2
$\begingroup$

In a recent article[1], John Baez is quoted as making a nice point about how non-commutativity is common in the world around us, whereas non-associativity is not:

[...] while it’s very easy to imagine noncommutative situations—putting on shoes then socks is different from socks then shoes—it’s very difficult to think of a nonassociative situation.” If, instead of putting on socks then shoes, you first put your socks into your shoes, technically you should still then be able to put your feet into both and get the same result. “The parentheses feel artificial.”

This thus begs the question: are there any examples of non-associative processes in nature/the physical universe? Note that I emphasize that it be an actual physical process (like putting on shoes) and not just a description of reality (cause there are plenty of examples like that to be found already on the wikipedia page [2], which are somehow artefacts of the language we use to describe things).

I think half the struggle is to figure out how to ask the question in physical terms. It makes sense to ask for two physical processes A and B such that 'first A then B' does not give the same outcome as 'first B then A', hence giving an example of non-commutativity. I don't quite know how to phrase the desired non-associative property for three physical processes A, B and C in a purely physical language. I guess my question is perhaps: are there parentheses in nature?

[1] https://www.wired.com/story/the-peculiar-math-that-could-underlie-the-laws-of-nature/

[2] https://en.wikipedia.org/wiki/Associative_property#Non-associative_operation

$\endgroup$
3
  • 1
    $\begingroup$ Function composition is associative, so when you talk about a sequence of physical processes, I think it will hard to come up a non-associative example. However, octonions have been in the news lately. $\endgroup$
    – saulspatz
    Aug 14, 2018 at 20:55
  • $\begingroup$ @saulspatz Indeed: Ref.[1] in my post is a reprint of the quanta article you link to :) $\endgroup$ Aug 14, 2018 at 21:17
  • 1
    $\begingroup$ So it is. I still stand by the first sentence in my comment, though. $\endgroup$
    – saulspatz
    Aug 14, 2018 at 21:19

3 Answers 3

1
$\begingroup$

The closest I can think of is when you put wheat flour into a hot sauce to make it thicker. If you put the dry flour directly into the hot sauce, there will easily be lumps, but if you first put the flour into cold water or milk, stir it, and then put that batter into the sauce it will work better. Thus, $$ (\text{cold fluid} + \text{flour}) + \text{hot fluid} \neq \text{cold fluid} + (\text{flour} + \text{hot fluid}) $$

$\endgroup$
1
$\begingroup$

Reproduction is non-associative. e.g., if you have three people, Alice, Bob and Clara, then a child of Alice and Bob reproducing with Clara is not equivalent to Alice reproducing with a child of Bob and Clara:

$$(\text{Alice} + \text{Bob}) + \text{Clara}) \neq \text{Alice} + (\text{Bob} + \text{Clara}).$$

In the first case the child has genes that are 25% Alice, 25% Bob, and 50% Clara, while in the second case the child is 50% Alice, 25% Bob, 25% Clara.

$\endgroup$
0
$\begingroup$

John Baez's comparison between commutativity and putting on shoes and socks seems to me to be fundamentally flawed. Commutativity does not involve any temporal ordering. Combining together A and B is a single event, the same as combining B and A.

Physically, you could consider the two entities (shoes, socks etc) as approaching from different directions. The cosmological principle tells us that the result should be the same in each case, due to isotropy. In this respect, the universe should be considered to be commutative.

However, associativity does involve ordering. In the case of shoes and socks, the feet are an extra factor. Combining the foot with a sock and enclosing the combination in a shoe gives a different result from combining a foot with a shoe and enclosing them in a sock.

This illustrates that the physical universe is commutative and non-associative, exactly the opposite of John Baez's assertion.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .