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Lambda-CDM theory - Woo or not?

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Errrr.... we were using the argument that it can be repulsive as well as attractive to demonstrate that your understanding of its physics was deeply lacking.

My explanation of this process allows for this process to be either repulsive or attractive depending on the specific geometry and circumstances whereas your definition does not. I'm afraid it's your understanding that is deeply lacking, not mine.
 
Please explain how yours does so then, and demonstrate that the pressure in your explanation is positive using the dE/dV definition.
 
http://physicsworld.com/cws/article/news/32380

The quantum Casimir effect comes about because a vacuum always contains fluctuating electromagnetic fields. Normally these fluctuations are roughly the same everywhere, but two close conducting surfaces set “boundary conditions” that limit the number of allowed field frequencies between them. Only waves that can fit multiples of half a wavelength between the surfaces resonate, leaving non-resonating frequencies suppressed. The result is that the total field inside a gap between conductors cannot produce enough pressure to match that from outside, so the surfaces are pushed together.

The pressure between the plates is "less than" the "pressure" on the outside of the plates so the plates are *pushed together* because of this "pressure difference". No surface area experiences "negative pressure", just "less pressure" than the other side of the plate.

This explanation is completely consistent with everything I've said, and completely consistent with the explanations and drawings in the WIKI article.

300px-Casimir_plates.svg.png


The small blue arrows represent the "pressure" on the inside surfaces of the plates, whereas the larger blue arrows represent the *greater pressure* on the outside of the plates that pushes them together. As the author puts it:

The result is that the total field inside a gap between conductors cannot produce enough pressure to match that from outside, so the surfaces are pushed together.
 
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So in order for there to be a repulsive force there must be more pressure between the plates in some geometries than is outside - how does this happen, when the conductive plates can only restrict the number of possible modes?
 
Are you referring to the dependence on boundary conditions I specifically mentioned or did you miss that as well?

You're still ignoring the key issue. There is absolutely no "negative pressure" involved in this process. The experiments always have "pressure" in them, both at the atomic level (due to atoms) and QM level. There is no negative pressure involved, just a "pressure difference" between the two regions. The "pressure" inside the plates is "less than" the pressure outside of the plates. You've confused the idea of "lower pressure" with "negative pressure".
 
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You're still ignoring the key issue. There is absolutely no "negative pressure" involved in this process. The experiments always have "pressure" in them, both at the atomic level (due to atoms) and QM level. There is no negative pressure involved, just a "pressure difference" between the two regions. The "pressure" inside the plates is "less than" the pressure outside of the plates. You've confused the idea of "lower pressure" with "negative pressure".

You still continue to ignore the significance of zero point energy and as a result confuse that energy as resulting in some kind of “pressure”. Lower then zero is by definition negative, in energy or pressure.

Please quantify your always positive QM “pressure”, as I have said before it would be a revolution in energy production as well as do away with conservation of energy and Newton’s third law.
 
Hi Micheal Mozina: Have you read up on pressure in a physics textbook yet?

In that case there are these outstanding questions.
Outstanding questions for MM from me:
First asked 26 March 2009
I know that these are absurdly simple questions that a high school student could answer but that should just make it easier for you :biggrin:.



Here is a simpler situation: Consider these 2 scenarios
  1. A force F pushes on a surface that has an area of A.
  2. A force F pulls on a surface that has an area of A.
What is the pressure in these 2 scenarios?
If you do not know what pressure is or cannot use the standard defeinition of pressure that then have a guess at:
Is the pressure positive or negative in each of the 2 scenerios?
First asked 1 April 2009
Now prove your assertion (that the Casimir effect is air pressure) by showing the the air pressure between 2 parallel plates exerts a pressure that varies as the fourth power of the distance betwen the plates (as shown experimentally).
For a genius like you this should be simple. But given your track record with questions I will timestamp this question.

I will add the outstanding questions that other posters have asked:



Asked 2 April 2009 (and many times before)
  • What is your definition of pressure (citations please) that only allows pressure to be positive?
  • What is the error in the derivation of the pressure of the Casimir effect that leads to the pressure being negative?
  • And a bonus question: Why do scientists actually measure a negative pressure?
    (since you have problems with infinity as in your recent posts, replace it by 101010 pascals!)
    Originally Posted by sol invictus
    Here's a paper measuring Casimir pressure. Note the sign of the result (see e.g. Figure 1), and the functional form (which shows that the pressure tends to minus infinity as the cavity shrinks - which in MM's world means the pressure outside must be +infinity).
 
You still continue to ignore the significance of zero point energy

There is no "zero point" in the first place. The universe is simply full of quantum energy, even a vacuum without atoms.

and as a result confuse that energy as resulting in some kind of “pressure”.

You are simply confusing "lower pressure" with "negative pressure". There is "positive pressure" on both sides of both plates, just more of that pressure on the outside than on the inside. There is no area that experiences "negative pressure", just "lower pressure" and "higher pressure".

Lower then zero is by definition negative, in energy or pressure.

It is not "zero" anywhere in the chamber, ever. It can't be. There is energy flowing through the chamber at all times.

Please quantify your always positive QM “pressure”, as I have said before it would be a revolution in energy production as well as do away with conservation of energy and Newton’s third law.

Huh? It does not do away with any sort of conservation of energy laws or any other laws of nature. What are talking about? QM simply describes the energy *already in the system*.
 
There is no "zero point" in the first place. The universe is simply full of quantum energy, even a vacuum without atoms.

Well, then it should be a simple matter for you to harness or at least measure your omnipresent “quantum energy”, please let us know when you get that figured out.


You are simply confusing "lower pressure" with "negative pressure". There is "positive pressure" on both sides of both plates, just more of that pressure on the outside than on the inside. There is no area that experiences "negative pressure", just "lower pressure" and "higher pressure".

As I said a "lower pressure" then zero is by definition a "negative pressure".


It is not "zero" anywhere in the chamber, ever. It can't be. There is energy flowing through the chamber at all times.

Again please let us know when you can measure or tap into this "pressure" or “energy flowing through the chamber at all times” of yours.


Huh? It does not do away with any sort of conservation of energy laws or any other laws of nature. What are talking about? QM simply describes the energy *already in the system*.

Just what energy would that be? Certainly not the zero point energy or vacuum energy, as you proclaim “It can't be”. So you will have to explain your own simple QM description of “the energy *already in the system*”.
 
You're still ignoring the key issue. There is absolutely no "negative pressure" involved in this process. The experiments always have "pressure" in them, both at the atomic level (due to atoms) and QM level. There is no negative pressure involved, just a "pressure difference" between the two regions. The "pressure" inside the plates is "less than" the pressure outside of the plates. You've confused the idea of "lower pressure" with "negative pressure".

I don't think it's particularly fair to end a line of argument like that and takes us several steps back without dealing with the criticisms levelled.
 
You mean the ones that say PV=nRT?

The physics textbook I am teaching out of shows pV = nRT on page 621. Let's look at that more closely, shall we?

"... the ideal gas equation pV = nRT which is based on experimental studies of gas behavior". Hmm, that doesn't say that that is the definition of pressure, it says that's the particular form of the pressure you get when your force arises from the kinetic behavior of an ideal gas. Is that all it says? No, on page 648 we find dW = p dV as a fundamental aspect of the first law of thermodynamics, and a note on page 647 that dW = -dE.

Let's try another textbook: Kittel and Kroemer, "Thermal Physics" (Damn, I borrowed this from my officemate years ago and should give it back ...), usually used for graduate students. Page 66:

"p_s = -dE_s/dV is the pressure on a system in state s"

Hmm, MM, that doesn't say PV = nRT. Try Reif, page 153:

"furthermore, the work done on the system when its volume is changed by an amount dV in the process is simply given by dW = p dV. Hence one obtains the fundamental thermodynamic relation T dS = dE + p dV"

Fundamental, eh? Huang, page 23:

"In an infinitesimal reversible transformation it is easily verified that dA = -P dV - S dT. From this follow the relations P = -dA/dV ... "

Remember that the Casimir Effect was discovered by, measured by, and studied by the same sort of people who write these textbooks. For example, the (very complex) predictions for the Casimir force between nonflat plates were first computed by Mehran Kardar, who recently published two statistical mechanics textbooks, one for particles and one for fields. (And by Bob Jaffe, one of the founding fathers of modern nuclear theory.) And before you announce that theorists live in some castle-in-the-sky and you're talking about reality, let me tell you that they embarked on this calculation because nanomechanical engineers wanted better computation tools for the Casimir forces on their little nano-gears and cantilevers. And here you are, MM, claiming that the Casimir Effect is misexplained because its proponents don't understand pressure as well as you do? Because you've thought really hard about the arrows in a Wikipedia diagram and you read PV=nRT in an intro physics textbook somewhere? Really, MM?
 
Well, then it should be a simple matter for you to harness or at least measure your omnipresent “quantum energy”, please let us know when you get that figured out.




As I said a "lower pressure" then zero is by definition a "negative pressure".




Again please let us know when you can measure or tap into this "pressure" or “energy flowing through the chamber at all times” of yours.




Just what energy would that be? Certainly not the zero point energy or vacuum energy, as you proclaim “It can't be”. So you will have to explain your own simple QM description of “the energy *already in the system*”.
(bold added)

But if you missed the quantitative revolution - as MM seems to have done - then it is (very likely) hard to downright impossible to see that "there is no negative pressure" (in the Casimir experiment) leads inexorably to "there is a vast store of essentially free energy waiting to be tapped" or, perhaps, "energy is not conserved".
 
You mean the ones that say PV=nRT?

Yes, those ones. Now, try actually consulting one of those textbooks. Look in the index for the word "pressure". Find the first instance of a reference to pressure. You will likely find that they define the term when they first use it.

Now, is that definition the ideal gas law? Not in any textbook I've seen.

We've been over this before, Michael. I've given you multiple textbook definitions of pressure. NONE of them rely on the ideal gas law to define pressure. It would be amazingly stupid to do so, since it's an equation of state whose entire usefulness comes from quantifying the relationship between previously defined quantities. Which is why nobody BUT you ever tries to do so. And why nobody believes that you've actually bothered to read about pressure in a physics textbook.
 
(bold added)

But if you missed the quantitative revolution - as MM seems to have done -

Evidently the "quantitative revolution" of elves and fairies is the only thing I "missed". It's fine when to quantify something when you can demonstrate it actually exists outside of your head. Since you folks can't do that, I'm afraid your "quantitative revolution" is a "religious revelation", not a scientific revolution.

then it is (very likely) hard to downright impossible to see that "there is no negative pressure" (in the Casimir experiment) leads inexorably to "there is a vast store of essentially free energy waiting to be tapped" or, perhaps, "energy is not conserved".

I have never once suggested that energy conservation laws are being violated. That is another perfect example of you "spinning" my words and building your own strawman. Typical creationist tactics.

The fact a "vacuum" has energy in it is no revelation to anyone except evidently to you folks.
 
Well, then it should be a simple matter for you to harness or at least measure your omnipresent “quantum energy”, please let us know when you get that figured out.

They already know it's there. Harnessing it is something else altogether.

As I said a "lower pressure" then zero is by definition a "negative pressure".

It's not "lower than zero", it "lower than the outside (of the plates) pressure, and *greater than zero*.

Hell, even when I provide you folks to links that agree with my explanation, diagrams that agree with my explanation and the whole nine yards, you all sit around in pure denial of simple fact. The very "best" vacuum on Earth contains atoms in it. The very best vacuum on Earth therefore has a "pressure" that is "greater than" zero. The best vacuums on Earth contain all sort of neutrinos and flowing energy in them. There is never any area of any vacuum that contains "zero" pressure, so there is no way you will ever get "less than" zero pressure inside a chamber that has pressure in it. This whole discussion amounts to pure denial on your part since the whole thing takes place in a positive pressure environment.
 
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