Howell - "Unconnected" (particle, observer) reference frames (RFp) & (RFo)
www.BillHowell.ca initial draft ?date?
Summary :
/*_file_insert_path "$d_Lucas/context/waiver, copyright.txt
Table of Contents
IV. "Unconnected" (particle, observer) reference frames (RFp) & (RFo)
22Jan2016 - Note that this Section ""Unconnected" (particle, observer) reference frames" is way out of date and uses older, simpler symbols.
It has to be redone, even though it is not of direct use for the review of Lucas's book.
This chapter may not be terribly useful for Lucas's book, but it was of interest to me. Initially this was because I was uncertain about the symbols and setups of reference frames in Lucas's book, but now it is still of interest as it would be useful for systmes with many [particles, systems] moving in a region, in a manner that one cannot use the simple expressions of Chapter II.
IV.2 Q(t) as the "hard link" between reference frames
Introduction
Figure "[Q(t),vov(t)]"
[Q(t),vov(t)] provide the "hard link" between observer and particle reference frames. As previously noted, for Chapter 4 vo is constant, so a convenient "hard link" is [Q0,vo], where Q0 = Q(t=0).
Expressions involving ro*vs*cosQ(t) will come up later, so it's convenient to derive expressions for [Q(t),cosQ(t)] now. Given that v is constant, the angle Q at any time t may be calculated knowing it's value at t=0, which I'll call Q0, and r at t=0. I need [r(t),rs(t),Q(t)] in terms of [r(t=0),Q(t=0),v,t]
??? Show Figure for Q(t) ???
Referring to the Figure above, consider the line of trajectory of the particle. Its direction is vh (from the velocity of the particle), but its 2D location on the unchanging plane of [r,v] relative to the observer must also be determined, as specified by [r(t=0),Q(t=0)]. Of course, the particle is always on the trajectory, but it's position at any time t is also needed. Here, the observer is considered to be stationary in his frame of reference.
A vector from the observer intercepts the line of trajectory at right angles at some time t (positive or negative time). We can use this diagram to help visualise the expression of each "dependent" variable from [ro(t),rocs(POIp,t),Q(t)] in terms of the "independent" variables [ro(t=0),Q(t=0),v,t]. Note that all independent variables except "t" are constant.
The notiation rs0 = rs(t=0) is used to simplify expressions. Perhaps 3 cases at t=0 for development : Q(t=0) is [acute, perpendicular, obtuse], but a general solution is likely possible. This initial analysis is for an acute angle O (theta, not zero).
Note that :
1. The [vov, rov] plane (same plane as [vov,rpv] DOES NOT rotate as the particle moves in the direction of vov, such that a fixed 2D plane (fixed in RFo and RFp) links the two frames of reference.
2. [Q(t),ro ,rs,xo,yo,rp] are functions of time, but not [Op,Pp, tp(t),v,xp(t),yp(t)] (nor of course the coordinate systems basis).
Symbol list
As per the "Introduction, basics" section,
PTox is the point defined by intersection of the particle's line of trajectory, vh, and a line perpendicular to that trajectory and that passes through the observer
Dv(t) is the distance at time t from the particle to PTox, which is along the line of trajectory of the particle vh
Dv0 = Dv(t=0)
Dw is the distance from the observer to PTox, which is perpendicular to the line of trajectory of the particle vh, and is the closest observer-particle distance.
tx is the time at which the particle would be at the PTox
Expressions for the basic variables
Dw = ros0*sinQ0
Dv0 = ros0*cosQ0
tx = Dv0/vs
= ros0*cosQ0/vs
Dv(t) = vs*t + Dv0 = vs*(t + tx)
rocs(POIp,t) = [Dw^2 + Dv(t)^2]^(1/2)
= [(ros0*sinQ0)^2 + (vs*(t + tx))^2]^(1/2)
= ros0*[sin^2Q0 + (vs*(t + tx)/ros0)^2]^(1/2)
Q(t) = arctan(Dw/rocs(POIp,t))
= arctan(ros0*sinQ0/rocs(POIp,t))
Polar expression for rocs(POIp,t)
rocs(POIp,t) = not convenient for now - maybe later
ro(t) = rocs(POIp,t) is in the direction Q(t)
= rs0*sinQ0/cos^2Q0/sin(Q(t))*Qh(t)
where Qh(t) is a unit vector in the direction of Q(t)
However, it's easier to use Cartesian coordinates in the [r,v] plane, centered on point on v trajectory perpendicular to the observer position.
ro(t) = Dv(t)*vh Dw*wh
= vs*(t + tx )*vh Dw*wh
ro(t) = vs*(t + rs0/vs*cosQ0)*vh rs0*sinQ0*wh
For the case where the observer and particle coordiante origins coincide at time t=0, with the observer x-axis colinear with vh, cosQ0 = 1, [sinQ0,ros0,Dw,Dv0] are all zero then :
r_coincident_observerFrame(t) = vs*t*vh 0*wh = v*t
Summarizing
2) Qx = π/2
3) Dv0 = ros0*cosQ0
4) Dw = ros0*sinQ0
5) tx = Dv0/vs = ros0*cosQ0/vs
6) Dv(t) =
7) rocs(POIp,t) = ros0*[sin^2Q0 + (vs*(t + tx)/ros0)^2]^(1/2)
8) ro(t) = (vs*t + rs0*cosQ0)*vh rs0*sinQ0*wh
8a) r_coincident_observerFrame(t) = v*t
9) Q(t) = arctan(ros0*sinQ0/rocs(POIp,t))
Derivatives of [ro(t),rocs(POIp,t)] expressed in terms of Q(t)
For ro(t) using :
8) ro(t) = (vs*t + rs0*cosQ0)*vh rs0*sinQ0*wh
dp(dt : ro(t))
= dp(dt : (vs*t + rs0*cosQ0)*vh rs0*sinQ0*wh )
= dp(dt : (vs*t + rs0*cosQ0)*vh 0*wh
= vs*vh 0*wh
= v
therefore, as expected :
10) dp(dt : ro(t)) = v
For rocs(POIp,t) using :
7) rocs(POIp,t) = ros0*[sin^2Q0 + (vs*(t + tx)/ros0)^2]^(1/2)
In the following derivation, ros is shown as rocs(POIp,t) for brevity.
dp(dt : rocs(POIp,t))
= dp(dt : ros0*[sin^2Q0 + (vs*(t + tx)/ros0)^2 ]^(1/2) *vh)
= (1/2)*ros0*[ sin^2Q0 + (vs*(t + tx)/ros0)^2 ]^(-1/2)*vh*
[ dp(dt : sin^2Q0) + dp(dt : vs*(t + tx)/ros0)^2)]
= (1/2)*ros0^2/rocs(POIp,t)*vh*dp(dt : vs*(t + tx)/ros0)^2)
= (1/2)*ros0^2/rocs(POIp,t)*vh*(vs/ros0)^2*dp(dt : (t + tx)^2
= (1/2)*ros0^2/rocs(POIp,t)*vh*(vs/ros0)^2*2*(t + tx)*dp(dt : (t + tx))
= /rocs(POIp,t)*vh *vs^2 *(t + tx)*dp(dt : (t + tx))
*where
5) tx = Dv0/vs = ros0*cosQ0/vs
*which is a constant, so dp(dt : (t + tx)) = 1, and
= vs^2/rocs(POIp,t)*(t + tx)*vh
10a) dp(dt : rocs(POIp,t)) = vs^2*(t + tx)/rocs(POIp,t)*vh
For the case where the observer and particle coordinate origins coincide at time t=0, with the observer x-axis colinear with vh, cosQ0 = 1, [sinQ0,ros0,Dw,Dv0] are all zero then :
dp(dt : rocs(POIp,t))_coincident_observerFrame(t) = vs^2*(t + tx)/rocs(POIp,t)
where under these conditions
7) rocs(POIp,t) = ros0*[sin^2Q0 + (vs*(t + tx)/ros0)^2]^(1/2)
becomes
rocs(POIp,t) = ros0*[vs*(t + tx)/ros0)^2]^(1/2)
rocs(POIp,t) = ros0 *vs*(t + tx)/ros0
rocs(POIp,t) = vs*(t + tx)
therefore
dp(dt : rocs(POIp,t))_coincident_observerFrame(t)
= vs^2*(t + tx)/vs*(t + tx)
= vs
10b) dp(dt : rocs(POIp,t))_coincident_observerFrame(t) = vs
As expected!
IV.3 Norm of ro = |rocv(POIp,t)|
Norm of ro
Note that the norm of a vector such as r is often depicted as ||r||, but here I follow Lucas's usage of |r|.
The expressions shown below are for the OBSERVER frame of reference. For the PARTICLE frame of reference, simply replace the qualifier "o" (small case letter "o") with "p" (small case letter "p").
??? I need to expand these ???
In spherical coordinates :
11) |ro(t)| = rocs(POIp,t)
In Cartesian coordinates :
12) |ro(t)| = [ |xo(t)|^2 + |yo(t)]^2 + |zo(t)|^2 ]^(1/2)
In [vh,wh] coordinates :
The following expressions do NOT apply to the particle frame of reference :
In [vh,wh] coordinates : repeating (8) and (15) (see next subsection) :
8) ro(t) = (vs*t + rs0*cosQ0)*vh + rs0*sinQ0*wh
15) |x| = (x dotProd x)^(1/2)
therefore
13) |ro(t)| = [ (vs*t + rs0*cosQ0)^2 + (rs0*sinQ0)^2 ]^(1/2)
Noting that rs0/cosQ0 ??????????????????
For the case where the observer and particle coordiante origins coincide at time t=0, with the observer x-axis colinear with vh, cosQ0 = 1, [sinQ0,ros0,Dw,Dv0] are all zero then :
8a) r_coincident_observerFrame(t) = v*t
13a) |ro(t)|_coincident_observerFrame = vs*t - rs0
IV.4 Derivative of |rocv(POIp,t)|
27Sep2015 inital using Lucas's formulations with (r - v*t)
29Sep2015 using Howell's formulation with r (abandoned)
30Sep2015 using a general (standard?) approach
04Jan2016 Note that this was used for trials as a substitute for Norm of (ro - vo*t) = |ro - vo*t|, but is retained here anyways
Kahan's formulation
http://www.eecs.berkeley.edu/~wkahan/MathH110/NormOvrv.pdf
p21h0.2
d||z|| = u_T dotProd dz / ||z||
where u_T is the linear functional dual to z wrt ||...||
z,u are vectors, and p16h0.7
||x||2 = (sum(|xi|^2))^(1/2)
or Kreyszig 1972 p200 Eqn (2)
15) |x| = (x dotProd x)^(1/2)
I will use |...| notation instead of ||...|| to denote the vector norm
Also - Does d||z|| have a DIRECTION? I wouldn't think so, as ||z|| is a scalar itself.
From http://www.math.utah.edu/~cherk/teach/12calcvar/210reminder.pdf
The vector derivative of the euclidean norm |a| of a nonzero vector a
is a row vector b,
b = d(da : (|a|^2)^(1/2) ) = a_T/(|a|^2)^(1/2) = a_T/|a|
Observe that b is codirected with a and has unit length.
But in my work here, I will be taking a scalar derivative (dp(dt : ...),
so Cherk's comment isn't applicable.
Taking the derivative of |ro(t)|:
d( |ro(t)|) = ro_T(t) dotProd d ( ro(t)) /|ro(t)|
dp(dt: |ro(t)|) = ro(t)_T dotProd dp(dt : ro(t)) /|ro(t)|
or, dropping the transpose notation to be the same as Lucas's book :
16) dp(dt : |ro(t)|) = ro(t) dotProd dp(dt : ro(t)) /|ro(t)|
From earlier
8) ro(t) = (vs*t + rs0*cosQ0)*vh rs0*sinQ0*wh
10) dp(dt : ro(t)) = v
therefore
ro(t) dotProd dp(dt : ro(t))
= ((vs*t + rs0*cosQ0)*vh rs0*sinQ0*wh )
dotProd
( vs*vh 0*wh)
= vs*(vs*t + rs0*cosQ0) + 0
= vs*(vs*t + rs0*cosQ0)
therefore
17) ro(t) dotProd dp(dt : ro(t)) = vs*(vs*t + rs0*cosQ0)
Subbing (17) into (16)
18) dp(dt : |ro(t)|) = vs*(vs*t + rs0*cosQ0)/|ro(t)|
where |ro(t)| can be calculated from :
13) |ro(t)| = [ (vs*t + rs0*cosQ0)^2 + (rs0*sinQ0)^2 ]^(1/2)
For the case where the observer and particle coordiante origins coincide at time t=0, with the observer x-axis colinear with vh, cosQ0 = 1, [sinQ0,ros0,Dw,Dv0] are all zero then :
dp(dt : |ro(t)|)_coincident_observerFrame
= vs*(vs*t + rs0*cosQ0)
/[ (vs*t + rs0*cosQ0)^2 + (rs0*sinQ0)^2 ]^(1/2)
= vs*(vs*t + rs0) / (vs*t + rs0)
= vs as expected
19) dp(dt : |ro(t)|)_coincident_observerFrame = vs
Derivatives of |rocv(POIp,t)|^n
From earlier :
8) ro(t) = (vs*t + rs0*cosQ0)*vh rs0*sinQ0*wh
10) dp(dt : ro(t)) = v
13) |ro(t)| = [ (vs*t + rs0*cosQ0)^2 + (rs0*sinQ0)^2 ]^(1/2)
18) dp(dt : |ro(t)|) = vs*(vs*t + rs0*cosQ0)/|ro(t)|
therefore
dp(dt : |ro(t)|^(-3))
= (-3)*|ro(t)|^(-4)*vs*(vs*t + rs0*cosQ0)/|ro(t)|
= (-3)*|ro(t)|^(-5)*vs*(vs*t + rs0*cosQ0)
Likewise :
dp(dt : |ro(t)|^(+1)) = (+1)*|ro(t)|^(-1)*vs*(vs*t + rs0*cosQ0)
dp(dt : |ro(t)|^(-3)) = (-3)*|ro(t)|^(-5)*vs*(vs*t + rs0*cosQ0)
dp(dt : |ro(t)|^(-5)) = (-5)*|ro(t)|^(-7)*vs*(vs*t + rs0*cosQ0)
dp(dt : |ro(t)|^(-7)) = (-7)*|ro(t)|^(-9)*vs*(vs*t + rs0*cosQ0)
For the case where the observer and particle coordiante origins coincide at time t=0, with the observer x-axis colinear with vh, cosQ0 = 1, [sinQ0,ros0,Dw,Dv0] are all zero then :
Taking the results above for dp(dt : |ro(t)|^n)
# dp(dt : |ro(t)|^(+1)) = (+1)*|ro(t)|^(-1)*vs^2*t
# dp(dt : |ro(t)|^(-3)) = (-3)*|ro(t)|^(-5)*vs^2*t
# dp(dt : |ro(t)|^(-5)) = (-5)*|ro(t)|^(-7)*vs^2*t
# dp(dt : |ro(t)|^(-7)) = (-7)*|ro(t)|^(-9)*vs^2*t
# Remember - this is ONLY for [observer, particle] coordinate systems
# that share a common origin at time t=0, with the x-axis pointing
# in the same direction as v.
Questions concerning the Derivative of |rocv(POIp,t)|
However, for later consideration during re-analysis :
• If Q replaces all sinOp - then the complexity problem may just have a different symbol, but remains the same problem?
• how are the integration limits set?
• For a diffuse charge, shouldn't phi also play a role?
• Q is "planar", Oo "latitudinal", and phi "longitudinal" - integration still a challenge?
IV.5 Norm of (ro - vo*t) = |ro - vo*t|
Norm of (ro - vo*t)
Given the constant-particle-velocity situation for Chapter 4, the Galilean transformation is a constant :
from 1) Measure(ro(t) - vo*t) = Measure(ro(t=0))
= Measure(rp(t=0)) IFF observer's refFrame origin coincident with particle's refFrame at t=0
which is a constant, and where "Measure" could be [rs,E,B,F, etc].
Simply put, the norm of (ro - vo*t) is ros(t=0) which is a constant :
20) |ro - vo*t| = ros(t=0)
IV.6 Derivative of |ro - vo*t|
See IV.4 :
27Sep2015 inital using Lucas's formulations with (r - v*t)
29Sep2015 using Howell's formulation with r (abandoned)
30Sep2015 using a general (standard?) approach
Newer :
04Jan2016 using (ro - vo*t)
As (ro - vo*t) is a constant (vector) for a given POI (Point Of Interest), its norm is as well, and the derivative is zero.
21) dp(dt : |ro - vo*t|) = 0
But to check consistency here I'll follow Kahan's formulation, as per Section IV.4.
Kahan's formulation
http://www.eecs.berkeley.edu/~wkahan/MathH110/NormOvrv.pdf
p21h0.2
d||z|| = u_T dotProd dz / ||z||
where u_T is the linear functional dual to z wrt ||...||
z,u are vectors, and p16h0.7
||x||2 = (sum(|xi|^2))^(1/2)
or Kreyszig 1972 p200 Eqn (2)
15) |x| = (x dotProd x)^(1/2)
I will use |...| notation instead of ||...|| to denote the vector norm
Also - Does d||z|| have a DIRECTION? I wouldn't think so, as ||z|| is a scalar itself.
Taking the derivative of |ro - vo*t|:
d( |ro - vo*t|) = (ro - vo*t)_T dotProd d (ro - vo*t)
/|ro - vo*t|
dp(dt: |ro - vo*t|) = (ro - vo*t)_T dotProd dp(dt : ro - vo*t)
/|ro - vo*t|
or, dropping the transpose notation to be the same as Lucas's book :
22) dp(dt: |ro - vo*t|) = (ro - vo*t) dotProd dp(dt : ro - vo*t)
/|ro - vo*t|
Given that ro is to a point of interest (POI) moving with the particle, such that :
23) dp(dt : ro) = -vo
then we can say
24) dp(dt : ro - vo*t) = dp(dt : ro) - dp(dt : vo*t) = 0
and from (22) :
25) dp(dt: |ro - vo*t|) = (ro - vo*t)_h dotProd 0 = 0
IV. Special integrals adapted to unconnected reference frames
IV.1 Summary
# see "Howell - Symbols for Bill Lucas, Universal Force.odt"
Recurring integrals from http://integral-table.com
# see "Howell - Symbols for Bill Lucas, Universal Force.odt" :
∫(dQ, 0 to Qf : sin^3Q ) = - 3/4*cosQ + cos(3*Q)/12
∫(dQ, 0 to Qf : sin^5Q ) = - sin^4Q*cosQ/5 - 3/5*cosQ + cos(3*Q)/15
∫(dQ, 0 to Qf : sinQ*cosQ ) = sin^2Q/2
∫(dQ, 0 to Qf : cosQ*sin^nQ ) = sin^(n + 1)Q/(n + 1) for n>0
∫(dQ, 0 to Qf : sinQ*cos(3*Q) ) = cos(2*Q)/4 - cos(4*Q)/8
∫(dQ, 0 to Qf : (cosQ - 1)*sinQ) = sin^2Q/2 + cosQ
∫(dQ, 0 to Qf : (cosQ - 1)*sin^2Q) = ??
∫(dQ, 0 to Qf : (cosQ - 1)*sin^3Q ) = sin^4Q/4 + 3/4*cosQ - cos(3*Q)/12
∫(dQ, 0 to Qf : (cosQ - 1)*sin^5Q ) = + sin^6Q - sin^4Q*cosQ/5 - 3/5*cosQ + cos(3*Q)/15
I don't have these :
∫(dQ, 0 to Qf : sinQ*cos(3*Q) ) -> dunno
∫(dQ, 0 to Qf : (cosQ - 1)*sin^5Q ) -> need to decifer sin^5Q
Literature reults
from http://integral-table.com
∫(dx : cos(a*x)*sin(b*x) = cos((a - b)*x)/2/(a - b) - cos((a + b)*x)/2/(a + b)
∫(dx : sin^2(a*x) ) = -3*cos(a*x)/4/a + cos(3*a*x)/12/a
from people.rit.edu search "table of integrals"
http://people.rit.edu/pnveme/pigf/Integrals/int_tables_trig_eqn.gif
∫(dx : sin^n(ax))
= - sin^(n-1)(a*x)*cos(a*x)/n/a + (n - 1)/n*∫(dx : sin^(n - 2)(a*x) )
∫(dx : sin^2(a*x) ) = x/2 - sin(2*a*x)/4/a
∫(dx : sin^n(a*x) ) = -cos(a*x)/a *F[2,1 : 1/2, (1 - n)/2, 3/2, cos^2(a*x)]
where F(2,1 : ) is the ??? function
∫(dQ, 0 to Qf : sin^3Q ) = -3*cosQ/4 + cos(3*Q)/12
∫(dQ, 0 to Qf : sin^4Q )
= - sin^3Q*cosQ/4/1 + 3/4*∫(dQ : sin^2*Q) )
∫(dQ, 0 to Qf : sin^5Q )
= - sin^(5-1)Q*cosQ/5 + (5 - 1)/5*∫(dQ : sin^(5 - 2)Q )
Subbing for sin^3Q
= - sin^4Q*cosQ/5 + 4/5*( -3/4*cosQ + cos(3*Q)/12 )
= - sin^4Q*cosQ/5 + ( -3/5*cosQ + cos(3*Q)/15 )
= - sin^4Q*cosQ/5 - 3/5*cosQ + cos(3*Q)/15
http://www.themathpage.com/aCalc/inverse-trig.htm
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions
d(dz : arccot(z) ) = -1/(1 + z^2)
http://www.math.brown.edu/UTRA/trigderivs.html
d(dz : arccot(x) ) = -1/(1 + z^2)
http://www.themathpage.com/aCalc/inverse-trig.htm
same thing - but great explanations!
https://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions
d(dz : cosecant(z) ) = -cosec(x)*cotan(x)
http://www.math.brown.edu/UTRA/trigderivs.html
d(dz : cosecant(x) ) = - cosecx*cotanx
http://www.themathpage.com/aCalc/sine.htm
d(dz : cosecant(x) ) = - cosec(x)*cotan(x)
http://www.themathpage.com/aCalc/inverse-trig.htm
d(dx : arctan(x)) = (1 + x^2)^(-1)
d(dx : arcsin(x)) = (1 - x^2)^(-1/2)
d(dx : arccos(x)) = (-1)*(1 - x^2)^(-1/2)
Howell's derivations :
∫(dQ, 0 to Qf : cosQ*sin^5Q )
try
d(dQ : sin^6Q) = 6*sin^5Q*cosQ
therefore
∫(dQ, 0 to Qf : cosQ*sin^5Q ) = sin^6Q/6
∫(dQ, 0 to Qf : (cosQ - 1)*sinQ )
= + ∫(dQ, 0 to Qf : cosQ*sinQ ) + ∫(dQ, 0 to Qf : -sinQ )
= + sin^2Q/2 + cosQ
∫(dQ, 0 to Qf : (cosQ - 1)*sin^2Q)
= ??
∫(dQ, 0 to Qf : (cosQ - 1)*sin^3Q )
= ∫(dQ, 0 to Qf : cosQ*sin^3Q ) + ∫(dQ, 0 to Qf : -sin^3Q )
= sin^4Q/4 from above - (-3/4*cosQ + cos(3*Q)/12_ above
= sin^4Q/4 + 3/4*cosQ - cos(3*Q)/12
∫(dQ, 0 to Qf : (cosQ - 1)*sin^5Q )
= + ∫(dQ, 0 to Qf : cosQ*sin^5Q ) see below
+ ∫(dQ, 0 to Qf : - sin^5Q ) see above
= + sin^6Q - sin^4Q*cosQ/5 - 3/5*cosQ + cos(3*Q)/15
∫(dQ, 0 to Qf : sinQ*cos(3*Q) ) = cos((3 - 1)*Q)/2/(3 - 1) - cos((3 + 1)*Q)/2/(3 + 1)
∫(dQ, 0 to Qf : sinQ*cos(3*Q) ) = cos(2*Q)/2/2 - cos(4*Q)/2/4
∫(dQ, 0 to Qf : sinQ*cos(3*Q) ) = cos(2*Q)/4 - cos(4*Q)/8
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