3 ms·

Casimir Effect – What causes this force?

peter_d_sherman 4y ago
I've always wondered exactly how electronic Capacitors work... When I read various "how does a capacitor work" articles, I always see such words and terms as electrons, electric charge, electric field, and charged particles, i.e. From https://www.birmingham.ac.uk/teachers/study-resources/stem/physics/capacitors.aspx https://www.birmingham.ac.uk/teachers/study-resources/stem/p... -- my first Google search result: >"a capacitor is a circuit component that temporarily stores electrical energy through distributing charged particles on (generally two) plates to create a potential difference" Well, that understanding is not wrong -- but is it the most detailed, most nuanced one? You see, we have two very similar patterns here: We know, on the one hand, that electric capacitance is a phenomena that occurs between two metal plates in an electric circuit -- and that this capacitance increases not linearly, but exponentially -- as the space between the plates decreases linearly... We know, on the other hand, that two metal plates (according to the Casimir Effect) -- will block increasingly lower frequency waves -- as the space between the plates decreases... So, are these two phenomena -- related? ? Or are they unrelated? ? Could Science conclusively prove -- or disprove (an irrefutable negative proof) -- a relationship between these two phenomena occuring relative to the space between two plates of metal? ? You see, I don't know the answer to that... But I submit that both phenomena occur between plates of metal relative to the space between them... this we know... So, are these two phenomena related, or are they unrelated? We as of yet don't know...
- gus_massa 4y ago
  > and that this capacitance increases not linearly, but exponentially The formula is C = εA/d, where d is the distance between the plates. https://en.wikipedia.org/wiki/Capacitor#Parallel-plate_capacitor https://en.wikipedia.org/wiki/Capacitor#Parallel-plate_capac...
  - peter_d_sherman 4y ago
    >and that this capacitance increases not linearly, but exponentially Well, perhaps I have the "but exponentially" part wrong ("I stand corrected!") -- but it certainly would seem NOT to be linear... Let us suppose that we set εA to 16, that is, εA = 16. And let us start with a distance of 16 (d = 16), and compute C for multiple values as we successively halve the distance... C[1] = εA = 16, d=16 = 16/16 = 1 C[2] = εA = 16, d=8 = 16/8 = 2 (difference between previous value = 1) C[3] = εA = 16, d=4 = 16/4 = 4 (difference between previous value = 2) C[4] = εA = 16, d=2 = 16/2 = 8 (difference between previous value = 4) C[5] = εA = 16, d=1 = 16/1 = 16 (difference between previous value = 8) As we can clearly see -- this is definitely not a linear relationship... If it were, the difference would be constant per unit of change of distance, and it is not. But, let's be sure by trying single step difference of 1 for the distance: εA = 16, d=16 = 16/16 = 1 εA = 16, d=15 = 16/15 = 1.06666666667 εA = 16, d=14 = 16/14 = 1.14285714286 Nope! Not linear! So perhaps "exponentially" is the wrong term -- but halving the distance between plates gets more capacitance, and halving it again gets even more! But it is definitely not a linear relationship... (Also, in theory, if the distance between plates in a capacitor could be made infinitely small (assuming that dieletric breakdown wouldn't be a problem -- we are dealing with a purely theoretical capacitor here!) -- then in theory, the Capacitance of such a device -- would be infinite... That's simply because if an equation of the form 1/x governs a relationship, then as x approaches 0, 1/x approaches infinity...) Anyway, I thank you for your correction!
    - gus_massa 4y ago
      Nitpicking: The correct name is https://en.wikipedia.org/wiki/Proportionality_(mathematics)#Inverse_proportionality https://en.wikipedia.org/wiki/Proportionality_(mathematics)#... --- TL;DR: Back to your original question, I guess the answer is YES, but please read the rest of my comment, because it's a surprising YES. --- Back to your original question, it's a weird question. I need like 24 hours to understand it and a few hour to think a good answer. If you try it IRL, give the other person a link to the video, my comment, and 24 or 48 hours to think about it. This is not my area, so I hope my answer is correct. I'm 90% sure it's correct, but for the record I'll only claim that I'm only 70% sure it is correct. Take all the rest with a small grain of salt. --- As I said before, C = εA/d. The force in each plate if F=Q^2/εA=εAV^2/d^2 [verify]. So if you have charged capacitor and the voltage V is constant, the force goes to infinity if the distance d goes to 0. So a very small d creates a very strong capacitor with a lot of force between the plates when they are charged. If the capacitos has no charge, then F=εA0^2/d^2=0 and there is no force. Anyway, in a real capacitor, there is thermal noise. The noise is different in each plate, so you get F=εA(V+noise_1)*(V+noise_2)/d^2 where noise_1 in the noise in one plate and noise_2 in the noise in the other plate. So even if V=0 you get F=εA\noise_1*noise_2/d^2. In the easy model, both noises are not correlated, so the average is 0. My guess is that there is a correlation and when one plate is slightly positive the other is most of the times slightly negative, and you get a average force that is not 0. This is tiny when the plates are far away, but I guess the correlation is greater when the plates are very close. Also, when d is very small, makes everything else much bigger. Note that this is not the Casimir effect. This is thermal noise, that is a classic effect cause by the movements of the parts of the capacitor and the molecules around it. This can be reduced if you put the capacitor inside something that is very cold like liquid air (or liquid helium if you have more money and need less noise). --- The Casimir effect is a quantum effect. You can't avoid it even putting the capacitor at very low temperatures. A simple way to imagine quantum mechanics is that it add uncertainty everywhere, so you again get V+"noise", but it's not the usual thermal noise, but some quantum noise. The problem with this is that you can imagine that you can use a tiny multimeter to measure this. The real thing is not that the V changes from time to time. The real effect is some weird algebraic properties and superposition. You can imagine that is like the classic effect where the V actually changes an get a good intuition in some experiments, but in others experiments you get the wrong result. Quantum Mechanics is actually more weird. So now F=εA*"noise_1"*"noise_2"/d^2, but it's not real noise, just a bad approximation to the quantum fluctuations. Again, the values of both "noises" are correlated, and you get an average that is not 0. The amount of correlations is bigger when they are close and 0 far away. Also the d^2 in the denominator makes this force almos 0 when they are far away, and big when they are very close. I guess this is what they call in the video "Van der Waals" forces. I've seen the Van der Waals forces for molecules, but I guess they can be extended for some (almost?) ideal capacitor plates. --- Again, this is not my area, so I hope my answer is correct. Take all the rest with a small grain of salt. I'm using regularly quantum mechanics, but for smaller system like molecules. A capacitor is too big for me.

I've always wondered exactly how electronic Capacitors work... When I read various "how does a capacitor work" articles, I always see such words and terms as electrons, electric charge, electric field, and charged particles, i.e. From https://www.birmingham.ac.uk/teachers/study-resources/stem/physics/capacitors.aspx https://www.birmingham.ac.uk/teachers/study-resources/stem/p... -- my first Google search result: >"a capacitor is a circuit component that temporarily stores electrical energy through distributing charged particles on (generally two) plates to create a potential difference" Well, that understanding is not wrong -- but is it the most detailed, most nuanced one? You see, we have two very similar patterns here: We know, on the one hand, that electric capacitance is a phenomena that occurs between two metal plates in an electric circuit -- and that this capacitance increases not linearly, but exponentially -- as the space between the plates decreases linearly... We know, on the other hand, that two metal plates (according to the Casimir Effect) -- will block increasingly lower frequency waves -- as the space between the plates decreases... So, are these two phenomena -- related? ? Or are they unrelated? ? Could Science conclusively prove -- or disprove (an irrefutable negative proof) -- a relationship between these two phenomena occuring relative to the space between two plates of metal? ? You see, I don't know the answer to that... But I submit that both phenomena occur between plates of metal relative to the space between them... this we know... So, are these two phenomena related, or are they unrelated? We as of yet don't know...

gus_massa 4y ago
> and that this capacitance increases not linearly, but exponentially The formula is C = εA/d, where d is the distance between the plates. https://en.wikipedia.org/wiki/Capacitor#Parallel-plate_capacitor https://en.wikipedia.org/wiki/Capacitor#Parallel-plate_capac...
- peter_d_sherman 4y ago
  >and that this capacitance increases not linearly, but exponentially Well, perhaps I have the "but exponentially" part wrong ("I stand corrected!") -- but it certainly would seem NOT to be linear... Let us suppose that we set εA to 16, that is, εA = 16. And let us start with a distance of 16 (d = 16), and compute C for multiple values as we successively halve the distance... C[1] = εA = 16, d=16 = 16/16 = 1 C[2] = εA = 16, d=8 = 16/8 = 2 (difference between previous value = 1) C[3] = εA = 16, d=4 = 16/4 = 4 (difference between previous value = 2) C[4] = εA = 16, d=2 = 16/2 = 8 (difference between previous value = 4) C[5] = εA = 16, d=1 = 16/1 = 16 (difference between previous value = 8) As we can clearly see -- this is definitely not a linear relationship... If it were, the difference would be constant per unit of change of distance, and it is not. But, let's be sure by trying single step difference of 1 for the distance: εA = 16, d=16 = 16/16 = 1 εA = 16, d=15 = 16/15 = 1.06666666667 εA = 16, d=14 = 16/14 = 1.14285714286 Nope! Not linear! So perhaps "exponentially" is the wrong term -- but halving the distance between plates gets more capacitance, and halving it again gets even more! But it is definitely not a linear relationship... (Also, in theory, if the distance between plates in a capacitor could be made infinitely small (assuming that dieletric breakdown wouldn't be a problem -- we are dealing with a purely theoretical capacitor here!) -- then in theory, the Capacitance of such a device -- would be infinite... That's simply because if an equation of the form 1/x governs a relationship, then as x approaches 0, 1/x approaches infinity...) Anyway, I thank you for your correction!
  - gus_massa 4y ago
    Nitpicking: The correct name is https://en.wikipedia.org/wiki/Proportionality_(mathematics)#Inverse_proportionality https://en.wikipedia.org/wiki/Proportionality_(mathematics)#... --- TL;DR: Back to your original question, I guess the answer is YES, but please read the rest of my comment, because it's a surprising YES. --- Back to your original question, it's a weird question. I need like 24 hours to understand it and a few hour to think a good answer. If you try it IRL, give the other person a link to the video, my comment, and 24 or 48 hours to think about it. This is not my area, so I hope my answer is correct. I'm 90% sure it's correct, but for the record I'll only claim that I'm only 70% sure it is correct. Take all the rest with a small grain of salt. --- As I said before, C = εA/d. The force in each plate if F=Q^2/εA=εAV^2/d^2 [verify]. So if you have charged capacitor and the voltage V is constant, the force goes to infinity if the distance d goes to 0. So a very small d creates a very strong capacitor with a lot of force between the plates when they are charged. If the capacitos has no charge, then F=εA0^2/d^2=0 and there is no force. Anyway, in a real capacitor, there is thermal noise. The noise is different in each plate, so you get F=εA(V+noise_1)*(V+noise_2)/d^2 where noise_1 in the noise in one plate and noise_2 in the noise in the other plate. So even if V=0 you get F=εA\noise_1*noise_2/d^2. In the easy model, both noises are not correlated, so the average is 0. My guess is that there is a correlation and when one plate is slightly positive the other is most of the times slightly negative, and you get a average force that is not 0. This is tiny when the plates are far away, but I guess the correlation is greater when the plates are very close. Also, when d is very small, makes everything else much bigger. Note that this is not the Casimir effect. This is thermal noise, that is a classic effect cause by the movements of the parts of the capacitor and the molecules around it. This can be reduced if you put the capacitor inside something that is very cold like liquid air (or liquid helium if you have more money and need less noise). --- The Casimir effect is a quantum effect. You can't avoid it even putting the capacitor at very low temperatures. A simple way to imagine quantum mechanics is that it add uncertainty everywhere, so you again get V+"noise", but it's not the usual thermal noise, but some quantum noise. The problem with this is that you can imagine that you can use a tiny multimeter to measure this. The real thing is not that the V changes from time to time. The real effect is some weird algebraic properties and superposition. You can imagine that is like the classic effect where the V actually changes an get a good intuition in some experiments, but in others experiments you get the wrong result. Quantum Mechanics is actually more weird. So now F=εA*"noise_1"*"noise_2"/d^2, but it's not real noise, just a bad approximation to the quantum fluctuations. Again, the values of both "noises" are correlated, and you get an average that is not 0. The amount of correlations is bigger when they are close and 0 far away. Also the d^2 in the denominator makes this force almos 0 when they are far away, and big when they are very close. I guess this is what they call in the video "Van der Waals" forces. I've seen the Van der Waals forces for molecules, but I guess they can be extended for some (almost?) ideal capacitor plates. --- Again, this is not my area, so I hope my answer is correct. Take all the rest with a small grain of salt. I'm using regularly quantum mechanics, but for smaller system like molecules. A capacitor is too big for me.