Does the sequence of a deck of cards have significance?

AI Thread Summary
The sequence of a deck of cards does not have intrinsic thermodynamic significance when considered in isolation. Shuffling the deck increases the observer's entropy due to the cognitive effort required to track the new order, rather than changing the entropy of the deck itself. The discussion draws parallels between the entropy of a deck of cards and that of a coin, emphasizing that the entropy of the coin remains unchanged as a closed system, regardless of whether it is observed. The act of observation and memory storage introduces additional entropy related to the observer's cognitive processes. Ultimately, the increase in entropy is attributed to the observer's engagement rather than the physical state of the card deck or coin.
Grinkle
Gold Member
Messages
819
Reaction score
236
If I open a new deck of cards the sequence is known.

If I shuffle them, the sequence is unknown. If I then memorize the sequence, it is again known.

State 1: New deck
State 2: Shuffled deck
State 3: Shuffled deck after I have memorized the sequence
State 4: Re-shuffled deck

I have computed the entropy difference of the deck sequence between state 1 and state 2. Many of us will have done the same as an introductory exercise in the second law of thermodynamics.

If I skip step 3 then it seems to me I must say that state 2 has equivalent entropy to state 4, but if I don't skip step 3 then it seems that delta entropy (1->2) must equal the delta entropy (2->4). Entropy really is a thermodynamic property - it is not just a mind game. Is there any way to conceptually relate the entropy of the sequence of a deck of cards to the entropy in a container of gas so they both have the same 'amount' of physical significance? One seems to be a state of mind and the other seems to be a thermodynamic property, but both share a common mathematical model.
 
Physics news on Phys.org
If you skip step 3 then you are talking about a completely different system. With it you are talking about the entropy of the deck plus the memory storage. Without it you are talking about the entropy of the deck alone.
 
Hmmm.

Is it well-formed / sound to discuss the sequence itself as having increased in entropy by having shuffled the deck - the exercises I have done have never asked about the energy dissipated by the observer. The system has always been described as a deck of cards.

One could also just think of a penny being heads or tails, the question is the same, I think. If I flip a coin with my eyes closed, so I don't know the start or end state, and my clone in another universe does the same with eyes open, is the penny's-state entropy change different in the two universes? Can one analyze purely the state of the penny (heads vs tail up) as a bounded system?

It may be that my problem comes from a poorly constructed exercise in my junior year thermo textbook.
 
For an idealized coin, if you don't look at the coin, the coin is in a mixed state with entropy of S = k ln 2. If you look at the coin, then the entropy of the coin becomes 0. If you randomize the coin, the entropy goes back to k ln 2.

The entropy of the coin can go down, but that's because the coin isn't a closed system.

Another way of thinking of it is that you aren't talking about the same coin after you look at it. You have shifted into a subset of worlds, so you are now referring to a subset-coin, which is just "part" of the original coin. In the many worlds picture, initially there are two worlds, one with heads and one with tails. If you look at the coin and then flip the coin, there are 4 worlds:
a. you remember seeing H, coin is in state 1
b. you remember seeing H, coin is in state 2
c. you remember seeing T, coin is in state 1
d. you remember seeing T, coin is in state 2
The total entropy is k ln 4 since there are 4 possibilities. But from your point of view, you never observe yourself splitting, so it looks like there are only two possible worlds. If you don't look at the coin before flipping the coin, there are just 2 worlds.
a. coin is in state 1
b. coin is in state 2
The world doesn't split when you flip the coin because coin's previous state interferes with itself.
 
DaleSpam said:
f you skip step 3 then you are talking about a completely different system. With it you are talking about the entropy of the deck plus the memory storage. Without it you are talking about the entropy of the deck alone.

Khashishi said:
The entropy of the coin can go down, but that's because the coin isn't a closed system.

This has been helpful food for thought.

I think all this implies that the answer to my initial question is the sequence of a deck of cards in isolation does not have any thermodynamic significance.

The entropy of a coin all by itself as a closed system cannot change.

So I think it must be the case that the same remains true of the coin per se if one adds an observer. The entropy of the observer will change as the observer does whatever work is required (brain burning calories etc) to attach symbolic meaning to the different possible states of the previously closed system being observed. Using entropy to describe the count of these interpreted states is a bookeeping mechanism, I think. Its misleading to think that the entropy of a deck of cards increases if the deck is shuffled. What increases is the entropy of the person thinking about what it means to shuffle cards and thinking about the different order of the cards, because the act of keeping track of the symbolism in the ordering of the deck takes computing power. If a blindfolded man flips a coin the entropy of the universe increases less than if a watching man does the same thing and comprehends heads vs tails after each flip, and the entropy difference is all in the brain activity of the watching man doing the comprehension. The coin is the same in both situations and its entropy is not changing in either situation.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top