# The temperature of acceleration (Unruh radiation)

• marcus
In summary: It was easy enough to do it over.So a mass of half a million metric tons (0.5 Mm) is about right. That is what I assumed in the previous post, saying a thousand solar masses. It is 550 solar masses. A 550 solar mass hole would be 0.43 micron diameter and would have Hawking radiation at a temperature of boiling water.In summary, the Unruh temperature was first discovered in 1976 by Bill Unruh at British Columbia, who realized that an accelerating observer would experience a temperature due to acceleration. This concept has been experimentally investigated at SLAC and other places. The formula for the temperature associated with acceleration is kT = hbar a/2&pi
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

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

The formula for the temperature associated with acceleration a is

kT = hbar a/2&pi;c

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 2&pi;
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 Hawking formula for black hole temperature which depends in the same way on the surface gravity (acceleration at the event horizon).

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

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?

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.

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

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).

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).

Last edited by a moderator:
Originally posted by Alexander

bravo

I got (1/24&pi;)E30 Planck mass but did not
work it out in microns, can't 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 (1/24&pi;)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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## 1. What is the temperature of acceleration?

The temperature of acceleration, also known as Unruh radiation, is a theoretical phenomenon in physics that describes the temperature experienced by an accelerating observer in empty space. It is based on the idea that a constant acceleration in a vacuum will create a background of thermal radiation, similar to the radiation emitted by a hot object.

## 2. How is the temperature of acceleration measured?

The temperature of acceleration is a theoretical concept and cannot be directly measured. However, it can be calculated using the Unruh effect, which is a mathematical formula that relates the acceleration of an observer to the temperature they experience. This calculation is based on the principles of quantum field theory.

## 3. What factors affect the temperature of acceleration?

The temperature of acceleration is affected by the acceleration of the observer, the duration of the acceleration, and the type of particle being observed. Different types of particles will experience different temperatures of acceleration, with lighter particles experiencing higher temperatures.

## 4. Is the temperature of acceleration a constant value?

No, the temperature of acceleration is not a constant value. It is dependent on the acceleration of the observer, and as the acceleration changes, so does the temperature. This means that an observer who is constantly accelerating will experience a range of different temperatures over time.

## 5. What are the implications of the temperature of acceleration?

The temperature of acceleration has important implications for our understanding of gravity and the nature of space-time. It also has potential applications in fields such as cosmology and quantum computing. However, since it is a theoretical concept, further research and experimentation are needed to fully understand its implications.

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