The temperature of acceleration (Unruh radiation)

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marcus

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In 1976 Bill Unruh at British Columbia realized that
acceleration has an intrinsic temperature---an accelerating observer will be bathed in radiation at that temperature.
He published his finding in Physics Reviews Series D vol 14 page 870 and following.

the Unruh temperature has been experimentally investigated at SLAC, and possibly elsewhere

here is a SLAC press release about the work of P. Chen
on Unruh radiation

http://www.slac.stanford.edu/slac/media-info/20000605/chen.html

The formula for the temperature associated with acceleration a is

kT = hbar a/2πc

Here is an example in planck units c=G=hbar=k=1

Suppose the acceleration is 18E-30
then to get the temperature we need to divide by 2π
That gives about 3E-30---a bit hotter than the familiar boiling point of water.

So if one accelerates a teakettle of water at 18 x 10^-30 planck, it will soon boil and one can brew tea with it.

All that is necessary to find the temp is to divide by 2pi.

this partily in response to dav2008, who in another thread mentioned the planck unit of acceleration

Unruh's formula is similar to Stephen Hawkings formula for black hole temperature which depends in the same way on the surface gravity (acceleration at the event horizon).
 

Alexander

How do you accelerate a teakettle at such tremendous acceleration 10E-30?
 

marcus

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Originally posted by Alexander
How do you accelerate a teakettle at such tremendous acceleration 10E-30?
A thoughtful question suggesting that you like tea.
It is a practical problem that would definitely have to be solved
before one could brew tea by this method.
18E-30 is almost exactly E22 gee----more precisely 1.02E22 times the earth sealevel gravity norm.

Do you happen to know what size of black hole would
have a Hawking radiation of the right temperature to
boil water? I suppose it would be extremely small!

Does anyone have a suggestion as to how to solve
the practical problem posed by Alexander about the
acceleration of teapots?
 

marcus

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the gravity feels like home

Just thought of a related question. What size ordinary black hole has a familiar earth-like gravity at its event horizon?


This is easiest to solve in planck c=G=hbar=1 terms. And then, to interpret the answer one should know that E38 planck lengths make a mile. And that one solar mass is about E38 planck mass.

(more exactly the solar mass is 0.93E38 planck but E38 is close enough)

Earth sealevel gravity is around 18E-52 planck----that's all you have to know to work the problem in planck.


There is this useful universal constant force F and the surface gravity of a black hole is F/4m, where m is the mass of the hole.
You know dividing a force by a mass gives an acceleration, so I'm really just telling you there is a 4 in the formula.

So we just have to solve this for m

F/4m = 18E-52

F is the unit force so it is just equal to one

1/m = 72E-52

That means that m = 1.4E50 planck

And its Schw radius is 2.8E50 planck.

So we can say that in miles, radius is 2.8E12 miles
or 2.8 trillion miles

And in "solar" masses of E38 planck the hole is 1.4 trillion solar masses.

It is not huge, just a thousand times larger than the solar system, but it is really massive----like a whole milkyway galaxy.

The nice thing about this is the simplicity of the arithmetic, just had to multiply by 4 and find one over something. The hard thing (which is not all that hard) is needing to know stuff like E38 is a mile and roughly a solar mass.
 

Alexander

So, how to boil a teakettle (by acceleration)?
 

marcus

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Originally posted by Alexander
So, how to boil a teakettle (by acceleration)?
tie a long rope to the handle of the teakettle
and lower it down near the surface of a black hole
which has the right acceleration of gravity at
its event horizon


it will probably have to be a very tiny teakettle
to match the small size of the black hole

or, if one only has a normal teakettle one must
gather together many black holes to make
an array of them the same size as the bottom
of the teakettle, like a stove burner.

I was thinking you might answer the question.
Please you tell me. What mass of black hole
has Hawking temperature to boil water?
That is, around 3E-30 planck temperature (if
you use planck units) or roughly 400 kelvin (if you use
kelvin).
 

Alexander

Originally posted by marcus

Do you happen to know what size of black hole would
have a Hawking radiation of the right temperature to
boil water? I suppose it would be extremely small!

Size? About a micron (2R~1.1x10-6m), or 0.4 Yg in mass (which is about mass of 0.5 Mm asteroid).
 
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marcus

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Originally posted by Alexander
Size? About a micron (2R~1.1x10-6m).
bravo

I got (1/24π)E30 planck mass but did not
work it out in microns, cant check this at the moment

I must go to an appointment so must continue
this very pleasant discussion later

Now I'm back. That result for the mass means the
Schw length (halfradius) is (1/24π)E30 planck length

So in metric terms, Schw radius is 0.43 micron, and diameter (for comparison with what you calculated) is 0.86 micron.
You got diameter about 1.1 micron which is in essential agreement. The difference is probably due to the fact that
I used a temperature slightly hotter than conventional boiling because I wanted a round number input that would get the kettle boiling quickly.

I used a temperature of 3E-30, while conventional boiling in planck terms is 2.63E-30. You probably used conventional boiling temperature 373 kelvin. So there would be a factor of 2.63/3 between the answers having to do with the hotter temperature that I assumed. Yeah, if I use boiling I get a diameter of right close to one micron.
 
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