Unruh Hawking Black hole temperature.

Labarbapostiza

https://en.wikipedia.org/wiki/Unruh_effect

The Unruh temperature, derived by William Unruh in 1976, is the effective temperature experienced by a uniformly accelerating detector in a vacuum field. It is given by[4]

T = [LATEX]\frac{\hbar a}{2\pi c k_\text{B}}[/LATEX],
where a is the local acceleration, [LATEX]k_\text{B}[/LATEX] is the Boltzmann constant, [LATEX]\hbar[/LATEX] is the reduced Planck constant, and c is the speed of light. Thus, for example, a proper acceleration of 2.5 × 1020 m·s−2 corresponds approximately to a temperature of 1 K.

The Unruh temperature has the same form as the Hawking temperature [LATEX]T_\text{H} = \hbar g/(2\pi c k_\text{B})[/LATEX] of a black hole, which was derived (by Stephen Hawking) independently around the same time. It is, therefore, sometimes called the Hawking–Unruh temperature.[5]

This has been bothering me for a while. Specifically Unruh's temperature, not Hawking. (Hawking's temperature bothers me too. But that's another story.)

The formulas are both the same. Unruh imagines that if you accelerate into a 0 zero kelvin space, you'll experience a temperature. Photons. His formula is T = ka (I've compacted all the constants into k).

Hawking's formula and Unruh's look identical. T = ka , T = Kg (where g is gravity).......But, they're not actually identical. For Hawking's Blackhole to have a positive temperature, the equivalent energy must fall into the blackhole. So, Hawking's temperature is in fact T = Kg - Kg, (That's T for Hawking's vacuum temperature).

Unruh doesn't take account for the creation of anti particles.

And looking at Unruh's equation, it appears to me, that and acceleration of force of just a Planck would be met with an infinite force of photons. Lot's and lots.

For existence in this universe to be possible, Uruh's equation must be

T = Ka - Ka = 0, where the minus Ka is the negative energy of the vacuum.

I think Hawking's equation is also wrong. (at least on the Unruh section of Wikipedia)

If Hawking's equation is correct. Then a black hole would evaporate in a ultra blinding flash of light, the instant it reached its' Swarzchild radius.

If Blackhole radiation does it exist, then it has to be a very small factor of energy. There are a number of other reasons the formulas appear to me to be wrong.

Waddya think?

Labarbapostiza

Any opinions?

Labarbapostiza

Labarbapostiza wrote: »

Any opinions?

I've some research and the WIKIPEDIA entry is Wrong.....So, I, and Stephen Hawking's are correct.

Hawking's formula is in fact

[LATEX]\frac{\hbar\kappa }{2\pi k} \approx 10^{-6}\left ( \frac{M_{o}}{M} \right )K^{o}[/LATEX]

Which makes sense, as it's a much smaller amount of energy.

Here's Hawking's original paper. It's less confusing than second or third hand explanations I've had. And of course the equation is correct.

http://digidownload.libero.it/steve.little/Articoli/Hawking-1975.pdf

There's an easier explanation here, by John Baez,

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html



So......Why didn't anyone just point out to me Wikipedia was wrong?

Morbert

Wikipedia isn't wrong.

Morbert

Labarbapostiza wrote: »

https://en.wikipedia.org/wiki/Unruh_effect



This has been bothering me for a while. Specifically Unruh's temperature, not Hawking. (Hawking's temperature bothers me too. But that's another story.)

The formulas are both the same. Unruh imagines that if you accelerate into a 0 zero kelvin space, you'll experience a temperature. Photons. His formula is T = ka (I've compacted all the constants into k).

Hawking's formula and Unruh's look identical. T = ka , T = Kg (where g is gravity).......But, they're not actually identical. For Hawking's Blackhole to have a positive temperature, the equivalent energy must fall into the blackhole. So, Hawking's temperature is in fact T = Kg - Kg, (That's T for Hawking's vacuum temperature).

Unruh doesn't take account for the creation of anti particles.

And looking at Unruh's equation, it appears to me, that and acceleration of force of just a Planck would be met with an infinite force of photons. Lot's and lots.

For existence in this universe to be possible, Uruh's equation must be

T = Ka - Ka = 0, where the minus Ka is the negative energy of the vacuum.

I think Hawking's equation is also wrong. (at least on the Unruh section of Wikipedia)

If Hawking's equation is correct. Then a black hole would evaporate in a ultra blinding flash of light, the instant it reached its' Swarzchild radius.

If Blackhole radiation does it exist, then it has to be a very small factor of energy. There are a number of other reasons the formulas appear to me to be wrong.

Waddya think?

You'll have to be clearer about your reasoning. What for example, is "acceleration of force of just a Planck". The Planck force is approximately 10^44 Newtons. The Unruh temperature, however, is a function of acceleration, not force.

Labarbapostiza

Morbert wrote: »

Wikipedia isn't wrong.

So, are you saying Hawking is wrong. Because the equation used in the equation section of the Wiki page is not the same formula as in his paper.

And reading on, there are paragraphs that simply do not give the same explanation as Hawking's.

I haven't had a chance to read Unruh's paper.

I'm going to have another look at it, but as far I can see the equations in the equation section are wrong.

Morbert

Labarbapostiza wrote: »

So, are you saying Hawking is wrong. Because the equation used in the equation section of the Wiki page is not the same formula as in his paper.

I don't see how it's different. The surface gravity [LATEX]\kappa[/LATEX] is the same as [LATEX]g[/LATEX].

I.e.

[LATEX]\frac{\hbar\kappa}{2\pi k} = \frac{\hbar g}{2\pi k}[/LATEX]

Labarbapostiza

Morbert wrote: »

I don't see how it's different. The surface gravity [LATEX]\kappa[/LATEX] is the same as [LATEX]g[/LATEX].

I.e.

[LATEX]\frac{\hbar\kappa}{2\pi k} = \frac{\hbar g}{2\pi k}[/LATEX]

You're right, k and g are both acceleration. What's wrong, is [LATEX]T = \frac{\hbar acceleration}{2\pi k}[/LATEX] doesn't represent the energy of the vacuum correctly.

I'm a little tired so I won't dig out the equation (at least this minute) that shows more correctly what's in the vacuum, but I'll say something about some of the misunderstandings. (I will dig out the formulas - just my eyes are strained, so not for this post).

The most basic principle of the vacuum is something that is deceptively simple. Does 0 = 0, or does 0 = 1 - 1. Both statements are correct, but if you're thinking about the vacuum you must always think of it in terms of 0 = 1 - 1, otherwise you get very strange results. If 0 = 1 - 1, does 0 = [LATEX]\psi - \psi[/LATEX].....Right it does, but if you get uncertainties 0 isn't quite absolutely zero.

Okay, how this formula [LATEX]T = \frac{\hbar acceleration}{2\pi k}[/LATEX], gives you a response of infinite energy with just a little motion.

If you imagine the electromagnetic field in the vacuum as 0 = 0, and then you accelerate an electron into this vacuum, and due to relativity, a photon is raised out of the zeroth energy level into the first, and higher depending on how much you accelerate the electron.

This is completely wrong.

First, how many photons are in the zeroth energy state before you accelerate. You could say none......but you could fit an infinite quantity of nothing in nothing.

Let's just say a zillion photons are <i>not</i> in the zeroth energy state. Just with the tiniest acceleration into this kind of vacuum, you'd raise zillions of photons into the next energy state. You're not getting the energy of one low energy photon, you're getting zillions. This is obviously not our universe.

Now, a slightly different vacuum, where you always think in terms of 0 = 1 -1. When you accelerate into it, a zillion photons are raised out of the zeroth state, but crucially, a zillion photons are also raised, or you could say lowered into a negative energy state. And viola, they cancel...........(cancel nearly completely - but that's not completely straightforward either.)

So, this formula [LATEX]T = \frac{\hbar g}{2\pi k}[/LATEX] for getting the Hawking temperature is wrong. The temperature from that formula would be infinite.

The whole thing gets more and more interesting, the more you think about it.

Morbert

You said Wikipedia is wrong and that "the equation used in the equation section of the Wiki page is not the same formula as in [Hawking's] paper". But they are the same, as I pointed out in my last post.

Labarbapostiza

Morbert wrote: »

You said Wikipedia is wrong and that "the equation used in the equation section of the Wiki page is not the same formula as in [Hawking's] paper". But they are the same, as I pointed out in my last post.

Yep....kinda.... [LATEX]T = 10^{-6}\left ( \frac{M_{o}}{M} \right )K^{o}[/LATEX] is more what caught my eye as being correct.

[LATEX]T = \frac{\hbar\kappa }{2\pi k} [/LATEX] is not correct

[LATEX]\frac{\hbar\kappa }{2\pi k} \approx 10^{-6}\left ( \frac{M_{o}}{M} \right )K^{o}[/LATEX]...I'm not going to say the left hand of the equation is wrong, it just depends on a few things.

The black hole, and the vacuum are not black bodies. I'm pretty sure Hawking's wasn't confused, but he just wanted to use the word temperature and the Boltzmann constant in the paper. The theoretical vacuum energy, is a quantity, that is expected not to self-annihilate. Maybe it has some unusual properties, or maybe it's not even there, which would be a shame.