If Netflix 'Roma' Wins Best Picture It Will Be Curtains For The Cinema - And Non-Cord Cutters

Among the movies competing for 'Best Picture' at this year's Academy Awards tonight is an aberration entitled "Roma".  It is an aberration because:  a) It was produced by Netflix, which is not a movie company - indeed, it is a streaming service, and b) This service is subsuming the rest of the entertainment industry with its $12 b a year spending, not to mention overloading the landscape with so much fare it's creating "option paralysis"

In the latter case, we find (WSJ., 'Entertainment Overload', Exchange, p. B1, Feb. 22-23):: "Audiences have been overwhelmed by too many choices"  leading to what Einstein once called "option paralysis". (One reason Einstein opted to wear the same sport coat and trousers every day). The effect of this choice overload inevitably leads to fewer choices being made over time, given the incipient stress accompanying each decision.   

Of course, this isn't new and was originally put forward in the book, 'The Paradox of Choice', by Brian Schwartz, an emeritus professor of psychology at Swarthmore College. The basic assumption is that humans are rational agents and, when offered a set of choices - say for a product- will end up making the optimum one. But Schwartz showed this isn't the case and when humans are offered a surfeit of choice they punt! They either don't make any purchase, or they opt for the first one that comes to mind or in their field of vision.

In the entertainment sphere, including television and  film, that means fewer choices made, fewer movies viewed-  and it could also translate to the streaming screen as well.  In the WSJ piece, Schwartz observes:  "If studios end up spending millions of dollars producing too much new programming they could end up in a situation where they give people everything under the sun and they end up just watching what they watched before."

"Too many choices, America! Not good!"   to quote the late George Carlin in one of his last riffs at the  Beacon Theater in NYC.  Expanding on his premise, Dr Schwartz also noted that an influx of options makes the ultimate choice more an expression of personal identity. In other words, the excess of  choices even in the entertainment sphere, continues to balkanization into different tribes of political (and cultural) identity.

Thus, when there are only five movies to choose from at a multiplex people tend not to put too much stock into their decision.  But increase the choice load to 30 movies it becomes a question of: "Am I the kind of person who would watch this kind of movie?"

Given Netflix  produced "dozens of original films in 2018" and plans even more this year, the streaming service is a primary contributor to overchoice and option paralysis. Worse, it is responsible for many of the younger demographic going on this cord cutting binge which -frankly  - appalls those of us (like wifey and myself) who love to watch films on the big screen, either in the cinema (like 'Alita- Battle Angel' which we saw three days ago in IMAX 3D), or on our 55" TV (where we watch films we missed on the cinema screen).   

Ordinarily we wouldn't give a damn about the cord cutting, streaming cosmos, but the lemming- like abandonment is now forcing higher subscription costs on those of us who remain, as well as ever creeping attrition of viewing choices.  Recent WSJ forecasts in the 'Business &  Investing' pages are also not sanguine about the future of cable TV or Direct TV, where diminishing returns seems to be the byword.

The cord cutting phenomenon is now generating a momentum for streaming on tiny screens which is pathological as well as dumbfounding and aberrant.  This is given so much more vivid entertainment experience is accessible on large screens - especially in theaters.   This is also why - according to the  aforementioned WSJ piece:

"Some Academy members consider 'Roma' an existential threat to a movie business already struggling to get people into theaters for anything but big budget franchise pictures."   

So bottom line these Academy members are Never -Netflixers,  like wifey and myself.  Naturally then, it is logical to think if a streaming flick like Roma wins 'Best Picture'  (god forbid)  it will further accelerate cord cutting, more Netflix viewers, more streaming, which will vastly fuel even more Netflix fare - and more choice overload and option paralysis.   With ever more viewers prone to more option paralysis expect really good, standout screen choices to collapse in lieu of cheaper, easier to produce schlock.

Both cinema choices, quality, not to mention large screen, original TV movies (such as produced by HBO, Showtime) stand to suffer, as a race to the bottom ensues.

But most importantly, it would likely sever the connection between motion pictures and cinema (theater going) forever.  Henceforth, the great majority of people opting to never go out to the cinema but just remain in their personal cocoons streaming flicks on tiny screens. An "atrocity", as Janice so aptly puts it.   I wouldn't use that strong a term, but it's at least aberrational. 

Yes, Roma deserves to win an award:

 'The Best Motion Picture On A Streaming Service'. 

But emphatically not an Oscar, sorry!

------
 My own top picks:

Best Picture:  'Green Book'

Best Director: Spike Lee, 'Blackkklansman'

Best Actor: Viggo Mortensen ('Green Book')

Best Actor in Supporting Role: Mahershala Ali ('Green Book')

Best Actress:  Glenn Close ('The Wife')

Best Actress in  Supporting Role:  Rachel Weisz ('The Favorite')
----------------------

Update:  "Green Book" wins - and a petulant pseudo-critic from the LA Times whines.  That would be Justin Chang. I haven't the time (or inclination) to douse all his hucksterish codswallop but suffice it to say I disagree with him on nearly every argument he makes, especially that this film was the "worst Best Picture award in ten years"  (Since 'Crash')   Those who want can read his balderdash here:

https://www.latimes.com/entertainment/movies/la-et-mn-oscars-green-book-worst-best-picture-winner-20190224-story.html

Perhaps the best (and most succinct) takedown of this smug turkey came from a commenter on the LA Times site,  Wendy:

"Despite all of the dressing down of this movie in this article, I was thoroughly entertained by this movie, and believe it deserved all the accolades. Mr Chang sounds bitter like he lost a bet or something!"

Selected Questions- Answers From All Experts Astronomy Forum (Celestial Mechanics)

Question: I am interested in how spacecraft get to other planets and how astronomers are able to predict future planet positions to do this. So basically can you provide some background details on what makes up celestial mechanics?  (P.S. I am a math major so don't hold back on math details! :) ) - Barbara, Santa Monica CA

Answer:

The name "celestial mechanics"  conveys exactly what the subject embodies: Mechanics applied to the dynamics of celestial objects. A typical diagram from my own Celestial Mechanics notes when I took it at Univ. of South Florida, is shown below:














The diagram shows assorted "orbital elements" for a planet of mass m2 (the Sun is m1) and this can be used as a basis for orbital energy analysis and also to predict future positions. Energy constants in celestial mechanics are very useful for quickly coming to terms with specific properties of an orbit such as shown in the accompanying sketch- designating a generic orbit in x-y-z space. In the diagram, w   is the argument of the perihelion, W is the longitude of the ascending node, f is the true anomaly and i is the inclination of the orbit. The critical or key parameter here is h, the angular momentum vector for the orbiting system. It may be useful here to refer to the diagram (b) in the figure used for angular momentum vector (z) in an atomic system:

Getting specific, assuming r and r' are r (radius vector) and d r/dt, respectively, the magnitude h, of the angular momentum vector is:

h = r x r’ =

(y z’ - z y’)

(z x’ - x z’) = (c1 c2 c3)

(x y’ - y z’)

so (r x r’) = (c1/ h, c2/ h, c3/h)

and inserting variables one finds:

c1/ h = sin W sin (i)

c2/ h = - cos W sin (i)

c3/h = cos(i)

Now since the inclination of Earth's orbit to the ecliptic  (i) is known (23.5 deg) and therefore cos(i) can be determined, then sin(i) can be as well. Also h can be determined, since: h = c3 / cos(i) = (GMm a (1 – e ²) ^{1 /2} where all the constants are known (a = semi-major axis of orbit, e = eccentricity of orbit)

(The energy equation is: ½V ² - m  /r = c, and the c's - energy integration constants- are  found as shown below)  Also, h can be determined, since: h = c3 / cos(i) .  (Also h =  [c1 ² +  c2 ²   +  c3 ²]  ^½)   We also know  W =  11.26 deg.

To fix ideas, W  is one of the key orbital elements (difference between mean anomaly  M and argument of the perihelion,  w )  where M can be obtained from a table based on observations, and w can be obtained using a Fourier expansion of the mean anomaly, M, e.g.:

 w  = M + (2e – e³ / 4) sin M + 5 e²/4 sin 2M + ... etc.

Once W is known, and c1 and c2 and c3 are known, the astronomer is ready to compute the position of a planet, say Jupiter, forty or so years in the future.  The actual calculations are formidable but can be found in a good celestial mechanics text, or using the orbital element formulae can be worked out using a computer.

The energy integration constants  (c1, c2, c3) factor into a host of different problems but let's consider just one relatively simple application, say to compute a planet's velocity at the perihelion and aphelion points of orbit.  The diagram below can be used:


Here,   the points A and P denote the aphelion (farthest point) and perihelion (closest point) to the Sun (S), respectively. We let   V_A, V_P  be the respective velocities at those orbital extrema. As may be deduced here, points A and P are the only ones in the whole orbit for which the velocities are truly tangential or at right angles to the radius vectors for those positions. Consequently, we can write: 

V = (2π/T) r

where r is the radius vector at the point, and T is the period. 

 If Kepler's 2nd (equal areas)  law holds at every point (i.e. equal areas swept out in equal intervals of time) we also have:

r² (2π/T) = h 

where 'h' is a constant ('specific relative angular momentum') which is twice the rate of area description (i.e. by the radius vector). Thus, if the radius vector is r1, then h = 2A1, when A1 = π(r1)². Hence, at aphelion and perihelion only we have: 

V = h/r    for which r = a(1   +  e)  OR:   r =  a (1  -  e)

For the perihelion velocity we have:

V_P = h/ a(1 - e)

where a is the semi-major axis, and e is the eccentricity.

For the velocity at aphelion: 

V_A = h/ a (1 + e)

Then the ratio of velocities is:

(V_P/V_A) = (1 + e)/ (1 - e)

The correct energy equation can be written:

½V² - m  / r =   a 

where a  is an energy integration constant.   More conventionally it is written  without  the energy constant:


V ² = m (2/r - 1/a)


which is also known as the "vis viva" equation, one of the most important in celestial mechanics. 

Since for any bound system of masses m1 and m2, m = G (m1 + m2), where G is the Newtonian gravitational constant (G = 6.7 x 10^-11 Nm²/kg²) then if we know V_P and V_A, along with a and e, we can compute  a, viz. 

a      = ½V_P² - m /a(1 - e) 

at perihelion, and 

a = ½V_A² - m /a(1 +  e) 

at aphelion. 

Consider for the Earth-Sun system :

m  = 1.33 x 10²⁰ Nm²/kg

We already know G and m1= 1.99 x 10³⁰ kg (Sun's mass) and m2 = 6.4 x 10²⁴ kg, (Earth's mass)

Also: a (semi-major axis)  = 1.496 x 10¹¹ m

Then h = + [m a(1 - e²)]^½ =    4.46 x 10¹⁵ N-m/kg = 

4.46 x 10¹⁵ J/kg 

The energy constant a  =   - m/ 2a   for an elliptical orbit

So:   

a  =    - (1.33 x 10²⁰ Nm²/kg) / 2 (1.496 x 10¹¹ m)  =   

-4.45 x 10⁸ m²/s² 

The eccentricity of the orbit e, can now be obtained from:

e =   [1   +   (2 h ²  a )/ m ² ] ^½ =

[1   +    2(4.46 x 10¹⁵ J/kg ) ² (-4.45 x 10⁸ m²/s²)/ (1.33 x 10²⁰ Nm²/kg)² ] ^½ =

0.016

(Which is confirmed on consulting a planetary data table)

What about the velocities at perihelion and aphelion?

Since we have obtained h and e,    the velocity at perihelion is easy to calculate, e.g. 

V_P = h/ a(1- e) =

4.46 x 10¹⁵ J/kg / [1.496 x 10¹¹ m(1 - 0.016)]

V_P = 3.03 x 10⁴ m/s  = 30, 300 m/s

and the velocity at aphelion can be obtained using:

V_A = h/ a(1 + e) =

4.46 x 10¹⁵ J/kg / [1.496 x 10¹¹ m(1 + 0.016)]

V_A = 2.93 x 10⁴ m/s    = 29, 300 m/s

Let's take another illustration from the Pluto-Charon system.   The orbit of Charon with respect to Pluto, has an eccentricity e = 0.0020. The semi-major axis of the orbit is 19, 450 km. The mass of Pluto = 1.27 x 10²² kg and the mass ratio (Charon to Pluto) is found to be m(c)/m(P) = 0.12.  From this  data, find: 

a) The mass of Charon

b) The ratio of the velocity of Charon at perihelion to aphelion

c) The period of Charon, and its velocity 

a) The mass is found using the mass ratio. Since m(c)/m(P) = 0.12 and m(P) =1.27 x 10²² kg , then:

m(C) = 0.12 (1.27 x 10²² kg ) = 1.52 x 10²¹ kg

b) This ratio is obtained from:

(V_P/V_A)  = (1 + e)/ (1 - e)

and we know, e = 0.0020, so:

(V_P/V_A)  = (1 + 0.0020)/ (1 - 0.0020) =

 (1.0020)/ (0.998) = 1.004

Therefore:  (V_P/V_A)  = 1.004

c) The period is obtained from: T = 2π (a³/m)^½

where a = 19, 450 km = 1.94 x 10⁷ m

For preserving unit consistency this is one time we decline to use units of AU and years, and instead use the distance in meters to conform with the required units for G. Then: T =

2π [(1.94 x 10⁷ m)³/[6.7 x 10^-11 Nm²/kg²) (1.422 x 10²² kg)]^½

T = 5.5 x 10⁵ s = 6.37 days


The velocity is:

V = 2πa/T = 2π( 1.94 x 10⁷ m )/(5.5 x 10⁵ s)=

 221.6 ms^-1


This is all meant to show the power of celestial mechanics at even a basic level.   Of course, much much more can be done such as determining the geocentric coordinates of Jupiter on July 20, 2048.  But that exercise is far beyond the scope of this response and requires knowing not only all the relevant orbital parameters of each planet (Earth, Jupiter) but also their mean motions and then using the equations needed such as the expansion of mean anomaly, e.g.

w  = M + (2e – e³ / 4) sin M + 5 e²/4 sin 2M + ...

Precision targeting of a planet for future spacecraft arrival, also necessitates computation of likely perturbations such as shown below:

Perturbations are important because they show how the positions of some given planet are influenced by the gravitational attraction of another. Unless one then reckons the differential location into computations, he will find a spacecraft destined for some destination planet will likely miss it entirely.

In the illustration shown  above we seek to quantitatively estimate the effect of Jupiter on the orbital position of Earth. The respective masses are: m1(Sun), m2(Earth) and m3 (Jupiter) and we assign relative radius vectors that approximate to the actual distance in AU or astronomical units where 1 AU = 149 million kilometers. Thus, r = 1.0 and r3 (for Jupiter) = 5.  Also, D defines the separation between Earth and Jupiter and is the key parameter for estimating the magnitude of the perturbation.

The angle S separating the r and r3 vectors can be anything but for working purposes we use S = 120 degrees, which yields a value: cos(S) = cos (120) =      - ½. This will be useful for computing the first three Legendre polynomials which factor into the kind of generic perturbations I'm looking at in this example.

The first three Legendre polynomials for this context are:

P 0  = 1

P 1 = cos (S)

P 2 = ½ (3 cos ² (S) – 1)

Note that by comparing  P 1  above to  P 1 (x) earlier we can easily discern:

x <-> cos (S)

In other words, x in the Rodrigues’ formula,  e.g.

d/dx [(1 – x ² ) d/dx P n  (x)] + n(n + 1)P n (x)

is now replaced by cos (S) for the purposes of this perturbation estimate.

As we are perturbing an inner planet by an outer one, we are expanding (1/ D) using the Legendre polynomials. Computing the Legendre polynomials for an Angle S = 120 deg, one finds: P 0 = 1,  P 1 = - ½ and  P 2 = -1/8

The perturbing function is then easily computed, viz.:
  
1/D  = 1/ {1 + (1/5) 2    – 2(1/5) (-½)} ^1/2  

And   1/D = 0.18

 is now replaced by cos (S) for the purposes of this perturbation estimate.  

Again, this is all intended as a general illustration of how celestial mechanics works in the situation of sending a spacecraft to a plant.

Solutions To Analytic Geometry Revisited (1)

1)Find the coordinates of the center of each for the following circles and the radius r. Sketch each of the circles

a)  x ²    +    y ²  - 2 y   =  3

b) 2x ²    +  2  y ²   + x  + y = 0

c) x ²    +    y²  + 2x  = 8

Solutions:

a)  x  ²   +    y ²  - 2 y   =  3

We have: A = 1, D = 0,  E = -2  and F = -3

Then the center is at: 

(h, k) =    (-D/ 2A),   (-E/ 2A)  =   (0,  1)

r ²     =   D ²  +    E  ²      -  (4AF) /  4 A ²    

So: 

r ²     = (0 ²   +    (-2) ²      - 4(-3_) /  4 (1) ²    

r ²     =   16/ 4   = 4  so radius  r  = 2

Sketch of the resulting circle:

b) 2x ²    +  2  y ²   + x  + y = 0

Here: A = 2,   C= 2,  D = 1,  E = 1,  F = 0

Center is at:

(h, k) =   (-D/ 2A),   (-E/ 2A)  =   (- 1/4,  -1/4)

Radius:

r ²       =   D ²    +    E ²       - 4  (AF) /  4 A ²    

So: 

r ²     = (1 ) ²    +    (1) ²      - 4(0) /  4 (2) ²    

r ²     =   2    -  0   =   2  

So:  r =   Ö 2   (Sketch below)

c) x ²    +    y ²  + 2x  = 8

We have: A = 1,  D= 2, E = 0, F = -8

Center is at:

(h, k) =   (-D/ 2A),   (-E/ 2A)  =   (- 2/ 2,  0 ) = (-1, 0)

Radius:

r ²      = (2 ²     - 4(-8)) /  4 A ²    

So: 

r ²     =    36/ 4   =  9

So; r =   Ö  9   =    3

The circle is as shown below

2) For ellipse:    We have:   A = 9,  B= 0, C = 4,  D= 36,  E =  -8 and F = 4

Then we get:  9 x ²  +   4  y ²   + 36 x   - 8 y + 4 = 0

With appropriate algebraic manipulation we get:

(x + 2) ² / 4  +   (y - 1 ) ²   / 9 = 1

Graphing yields:

 

3) Determine the equation to which:

x²  +   xy  +  y²    =  1

reduces when the axes are rotated to eliminate the cross product term. Sketch the resulting curve. 

Solution:

The given equation has: A = B = C = 1

Choose Θ  according to: cot 2Θ  =   (A  - C)/ B  =   (1 - 1) / 1 = 0/1 = 0

Then:  2Θ  =  90 deg     so: Θ  =  45 deg    

Then:  x  =   (x'   -   y')   / Ö 2       

y = (x'   +   y' )  /  Ö 2  
 
And, after some algebra:


3(x') ²  +  ( y') ²    =  2

Divide through by 2:

3(x') ² / 2   +  ( y') ²  / 2  =  1

This curve defines an ellipse which is shown below: