Homoeopathy: Help explain it to our credulous parliamentarians!

[kyleain]

ok well u c, homopathy is when homosexuals are pathy n yeah ugh, i have post 12 stupid more posts to type urls ugh, yeah thats really preventing spam

 

[fls]

ok well u c, homopathy is when homosexuals are pathy n yeah ugh, i have post 12 stupid more posts to type urls ugh, yeah thats really preventing spam

Are you preparing to spam us once you reach 15?

I see you have spent the last hour dropping disparaging remarks into random threads, in order to get your post count up.

Linda

 

[jimbob]

ok well u c, homopathy is when homosexuals are pathy n yeah ugh, i have post 12 stupid more posts to type urls ugh, yeah thats really preventing spam

As I read this I see:

kyleain
Suspended

But that might prevent spam.

 

[Monza]

Please feel free to check my math and let me know of any errors. But here is what I came up with...

OK, so let's assume we have 1 liter of water. The density of water is 1 kg/m^3. There are .1 m^3/liter. So we have 100 g/liter.

The molecular weight of water is 18 grams (16 + 2x1). How many molecules of water are in 1 liter?

N = 100g / [18g/(6.02x10^23)] = 33.4 x 10^23 molecules in 1 liter of water

The most dilution you could have is one molecule of substance out of the 33.4 x 10^23 molecules in the liter of water. This is about a 12C dilution. By definition, it is impossible to dilute this volume any more. If you have any dilution greater than 12C then you will need a larger volume of water (i.e. more water molecules) to get the ratios right.

Looking at a common 30C dilution, you would need at least 1 x 10^60 molecules of water. How much is that?

m = 1 x 10^60 x [18g/(6.02x10^23)] = 2.99 x 10^37 grams

V = m/p = 2.99 x 10^37 grams / (100 g/liter) = 2.99 x 10^35 liters

Let's calculate how big a sphere of water is needed to equal 2.99 x 10^35 liters.

V = 4/3*pi*r^3
2.99 x 10^35 liters x .1 m^3/liter = 4/3*pi*r^3

Solving for r...

r = [(2.99 x 10^32 x 3/4) / pi]^(1/3)
r = 4.15 x 10^10 m
d = 2r = 8.30 x 10^10 m

This is equal to a sphere of water 83 million kilometers in diameter!

So how can someone sell you a vial of homeopathic remedy with 24C or 30C or 200C dilutions? The fact is, there is not a single molecule of the original substance in the vial.

 

[rjh01]

One issue. 1 Kg water = 1 Litre of water = 1,000 grams

 

[Monza]

Aghh! I knew I'd screw up a decimal point somewhere. OK, here is the fixed version. Again, if you see an error let me know.


OK, so let's assume we have 1 liter of water. The density of water is 1000 kg/m^3. There are .001 m^3/liter. So we have 1000 g/liter.

The molecular weight of water is 18 grams (16 + 2x1). How many molecules of water are in 1 liter?

N = 1000g / [18g/(6.02x10^23)] = 3.34 x 10^25 molecules in 1 liter of water

The most dilution you could have is one molecule of substance out of the 3.34 x 10^25 molecules in the liter of water. This is about a 12C dilution. By definition, it is impossible to dilute this volume any more. If you have any dilution greater than 12C then you will need a larger volume of water (i.e. more water molecules) to get the ratios right.

Looking at a common 30C dilution, you would need at least 1 x 10^60 molecules of water. How much is that?

m = 1 x 10^60 x [18g/(6.02x10^23)] = 2.99 x 10^37 grams

V = m/p = 2.99 x 10^37 grams / (1000 g/liter) = 2.99 x 10^34 liters

Let's calculate how big a sphere of water is needed to equal 2.99 x 10^34 liters.

V = 4/3*pi*r^3
2.99 x 10^34 liters x .1 m^3/liter = 4/3*pi*r^3

Solving for r...

r = [(2.99 x 10^31 x 3/4) / pi]^(1/3)
r = 4.15 x 10^9 m
d = 2r = 8.30 x 10^9 m

This is equal to a sphere of water 8.3 million kilometers in diameter!

 

[jimbob]

Now Monza, for a real laugh, work through the calculations for a 200C solution (I did it for a 100X solution....

 

[Monza]

OK, jimbob...

200C = 100^200 = 10^400

So a 200C dilutions means there is one part of active ingredient for every 10^400 parts of solution (let's assume water). By the way, that number is a 1 followed by 400 zeros!

m = 1 x 10^400 x [18g/(6.02x10^23)] = 2.99 x 10^37 grams

V = m/p = 2.99 x 10^37 grams / (1000 g/liter) = 2.99 x 10^374 liters

Let's calculate how big a sphere of water is needed to equal 2.99 x 10^374 liters.

V = 4/3*pi*r^3
2.99 x 10^374 liters x .001 m^3/liter = 4/3*pi*r^3

Solving for r...

r = [(2.99 x 10^371 x 3/4) / pi]^(1/3)
r = 4.15 x 10^123 m
d = 2r = 8.30 x 10^123 m

1 light-year is equal to 9.461 x 10^15 m, so...

8.30 x 10^123 m / (9.461 x 10^15 m) = 8.77 x 10^107 light-years

The Milky Way galaxy is only 1 x 10^5 light-years in diameter. The known universe is 9.3 x 10^10 light-years across. So we're looking at something about 4800 times larger than the known universe.

Homeopathy claims that the higher the dilution, the more potent the substance. That is, a 30C dilution is more effective than a 10C dilution. So a 200C dilution is even more effective and I would assume, more expensive.

Keep in mind that the calculation above is for the absolute minimum amount of water needed to achieve a 200C dilution. Since the smallest amount of a substance is a single molecule of that substance, in order to get a ratio of 1:1x10^400 you must have as a minimum a sphere of water 8.77 x 10^107 light-years in diameter. Ignore whether or not Homeopathy works. The next time you buy a 200C homeopathic dilution, if the store clerk hands you over a container smaller than 8.77 x 10^107 light-years in diameter, by definition you are getting a dilution less than 200C which is less effective. You are getting ripped off!

 

[steenkh]

Maybe I could draw your attention to this post by MRC_Hans where he has made these sort of calculations:

Here is a short list of potencies where the volume of water to one drop (0.1 ml) is related to something in the real world:

6C = 4 Olympic class swimming-pools.
13C = 3/4 of all Earth's oceans.
16C = Sphere the size of Earth.
19C= 10 Spheres with same diameter as our Solar System
22C = 1 cubic light-year
26C = 1 Sphere with same diameter as the Milky Way

The very popular 30C equals one drop of mother tincture in nearly a hundred million galaxy-sized balls of water .

 

[jimbob]

OK, jimbob...

<snip>
The Milky Way galaxy is only 1 x 10^5 light-years in diameter. The known universe is 9.3 x 10^10 light-years across. So we're looking at something about 4800 times larger than the known universe.

Homeopathy claims that the higher the dilution, the more potent the substance. That is, a 30C dilution is more effective than a 10C dilution. So a 200C dilution is even more effective and I would assume, more expensive.

Keep in mind that the calculation above is for the absolute minimum amount of water needed to achieve a 200C dilution. Since the smallest amount of a substance is a single molecule of that substance, in order to get a ratio of 1:1x10^400 you must have as a minimum a sphere of water 8.77 x 10^107 light-years in diameter. Ignore whether or not Homeopathy works. The next time you buy a 200C homeopathic dilution, if the store clerk hands you over a container smaller than 8.77 x 10^107 light-years in diameter, by definition you are getting a dilution less than 200C which is less effective. You are getting ripped off!

It's fun isn't it..(the calculations I mean)...

ETA:



But it its the shaking and hitting with a leather glove that is important.

 

[Monza]

steenkh, thank you (and MRC_Hans) for the linked post. Very interesting.

jimbob, I have always heard the statement that in most homeopathic remedies there is not a single molecule of the original active ingredient. But until recently I have never just sat down and worked through a calculation myself. It is mind-boggling how insane 30C+ dilutions really are. What really gets me is that it is impossible to make certain dilutions. Even if homeopathy worked, at the very least someone is lying by stating it is diluted to 30C. This is something that can't be fixed by hitting it with a leather glove.

 

[jimbob]

Ignore whether or not Homeopathy works. The next time you buy a 200C homeopathic dilution, if the store clerk hands you over a container smaller than 8.77 x 10^107 light-years in diameter, by definition you are getting a dilution less than 200C which is less effective. You are getting ripped off!

I nominated this post

... BTW "sup" " /sup" in swuare brackets gives you superscripts:

8.77 x 10¹⁰⁷

 

[Lothian]

The next time you buy a 200C homeopathic dilution, if the store clerk hands you over a container smaller than 8.77 x 10^107 light-years in diameter, by definition you are getting a dilution less than 200C which is less effective. You are getting ripped off!

I wouldn't argue with your calculations but you only need* a litre of water to make a 5ml 200C dilution.

* As you showed; chemically an identical result could be made with just 5ml of water

 

[Lothian]

What really gets me is that it is impossible to make certain dilutions. Even if homeopathy worked, at the very least someone is lying by stating it is diluted to 30C.

No. Take one drop of ingredient add to 100 drops of water (c). Take one drop of this mixture and add it to 100 drops of water (2c). Take one drop of this and add it to 100 drops of water (3c).

You only need 3000 drops to make a 30C mixture*.


*I use the term mixture very loosely.

 

[Monza]

No. Take one drop of ingredient add to 100 drops of water (c). Take one drop of this mixture and add it to 100 drops of water (2c). Take one drop of this and add it to 100 drops of water (3c).

You only need 3000 drops to make a 30C mixture*.


*I use the term mixture very loosely.

That method only works if if an ingredient can be infinitely reduced in volume. It can't. An ingrediant has a finite point at which it cannot be divided any further and still be the same ingrediant. Once you get to one molecule of an ingredient, that's it. You can't divide it any further.

So somewhere around your 12th time of taking a drop of the mixture and adding another 100 drops of water, you start to run low on molecules of ingredient. So then the next time you take a drop of the mixture, it is not a mixture...it is only water. So when you add another 100 drops of water you are only adding water to water, not water to a mixture.

This was my point. You can run through the process of adding drops of water 30 or 100 times. But it is not a 30C or 100C mixture by definition.

 

[Monza]

Here is another way of looking at it.

Let's assume we have a swimming pool full of marbles. Half of the marbles are black and half are white. Further more, the marbles are evenly distributed. They are not randomly distributed, but evenly distributed. This is the same as a fully disolved solute in a solution.

Now you take a back-hoe and scoop out a bunch of marbles and put them into a new container. This container still has a 50/50 ratio of black marbles to white marbles. You want to dilute the black marbles so you add a back-how amount of white marbles. Now shake them up so that they are evenly distributed again.

Now you take a bucket and scoop out a bunch of marbles from the container. The marbles in the bucket have the same 1:3 black to white marble ratio. You want to dilute the black marbles further so you add another bucket full of white marbles. Once they are shaken up and the marbles are distributed evenly, you can take another scoop of this mixture that has the same 1:7 ratio of black to white marbles as the bucket.

This cannot go on indefinately. At some point, your scoop will only have one black marble. You cannot cut the marble in half or break it down into smaller sizes. So if you want to make a certain dilution of black marbles, you have no choice but to add more white marbles to this lone black marble.

A dilution is nothing more than a ratio of black marbles to white marbles. For example, a 30C dilution of black marbles is a ratio of 1 black marble for every 10⁶⁰ white marbles. Since you cannot have less than 1 black marble, in order to have a 30C dilution you need a minimum of 10⁶⁰ white marbles. If you have any number of white marbles less than this, then by definition you have a dilution less than 30C.

 

[Lothian]

This was my point. You can run through the process of adding drops of water 30 or 100 times. But it is not a 30C or 100C mixture by definition.

Ok

 

[steenkh]

Since you cannot have less than 1 black marble, in order to have a 30C dilution you need a minimum of 10⁶⁰ white marbles. If you have any number of white marbles less than this, then by definition you have a dilution less than 30C.

In homoeopathy, having "lost the marbles" is not just what is actually happening to their dilutions, but it is also a state of mind, the one leading to the other.

 

[Simplegreentinhouse]

Erm. I do not believe homeopathy works because, water molecules rotate every nine picoseconds and so doesn't really remember these so called "footprints".

Edit: It supposidly works by serial dilution. The water remembers the electromagnetic arrangement of the original content and is then serial diluted so that there is virtually none of this stuff left but the water crystals remember the arragement and the water can then use these properties to heal without actually having the extra chemical in it and can cheaply be mass produced. It really works on the placibo principle, because, however magnificant water is, it still rotates every 9 picoseconds and so is not going to easily store this arrangement. The only way would be to freeze at absolute zero temperature, but then how can one dilute something that is frozen? What I find an intersting discussion is the symetry/asymetry of ice crystals. If one were to play music near a glass of water and then freeze it, one could see the music waves in the crystals, and if the music were of one frequency and wavelength (or monotone) the crystals would be symetrical, yet if we were to play loud interfering and mixed music the crystals would be asymetrical.

 

[rjh01]

To record music in ice you would need to freeze it very quickly. As in a fraction of a second. And then I am not sure that this would even work. Remember that sound travels 330 meters eery second.