physical fact is that black holes are not static entities
in our Universe.  The best observation evidence is that they
have a definite "birth" and growth, and according to theory
advanced by Stephen Hawking, eventual dissipation.  Thus the
chronological extent of a black hole phenomenon would seem to
be finite, at least in reference to external coordinate time,
and not extended to infinity as the (necessarily simplified)
mathematical model suggests.


regards, mathtalk

5 stars Accepted and rated by kline on Tue 2 Dec 2008 - 7:26 am UTC:

Hi Mathtalk,

I have accepted your answer. It does not entirely agree with my own
conclusions, but I asked for your honest opinion and you provided that and
I appreciate your valued input. Some of your observations are particulary
interesting. I am working on a project that this question is just a small 
part of, and if I get stuck on the maths again, I will know who to go to.

Thanks

Uclue Researcher Answer clarification by Researcher mathtalk on Tue 2 Dec 2008 - 11:43 am UTC:

Thanks. I am racking my brain for mathematical trails that I may have
overlooked regarding alternative solutions.

I think there is a piece of the mathematical solution that we should fill
in here, namely the solutions t(α) that correspond to the case apogee R <
2m.

In other words we began following Kevin Brown's writeup by looking at the
solution for r(α) and τ(α) in the case R < 2m, and then worked through
the observation that despite the apparent singularity in the 2nd order
differential equation, the form of solution remains the same (and is
continuous) coming across from R > 2m.

The solutions t(α) corresponding to the latter are what were pictured in
the mathpages graph and discussed above, but I don't think any discussion
has been made yet about t(α) for R < 2m.  It seems likely that because the
singularity in coordinate time t at r = 2m refuses to fade away from the
calculations, the form of t(α) inside the event horizon may be different,
possibly involving tanh(α/2) instead of tan(α/2).  Just a speculation at
present...


regards, mathtalk

Request for clarification by kline on Sat 13 Dec 2008 - 2:24 am UTC:

Hi mathtalk,

> It seems likely that because the singularity in coordinate time t
> at r = 2m refuses to fade away from the calculations, the form of t(α)
> inside the event horizon may be different, possibly involving tanh(α/2)
> instead of tan(α/2). 

I see you have discovered some of the problems involved in the time aspect
of the motion equations. As I mentioned before, this question was part of a
larger project I was working on. I have largely completed that project now
and I would like you review my project and see if, for you it resolves any
of the problems. I posted the review request in the form of a new question.
I hope you have time to take a look at it.

Best regards,

Kev

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