Casimir effect and dimensional analysis

Click For Summary

Discussion Overview

The discussion revolves around the use of dimensional analysis to estimate the force between two plates in the context of the Casimir effect. Participants explore the relationship between the force, the area of the plates, and their separation, while debating the implications of their dimensional analysis approaches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a formula for the force as a function of area and separation, suggesting that the force scales as ##F \sim \frac{A}{L}##.
  • Another participant questions the validity of this formula, pointing out that it implies the force is independent of the distance between the plates.
  • There is a discussion about the role of dimensionful constants like ##\hbar## and ##c## in the dimensional analysis, with some participants arguing that these constants should be included in the analysis.
  • One participant expresses confusion about whether the choice of including ##\hbar## and ##c## in the dimensional analysis is arbitrary or justified based on the physics involved.
  • Another participant suggests that dimensional analysis is not an exact science and that educated guesses based on known physics are necessary.
  • There is a debate about the necessity of having three equations corresponding to the three fundamental dimensions (mass, length, time) to solve for the parameters in the analysis.
  • Participants discuss the implications of their dimensional analysis results and the need to revisit assumptions if the results are inconsistent.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the proposed formulas and the role of dimensional analysis. There is no consensus on the correct approach or the implications of the analysis, indicating ongoing disagreement and exploration of the topic.

Contextual Notes

Participants highlight that the dimensional analysis may neglect certain physical quantities and constants that could influence the results. The discussion reflects the complexity of applying dimensional analysis in the context of quantum-field-theoretic effects.

spaghetti3451
Messages
1,311
Reaction score
31
How can you use dimensional analysis to estimate the force between two plates, as a function of the area of the plates and their separation?
 
Physics news on Phys.org
This is my hunch:

Estimating the mass of the plates to be ##0.1\ \text{kg}## and the time taken to move ##1\ \text{m}## be ##100\ \text{years} \sim 10^{9}\ \text{s}##, the force is

##F=\frac{MA}{T^{2}L}=10^{-8}\frac{A}{L}##

in SI units.

What are your thoughts?
 
First, is this homework?

Second, dimensional analysis does not mean "make up some numbers and see what happens". It means to ask yourself questions like "if I double the distance between the plates, what happens to the force?". What you wrote down, for example, says the force is independent of distance between the plates. Does that seem reasonable to you?
 
Vanadium 50 said:
First, is this homework?

No, this is not homework.

Vanadium 50 said:
Second, dimensional analysis does not mean "make up some numbers and see what happens". It means to ask yourself questions like "if I double the distance between the plates, what happens to the force?". What you wrote down, for example, says the force is independent of distance between the plates. Does that seem reasonable to you?

Well, the force is independent of the separation of the plates in my formula. In fact, I have ##F\sim\frac{A}{L}##, where ##L## is the separation between the plates.
 
failexam said:
Well, the force is independent of the separation of the plates in my formula.

Which is why your formula is wrong.
 
Oh no, I meant 'dependent'. The formula has ##L##, the separation between the plates.
 
But the force between the plates does not go as 1/L. You're writing nonsense.
 
I'm using the following dimensional analysis:

##\displaystyle{[F]=\frac{kg\ m}{s^{2}}}##,

so that the force ##F## should scale as the length of some quantitiy in the problem. Now there are two parameters ##A## and ##L## in the problem. ##A## is the area between the plates and ##L## is the separation between the plates, so I found it reasonable to say that ##\displaystyle{F \sim \frac{A}{L}}##.

Where's my mistake?
 
The problem is that there may be dimensionful constants, such as ##\hbar## and ##c## involved. You realize that your analysis neglects the presence of the ##\mathrm{kg\, s}^{-2}## terms in the dimensions of the force - you can't just extract the length dimension and make a claim based on that alone (because other physical quantities, apart from the constants, can also be 'coupled' to the length dimension).
 
  • #10
Fightfish said:
The problem is that there may be dimensionful constants, such as ##\hbar## and ##c## involved. You realize that your analysis neglects the presence of the ##\mathrm{kg\, s}^{-2}## terms in the dimensions of the force - you can't just extract the length dimension and make a claim based on that alone (because other physical quantities, apart from the constants, can also be 'coupled' to the length dimension).

But the Casimir force cannot only depend on ##\hbar## and ##c##. I tried dimensional analysis using

##F=\hbar^{\alpha}c^{\beta}##

and I get three equations for the two variables ##\alpha## and ##\beta## by using dimensional analysis. The equations are inconsistent with each other.

Do you suppose that Newton's gravitational constant ##G## may also be involved?
 
  • #11
failexam said:
But the Casimir force cannot only depend on ##\hbar## and ##c##.
Of course it doesn't; it also depends on the separation of the plates and the area of the plates.
failexam said:
Do you suppose that Newton's gravitational constant ##G## may also be involved?
As a first guess (which turns out to be correct), we wouldn't naturally include it. At the end of the day, dimensional analysis is not an exact science - and we can only guess based on the supposed physics involved in the situation. In this case, the Casimir effect is thought to be due to the zero-point energy / (electromagnetic) vacuum fluctuations, and so it seems natural to include ##\hbar## and ##c##.
 
  • #12
Fightfish said:
Of course it doesn't; it also depends on the separation of the plates and the area of the plates.

Ah! I see!

But dimensional analysis also suggests that

##\displaystyle{F = \frac{kg m}{s^{2}}}##.

Are you saying that I should encode ##\displaystyle{\frac{kg}{s^{2}}}## into factors of ##\hbar## and ##c##?
 
  • #13
No, there is no reason to separate out that particular part of the dimension of the force. (Besides if you tried that, you'll find that it's impossible) Just use the dimensional analysis relation you did earlier but including all four quantities this time: ##[F] = [\hbar]^{\alpha} [c]^{\beta} [A]^{\gamma} [L]^{\delta}##. We intuitively expect the force to scale linearly with the surface area of the plate, so we can further take ##\gamma = 1## to simplify the situation.
 
  • #14
Fightfish said:
No, there is no reason to separate out that particular part of the dimension of the force. (Besides if you tried that, you'll find that it's impossible) Just use the dimensional analysis relation you did earlier but including all four quantities this time: ##[F] = [\hbar]^{\alpha} [c]^{\beta} [A]^{\gamma} [L]^{\delta}##. We intuitively expect the force to scale linearly with the surface area of the plate, so we can further take ##\gamma = 1## to simplify the situation.

But isn't this choice of equation ##[F] = [\hbar]^{\alpha} [c]^{\beta} [A]^{\gamma} [L]^{\delta}## completely arbitrary?

After all, F depends on mass, length and time. ##\hbar## and ##c## might often appear in some physical quantities due to the combinations ##Js## and ##ms^{-2}## which are commonly found in physical quantities in quantum-field-theoretic systems.

So, to me, ##[F] = [\hbar]^{\alpha} [c]^{\beta} [A]^{\gamma} [L]^{\delta}## looks like an arbitrary choice of equation to solve.
 
Last edited:
  • #15
Well it is arbitrary to some extent - but not completely arbitrary. Its just like an ansatz, an educated guess that we make based on what we know (or think we know) about the underlying physics of the phenomenon. There's certainly no guarantee that dimensional analysis will end up giving us right results especially if we chose the wrong starting quantities to consider.
 
  • #16
But then, are you by any chance using the fact that there are three units - mass, length, time - in force and so we need three equations for the separate units of mass, length and time. Now, this is a quantum-field-theoretic effect so that ##\hbar## and ##c## are natural choices to consider and then ##L## is something that we just intuitively expect the force to depend on. Is that it?
 
  • #17
failexam said:
But then, are you by any chance using the fact that there are three units - mass, length, time - in force and so we need three equations for the separate units of mass, length and time.
I wouldn't think this is a particularly important consideration - after all, certain quantities carry several dimensions, so we don't necessarily need three equations.
failexam said:
Now, this is a quantum-field-theoretic effect so that ##\hbar## and ##c## are natural choices to consider and then ##L## is something that we just intuitively expect the force to depend on.
Mainly this I would say. As with all guesswork formulations, if somehow we cannot get a reasonable solution for the powers, then its back to the drawing board to re-examine what other quantities we might have missed out and then include them in and repeat the analysis.
 
  • #18
Fightfish said:
I wouldn't think this is a particularly important consideration - after all, certain quantities carry several dimensions, so we don't necessarily need three equations.

But in this case,

##[F] = [\hbar]^{\alpha} [c]^{\beta} [L]^{\gamma}##

##kg ms^{-2} = (Js)^{\alpha} (ms^{-1})^{\beta} (m)^{\gamma}##

##kg ms^{-2} = (kgm^{2}s^{-1})^{\alpha} (ms^{-1})^{\beta} (m)^{\gamma}##

so that the equation decouples into three separate equations for ##kg##, ##m## and ##s##.

This shows that we need three parameters (in this case, they are ##\hbar##, ##c## and ##L##) since we need three powers to solve for from the three equations.

This is why I say that we need three parameters for this problem (in this case, they are ##\hbar##, ##c## and ##L##).
 
  • #19
What are your thoughts?
 
  • #20
I still don't think it is always necessary for there to be three quantities involved. After all, another way to write the force is in terms of mass x acceleration - that's two quantities. It's also possible that four or more quantities are required. Basically, there is no unique way to decompose one quantity in terms of other quantities, so it still depends on the choices that we make based on the physics involved.
 
  • #21
I see.
 
  • #22
So, I get

##\displaystyle{F \sim \frac{\hbar c}{L^{2}} \sim \frac{\text{197 eV nm}}{L^{2}}}##.

(I used ##\hbar##, ##c## and ##L## for the dimensional analysis.)

Does this look good?
 
  • #23
failexam said:
So, I get

##\displaystyle{F \sim \frac{\hbar c}{L^{2}} \sim \frac{\text{197 eV nm}}{L^{2}}}##.

(I used ##\hbar##, ##c## and ##L## for the dimensional analysis.)

Does this look good?
Nope, you left out the surface area ##A## of the plate, and also the result of the dimensional analysis should be left in terms of quantities, not expressed as numerical values or units.
 
  • #24
Fightfish said:
Nope, you left out the surface area ##A## of the plate, and also the result of the dimensional analysis should be left in terms of quantities, not expressed as numerical values or units.

Ok, so I expect the answer to be

##F \sim \frac{\hbar c}{A}##,

that is, the force is independent of the distance between the plates.

What do you think?
 
  • #25
Erm...I was under the impression that we earlier reasoned that the force should depend on both the surface area of the plate and the separation of the plates, and furthermore that the dependence on the surface area should be linear. So you should get ##[F] = \frac{\hbar c A}{L^4}##

(as you can see, both of your answers are also "valid" guesses via dimensional analysis - which shows how important the choice of quantities to consider is to the accuracy of the result that we get)
 
  • #26
Right, it makes sense that the force ##F## should scale as the area ##A## between the plates since the increasing the area increases the area for vacuum fluctuations of the electromagnetic field to occur.

Also, the force ##F## should scale as some inverse power of the separation ##L## between the plates since increasing the separation increases the distance that the virtual photons have to propagate to interact with each other. This sentence is rather vague but it gives a rough qualitative view of hy the force might scale as the inverse power of the separation between the plates.

What do you think?
 
  • #27
failexam said:
Right, it makes sense that the force ##F## should scale as the area ##A## between the plates since the increasing the area increases the area for vacuum fluctuations of the electromagnetic field to occur.
Yup, and we should expect that this scales linearly if the force is homogeneous across the plate.
failexam said:
Also, the force ##F## should scale as some inverse power of the separation ##L## between the plates since increasing the separation increases the distance that the virtual photons have to propagate to interact with each other.
Well, since you're taking a QFT course (I presume), it's best to be careful when talking about virtual particles haha. But yes, intuitively, the force should fall off with distance - and relatively quickly as well. You realize that this doesn't necessarily have to be a power law; it could very well be an exponential decay, but that will not be possible to obtain via dimensional analysis. In this case though, the proper calculations do indeed produce a result that agrees with that obtained via dimensional analysis, so all is nice and good.
 
  • #28
Fightfish said:
Yup, and we should expect that this scales linearly if the force is homogeneous across the plate.

Well, since you're taking a QFT course (I presume), it's best to be careful when talking about virtual particles haha. But yes, intuitively, the force should fall off with distance - and relatively quickly as well. You realize that this doesn't necessarily have to be a power law; it could very well be an exponential decay, but that will not be possible to obtain via dimensional analysis. In this case though, the proper calculations do indeed produce a result that agrees with that obtained via dimensional analysis, so all is nice and good.

Thanks.

My original question was about using dimensional analysis to ##\textit{estimate}## the force between the plates, as a function of the area ##A## of the plates and their separation ##L##.

Since I now have an expression for the force as a function of the area of the plates and their separation, won't it make sense to evaluate ##\hbar c## to ##\textit{estimate}## the force between the plates?
 
  • #29
Hmm...I'm not sure about that though - because our dimensional analysis approach completely leaves out any numerical factors or dimensionless constants, and these could affect the order of magnitude of the estimate significantly.
 
  • #30
Fightfish said:
Hmm...I'm not sure about that though - because our dimensional analysis approach completely leaves out any numerical factors or dimensionless constants, and these could affect the order of magnitude of the estimate significantly.

Thanks. I get the idea.

I was wondering what boundary conditions the electromagnetic field obeys at anyone of the plates. My hunch is that the field vanishes at the plates because the electromagnetic field on or within a (conducting) plate is zero. What do you think?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K