Microwaving Marshmallows and c

[jds10011]

Hi all,
I teach high school physics, and as usual I welcome the opportunity for someone out there to set me straight. A common (and somewhat silly) demo involves taking a microwave, removing the rotating tray, and microwaving a plate of marshmallows for a brief time to observe the irregular pattern of heating the microwave produces. By claiming the melted spots are anti-nodes, and measuring the distance between them, one theoretically observes half the wavelength of the standing wave set up within. Then, in combination with the frequency printed on the back of the unit (usually 2450 MHz), one can use v=f(lambda) to calculate the speed of light. Obviously this is circular reasoning, and I don't care to belabor that point a great deal, other than to say that it's a neat way to introduce a concept and kids like marshmallows; I'm very clear to the students about this. However, the problem is that I (and a number of others I've found online) consistently measure about 12 cm between melted spots, when it seems one ought to measure 6 cm if the math is to come out correctly. Care to let me know what I'm missing? I'm concerned this may be related to some debate I've heard regarding misconceptions about microwaves and resonance, although some more mundane misconception about the existence of reflectors/fans in the microwave (or how the magnetron works altogether) is also relatively likely.

Thoughts?

A typical description (from a journal) is here: http://www.physics.umd.edu/icpe/newsletters/n34/marshmal.htm

Thanks!

 

[Bill_K]

Is your microwave at rest?

 

[jds10011]

Is your microwave at rest?

Unless you mean something more complicated, I certainly hope so...

 

[G01]

There are two nodes per wavelength, one at 0 radians, one at $\pi$ radians. The next wavelength starts with a node at $2\pi$.

Thus, the distance between any two nodes, 6cm, is the distance between the nodes at 0 and $\pi$, or half the wavelength. This is where your missing factor of 2 comes from.

 

[Bill_K]

Seriously, I'd be interested to hear more details about your results. Do you get a square grid pattern? Is it well-defined? I would think that the microwave manufacturer does everything he can to reduce this effect and make the cooking as even as possible.

 

[epenguin]

Glad a teacher is pointing out the circular nature of the reasoning, there is too much of it in physics and chemistry teaching in my opinion and (distant) experience.

Did you get this from BBC TV? I guess it is OK as an illustration of the wave nature of the radiation (and standing waves?). It would be OK if you knew how they know the frequency (I don't). One figure that you are told, put through a calculation gives c agreeing with another figure that has been reported, though at least experiments to determine that are comprehensible.

I suppose this effect is why they make the plates rotate?

 

[vanhees71]

But he said, he measures 12cm, i.e., the wave length. That's indeed strange since the nodes should indeed have a distance of half the wave length since the energy density (or energy flow, i.e., the Poynting vector) of the em. field is what determines the heating, and that's $$u=\frac{1}{2} (\vec{E}^2+\vec{B}^2)$$. The square of a sine has half the wave-length of the sine itself and thus the nodes (or anti-nodes) should indeed have a distance of $$\lambda/2$$. Unfortunately, I've no marshmallows at hand to perform the experiment myself.

 

[jds10011]

Seriously, I'd be interested to hear more details about your results. Do you get a square grid pattern? Is it well-defined? I would think that the microwave manufacturer does everything he can to reduce this effect and make the cooking as even as possible.

On a plate of ~25 cm diameter, I routinely get three antinodes (i.e. very melted spots), rather exactly equidistant (think equilateral triangle) at about 12 cm. They're pretty obvious and the rest of the plate appears untouched. I've tried this with three microwaves (one quite old) at my disposal, all using the same frequency, and all with their turntables removed. Results are consistent.

I think the rotation of the plates is their only (rather poor) attempt to make cooking even (or assume you're only using it to heat more liquid items).

 

[jds10011]

There are two nodes per wavelength, one at 0 radians, one at $\pi$ radians. The next wavelength starts with a node at $2\pi$.

Thus, the distance between any two nodes, 6cm, is the distance between the nodes at 0 and $\pi$, or half the wavelength. This is where your missing factor of 2 comes from.

I think you misunderstand; I'm aware that I ought to get half the wavelength, and that this should be 6 cm, but instead I measure 12 cm between antinodes, as if half of them were missing.

 

[jds10011]

But he said, he measures 12cm, i.e., the wave length. That's indeed strange since the nodes should indeed have a distance of half the wave length since the energy density (or energy flow, i.e., the Poynting vector) of the em. field is what determines the heating, and that's $$u=\frac{1}{2} (\vec{E}^2+\vec{B}^2)$$. The square of a sine has half the wave-length of the sine itself and thus the nodes (or anti-nodes) should indeed have a distance of $$\lambda/2$$. Unfortunately, I've no marshmallows at hand to perform the experiment myself.

Since you mention it, if you (or anyone else) wants to gather data, I'm told thin chocolate bars, sliced cheese, a layer of chocolate chips, or similar also works for this. Make sure you only run it long enough to see the pattern emerge, not until you have a huge mess.

 

[G01]

I think you misunderstand; I'm aware that I ought to get half the wavelength, and that this should be 6 cm, but instead I measure 12 cm between antinodes, as if half of them were missing.

O ok. I see. Yeah that is strange. I'll have to think about it...

 

[JaredJames]

I assume you're sure of the frequency?

 

[skeptic2]

I find this puzzling as well. Could you give us the dimensions of the inside of the microwave. Are the dimensions related to the wavelength? For instance I would expect an anti-node to occur 1/4 wavelength from an interior side. Could they have designed the microwave so that the nodes in one dimension coincide with the anti-nodes in another?

 

[jds10011]

I assume you're sure of the frequency?

I don't have any way of measuring, just the manufacturer's marking on the attached label. Also, in some internet research, 2450 MHz seems to be the most standard frequency.

I suppose there might be some odd standard in how this is designated for this particular sort of device that we're unaware of?

 

[JaredJames]

I'm just wondering about interference.

 

[jds10011]

I find this puzzling as well. Could you give us the dimensions of the inside of the microwave. Are the dimensions related to the wavelength? For instance I would expect an anti-node to occur 1/4 wavelength from an interior side. Could they have designed the microwave so that the nodes in one dimension coincide with the anti-nodes in another?

OK, the two of those I tested that I've got easy access to have interior dimensions of 29 cm x 27 cm, & 33 cm x 32 cm (these measurements are width (front) x depth to rear of inside, if you're curious).

I suppose nodes and anti-nodes could cancel like that, but to what end? I imagined the manufacturer would want as many anti-nodes as possible, then spin the food to make these pass through as much of it as possible...

 

[RedX]

A microwave is 3-dimensional is it not? So if you travel a distance 12 in the x-direction, then you travel a phase of $$\pi$$, since you go from antinode to antinode:

$$(k_x)(12)=\pi$$

so that the wavenumber is

$$k_x=\frac{\pi}{12}$$

So $$\lambda=\frac{2 \pi}{\sqrt{(k_x)^2+(k_y)^2+(k_z)^2}} =\frac{2 \pi}{\sqrt{3(k_x)^2}}=\frac{2 \pi}{\sqrt{3(\frac{\pi}{12})^2}}= \frac{24}{\sqrt{3}}$$

 

[RedX]

eh I found a rectangular cavity applet:

http://www.falstad.com/embox/

My suggestion is to change "show field (tricolor)" into "show field magnitude" and change "oscillation speed" to its lowest setting.

I don't see any single mode patterns that give a triangle on a plate. You can select two modes at a time, and I guess they weight each one the same? So maybe you can get lucky and one of the possible linear combos with equal weight equals a flat triangle.

I don't even understand how you can get an answer like this. But I looked it up online and lots of people got the right answer to the speed of light assuming everything is just one-dimensional.

 

[jds10011]

A microwave is 3-dimensional is it not? So if you travel a distance 12 in the x-direction, then you travel a phase of $$\pi$$, since you go from antinode to antinode:

$$(k_x)(12)=\pi$$

so that the wavenumber is

$$k_x=\frac{\pi}{12}$$

So $$\lambda=\frac{2 \pi}{\sqrt{(k_x)^2+(k_y)^2+(k_z)^2}} =\frac{2 \pi}{\sqrt{3(k_x)^2}}=\frac{2 \pi}{\sqrt{3(\frac{\pi}{12})^2}}= \frac{24}{\sqrt{3}}$$

OK, this is very interesting. I'm having visions of an E&M class I took a while back. If I run this backward by assuming that I ought to get a wavelength of c/2450 MHz, I arrive at a value of 11 cm between anti-nodes, assuming I've run the calculations correctly (perhaps you'll check me?). As I'm not using a set of calipers, and the gooey clumps of melted marshmallows are not exact points, I'm willing to believe that this is what is actually occurring.

Can't say this really makes for a good explanation for a high school demo, though...

 

[RedX]

OK, this is very interesting. I'm having visions of an E&M class I took a while back. If I run this backward by assuming that I ought to get a wavelength of c/2450 MHz, I arrive at a value of 11 cm between anti-nodes, assuming I've run the calculations correctly (perhaps you'll check me?). As I'm not using a set of calipers, and the gooey clumps of melted marshmallows are not exact points, I'm willing to believe that this is what is actually occurring.

Can't say this really makes for a good explanation for a high school demo, though...

You can find youtube videos of people doing this experiment: you can view their experimental setup and see how yours is different, because they all consistently claim 6 centimeters. For now I'll just assume 6 centimeters is what I'd get if I did the experiment (even though you got 12) - I will actually do this experiment next weekend, as marshmellows ought to be on sale for easter. Here's something I found:

http://demonstrations.wolfram.com/MeasuringTheSpeedOfLightWithMarshmallows/

(note: for some reason, you have to delete the "?affilliate=1" at the end of the url in your browser, and refresh to see everything)

From the picture, in the x-direction hot spots are spaced 6 apart. In the y direction, hot spots would be spaced 6*sqrt(3) apart.

Therefore for me the best fit for a 30x30x20 microwave would be the mode:

$$\lambda=\frac{2 \pi}{\sqrt{(4\pi /30)^2+(3\pi /30)^2+(\pi/20)^2}}=11.5 \mbox{ cm}$$

where $$(k_x,k_y,k_z)=(4\pi/30,3\pi/30,\pi/20)$$ which corresponds to antinode spacing of:
$$(L_x,L_y,L_z)=(7.5,10,20)$$

That 7.5 is very different from 6, but you know what? That's just how I'd do things. Maybe I'm wrong. But I'd like to learn.

Does anyone know how a microwave selects which modes?

 

[Born2bwire]

I think the problem with this is that it would be a poor microwave if such an experiment worked (but it wouldn't surprise me if the cheaper varieties did demonstrate this). The first problem of course is that the nodal points will depend upon the dimensions of the microwave and the modes excited. Seeing as how a plate of marshmallows will only provide a two-dimensional plane of measurement, one may not have marshmallows that intersect a plane with nodal points (probably want to use big marshmallows that are at least 1 in or more tall). But given that a microwave is a large cavity then I think the problem may be that there would be too many nodal points.

The main reason is that many microwaves have a stirrer. A bad design would be to just let the waveguide exit out into a cavity and leave it at that. Some of the cavity modes may not be excited and thus we may see poor distributions of energy. However, a metal fan is often used at the top of the microwave so that it adds randomized reflections of the waves. This acts as a "stirrer" to continually excite different cavity modes despite the fact that the source may only directly excite a few of them.

So a decent microwave is going to excite many cavity modes and the marshmallows may intersect nodal points associated with multiple modes. If this were to work, I think you will probably want to use a small microwave to reduce the number of modes, a very cheap microwave that would rely on the turntable instead of adding a stirrer along with it, and using the regular sized marshmallows stood on end to make sure that they have a good chance of hitting a nodal point.

Oh, well, I think the actual thing to do is hit the ANTInodal points. These would correspond to the hot spots as we are really measuring the hot spots, not the cool spots.