Broadcast Industry Sold Out 1st Amendment Rights For Lucrative Deals? Evidently - According to Econ Prof Writing In WSJ

                       Kimmel tears into Dotard after his return to ABC last September.

It was Robert McChesney in his 'The Problem of the Media' who first pointed out that when media are taken over by corporate interests diverse viewpoints are the first to go.  In retrospect he might have added that the broadcast industry's aversion to diverse viewpoints - to the extent of knocking out competition - caused them to sign their own 'death warrants' vis-a-vis free speech. His original warning about corporate media networking remains trenchant today, i.e.

"Given this networking system it is safe to say that the media in the United States effectively represents the interests of corporate America and that the media elite are the watchdogs of what constitutes acceptable ideological messages, the parameters of news and information content, the general use of media resources."\

But now may need to be extended. According to a recent Wall Street Journal op-ed piece:

ABC vs. Trump and the Broadcasters’ Free-Speech Sellout - WSJ

It appears at the very inception of U.S. radio broadcasting, the industry sold out. According to the Journal:

"The First Amendment lawsuit ABC filed against the Federal Communications Commission looks straightforward. The network claims the commission is tying up speakers hostile to the president in red tape. With the licenses of eight ABC-owned TV stations in jeopardy—threatened by extensive DEI evaluations, years-early renewal reviews, and presidential condemnations of “fake news” and “media offenders”—the battle lines are drawn.

President Trump complains that the host of late night’s “Jimmy Kimmel Live” and the ladies of daytime’s “The View” are biased against him, in the pockets of Democrats, and hence should cost ABC the “privilege” of holding licenses the FCC issues according to the “public interest.”

Seems crystal clear, right?  After all, Jimmy Kimmel and ABC prevailed the last time the FCC top honcho Brendan Carr came down on them (last year) after a Kimmel segment following the Charlie Kirk killing.

But not so fast. As Clemson University professor of economics Thomas Hazlett goes on to point out:

"But there's a loophole.  While most media operate in a laissez-faire regime, terrestrial radio and television are licensed and their airwave access has at times been deemed a privilege not a right. Harvard Law's Laurence Tribe traces the anomaly to a technical error: 'a profound fallacy about spectrum scarcity' or as comedian George Carlin once put it: 'Radio and television are the only two parts of American life not protected by the free speech provisions of the First Amendment to the Constitution.'".

How can this be? Well, Robert McChesney would probably call it the greed of the corporate interests in the broadcast industry at the time. So greedy were they for lucrative free licenses and reduced competition they traded they traded away their free-speech rights. As Prof. Hazlett goes on to note:

"The broadcasters traded freedom of speech for free licensing and a lucrative, protective market.  As a result, upstart rivals (in the 40s, 50s) were suppressed - then cable and satellite TV in the 1960s were also suppressed. This while only a handful of viable stations were assigned more than 80 TV channels. In return, regulators gained clout over an industry and influence over content."

I warrant not many Americans, especially who cheered on Kimmel's campaign last year, are aware of this. McChesney surely is as he made reference in his book to the 1927 Radio Act.  As well as the fact that the public interest language in it was written by the National Association of Broadcasters.

As for Prof. Hazlett, he goes on to ask: How violent a violation of the First Amendment is America willing to countenance. Well, I suspect not much. 

Recall that after Disney (owner of ABC) pulled his show off the air last September.  This followed remarks he made about "the MAGA gang trying to score political points" off of the fatal shooting of far-right activist Charlie Kirk, a public backlash ensued.  To wit, millions of Kimmel-ABC viewers began a boycott of Disney programs, especially the costly streaming shows. The pain to profits was too much and Disney / ABC removed the suspension.

There' s good reason to believe ABC will succeed in its new battle with Trump and the FCC. As the prof writes: "Leave it to Donald Trump to up the ante radically and tweet out his worst."

Basically, highlighting an even worse free speech excess. So no surprise at the end Hazelett writes:

"Best of luck, ABC. May free speech soon, finally, be yours."

And I am confident it will, since most sane Americans have already extended that right to broadcasting, especially since last year's Jimmy Kimmel brouhaha. The critical takeaway from the Kimmel episode is that public push back is effective and is the key to preserving our free speech rights under the first amendment. Even if TV broadcasting wasn’t originally under that umbrella.

See Also:

Citizens' Backlash Beats Back The Right's "Cancel Culture" & Anti-Free Speech Thuggery - As Jimmy Kimmel Returns

And:

Disney Has Lost BILLIONS After Canning Jimmy Kimmel

And:

John Oliver DESTROYS Trump & ABC Over Kimmel Firing

And:

Television

‘Streisand Effect on Steroids’: Official Ratings for Kimmel’s Return Described As ‘Insane’

by Ailia Zehra | September 25, 2025 - 5:41am | permalink

— from Alternet

Late-night host Jimmy Kimmel’s return to "Jimmy Kimmel Live!" drew 6.2 million viewers on Tuesday night, nearly four times the show’s usual audience, according to preliminary Nielsen figures.

The episode was Kimmel’s first since ABC and its parent company, Disney, suspended the program last week following backlash over remarks he made about the man accused of killing conservative activist Charlie Kirk. The suspension sparked a national debate over free speech.

The New York Times noted in a report that more than 20 percent of ABC affiliates did not air the episode. Nexstar and Sinclair, two major station groups, said they would continue to pre-empt the show, though Nexstar said Wednesday it was in talks with Disney.

Online interest was also strong: his monologue drew more than 17 million views on YouTube within a day.

» article continues...

Reviewing Algebra II- Part Three: Solving & Simplifying Radical Equations & Expressions

 In Part One the focus was on relatively benign algebraic fractions and in Part Two the equally benign exponential equations (solved without use of logarithms). Now we will move on to the more challenging solution of radical equations, which may also involve surds. So a starting point is distinguishing radicals from surds.

Example: Solve the radical equation:

 Ö(5x - 11)   - Ö(x - 3)   = 4  

The equation contains a root (in Ö5  as a coefficient of x) which is both a radical and surd.

Solving the equation is straightforward.  We begin by isolating the radical on the left side, making it the subject, viz.

Ö(5x - 11)   =  Ö(x - 3)   + 4  

Next equate the squares of the members:

5x - 11   =  x - 3   + 8Ö (x - 3)   + 16

Isolating 8Ö (x - 3), making it the subject:

 8Ö (x - 3)   = 5x - 11  -  x + 3  -16

Simplify the right side and combine like terms:

 8Ö (x - 3)   =  4x -  24

Divide each side by 4: 

  2 Ö (x - 3) = x - 6 

 Squaring each side: 

4 (x - 3) = x ²  - 12x  + 36 

 Apply distributive law:

4x -  12 = x ²  - 12x  + 36 

Add 4x - 12 to each side:

x ²  - 16x  + 48 =  0

Factoring the left hand side:

(x - 12)  (x - 4)  =  0  

Setting x - 12 = 0 and solving:

x = 12

Doing the same for x -4:

x = 4

Substitute values back into original equation, i.e.: 

Ö(5x - 11)   =  Ö(x - 3)   + 4  

We obtain:

Ö(60 - 11)  -  Ö(12 - 3)   =  Ö49   -  Ö9  = 7 - 3  = 4 

Therefore 12 is one root.

But, checking x = 2: 

Ö(20 - 11)  -  Ö(4 - 3)   =  Ö9   -  Ö1 = 3 - 1  = 2

2  ≠ 4, so x= 12 is the only root

The method is longer than one might expect because squaring a radical-surd equation doesn't always result in an equivalent equation.

Practice Problems: 

Solve each for x and verify the root is correct.

1) Ö(2x - 3)   = 5

2)  Ö(x + 3)   + Öx   = 5

3)   3   - Ö(x + 1) =  0

4)   Ö(x + 5)   =   1 +  Öx

5)  Ö(x + 1)   + Ö(2x + 3) - Ö(8x + 1)    = 0

All Experts Redux: Heat, Temperature, The Origin Of The Sun's Energy - And Why Space Is Cold

                              Interior structure of the Sun

 Question -

We know the higher up a mountain you go the colder it is. We also know space is cold, so it must be the microwaves from the Sun that reach Earth, excite the atmosphere and hence make us warm.  Please help to clarify, sir.


Answer -

 With a surface temperature of 6,000 C (11,000 F) and a core temperature of 15 million Celsius, it is nuclear fusion in the Sun's core which is the generator of the radiant energy responsible for warming the Earth - via the infrared radiation (not microwaves.)

Indeed the extremely high core temperatur is needed in order to achieve the fusion of hydrogen nuclei into helium.  The primary fusion source is the proton -proton cycle i.e.

¹H + ¹H + e^- ®  ² H   + n + 1.44 MeV

² D   + ¹H ®  ³ He + g + 5.49 MeV

³ He + ³ He ®  ⁴ He + ¹H + ¹H  + 12.85 MeV

The top line shows two protons fusing to yield deuterium (heavy hydrogen) with a positron and neutrino (n) emitted, along with 1.44 MeV of energy. Empirical evidence of this reaction is obtained from gallium detectors, of the neutrinos given off, which are within 1-2% of what theoretical models predict.[1] In the second fusion reaction, the deuterium combines with a proton to give the isotope helium 3, along with a gamma ray (g) and 5.49 MeV energy. In the final fusion, two helium-3 nuclei combine to yield one helium-4 nucleus, along with two protons, and 12. 85 MeV energy. Note that the two ending product protons commence the cycle anew, so that the generation of nuclear energy is ongoing.  

Regarding colder temperatures with altitude experienced here on Earth, this has everything to do with the density of oxygen, nitrogen (e.g. air) molecules decreasing.  Hence, the higher one goes up a mountain the fewer molecules (matter) there are to absorb heat- or retain it.

What we call 'heat'  (internal energy) is really energy associated with the motion
of molecules in a medium.  The graphic shown below illustrates the relation of
heat energy to kinetic or motion energy in a gas contained within a cylinder. The
cylinder has a solid bottom at one end and a moveable piston at the other.

In this case the heat imparted by the Bunsen burner causes the molecules in the gas to move more rapidly - colliding with each other and causing the piston to move upward -expanding the gas volume. This changes the volume of the gas from V1 to V2 where V2 > V1.

Space is "cold" because there are so very few molecules of matter in it, being nearly a perfect vacuum.  The presence of so few molecules means very few collisions and little available internal energy.  Hence,  even if you try to heat a volume of space there will be little detectable difference.

Regarding the Sun, it radiates at all wavelengths of the electromagnetic spectrum - not just microwaves.

If you study the diagram of the EM spectrum below you will see that microwaves form only a relatively small region:


These are associated with longer wavelength EM radiation, as are radio waves. The shorter wavelengths (to the left) include what we call the "visible" band - as well as x-rays, UV (ultraviolet) and gamma rays.

The Sun actually shines with its maximum radiation emitted at about 5500 Angstroms (550 nm)   - or in yellow light. It radiates at all wavelengths but has the maximum at that band. This is depicted below:

The fact is the Sun's radiant energy is being transferred to the Earth through space by the process known as radiation - which can occur even in a vacuum.  But the fact the Sun's radiation (including as
heat, i.e. infrared radiation) can be transferred through space doesn't mean space itself will be hot - as  I've explained.

Space is 'cold' not because it absolutely lacks heat  but because its
density (of particles) is too low to have much quantity of heat, or
'thermal capacity'.

The thermal capacity, W, is defined by the amount:

W = mc

where m is the mass of the medium and c, the heat capacity. It is a measure of how difficult it is to increase the temperature of a medium by one degree (e.g. Kelvin).  Obviously, since space is a near-vacuum, m is near 0, and c is near 0, so little or no thermal capacity exists. What this means is that energy from the Sun can be transferred through space, without appreciably heating space.

Back to Earth:  If air is thin (low density), as it is at high altitude, i.e. the top of a mountain, there isn't enough mass present to enable or facilitate energy transfer to the medium. Hence,  it feels 'cold' to the human observer. But this has nothing to do with the Sun being 'cold' because it exists in space. (As I noted earlier.) Conversely, the more one descends in altitude, the greater the number of particles encountered, and the greater the atmosphere's retention of heat. (thermal capacity)

The fundamental point here is that no heat energy transfer occurs from a cooler body to a hotter- ONLY from a hotter to cooler. Hence, it follows the Earth must be the cooler body compared to the Sun

One further point:  One can have an enormous temperature - say for the Sun's corona- which is still not enough to burn anybody! In the case of the solar corona, we estimate a kinetic temperature of 2 million degrees K, yet if one could insert a finger into it, there'd be NO burn. Why is this?

Because the corona is essentially a vacuum containing very few particles (low thermal capacity). However, those few particles have very high velocities, so possess extremely high 'kinetic temperature'  - which is most of what the '2  million Kelvin' magnitude is about.

Revisiting Diagonalization of Tensors & Obtaining Principal Axes

 Basic tensor algebra is a crucial first step in getting to use tensor calculus and also a better grasp of differential geometry. (Stay tuned for Part 6 in the introductory series.) Let's begin by noting that just as in linear algebra, e.g.

• Math Revisited: Looking At Linear Algebra

 matrices factor into this.  I.e. every symmetric tensor will have the form (using dummy indices i, j):   

a _{i j} =

(a₁₁ ……a₁₂………a ₁₃)

(a₁₂….. a₂₂……….a₂₃)

( a₁₃……a₂₃………a₃₃)

The anti-symmetric or skew symmetric tensor will have the form (note how the order of the dummy indices changes):     a _{j I} =

(0 ……a₁₂………-a ₃₁)

(-a₁₂…..0……….a₂₃)

( a₃₁……- a₂₃……0)

Or effectively only three distinct components.

The Kronecker delta, which we already saw in Introduction to Differential Geometry, Part 1, is:

d _ik = 

(1….0…..0)

(0….1…..0)

(0….0… ..1)

Example 1:  Compute:  a_{i j}  d i _j   

If a_{i j}  is a second order tensor with matrix:

Then we have:

And we let   x  _i  and   y _j   be first order (rank 1) tensors (i.e. vectors) respectively given by:

x _i    =  (2, 1, 4)

y _j    =  (3, 7, -1)

a)Find:   a _ij  x _i     +  a _ji 

b)Find:  ( a_{i j}    -  2/3  d  _ij   )

Solutions:

a) From Mathcad matrix computations:

b)

Diagonalizing tensors is analogous to obtaining the eigenvalues for a matrix in linear algebra.  Hence, we need to extract the eigenvalue equations via diagonalization and obtain the distinct eigenvalues.   

Consider the object (solid tetrad) below for which we want the principal axes:

Given:

                                                     a _{i j} = 

With T' =  A × I × A^t

Where t denotes the transpose.  Then we obtain, T' =

To be diagonalized.  Writing this out:

(15 - l…..0……..0)

(0…….11- l ....-3Ö2)

(0……..-3Ö2…8- l )

This leads to a cubic equation with triple roots which are:

l1 = 15,  l2 =  5, and l3 =  14

Substituting l1 in the matrix we get:

Þ 

0c _x    +  0 + 0 = 0 

- 4c _y  -  Ö3  c _z / 2  = 0 

Ö3  c _y / 2    - 7 c _z / 2  = 0 

Þ 

c _y =  c _z  = 0,  c _y =  Ö3 c _z   / 4 ,  c _y =  -7c _z  / 3Ö2 

 

  = c₁  e1^ + 0 + 0   Þ   e1'^ = e1''^

Now we can  find the remaining two eigenvalues, l2 and l3, to diagonalize the matrix and obtain solutions in c _x, c _y and c _z.

Thereby obtain the principal axes in terms of: 

e^’_x,      e ^’_y and e^’ _z .

In particular that:

  e^' _z    = - 1/ Ö3   e1^ -  1/ Ö3   e2^ + 1/ Ö3   e3^

Similarly:

Using  the eigenvalue l2  = 5 we arrive at:

Þ   10c _x  + 0 + 0 =  0

       

        0 + 6  c _y   - Ö3 c _z / 2 =  0

         

        0 - 3 c _y / 2   + 3 c _z    = 0

For which:

c _y = c _z / Ö 2    and:

  

 = 0 e₁^  -  c _z  e₂^ / Ö 2 + c _z  e₃^''

Whence:

(c _z  ² / 2  +  c _z  ²)  = 1    Þ   c _z   =  Ö (2/ 3)

Further: 

e₂^'  =

[ 1/Ö 2 (Ö (2/ 3)] e_2''  +  Ö (2 e_3''/ 3)   =  e_2''  /Ö 3  + Ö (2/ 3) e_3'' 

Lastly, for  the eigenvalue  l3  =  14:

We arrive at:

Yielding:   

c _x =  0

- 3 c _y  - Ö 3c _z  /2    = 0

 0 -  Ö 3 c _y /2  + 3c _z     = 0

Then: c _y = -  Ö 2 c _z   and   c _z    =  1/ Ö 3

Þ   c   =

 -  Ö 2  c _z   e_2''   +  c _z   e_3''   =  -Ö (2/ 3) e₂^''  + Ö (1/ 3) e_3''  

On converting to the unprimed coordinate system:

  e^ _j =  å³ _{i= 1}   a _{i j}   e _{j '}             And:  e _i’ =  å³ _{j= 1}   a _{i j}   e _j

Recall:

e _x''    =  - (1/ Ö 2) e^₁   - (1/ Ö 2) e^₂   

So: 

e _x^'    =  e _x''    =  ( 1/ Ö 2) e ₁   - (1/ Ö 2) e ₂    + 0

Similarly:  

e^ _y''    =   (1/ Ö 2) e^ ₁   +   (1/ Ö 2) e^ ₂    + 0

And:  e _z''    =   e^ ₃     Þ   

e _y^' =

[(1/ Ö 3) (1/ Ö 2)] e^ ₁   + [(1/ Ö 3 )(1/ Ö 2)] e^₂  +   [Ö (2/ 3)] e₃^ 

=   e^₁  /Ö 6   +   e^ ₂ / Ö 6   +   2e^ ₃ / Ö 6  

Lastly:  e _z^'    =   (- 1/ Ö 3 ) e^₁  - (1/ Ö 3 ) e^ ₂   + e^₃/ Ö 3    

(The Principal axes are: e _x^' ,    e _y^'   and  e _z^' )

Suggested Problem:

In a certain rectangular coordinate system, the directions of whose axes are given by the unit vectors i, j and k, the inertia tensor of an object is given by:

I = K x= 

(1….0…..0)

(0….1…..1)

(0….1… ..1)

a) What are the principal moments of inertia of the object (the moments of inertia along the principal axis) relative to the origin of the above coordinate system?

b) What is the direction of the principal axis corresponding to the principal moment of inertia and equal to K?

c)If the origin of the above rectangular coordinate system is at the center of mass of the object and the total mass of the object is M, what is the change in the inertial tensor of the object if the rectangular coordinate system is displaced parallel to itself a distance r_o in the direction

 

 (1/ Ö2)j +  (1/Ö2)k?