Re: Deriving the Born Rule

[Bruce Kellett]

On Wed, May 13, 2020 at 3:30 PM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

> On 5/12/2020 10:08 PM, Bruce Kellett wrote:
>
> On Wed, May 13, 2020 at 2:06 PM 'Brent Meeker' via Everything List <
> everything-list@googlegroups.com> wrote:
>
>>
>> > Consequently, the amplitude multiplying any sequence of M zeros and
>> > (N-M) ones, is a^M b^(N-M). Again, differentiating with respect to 'a'
>> > to find the turning point (and the value of 'a' that maximizes this
>> > amplitude), we find
>> >
>> > |a|^2 = M/N,
>>
>> Maximizing this amplitude, instead of simply counting the number of
>> sequences with M zeroes as a fraction of all sequences (which is
>> independent of a) is effectively assuming |a|^2 is a probability
>> weight.  The "most likely" number of zeroes, the number that occurs most
>> often in the 2^N sequences, is is N/2.
>>
>
> I agree that if you simply look for the most likely number of zeros,
> ignoring the amplitudes, then that is N/2. But I do not see that maximising
> the amplitude for any particular value of M is to effectively assume that
> it is a probability.
>
>
> I think it is.  How would you justify ".. the amplitude multiplying any
> sequence of M zeros and (N-M) ones, is a^M b^(N-M)..." except by saying a
> is a probability, so a^M is the probability of M zeroes.  If it's not a
> probability why should it be multiplied into and expression to be maximized?
>


Trivially, without any assumptions at all. The original state has
amplitudes a|0> + b|1>. If you carry the coefficients through at each
branch, the branch containing a new |0> carries a weight a, and similarly,
the branch containing a new |1> carries a weight b. One does not have to
assume that these are probabilities to do this -- each repeated trial is a
branch point, so each is another measurement of an instance of the initial
state, so automatically has the coefficients present. I don't see anything
sneaky here.

As to the question as to why it should be maximised? Well, why not? I am
simply maximising the carried through coefficients to find if this has any
bearing on the proportion M of zeros. The argument for probabilities then
proceeds by means of the analogy with the traditional binomial case. I
agree, this may not count as a derivation of the Born rule for
probabilities, but it is certainly a good explication of the same.



> In any case though, I don't see the form of the Born rule as something
> problematic.  It's getting from counting branches to probabilities.
>


I think my issue here is that counting branches is not the thing to do,
because the branches are not in proportion to the coefficients (which turn
out to be probabilities). And counting branches for probabilities requires
the self-location assumption, and that is intrinsically dualist (as David
Albert points out).

 Once you assume there is a probability measure, you're pretty much forced
> to the Born rule as the only consistent probability measure.
>

I agree. And that is the argument Everett made in his 1957 paper -- once
you require additivity, the fact that states are normalised, screams for
the coefficients to be treated as probabilities. The squared amplitudes
obey all the Kolmogorov axioms, after all.

Bruce

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Re: Deriving the Born Rule

['Brent Meeker' via Everything List]

On 5/12/2020 10:08 PM, Bruce Kellett wrote:
On Wed, May 13, 2020 at 2:06 PM 'Brent Meeker' via Everything List 
> wrote:


On 5/12/2020 7:12 PM, Bruce Kellett wrote:

> If we now turn our attention to the quantum case, we have a
> measurement (or sequence of measurements) on a binary quantum state
>
>  |psi> = a|0> + b|1>,
>
> where |0> is to be counted as a "success", |1> represents anything
> else or a "fail", and a^2 + b^2 = 1. In a single measurement, we
can
> get either |0> or 1>, (or we get both on separate branches in the
> Everettian case). Over a sequence of N similar trials, we get a
set of
> 2^N sequences of all possible bit strings of length N. (These all
> exist in separate "worlds" for the Everettian, or simply represent
> different "possible worlds" (or possible sequences of results)
in the
> single-world case.) This set of bit strings is independent of the
> coefficients 'a' and 'b' from the original state |psi>, but if we
> carry the amplitudes of the original superposition through the
> sequence of results, we find that for every zero in a bit string we
> get a factor of 'a', and for every one, we get a factor of 'b'.

This is what you previously argued was not part of the Schroedinger
equation and was a cheat to slip the Born rule in.  It's what I
said was
Carroll's "weight" or splitting of many pre-existing worlds.


This is not what Carroll does. He looks at a single measurement, and 
boosts the number of components of the wave function so that all have 
the same amplitude. That, I argue, is a mistake.


>
> Consequently, the amplitude multiplying any sequence of M zeros and
> (N-M) ones, is a^M b^(N-M). Again, differentiating with respect
to 'a'
> to find the turning point (and the value of 'a' that maximizes this
> amplitude), we find
>
>     |a|^2 = M/N,

Maximizing this amplitude, instead of simply counting the number of
sequences with M zeroes as a fraction of all sequences (which is
independent of a) is effectively assuming |a|^2 is a probability
weight.  The "most likely" number of zeroes, the number that
occurs most
often in the 2^N sequences, is is N/2.


I agree that if you simply look for the most likely number of zeros, 
ignoring the amplitudes, then that is N/2. But I do not see that 
maximising the amplitude for any particular value of M is to 
effectively assume that it is a probability.


I think it is.  How would you justify ".. the amplitude multiplying any 
sequence of M zeros and (N-M) ones, is a^M b^(N-M)..." except by saying 
a is a probability, so a^M is the probability of M zeroes. If it's not a 
probability why should it be multiplied into and expression to be maximized?


In any case though, I don't see the form of the Born rule as something 
problematic.  It's getting from counting branches to probabilities.  
Once you assume there is a probability measure, you're pretty much 
forced to the Born rule as the only consistent probability measure.


Brent

When you do this, you can see by analogy that it is a probability, but 
one did not assume this at the start.


Bruce
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Re: Deriving the Born Rule

[Bruce Kellett]

On Wed, May 13, 2020 at 2:06 PM 'Brent Meeker' via Everything List <
everything-list@googlegroups.com> wrote:

> On 5/12/2020 7:12 PM, Bruce Kellett wrote:
>
> > If we now turn our attention to the quantum case, we have a
> > measurement (or sequence of measurements) on a binary quantum state
> >
> >  |psi> = a|0> + b|1>,
> >
> > where |0> is to be counted as a "success", |1> represents anything
> > else or a "fail", and a^2 + b^2 = 1. In a single measurement, we can
> > get either |0> or 1>, (or we get both on separate branches in the
> > Everettian case). Over a sequence of N similar trials, we get a set of
> > 2^N sequences of all possible bit strings of length N. (These all
> > exist in separate "worlds" for the Everettian, or simply represent
> > different "possible worlds" (or possible sequences of results) in the
> > single-world case.) This set of bit strings is independent of the
> > coefficients 'a' and 'b' from the original state |psi>, but if we
> > carry the amplitudes of the original superposition through the
> > sequence of results, we find that for every zero in a bit string we
> > get a factor of 'a', and for every one, we get a factor of 'b'.
>
> This is what you previously argued was not part of the Schroedinger
> equation and was a cheat to slip the Born rule in.  It's what I said was
> Carroll's "weight" or splitting of many pre-existing worlds.
>

This is not what Carroll does. He looks at a single measurement, and boosts
the number of components of the wave function so that all have the same
amplitude. That, I argue, is a mistake.

> >
> > Consequently, the amplitude multiplying any sequence of M zeros and
> > (N-M) ones, is a^M b^(N-M). Again, differentiating with respect to 'a'
> > to find the turning point (and the value of 'a' that maximizes this
> > amplitude), we find
> >
> > |a|^2 = M/N,
>
> Maximizing this amplitude, instead of simply counting the number of
> sequences with M zeroes as a fraction of all sequences (which is
> independent of a) is effectively assuming |a|^2 is a probability
> weight.  The "most likely" number of zeroes, the number that occurs most
> often in the 2^N sequences, is is N/2.
>

I agree that if you simply look for the most likely number of zeros,
ignoring the amplitudes, then that is N/2. But I do not see that maximising
the amplitude for any particular value of M is to effectively assume that
it is a probability. When you do this, you can see by analogy that it is a
probability, but one did not assume this at the start.

Bruce

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Re: Deriving the Born Rule

['Brent Meeker' via Everything List]

On 5/12/2020 7:12 PM, Bruce Kellett wrote:
The meaning of probability and the origin of the Born rule has been 
seem as one of the outstanding problems for Everettian quantum theory. 
Attempts by Carroll and Sebens, and Zurek to derive the Born rule have 
considered probability in terms of the outcome of a single experiment 
where the result is uncertain. This approach is can be seen to be 
misguided, because probabilities cannot be determined from a single 
trial. In a single trial, any result is possible, in both the 
single-world and many-worlds cases -- probability is not manifest in 
single trials.


It is not surprising, therefore, the Carroll and Zurek have 
concentrated on the basic idea that equal amplitudes have equal 
probabilities, and have been led to break the original state (with 
unequal amplitudes) down to a superposition of many parts having equal 
amplitude, basically by looking to "borrow" ancillary degrees of 
freedom from the environment. The number of equal amplitude components 
then gives the relative probabilities by a simple process of branch 
counting. As Carroll and Sebens write:


"This route to the Born rule has a simple physical interpretation. 
Take the wave function and write it as a sum over orthonormal basis 
vectors with equal amplitudes for each term in the sum (so that many 
terms may contribute to a single branch). Then the Born rule is simply 
a matter of counting -- every term in that sum contributes an equal 
probability." (arxiv:1405.7907 [gr-qc])


Many questions remain as to the validity of this process, particularly 
as it involves an implementation of the idea of self-selection: of 
selecting which branch one finds oneself on. This is an even more 
dubious process than branch counting, since it harks back to the 
"many-minds" ideas of Albert and Loewer, which even David Albert now 
finds to be "bad, silly, tasteless, hopeless, and explicitly dualist."


Simon Saunders (in his article "Chance in the Everett Interpretation" 
(in  "Many Worlds: Everett, Quantum Theory, & Reality" Edited by 
Saunders, Barrett, Kent and Wallace (OUP 2010)) points out that 
probabilities can only be measured (or estimated) in a series of 
repeated trials, so it is only in sequences of repeated trials on an 
ensemble of similarly prepared states that we can see how probability 
emerges. This idea seemed promising, so I came up with the following 
argument.


If, in classical probability theory, one has a process in which the 
probability of success in a single Bernoulli trial is p, the 
probability of getting M successes in a sequence of N independent 
trials is p^M (1-p)^(N-M). Since there are many ways in which one 
could get M successes in N trials, to get the overall probability of M 
successes we have to sum over the N!/M!(N-M)! ways in which the M 
successes can be ordered. So the final probability of getting M 
successes in N independent trials is


  Prob of M successes = p^M (1-p)^(N-M) N!/M!(N-M)!.

We can find the value of p for which this probability is maximized by 
differentiating with respect to p and finding the turning point. A 
simple calculation gives that p = M/N maximizes this probability (or, 
alternatively, maximizes the amplitude for each sequence of results in 
the above sum). This is all elementary probability theory of the 
binomial distribution, and is completely uncontroversial.


If we now turn our attention to the quantum case, we have a 
measurement (or sequence of measurements) on a binary quantum state


 |psi> = a|0> + b|1>,

where |0> is to be counted as a "success", |1> represents anything 
else or a "fail", and a^2 + b^2 = 1. In a single measurement, we can 
get either |0> or 1>, (or we get both on separate branches in the 
Everettian case). Over a sequence of N similar trials, we get a set of 
2^N sequences of all possible bit strings of length N. (These all 
exist in separate "worlds" for the Everettian, or simply represent 
different "possible worlds" (or possible sequences of results) in the 
single-world case.) This set of bit strings is independent of the 
coefficients 'a' and 'b' from the original state |psi>, but if we 
carry the amplitudes of the original superposition through the 
sequence of results, we find that for every zero in a bit string we 
get a factor of 'a', and for every one, we get a factor of 'b'.


This is what you previously argued was not part of the Schroedinger 
equation and was a cheat to slip the Born rule in.  It's what I said was 
Carroll's "weight" or splitting of many pre-existing worlds.




Consequently, the amplitude multiplying any sequence of M zeros and 
(N-M) ones, is a^M b^(N-M). Again, differentiating with respect to 'a' 
to find the turning point (and the value of 'a' that maximizes this 
amplitude), we find


    |a|^2 = M/N,


Maximizing this amplitude, instead of simply counting the number of 
sequences with M zeroes as a fraction of all sequences (which is 
independent of a) is 

Deriving the Born Rule

[Bruce Kellett]

The meaning of probability and the origin of the Born rule has been seem 
as one of the outstanding problems for Everettian quantum theory. 
Attempts by Carroll and Sebens, and Zurek to derive the Born rule have 
considered probability in terms of the outcome of a single experiment 
where the result is uncertain. This approach is can be seen to be 
misguided, because probabilities cannot be determined from a single 
trial. In a single trial, any result is possible, in both the 
single-world and many-worlds cases -- probability is not manifest in 
single trials.


It is not surprising, therefore, the Carroll and Zurek have concentrated 
on the basic idea that equal amplitudes have equal probabilities, and 
have been led to break the original state (with unequal amplitudes) down 
to a superposition of many parts having equal amplitude, basically by 
looking to "borrow" ancillary degrees of freedom from the environment. 
The number of equal amplitude components then gives the relative 
probabilities by a simple process of branch counting. As Carroll and 
Sebens write:


"This route to the Born rule has a simple physical interpretation. Take 
the wave function and write it as a sum over orthonormal basis vectors 
with equal amplitudes for each term in the sum (so that many terms may 
contribute to a single branch). Then the Born rule is simply a matter of 
counting -- every term in that sum contributes an equal probability." 
(arxiv:1405.7907 [gr-qc])


Many questions remain as to the validity of this process, particularly 
as it involves an implementation of the idea of self-selection: of 
selecting which branch one finds oneself on. This is an even more 
dubious process than branch counting, since it harks back to the 
"many-minds" ideas of Albert and Loewer, which even David Albert now 
finds to be "bad, silly, tasteless, hopeless, and explicitly dualist."


Simon Saunders (in his article "Chance in the Everett Interpretation" 
(in  "Many Worlds: Everett, Quantum Theory, & Reality" Edited by 
Saunders, Barrett, Kent and Wallace (OUP 2010)) points out that 
probabilities can only be measured (or estimated) in a series of 
repeated trials, so it is only in sequences of repeated trials on an 
ensemble of similarly prepared states that we can see how probability 
emerges. This idea seemed promising, so I came up with the following 
argument.


If, in classical probability theory, one has a process in which the 
probability of success in a single Bernoulli trial is p, the probability 
of getting M successes in a sequence of N independent trials is p^M 
(1-p)^(N-M). Since there are many ways in which one could get M 
successes in N trials, to get the overall probability of M successes we 
have to sum over the N!/M!(N-M)! ways in which the M successes can be 
ordered. So the final probability of getting M successes in N 
independent trials is


  Prob of M successes = p^M (1-p)^(N-M) N!/M!(N-M)!.

We can find the value of p for which this probability is maximized by 
differentiating with respect to p and finding the turning point. A 
simple calculation gives that p = M/N maximizes this probability (or, 
alternatively, maximizes the amplitude for each sequence of results in 
the above sum). This is all elementary probability theory of the 
binomial distribution, and is completely uncontroversial.


If we now turn our attention to the quantum case, we have a measurement 
(or sequence of measurements) on a binary quantum state


 |psi> = a|0> + b|1>,

where |0> is to be counted as a "success", |1> represents anything else 
or a "fail", and a^2 + b^2 = 1. In a single measurement, we can get 
either |0> or 1>, (or we get both on separate branches in the Everettian 
case). Over a sequence of N similar trials, we get a set of 2^N 
sequences of all possible bit strings of length N. (These all exist in 
separate "worlds" for the Everettian, or simply represent different 
"possible worlds" (or possible sequences of results) in the single-world 
case.) This set of bit strings is independent of the coefficients 'a' 
and 'b' from the original state |psi>, but if we carry the amplitudes of 
the original superposition through the sequence of results, we find that 
for every zero in a bit string we get a factor of 'a', and for every 
one, we get a factor of 'b'.


Consequently, the amplitude multiplying any sequence of M zeros and 
(N-M) ones, is a^M b^(N-M). Again, differentiating with respect to 'a' 
to find the turning point (and the value of 'a' that maximizes this 
amplitude), we find


    |a|^2 = M/N,

where we have taken the modulus of 'a' since a is, in general, a complex 
number. Again, there will be more than one bit-string with exactly M 
zeros and (N-M) ones, and summing over these gives the additional factor 
of N!/M!(N-M)!, as above.


If we now compare the quantum result over N measurements with the 
classical probability result for N independent Bernoulli trials, we find 
that the amplitude for M 

Re: Wolfram Models as Set Substitution Systems

[Philip Thrift]

But we know though, there is no real physical theory.

@philipthrift

On Tuesday, May 12, 2020 at 4:32:16 PM UTC-5, Lawrence Crowell wrote:
>
> My primary difficulty with this is not that this is a possibly useful 
> math-method, but that I have little physical sense of what this means. As 
> some combinatorics or paths or states this may have some utility, but this 
> to me is not terribly much a real physical theory.
>
> LC
>
> On Tuesday, May 12, 2020 at 3:13:05 AM UTC-5, Philip Thrift wrote:
>>
>>
>> *Wolfram Models as Set Substitution Systems*
>> https://github.com/maxitg/SetReplace
>>
>> cf. https://www.wolframphysics.org/
>>
>> Stephen Wolfram (Ph.D. in theoretical physics at the California Institute 
>> of Technology in 1979—at the age of 20): 
>>
>> “I’m disappointed by the naivete of the questions that you’re 
>> communicating.” 
>>
>>
>> https://www.scientificamerican.com/article/physicists-criticize-stephen-wolframs-theory-of-everything/
>>
>> “I don’t know of any others in this field that have the wide range of 
>> understanding of Dr. Wolfram,” Feynman wrote ( in 1981).
>>
>>
>> @philipthrift
>>
>

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Re: Wolfram Models as Set Substitution Systems

[Lawrence Crowell]

My primary difficulty with this is not that this is a possibly useful 
math-method, but that I have little physical sense of what this means. As 
some combinatorics or paths or states this may have some utility, but this 
to me is not terribly much a real physical theory.

LC

On Tuesday, May 12, 2020 at 3:13:05 AM UTC-5, Philip Thrift wrote:
>
>
> *Wolfram Models as Set Substitution Systems*
> https://github.com/maxitg/SetReplace
>
> cf. https://www.wolframphysics.org/
>
> Stephen Wolfram (Ph.D. in theoretical physics at the California Institute 
> of Technology in 1979—at the age of 20): 
>
> “I’m disappointed by the naivete of the questions that you’re 
> communicating.” 
>
>
> https://www.scientificamerican.com/article/physicists-criticize-stephen-wolframs-theory-of-everything/
>
> “I don’t know of any others in this field that have the wide range of 
> understanding of Dr. Wolfram,” Feynman wrote ( in 1981).
>
>
> @philipthrift
>

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Re: Wolfram Models as Set Substitution Systems

[ronaldheld]

Not  a favorable review from SA.   If he submitted peer reviewed documents 
years ago, things might be different today.
  Ronald

On Tuesday, May 12, 2020 at 4:13:05 AM UTC-4, Philip Thrift wrote:
>
>
> *Wolfram Models as Set Substitution Systems*
> https://github.com/maxitg/SetReplace
>
> cf. https://www.wolframphysics.org/
>
> Stephen Wolfram (Ph.D. in theoretical physics at the California Institute 
> of Technology in 1979—at the age of 20): 
>
> “I’m disappointed by the naivete of the questions that you’re 
> communicating.” 
>
>
> https://www.scientificamerican.com/article/physicists-criticize-stephen-wolframs-theory-of-everything/
>
> “I don’t know of any others in this field that have the wide range of 
> understanding of Dr. Wolfram,” Feynman wrote ( in 1981).
>
>
> @philipthrift
>

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Re: Human Challenge Trials

[Lawrence Crowell]

On Tuesday, May 12, 2020 at 9:10:04 AM UTC-5, John Clark wrote:
>
> On Tue, May 12, 2020 at 9:29 AM Lawrence Crowell  > wrote:
>
> > t'Rump has a hard time making words such as "and" and "the" truthful.
>>
>
> For Trump lying comes as naturally as breathing, if he said "hello" to me 
> I wouldn't believe him.
>  
>
>> > In China, S, Korea and Germany the pandemic is already rebounding as 
>> they have opened up.
>>
>
> South Korea reported its first case OF COVID-19 on January 20, the exact 
> same day the US did, but unlike the US within 2 days they had developed 
> their own test for the virus and started a massive program of testing. So 
> as of today only 10,936 people in South Korea have gotten sick from the 
> virus and 258 have died, but in the US 1,387,496 have gotten sick from 
> the virus and 81,937 have died from it. And in the first quarter the GDP of 
> South Korea declined by 1.4%, in the USA it declined by 4.8%. And today the 
> unemployment rate in the US is 14.7%, in South Korea It's 3.8%.
>
> > we have a clown for a President
>>
>
> I think you're being a little unfair to clowns, they make you want to 
> laugh, this jackass makes you want to cry. 
>
> John K Clark
>

In the end your final defense is humor. Jimmy Buffet got it right with, "If 
we couldn't laugh, we would all go insane."

A friend in Brazil has given me a load of Bolsonaro, who is a magnified 
version of t'Rump, and fortunately in a country that is in far less power 
situation than the US. If there were to be a God these people are ways this 
God has of pulling tricks or jokes on us.

https://www.youtube.com/watch?v=YcmLij7s1wg

LC

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Re: Human Challenge Trials

[John Clark]

On Tue, May 12, 2020 at 9:29 AM Lawrence Crowell <
goldenfieldquaterni...@gmail.com> wrote:

> t'Rump has a hard time making words such as "and" and "the" truthful.
>

For Trump lying comes as naturally as breathing, if he said "hello" to me I
wouldn't believe him.


> > In China, S, Korea and Germany the pandemic is already rebounding as
> they have opened up.
>

South Korea reported its first case OF COVID-19 on January 20, the exact
same day the US did, but unlike the US within 2 days they had developed
their own test for the virus and started a massive program of testing. So
as of today only 10,936 people in South Korea have gotten sick from the
virus and 258 have died, but in the US 1,387,496 have gotten sick from the
virus and 81,937 have died from it. And in the first quarter the GDP of
South Korea declined by 1.4%, in the USA it declined by 4.8%. And today the
unemployment rate in the US is 14.7%, in South Korea It's 3.8%.

> we have a clown for a President
>

I think you're being a little unfair to clowns, they make you want to
laugh, this jackass makes you want to cry.

John K Clark

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Re: Human Challenge Trials

[Lawrence Crowell]

On Tuesday, May 12, 2020 at 6:51:07 AM UTC-5, Bruno Marchal wrote:
>
>
> On 8 May 2020, at 18:58, John Clark > 
> wrote:
>
> If you want to do something drastic to end this virus nightmare there is 
> something we could do that would be far more effective than waiting for 
> herd immunity as well as being less ethically questionable, although I'm 
> sure some would still clutch their pearls in horror, I'm referring to Human 
> Challenge Trials. The idea is young healthy volunteers would be injected 
> with a experimental vaccine (or a placebo) and then deliberately infected 
> with the COVID-19 virus. This would dramatically speed up vaccine 
> development and save many thousands, perhaps millions, of lives; not to 
> mention stop the economy from collapsing into rubble. The death rate for 
> young healthy people who get COVID-19 is only about 1.4 deaths per 10,000 
> and the death rate for those who volunteer as kidney donors is 3 times 
> that, we accept one as being ethical why not the other?
>
>
> I agree. In fact I think that a government has nothing to say about which 
> medication or medical technic can be used. It is only you and your doctor 
> or shaman who can decide, and evaluate their risk. The government can 
> impose “warnings”, and impose the traceability of products, and many 
> things, but not a treatment, nor any particular medication, at least in 
> normal time. If people are OK to try a new medication, there is no problem, 
> as long as they get the information right, including the possible lack of 
> information. A government can impose medical prescription, though, but 
> again, that must be debatable, and only the physicians should be able to 
> provide informations if such a decision makes sense.
>
> Can a government imposes a vaccine? Yes, but again, only if there is some 
> regulating independent court to assess the need. If not, that becomes an 
> easy way to make money on diseases, like “big-pharma” already does a lot 
> (and that is a *big* current problem).
>
> During a pandemic, a government can impose both vaccine and quarantine, 
> but again, only if such decisions are well explained and limited in time. 
>
> About all this, the following video makes me worry a lot:
>
> "US makes 'big bet' on vaccine company that's never brought a vaccine to 
> market":
> https://www.youtube.com/watch?v=TddfLvAeTqQ
>
> It is so weird … that I am waiting for more information confirming this, 
> but that is an example of what should never been done. It is like putting 
> all eggs in the same basket which does not yet exist.
>
> Bruno
>
>
Donald t'Rump has no idea what he is doing. He even expresses confusion 
over what exactly a test does, where he has made statements suggesting he 
thinks it is preventative method. t'Rump operates by pure cronism, and his 
plugging of hydrochloriquine is because he bought a lot of stock in an 
company in India that makes this anti-malarial drug. It has to be mentioned 
that this sort of activity, buying stock and doing business, violates the 
emoluments clause of the Constitution. On top of it t'Rump has a hard time 
making words such as "and" and "the" truthful. Don-the-Con t'Rump has quite 
a cultish set of followers and it may require a serious damaging event to 
unseat him in the coming election. 

The idea of paying people for human trials to expedite the development of a 
vaccine is probably at least worth considering. Given all of these 
unemployed people out there a decent statistical sample space is possible, 
Also the emergency employment pool exists to do testing and contact 
tracing. There seems to be only tepid action along these lines. Reopening 
the economy would be better if it could be more micromanaged this way 
before a vaccine comes  In China, S, Korea and Germany the pandemic is 
already rebounding as they have opened up. In the US the same will occur in 
spades, and we have a clown for a President who has no clue about anything 
scientific.

LC
 

>
>
> As one ethicist put it:
>
>
> *"This is the trolley problem where the fat man wants to jump knowing his 
> chance of death is below 1% and our decision is whether to stop him."*
> Should volunteers to be infected with coronavirus to test vaccines? 
> 
>
> Human Challenge Trials—A Coronavirus Taboo 
> 
>
> John K Clark
>
>
>
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>  
> 

Re: Human Challenge Trials

[Bruno Marchal]

> On 8 May 2020, at 18:58, John Clark  wrote:
> 
> If you want to do something drastic to end this virus nightmare there is 
> something we could do that would be far more effective than waiting for herd 
> immunity as well as being less ethically questionable, although I'm sure some 
> would still clutch their pearls in horror, I'm referring to Human Challenge 
> Trials. The idea is young healthy volunteers would be injected with a 
> experimental vaccine (or a placebo) and then deliberately infected with the 
> COVID-19 virus. This would dramatically speed up vaccine development and save 
> many thousands, perhaps millions, of lives; not to mention stop the economy 
> from collapsing into rubble. The death rate for young healthy people who get 
> COVID-19 is only about 1.4 deaths per 10,000 and the death rate for those who 
> volunteer as kidney donors is 3 times that, we accept one as being ethical 
> why not the other?

I agree. In fact I think that a government has nothing to say about which 
medication or medical technic can be used. It is only you and your doctor or 
shaman who can decide, and evaluate their risk. The government can impose 
“warnings”, and impose the traceability of products, and many things, but not a 
treatment, nor any particular medication, at least in normal time. If people 
are OK to try a new medication, there is no problem, as long as they get the 
information right, including the possible lack of information. A government can 
impose medical prescription, though, but again, that must be debatable, and 
only the physicians should be able to provide informations if such a decision 
makes sense.

Can a government imposes a vaccine? Yes, but again, only if there is some 
regulating independent court to assess the need. If not, that becomes an easy 
way to make money on diseases, like “big-pharma” already does a lot (and that 
is a *big* current problem).

During a pandemic, a government can impose both vaccine and quarantine, but 
again, only if such decisions are well explained and limited in time. 

About all this, the following video makes me worry a lot:

"US makes 'big bet' on vaccine company that's never brought a vaccine to 
market":
https://www.youtube.com/watch?v=TddfLvAeTqQ 


It is so weird … that I am waiting for more information confirming this, but 
that is an example of what should never been done. It is like putting all eggs 
in the same basket which does not yet exist.

Bruno



> As one ethicist put it:
> 
> "This is the trolley problem where the fat man wants to jump knowing his 
> chance of death is below 1% and our decision is whether to stop him."
> 
> Should volunteers to be infected with coronavirus to test vaccines? 
> 
> 
> Human Challenge Trials—A Coronavirus Taboo 
> 
> 
> John K Clark
> 
> 
> 
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> To unsubscribe from this group and stop receiving emails from it, send an 
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> .
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>  
> .

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Wolfram Models as Set Substitution Systems

[Philip Thrift]

*Wolfram Models as Set Substitution Systems*
https://github.com/maxitg/SetReplace

cf. https://www.wolframphysics.org/

Stephen Wolfram (Ph.D. in theoretical physics at the California Institute 
of Technology in 1979—at the age of 20): 

“I’m disappointed by the naivete of the questions that you’re 
communicating.” 

https://www.scientificamerican.com/article/physicists-criticize-stephen-wolframs-theory-of-everything/

“I don’t know of any others in this field that have the wide range of 
understanding of Dr. Wolfram,” Feynman wrote ( in 1981).


@philipthrift

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