In light of new evidence, what might be behind the Casimir Effect?

[Demystifier]

The presence of the conductors at the boundary of the cavity restricts the quantum field modes that are available inside the cavity, as compared to a free vacuum with no restriction on available modes. The fewer modes that are available, the lower the ground state energy of the vacuum.

This is the standard narrative. The problem with it is that it perpetuates the misleading picture that Casimir effect originates from vacuum energy.

 

[peterbb]

This is the standard narrative. The problem with it is that it perpetuates the misleading picture that Casimir effect originates from vacuum energy.

What is your take on the mechanism of the Casimir effect?

 

[Demystifier]

What is your take on the mechanism of the Casimir effect?

https://arxiv.org/abs/1702.03291
http://thphys.irb.hr/wiki/main/images/2/2c/Casimir.pdf

 

[peterbb]

https://arxiv.org/abs/1702.03291
http://thphys.irb.hr/wiki/main/images/2/2c/Casimir.pdf

I read the abstract of the first and the whole pdf of the second. My mathematics and quantum physics are not advanced enough to be able to understand the proofs. But I am familiar with van der Waals. I can see how intuitively van der Waals can explain the effect. What are your thoughts given the findings of the paper cited in this thread (and, for the record, there is no light shining on it, otherwise it would not be an anomaly worth publishing - it is producing its own energy internally through some mechanism according to the IV curve).

 

[peterbb]

That's not what the specific passages I quoted say. They say there is an external light source.

I can see how one might come to the conclusion by taking a sentence out of context. If you read the first two paragraphs it places things in context. There is no external light source. Again, by ruling out all other possibilities, it is reasonable to conclude that the illumination is coming from within the device, likely via quantum fluctuations. That's why illumination and photo-injection is talked about extensively. There's no light shining on it from the outside. Then there would be no "anomaly".

So that means there is another external source of energy, namely, the power supply.

(for anyone who's reading this without context, please read my earlier reply where I explain the role of the power supply in IV graphs, and about the (V=0, I=0) intersection in detail).

Correct - there is an external source of energy, the power supply. That's needed to make an IV curve. With a non-power generating resistor, you expect the line to go through the origin. In this case, it takes 0.07mv to stop the current. This is not possible unless the device is producing its own energy somehow (or getting it from somewhere else - but the 8 culprits mentioned were accounted for - there may be others of course, and there is no light shining on it).

I've created a visual to explain this using the figure in the paper. Note how the green line passes through the origin when V=0. This is how a non-power generating device behaves. In our case, there is a positive current even when V is negative.

Again, it's not so much the I > 0 when V = 0, it's that there are data points within the 2nd quadrant, demonstrating that there is positive current when external voltage is applied against the flow of current.

 

[Demystifier]

What are your thoughts given the findings of the paper cited in this thread

I don't have much thoughts on it, the paper is purely experimental, while I'm a theorist.

 

[PeterDonis]

If you read the first two paragraphs it places things in context. There is no external light source.

I have read these repeatedly and I'm not getting that. I'm getting that there is an external light source. But even if I am wrong and you are correct, that's still not the primary issue. See below.

With a non-power generating resistor, you expect the line to go through the origin.

No, with a completely dissipating resistor, you expect the line to go through the origin. In other words, a resistor does not convert any input power into current. Apparently the devices used in this experiment do. But that doesn't mean they are generating power from nowhere. They are generating power, if they are doing so at all, from the input provided by the power supply. With no power supply hooked up, the thing just sits there and doesn't generate anything. At least, that would be my expectation--but the people who ran this experiment did not even think of running this obvious control experiment.

To really test what these experimenters are claiming to test, you would hook up a circuit to the device containing a tiny load, with voltage and current measurements attached to the load (or part of the load), but with no power supply at all, and see what happens. A device that can actually generate power "from the vacuum" would show a nonzero voltage and current under such conditions, which the load would use for whatever it does. I predict that the devices in this experiment would do no such thing. But in any case this is an obvious thing to do which these experimenters did not do. Again, that makes me highly skeptical that they are showing a real effect, rather than an artifact of the fact that their experimental setup contains a power supply.

 

[peterbb]

I have read these repeatedly and I'm not getting that. I'm getting that there is an external light source. But even if I am wrong and you are correct, that's still not the primary issue. See below.

In this paper there are detailed specs of all the relevant raw data to reproduce the experiment: the dimensions of the parts of the device, thickness, materials, even fabrication methods, and other extra data like refraction index.

If they were shining a light on it, they would include details about the spectrum of light, intensity, distance, angle, etc. You don't find any of this information in the paper, because they are not shining a light on it.

Further down the paper, below Fig 4:

For an MIM device in equilibrium, any carriers excited by ZPFs and injected into the insulator from the upper electrode would be balanced by ZPF-excited hot carriers from the base electrode. We hypothesize that the presence of the Casimir cavity breaks this balance, resulting in a change in the net injection rate and hence increased G. This imbalance is discussed in greater detail after the presentation of the results. Because cavities suppress zero-point electromagnetic modes having wavelengths greater than twice the cavity thickness, reducing the cavity thickness increases the range of suppressed wavelengths. We would therefore expect an increase in G for decreasing cavity thickness, a trend that can be seen in Fig. 3(b).

Examining the results of Fig. 3(b) further, we would expect wavelengths in the visible range to dominate the response because, for near-UV wavelengths <250nm, the PMMA becomes highly absorptive [40], and the infrared response is limited because the available vacuum energy density falls off with the cube of the wavelength. Although the vacuum energy density in ideal Casimir cavities varies inversely with the thickness cubed [7], we would not expect the conductance to exhibit such a cubic dependence because there are multiple energy-dependent mechanisms in play, including (i) variations of photoinjection yield with photon energy, as described by extensions of Fowler's theory of photoemission [41]; (ii) the interband transition threshold of Pd [42], which limits the transport of high-energy carriers; (iii) the energy-dependent reflectivity of the mirrors and absorptivity of the transparent dielectric; and (iv) the energy dependence of hot carrier scattering [43].

I hope this makes it really clear that the photo-injection the paper is referring to is explicitly from zero-point fluctuations in the electromagnetic field. They are not shining a light on it.

 

[peterbb]

No, with a completely dissipating resistor, you expect the line to go through the origin. In other words, a resistor does not convert any input power into current. Apparently the devices used in this experiment do. But that doesn't mean they are generating power from nowhere. They are generating power, if they are doing so at all, from the input provided by the power supply. With no power supply hooked up, the thing just sits there and doesn't generate anything. At least, that would be my expectation--but the people who ran this experiment did not even think of running this obvious control experiment.

To really test what these experimenters are claiming to test, you would hook up a circuit to the device containing a tiny load, with voltage and current measurements attached to the load (or part of the load), but with no power supply at all, and see what happens. A device that can actually generate power "from the vacuum" would show a nonzero voltage and current under such conditions, which the load would use for whatever it does. I predict that the devices in this experiment would do no such thing. But in any case this is an obvious thing to do which these experimenters did not do. Again, that makes me highly skeptical that they are showing a real effect, rather than an artifact of the fact that their experimental setup contains a power supply.

The IV curve is more informative and comprehensive than doing a power load test because it gives a broader insight as to its behaviour and characteristics, max power output (at which voltage), etc. Using these values you can calculate how it would behave with a resistor alone, and a lot more. You get the open-circuit voltage, short-circuit current, maximum power point, and others that would be more relevant if it were not linear. With a simple load test you get more of a "snapshot", but with IV curve you got a whole spectrum of information.

In this case it's linear so it's quite simple. Still, it gives a clearer picture, and we know that it's linear.

I think you're right that it would be more intuitive for people without less technical background to prefer to see a simple resistor and voltage measurement. I personally would like to see that as well, but it doesn't change the fact that that information can be garnered from the IV curve.

Thank you for staying in the conversation.

 

[PeterDonis]

The IV curve is more informative and comprehensive than doing a power load test

Perhaps it is for "standard" components. But this paper is claiming that the device in question is anything but a "standard" component. So appealing to the normal behavior and usage of an IV curve is not valid in this case. A claim like "this device can generate power from the vacuum with no energy input" should be tested directly. This test is not a direct test since there is energy input (even if we leave aside the light source question, there is a power supply).

 

[hutchphd]

The IV curve is more informative and comprehensive than doing a power load test because it gives a broader insight as to its behaviour and characteristics, max power output (at which voltage), etc.

In many circumstances the IV curve is used because the the slope is a immune from the issues of where the origin is exactly. It turns out that because of temperature dependent contact potentials and a host of other complications in layered structures finding these zeros is fraught. These folks seem to infer far more than is reasonable from the data presented.

 

[berkeman]

Since the discussion has died down, and since the paper is questionable, this thread is now tied off. If there are new papers with validations of the results by other groups, please PM a Mentor to either reopen this thread or allow a new thread to be started. Thanks folks.