On Jul 25, 10:27 pm, (Phillip Helbig---
remove CLOTHES to reply) wrote:
In article , Chalky





writes:
What is the Hawking temperature for a black hole of macho mass? (Yes, I
know,

You astonish me.

Wonders never cease! :-)

Please tell me both the mass of the macho,

I meant the mass range the various observational projects (MACHO, OGLE,
MOA etc) were/are sensitive to. This is roughly between a millionth of
a solar mass and a solar mass. A solar mass has a Hawking temperature
of about 60 nK. (Thus, it would absorb more energy from the CMB than it
emits, causing it to increase in mass and thus decrease in temperature.)
The temperature is inversely proportional to the mass. In fact, it is
one of the great formulae:

T = \frac{\hbar c^{3}}{(8\pi^GMk}

According tohttp://casa.colorado.edu/~ajsh/hawk.html,

kT = hbar g / (2 pi c) where g = G M / R^2 and k is Boltzmann's
constant.

This is more logical, since it gives increasing temperature (and
intensity) with increasing mass. According to you, this relationship
is inverted.

[Mod. note: I think you'll find that R is a function of M. Indeed, the
web page says

`Bigger black holes are colder and dimmer: the Hawking temperature
is inversely proportional to the mass'

-- mjh]

(for a Schwarzschild black hole) where T is the temperature, hbar is the
reduced Planck constant, c is the speed of light, G is the gravitational
constant, M is the mass and k is Boltzmann's constant.

Thus, even at the bottom of the "MACHO mass range", the temperature is
only about 60 mK, still much less than that of the CMB.

I think I will trust my intuition andhttp://casa.colorado.edu, thank
you.

As the moderator's note shows, it is clear where your intuition is
taking you.

I quote one of the most famous formulae in all of physics, and point out
that the temperature is inversely proportional to the mass. Your
intuition indicates the opposite and you prefer to go with that. Thus,
don't be surprised if people are taking you less and less seriously with
every post of yours they read. One can debate about how observations
are interpreted, about whether perhaps something has been overlooked
etc, but questioning formulae derived from first principles (note:
Stephen Hawking is a theorist, not an experimentalist) or, worse,
stating that they are wrong, is tantamount to leaving scientific
discussion altogether.- Hide quoted text -

Two extremes:
1) Question nothing and swallow everything (AKA parrot learning).
2) Question everything, and build only on answers that make sense.

You will find it is scientists in the second category who come up with
the most revolutionary breakthroughs: Copernicus, Newton, and
Einstein, for example, and scientists in the first category who then
learn those findings parrot fashion, once they have become accepted by
the scientific community at large.

If you fell into the second category, you would have noticed that the
formula seemed counter-intuitive, and then worked out why, as I did:
the operative gravitational force is the tidal force near the
Schwartzchild radius, which is larger for smaller masses, because the
Schwartzchild radius is much smaller.

The fact that you did not explain this, but instead criticised me for
asking questions, suggests your affinities lie closer to the first
extreme.

Please note that my response of Jul 25, 9:53 am, confirms that I had
worked this out for myself, and, consequently, accepted the formula,
long before your posting of Jul 25, 10:27 pm, ridiculing me for
needing to know why the formula was as it was.

Please stick to the point, and respond to my posting of Jul 25, 9:53
am.

Chalky

[Mod. note: Again, can I remind participants that we are supposed to
be discussing the science, not each other -- mjh]