Is there a symmetry in the Unruh effect?

[Heidi]

Hi Pf
I read that the accelerated Unruh observer is in a thermal bath of particles. For him the mean value of his occupation number N of the inertial vacuum V0 is not null. <V0 N V0> is greater than 0.
He is not in his vacuum V'. Suppose that he travels with a box. he cools it and leaves it empty.
What will see the inertial observer ? an accelerated box of course , but how will he see the vacuum V' in it ?
thanks

 

[mitochan]

AS for the title of OP " Is there a symmetry in the Unruh effect ? " I take it as symmetry of radiation that people of an accelerated box observe. They observe event horizon plane below their feet. Unruh radiation come from it to them straight up say along z axis.　　I would appreciate person who knows well could enlighten us.

 

[Heidi]

I can tell it another way
does <V0 H' V0> greater 0 imply
<V' H V'> greater 0?
Vo and V' are the vacuaa of the inertial and accelerated observers.
H and H' their hamiltonian.

 

[PeterDonis]

Suppose that he travels with a box. he cools it

How does he cool the box?

 

[Heidi]

he puts it in his accelerated fridge of course. ;)

 

[PeterDonis]

he puts it in his accelerated fridge of course. ;)

Joking aside, what happens when he does this? Hint: in order for the temperature of the box to change, some energy has to leave the box and go somewhere else. How does that happen? Where does it go?

 

[Heidi]

i do not say that the accelerated observer suppress the particles everywhere just in a given region. Here on Earth we are in a gravity field and we can get empty cavities .
i asked a question in post 3.
Have you an answer to it?

 

[PeterDonis]

i do not say that the accelerated observer suppress the particles everywhere just in a given region.

I have no idea what "suppress the particles" refers to.

Here on Earth we are in a gravity field and we can get empty cavities .

I have no idea what this means.

 

[PeterDonis]

does <V0 H' V0> greater 0 imply
<V' H V'> greater 0?
Vo and V' are the vacuaa of the inertial and accelerated observers.
H and H' their hamiltonian.

This question is unanswerable because the expressions you are using are not well-defined. The two quantum field theory constructions, "inertial" and "accelerated", are not unitarily equivalent, so there is no way to evaluate the expectation values you describe.

 

[Heidi]

This is exactly the sense of my question!
Are inequivalent representations a great problem if they describe the universe for every observer or can we accept it if they describe what the two different observers measure?

Maybe you think that the Unruh effect does not exist?

edit the two inequalities are not for the same observer

 

[PeterDonis]

This is exactly the sense of my question!

That was not at all evident from any of your previous posts. If you wanted to ask about unitarily inequivalent representations, you could have just asked about unitarily inequivalent representations.

Are inequivalent representations a great problem if they describe the universe for every observer or can we accept it if they describe what the two different observers measure?

That depends on which physicist you ask. Some think it's a problem, some don't.

Maybe you think that the Unruh effect does not exist?

I have never said any such thing.

the two inequalities are not for the same observer

If you mean the two inequivalent representations are not for the same observer, yes, of course not. Furthermore, the two observers must have different proper accelerations (different in magnitude or direction or both) for their representations to be inequivalent. Any two inertial observers, even if they are in relative motion, will have unitarily equivalent representations.

 

[Heidi]

I agree with all that.
what abour the question about the inequalities in post 3?
thanks

 

[PeterDonis]

what abour the question about the inequalities in post 3?

See post #9.

 

[Heidi]

Please read eq 4 in this article
you will find my first inequality (it is well defined).
the question was about <0_{unruh} |a`^\dagger a| 0_{unruh}> the mean value of the occupation number (of the inertial observer ) in the unruh vacuum.

 

[PeterDonis]

Please read eq 4 in this article
you will find my first inequality (it is well defined).

Equation (4) in that article is not either of the inequalities you wrote in post #3.

the question was about <0_{unruh} |a`^\dagger a| 0_{unruh}> the mean value of the occupation number (of the inertial observer ) in the unruh vacuum.

That's not what Equation (4) in the article gives. In that equation, the vacuum state $|0\rangle$ is the Minkowski (inertial) vacuum, and the ladder operators $b_k$, $b^\dagger_k$ are operators in the inertial QFT construction, as shown by Equation (3) where they are written in terms of the "inertial" ladder operators $a_j$, $a^\dagger_j$. Your inequalities mix the constructions (either "inertial" vacuum with "accelerated" Hamiltonian, or accelerated vacuum with inertial Hamiltonian).

Also note the cautionary statement in the article after Equation (4):

Insofar as $b^\dagger_k b_k$ can be regarded as the operator representing the observable "number of particles in mode $k$"

That is an implicit reference to the fact that those operators are not operators in a true "accelerated" QFT construction. A true "accelerated" QFT construction would be unitarily inequivalent to the "inertial" construction. That means it is not possible to write down true "accelerated" operators in terms of "inertial" operators. Such an expression would have to be invertible, and as noted in the second paragraph after Equation (3) in the article, the transformation expressed in Equation (3) is not invertible if you only consider the QFT construction in the quadrant $z > |t|$, which is what a true "accelerated" construction would be. To invert Equation (3), as the article notes, you have to use $b_k$ operators from the quadrant $z < - |t|$ as well.

 

[PeterDonis]

the question was about <0_{unruh} |a`^\dagger a| 0_{unruh}> the mean value of the occupation number (of the inertial observer ) in the unruh vacuum.

I would advise reading the last paragraph of the "Model detector" section of the article you referenced, in particular the italicized part, which shows that this is not the question that the article is answering.

 

[PeterDonis]

I would advise reading the last paragraph of the "Model detector" section of the article you referenced, in particular the italicized part, which shows that this is not the question that the article is answering.

I would also advise thinking very carefully about how you would physically realize the "accelerated vacuum" state you have referred to. You handwaved it by saying "cool the box", but as my question in response to that statement of yours was trying to tell you, it is nowhere near that simple. You need to actually describe a physical process that will cool the box, and describe the effect of that process on the quantum field as well as the box. Your description needs to take into account the fact that the box starts out immersed in a thermal bath of particles because it is accelerated and the quantum field starts out in the inertial vacuum state.

Note also that the very fact that the box is accelerated implies that the answer to the title question of this thread is "no": an observer moving with the box feels nonzero proper acceleration, whereas an inertial observer does not. So there is an obvious asymmetry between them.

 

[Heidi]

thank you for these answers. I will read them carefully.

 

[Vanadium 50]

As I understand the question, it is as follows:

Rocket-man carries a box of vacuum with him as he accelerates. Because he is accelerating, the box contains a thermal bath of particles at temperature T. He cools the box (somehow - perhaps sorcery) so it is now at T2 < T. Does a non-accelerated observer think the box contains vacuum?

That answer is clearly "no". All observers agree that the box's state is not what it was pre-cooling. Because of the uniqueness of the vacuum, the non-accelerated observer must determine the box is not in the vacuum state, therefore <N> is not zero.

 

[PeterDonis]

Rocket-man carries a box of vacuum with him as he accelerates.

Just to be clear, the box is not present in the standard Unruh effect scenario. In that scenario, the vacuum is the space all around the rocket. To an inertial observer, this space contains zero particles; a particle detector carried by the inertial observer has zero probability of detecting any particles. However, to rocket-man, the space all around the rocket does contain particles: a particle detector carried by rocket-man has a nonzero probability of detecting particles, and if you work out the details, it turns out that rocket-man's detector tells him he is immersed in a thermal bath of particles at the Unruh temperature.

The OP appears to be imagining a box that starts out containing empty space before rocket-man accelerates. But none of the OP's posts and none of the references the OP has given say anything at all about any box carried by rocket-man, let alone about what happens to the box as rocket-man accelerates (does it actually contain a thermal bath of particles at the Unruh temperature? the usual Unruh effect calculations don't say that), or how rocket-man is supposed to cool the box (which is why I asked the OP to think carefully about what actual physical process would be involved, since none of the references talk about that).

 

[PeterDonis]

Because of the uniqueness of the vacuum

Actually, one of the points of the standard derivation of the Unruh effect is supposed to be that there is not a "unique" vacuum state. Which state is the vacuum is observer-dependent. In flat spacetime, all inertial observers will agree on which state is the vacuum (even if they are in relative motion), but accelerated observers will not agree.

In fact, because the quantum field theory constructions for inertial and accelerated observers are unitarily inequivalent, it's not even clear that there is any well-defined notion of "state" that is common to both of them. In other words, it's not even clear that there is any well-defined representation at all of the inertial "vacuum" state in the accelerated QFT construction, or vice versa.

 

[Heidi]

I think that we have not to worry about how the box is cooled and how the accelerated observer succeed to get "his" vacuum in the box.
Read Peskin and Schroeder chapter 2 (creation and annihilation). before t = 0 the space time is in the vacuum state. Next there is a non null classical source f(x,t) between t=0 and t=1.
And then after t=1, the inertial observer who saw no particle before t=0 sees particles after t=1.
I think that nobody wrote the authors to get the details about the device that turned f(x) on and off.

 

[PeterDonis]

I think that we have not to worry about how the box is cooled and how the accelerated observer succeed to get "his" vacuum in the box.

I think you are waving your hands instead of doing the actual work.

Read Peskin and Schroeder chapter 2

In that example they did the actual work; they defined, mathematically, what the source was and derived, mathematically, what its effect was. You have done none of that.

I think that nobody wrote the authors to get the details about the device that turned f(x) on and off.

That's because nobody needed to; the physical effects involved were already completely specified. You have done no such thing. Waving your hands and saying "the box cools" or "the vacuum inside the box gets accelerated" is not the same as actually doing the math.

 

[PeterDonis]

The OP question has been answered. Thread closed.