Richard Hertz wrote:

> On Tuesday, December 7, 2021 at 5:17:28 PM UTC-3, Thomas 'PointedEars'
> Lahn wrote:
>> Δt' = γ (Δt − v/c² Δx).
>>
>> Now let the events e₁ and e₂ be simultaneous in the unprimed frame, i.e.
>>
>> Δt = 0.
>>
>> Then we have
>>
>> Δt' = γ (0 − v/c² Δx) = −γ v/c² Δx.
>>
>> We know that v ≠ 0 and (therefore) γ ≥ 1. This means that if the events
>> are spatially separated, i.e. Δx ≠ 0, then Δt' ≠ 0. But that means that
>> in this case the events e₁ and e₂, which are simultaneous in the unprimed
>> frame, are NOT simultaneous in the primed frame.
>
> <snip>
>
> 1) Nice way to write equations. I bet that you are a teacher or professor,
> and a good one.

Thank you. I am only a (university) student for now; but I like to explain,
I frequently get positive responses to my explanations, and I have
considered taking up tutoring in mathematics and physics as a paid
occupation (again).
> 2) I've left only that part and your comment, because for me it represent
> the heart of the illusion of relativity.

It is not an illusion; it can be measured, and is therefore real:

<https://www.youtube.com/playlist?list=PLCfRa7MXBEspw_7ZSTVGCXpSswdpegQHX>
> Mathematically speaking, it's a correct derivation. It was Voigt, 17 years
> before Einstein's relativity, who invented such term for "local time"

No, “local time” was only introduced by Lorentz, 8 years after Voigt.

<https://en.wikipedia.org/wiki/History_of_Lorentz_transformations#Voigt_(1887)>

> Now, SIMULTANEITY is a "perceptual" concept […]

Not only.

PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)

Le 08/12/2021 à 02:12, Thomas 'PointedEars' Lahn a écrit :

> u(v, u') = (v + u')/(1 + v u'/c²).

This is longitudinal addition speed.

Here is perpendicular addition speed ---> u(v,u') =
sqrt[v²+u'²-vu'/c²]

But it's better to give the general addition and not only this two cases.

u(v,u') = sqrt[(v+cosµ.u')²+(sinµ.u')²(1-v²/c²)] / 1+cosµ.v.u'/c²

If we take µ = 0 ° and µ = 90 °, we obviously find the two formulas
already written.

In manual writing :<http://news2.nemoweb.net/jntp?sEnkWhe7T_OBHYfsgCnHjq3hMwM@jntp/Data.Media:1>

> Δx' = γ (Δx − v²/c² Δx)
> = Δx γ (1 − v²/c²)
> = Δx γ/γ²
> = Δx/γ.

No. The real equation is not that.

The real equation is Δx' = Δx.sqrt(1-v²/c²)/1+cosµ.v/c

Relativists do not take spatial anisocheony into account, and so must be.
They only take the contraction of distances by the effect of the second
degree, but not the other. However, this effect is not only apparent and
due to the speed of light (as they believe) but real.

>
> This means that the distance to the destination is shorter for the traveler
> the faster their relative speed relative to origin or destination, and

It's nice. BUT it's false. They make a catastrophic mistake of thinking.

It's not l'=l.sqrt(1-v²/c²) but l'=l.sqrt(1-v²/c²)/1+cosµ.Vo/c the
real and natural equation.

>
> [A textbook example: If you have a fast enough starship (that is not a
> warp ship, so all within the bounds of special relativity), you can
> reach the center of the Milky Way, ≈ 37'000 ly away from here, within a
> human lifetime (say 70 years) while traveling at *subluminal* speeds
> (v ≈ < c). The only problem is that for everybody not traveling with
> you it will take more than 37'000 years there and back, so when you
> come back, you have to rely on that the knowledge of your journey was
> preserved.]
>

Yes.

You are obviously absolutely right.

That is what I am saying.

But you use for that: t '= t / sqrt (1-v² / c²)

This is both true and false in the sense that it gives the average
calculation. The average.

At 0.8c, for example; we will have an 18-year-old brother, and the other
30, on a course of 24 light years.

That is quite right.

Where I no longer get along with relativists is when he thinks that this
average is a continuous average.

Take the students of a class, their average weight is 58 kgs.

I agree with that.

But that does not mean that everyone is 58 kgs.

This is what happens when a rocket leaves for space. The average will be
as you say. But if I break the outward and return trip into two parts, the
two measures will not be half of this average, but t '= t. (1 + cosµ.v /
c) / sqrt (1-v² / c²) either on the outward journey t '= 3 years and on
the return trip t' = 27 years.

t = 9 years + 9 years = 18 years in the rocket.

> your delusions of grandeur.

Oh, no, sir.

LOL, I assure you that I am the most humble of men, and that is moreover
what my entourage often confirms to me.

> PointedEars

Thank you for your very detailed answers.

It's always great to chat with someone about ideas, not people.

Even if we do not agree on several ideas and several implications of
special relativity.

R.H.

Le 08/12/2021 à 02:12, Thomas 'PointedEars' Lahn a écrit :

> u(v, u') = (v + u')/(1 + v u'/c²).

This is longitudinal addition speed.

Here is perpendicular addition speed ---> u(v,u') =
sqrt[v²+u'²-vu'/c²]

But it's better to give the general addition and not only this two cases.

u(v,u') = sqrt[(v+cosµ.u')²+(sinµ.u')²(1-v²/c²)] / 1+cosµ.v.u'/c²

If we take µ = 0 ° and µ = 90 °, we obviously find the two formulas
already written.

In manual writing :

<http://news2.nemoweb.net/jntp?1QLzbGm60wNMEbnu8RqF1WCkfs8@jntp/Data.Media:1>

> Δx' = γ (Δx − v²/c² Δx)
> = Δx γ (1 − v²/c²)
> = Δx γ/γ²
> = Δx/γ.

No. The real equation is not that.

The real equation is Δx' = Δx.sqrt(1-v²/c²)/1+cosµ.v/c

Relativists do not take spatial anisocheony into account, and so must be.
They only take the contraction of distances by the effect of the second
degree, but not the other. However, this effect is not only apparent and
due to the speed of light (as they believe) but real.

>
> This means that the distance to the destination is shorter for the traveler
> the faster their relative speed relative to origin or destination, and

It's nice. BUT it's false. They make a catastrophic mistake of thinking.

It's not l'=l.sqrt(1-v²/c²) but l'=l.sqrt(1-v²/c²)/1+cosµ.Vo/c the
real and natural equation.

>
> [A textbook example: If you have a fast enough starship (that is not a
> warp ship, so all within the bounds of special relativity), you can
> reach the center of the Milky Way, ≈ 37'000 ly away from here, within a
> human lifetime (say 70 years) while traveling at *subluminal* speeds
> (v ≈ < c). The only problem is that for everybody not traveling with
> you it will take more than 37'000 years there and back, so when you
> come back, you have to rely on that the knowledge of your journey was
> preserved.]
>

Yes.

You are obviously absolutely right.

That is what I am saying.

But you use for that: t '= t / sqrt (1-v² / c²)

This is both true and false in the sense that it gives the average
calculation. The average.

At 0.8c, for example; we will have an 18-year-old brother, and the other
30, on a course of 24 light years.

That is quite right.

Where I no longer get along with relativists is when he thinks that this
average is a continuous average.

Take the students of a class, their average weight is 58 kgs.

I agree with that.

But that does not mean that everyone is 58 kgs.

This is what happens when a rocket leaves for space. The average will be
as you say. But if I break the outward and return trip into two parts, the
two measures will not be half of this average, but t '= t. (1 + cosµ.v /
c) / sqrt (1-v² / c²) either on the outward journey t '= 3 years and on
the return trip t' = 27 years.

t = 9 years + 9 years = 18 years in the rocket.

> your delusions of grandeur.

Oh, no, sir.

LOL, I assure you that I am the most humble of men, and that is moreover
what my entourage often confirms to me.

> PointedEars

Thank you for your very detailed answers.

It's always great to chat with someone about ideas, not people.

Even if we do not agree on several ideas and several implications of
special relativity.

R.H.

--
Ce message a été posté avec Nemo : <http://news2.nemoweb.net/?DataID=1QLzbGm60wNMEbnu8RqF1WCkfs8@jntp>

Le 08/12/2021 à 02:12, Thomas 'PointedEars' Lahn a écrit :

> u(v, u') = (v + u')/(1 + v u'/c²).

This is longitudinal addition speed.

Here is perpendicular addition speed ---> u(v,u') =
sqrt[v²+u'²-vu'/c²]

But it's better to give the general addition and not only this two cases.

u(v,u') = sqrt[(v+cosµ.u')²+(sinµ.u')²(1-v²/c²)] / 1+cosµ.v.u'/c²

If we take µ = 0 ° and µ = 90 °, we obviously find the two formulas
already written.

In manual writing :

<http://news2.nemoweb.net/jntp?iaioLATGdLgEZBgyOELPUPE-uOw@jntp/Data.Media:1>

> Δx' = γ (Δx − v²/c² Δx)
> = Δx γ (1 − v²/c²)
> = Δx γ/γ²
> = Δx/γ.

No. The real equation is not that.

The real equation is Δx' = Δx.sqrt(1-v²/c²)/1+cosµ.v/c

Relativists do not take spatial anisocheony into account, and so must be.
They only take the contraction of distances by the effect of the second
degree, but not the other. However, this effect is not only apparent and
due to the speed of light (as they believe) but real.

>
> This means that the distance to the destination is shorter for the traveler
> the faster their relative speed relative to origin or destination, and

It's nice. BUT it's false. They make a catastrophic mistake of thinking.

It's not l'=l.sqrt(1-v²/c²) but l'=l.sqrt(1-v²/c²)/1+cosµ.Vo/c the
real and natural equation.

>
> [A textbook example: If you have a fast enough starship (that is not a
> warp ship, so all within the bounds of special relativity), you can
> reach the center of the Milky Way, ≈ 37'000 ly away from here, within a
> human lifetime (say 70 years) while traveling at *subluminal* speeds
> (v ≈ < c). The only problem is that for everybody not traveling with
> you it will take more than 37'000 years there and back, so when you
> come back, you have to rely on that the knowledge of your journey was
> preserved.]
>

Yes.

You are obviously absolutely right.

That is what I am saying.

But you use for that: t '= t / sqrt (1-v² / c²)

This is both true and false in the sense that it gives the average
calculation. The average.

At 0.8c, for example; we will have an 18-year-old brother, and the other
30, on a course of 24 light years.

That is quite right.

Where I no longer get along with relativists is when he thinks that this
average is a continuous average.

Take the students of a class, their average weight is 58 kgs.

I agree with that.

But that does not mean that everyone is 58 kgs.

This is what happens when a rocket leaves for space. The average will be
as you say. But if I break the outward and return trip into two parts, the
two measures will not be half of this average, but t '= t. (1 + cosµ.v /
c) / sqrt (1-v² / c²) either on the outward journey t '= 3 years and on
the return trip t' = 27 years.

t = 9 years + 9 years = 18 years in the rocket.

> your delusions of grandeur.

Oh, no, sir.

LOL, I assure you that I am the most humble of men, and that is moreover
what my entourage often confirms to me.

> PointedEars

Thank you for your very detailed answers.

It's always great to chat with someone about ideas, not people.

Even if we do not agree on several ideas and several implications of
special relativity.

R.H.

--
Ce message a été posté avec Nemo : <http://news2.nemoweb.net/?DataID=iaioLATGdLgEZBgyOELPUPE-uOw@jntp>

On Tuesday, December 7, 2021 at 7:47:17 PM UTC-6, Thomas 'PointedEars' Lahn wrote:

> Only your broken browser (probably broken by your broken “Web Protection by
> Bitdefender”, the so-called “internet security software”, among actual
> Internet security experts better known as “snake oil”) does not.

Bitdefender does very well in comparative tests against other antivirus
programs. A number of reviewers note that it -does- seem to have a
somewhat elevated tendency to report false positives (for example the
following https://www.tomsguide.com/news/bitdefender-vs-kaspersky-antivirus)
but it always ranks quite well.

No antivirus program has a perfect score. Does yours?

Richard Hachel wrote:

> Le 08/12/2021 à 02:12, Thomas 'PointedEars' Lahn a écrit :
>> u(v, u') = (v + u')/(1 + v u'/c²).
>
> This is longitudinal addition speed.
>
> Here is perpendicular addition speed ---> u(v,u') =
> sqrt[v²+u'²-vu'/c²]

I will get back to that later.
>> Δx' = γ (Δx − v²/c² Δx)
>> = Δx γ (1 − v²/c²)
>> = Δx γ/γ²
>> = Δx/γ.
>
> No. The real equation is not that.
>
> The real equation is Δx' = Δx.sqrt(1-v²/c²)/1+cosµ.v/c

No, it is not as it is missing the temporal difference in the unprimed
frame, Δt.
> Relativists do not take spatial anisocheony into account,

There is no such thing.

PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven

Le 08/12/2021 à 18:43, Thomas 'PointedEars' Lahn a écrit :
> Richard Hachel wrote:
>
>> Le 08/12/2021 à 02:12, Thomas 'PointedEars' Lahn a écrit :
>>> u(v, u') = (v + u')/(1 + v u'/c²).
>>
>> This is longitudinal addition speed.
>>
>> Here is perpendicular addition speed ---> u(v,u') =
>> sqrt[v²+u'²-vu'/c²]
>
> I will get back to that later.
>
>>> Δx' = γ (Δx − v²/c² Δx)
>>> = Δx γ (1 − v²/c²)
>>> = Δx γ/γ²
>>> = Δx/γ.
>>
>> No. The real equation is not that.
>>
>> The real equation is Δx' = Δx.sqrt(1-v²/c²)/1+cosµ.v/c
>
> No, it is not as it is missing the temporal difference in the unprimed
> frame, Δt.
>
>> Relativists do not take spatial anisocheony into account,
>
> There is no such thing.
>
>
> PointedEars

Well.
To fully understand people, you must first read them.
What do I say? I say that the Lorentz transformations are only valid when
O and O 'intersect, and that, outside of that, mine (more general) must be
used.
I repeat, the eqaution you give is correct, but it only applies for an
object observed in purely transverse displacement.
In this case, we therefore have, for example, a 30-meter rocket which will
only measure 24 to 0.6c, or 18 to 0.8c.
We agree on that.
If the displacement is transverse, then µ = 90 ° and cosµ = 0.
Obviously, we get l '= l.sqrt (1-v²/c²).
But in my own relativity, we must not limit ourselves to that, and we must
give the complete equation.
l '= l.sqrt (1-v²/c²)/1+cosµ.v/c
We can therefore see that the same rocket will not have the same length
for whom to see it flees longitudinally for example. cosµ = 1
Or a rocket that will appear even shorter. 10 meters at 0.8c, and 15
meters at 0.6c.
If the rocket approaches in the line of sight, then cosµ = -1.
The rocket then appears much longer. 90 meters at 0.8c, 60 at 0.6c.
These effects are real and not due to an apparent Doppler effect.
Relativistic physicists obviously find the same values ​​as me. It is
not on these values ​​that I base myself, since we all have the same
ones. I am basing myself on the ideology: with me, all of this is real,
not just the transverse Doppler effect which he calls the gamma factor.
The longitudinal Doppler effect is with me, just as real.

<http://news2.nemoweb.net/jntp?TN1OmI2wHNhTuOw8DyCWygpNq4w@jntp/Data.Media:1>

<http://news2.nemoweb.net/jntp?TN1OmI2wHNhTuOw8DyCWygpNq4w@jntp/Data.Media:2>

R.H.

--
Ce message a été posté avec Nemo : <http://news2.nemoweb.net/?DataID=TN1OmI2wHNhTuOw8DyCWygpNq4w@jntp>

Richard Hachel wrote:

> What do I say? I say that the Lorentz transformations are only valid when
> O and O 'intersect, […]

And that is still wrong, as has been proved to you /ad nauseam/.

That you are insisting that it would not be so is a strong indication that
you are mentally ill.

PointedEars
--
Heisenberg is out for a drive when he's stopped by a traffic cop.
The officer asks him "Do you know how fast you were going?"
Heisenberg replies "No, but I know where I am."
(from: WolframAlpha)

Prokaryotic Capase Homolog wrote:

> On Tuesday, December 7, 2021 at 7:47:17 PM UTC-6, Thomas 'PointedEars'
> Lahn wrote:
>> Only your broken browser (probably broken by your broken “Web Protection
>> by Bitdefender”, the so-called “internet security software”, among actual
>> Internet security experts better known as “snake oil”) does not.
>
> Bitdefender does very well in comparative tests against other antivirus
> programs.

What was described here is NOT what an antivirus is supposed to do.

> A number of reviewers note that it -does- seem to have a
> somewhat elevated tendency to report false positives (for example the
> following
> https://www.tomsguide.com/news/bitdefender-vs-kaspersky-antivirus)

In the case here it appears to be not just a false positive: The Web browser
or the "internet security software" (probably the latter) *falsely*
*redirects* requests to “http:” to “https:” (even if the server side does
not) as if that would provide additional security (it doesn’t). And then of
course the Web browser notifies the user that there is no security
certificate; as the Web site was not intended to be accessed via HTTP over
SSL/TLS in the first place, there *cannot* *be* a security certificate.

It is snake oil for people without a minimum clue about IT security.

> but it always ranks quite well.

You, too, don’t know what you are talking about.
> No antivirus program has a perfect score. Does yours?

No, but it (ClamAV) does not do such nonsense, and works quietly in the
background, only scanning files when needed. I had almost forgotten that it
was even installed.

That may have to do with the fact that I know what I am doing: I am avoiding
inherently b0rked operating systems, I have configured my operating system
so that no services are running and no ports are open that are not needed,
and so that software only has the privileges that it needs (on a case-by-
case basis); and I am not clicking on everything that is not in the trees on
the count of 3.

HTH

PointedEars
--
Q: What did the nuclear physicist order for lunch?
A: Fission chips.

(from: WolframAlpha)

On 12/8/21 6:42 PM, Thomas 'PointedEars' Lahn wrote:
> Prokaryotic Capase Homolog wrote:
>> [...]
> In the case here it appears to be not just a false positive: The Web browser
> or the "internet security software" (probably the latter) *falsely*
> *redirects* requests to “http:” to “https:”

You guys are talking past each other.

Redirecting http: to https: does not provide additional security in the
sense of antivirus/antimalware efforts. But it does provide additional
security in the sense of preventing others from reading what you are
doing (specifically, bad guys in the various networks and ISPs your
traffic traverses, especially your local WiFi).

Such redirection is not performed by bitdefender, or any other antivirus
program; it is performed by the browser. My browser, Firefox, does not
do it, but it does indicate that http: connections are "not secure". I
still have a number of web sites that use http: (no httos), but there is
nothing a bad guy could gain by reading their traffic.

Tom Roberts

Tom Roberts <tjroberts137@sbcglobal.net> wrote:

> On 12/8/21 6:42 PM, Thomas 'PointedEars' Lahn wrote:
> > Prokaryotic Capase Homolog wrote:
> >> [...]
> > In the case here it appears to be not just a false positive: The Web browser
> > or the "internet security software" (probably the latter) *falsely*
> > *redirects* requests to "http:" to "https:"
>
> You guys are talking past each other.
>
> Redirecting http: to https: does not provide additional security in the
> sense of antivirus/antimalware efforts. But it does provide additional
> security in the sense of preventing others from reading what you are
> doing (specifically, bad guys in the various networks and ISPs your
> traffic traverses, especially your local WiFi).
>
> Such redirection is not performed by bitdefender, or any other antivirus
> program; it is performed by the browser. My browser, Firefox, does not
> do it, but it does indicate that http: connections are "not secure". I
> still have a number of web sites that use http: (no httos), but there is
> nothing a bad guy could gain by reading their traffic.

Summary: It is a swindle,

Jan

Tom Roberts wrote:

> On 12/8/21 6:42 PM, Thomas 'PointedEars' Lahn wrote:
>> Prokaryotic Capase Homolog wrote:
>>> [...]
>> In the case here it appears to be not just a false positive: The Web
>> browser or the "internet security software" (probably the latter)
>> *falsely* *redirects* requests to “http:” to “https:”
>
> You guys are talking past each other.
>
> Redirecting http: to https: does not provide additional security in the
> sense of antivirus/antimalware efforts.

If the *server* does not do it, then it does not provide additional security
in any way.

> But it does provide additional security in the sense of preventing others
> from reading what you are doing (specifically, bad guys in the various
> networks and ISPs your traffic traverses, especially your local WiFi).

No offense meant, but you should stick to your area of expertise or educate
yourself. Before I was studying Physics, I earned a degree in computer
science while having a long career in that field, with some experience in IT
security as well (I remember that I gave a talk on OWASP, for example).
Therefore I know that what you just said is pure bullshit:

Without a valid security certificate there is *no authentication*, which is
precisely why browsers indicate that if one uses an “https:” URI and there
is either no certificate or it is in some way invalid (usually: expired,
registered to a different domain, or not recognized because the required CA
certificates were not installed and activated).
> Such redirection is not performed by bitdefender, or any other antivirus
> program; it is performed by the browser.

Unlikely; I have yet to see any Web browser do that, and I know *a lot* of
them. Iff Bitdefender did not do it, then it is more likely that the
PEBKAC: Paul thought it would be a good idea to use “https:” instead of
“http:” and, then he ran into Bitdefender’s blocking.

He/you can try the same with my Web site, <http://PointedEars.de/>, which
*also* does not have a security certificate (yet).

PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)

On 12/10/21 6:52 PM, Thomas 'PointedEars' Lahn wrote:
> Tom Roberts wrote:
>> Redirecting http: to https: does not provide additional security in the
>> sense of antivirus/antimalware efforts.
>> But it does provide additional security in the sense of preventing others
>> from reading what you are doing (specifically, bad guys in the various
>> networks and ISPs your traffic traverses, especially your local WiFi).
>
> No offense meant, but you should stick to your area of expertise

This is my area of expertise. In addition to being a physicist, I am a
highly experienced software developer, including networking.

> Without a valid security certificate there is *no authentication*,

Right. And there also is no https unless the user manually accepts their
browser notification of an invalid certificate.

>> Such redirection is not performed by bitdefender, or any other antivirus
>> program; it is performed by the browser.
>
> Unlikely; I have yet to see any Web browser do that,

I haven't either, but a browser could do so. The implication was that it
happened, and the only possibility is their browser. To do this in a
browser would have to be smart and not redirect if https was refused by
the server. There was discussion about doing this automatically, but I
have not kept current on its status.

Tom Roberts

Thomas 'PointedEars' Lahn wrote:

>> Such redirection is not performed by bitdefender, or any other
>> antivirus program; it is performed by the browser.
>
> Unlikely; I have yet to see any Web browser do that, and I know *a lot*
> of them. Iff Bitdefender did not do it, then it is more likely that the
> PEBKAC: Paul thought it would be a good idea to use “https:” instead of
> “http:” and, then he ran into Bitdefender’s blocking.

idiot, an antivirus crap is higher level thing, AFTER the data is been
decoded. The restricted sites, encrypted communication or not, are
browser business. But the shift from https to http, including DNS over
https is again a browser business done automatically. You are fucking
idiot in many areas.

Tom Roberts wrote:

> On 12/10/21 6:52 PM, Thomas 'PointedEars' Lahn wrote:
>> No offense meant, but you should stick to your area of expertise
>
> This is my area of expertise. In addition to being a physicist, I am a
> highly experienced software developer, including networking.
>
>> Without a valid security certificate there is *no authentication*,
>
> Right. And there also is no https unless the user manually accepts their
> browser notification of an invalid certificate.

both wrong. You accept their invalid certificate, not only the
notification. I suspect you guys have to go inside and delete a lot of
certificates.

Tom Roberts wrote:

> On 12/10/21 6:52 PM, Thomas 'PointedEars' Lahn wrote:
>> Tom Roberts wrote:
>>> Redirecting http: to https: does not provide additional security in the
>>> sense of antivirus/antimalware efforts.
>>> But it does provide additional security in the sense of preventing
>>> others from reading what you are doing (specifically, bad guys in the
>>> various networks and ISPs your traffic traverses, especially your local
>>> WiFi).
>>
>> No offense meant, but you should stick to your area of expertise
>
> This is my area of expertise. In addition to being a physicist, I am a
> highly experienced software developer, including networking.

There is a difference between a computer scientist and a software developer.
Anyhow, then I suggest that you review and update your knowledge in that
area.

>> Without a valid security certificate there is *no authentication*,
>
> Right. And there also is no https unless the user manually accepts their
> browser notification of an invalid certificate.

In particular, if the security certificate is not only not valid, but non-
existent, there is zero added security, too; only user hassle. Therefore:

>>> Such redirection is not performed by bitdefender, or any other antivirus
>>> program; it is performed by the browser.
>>
>> Unlikely; I have yet to see any Web browser do that,
>
> I haven't either, but a browser could do so. […]

And if my grandmother would have had wheels, she would have been a
bandwagon.

<facepalm/>

PointedEars
--
Q: Why is electricity so dangerous?
A: It doesn't conduct itself.

(from: WolframAlpha)