Determine how many microstates and macrostates (Thermodynamics)

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SUMMARY

This discussion focuses on determining the number of microstates and macrostates for a two-state paramagnet with N = 40. The correct calculation for microstates is given by the formula ##2^{40}##, not ##40^2##. The number of macrostates is established as 41, corresponding to the different total magnetic dipole moments possible, ranging from 0 heads (all tails) to 40 heads (all heads). The macrostate is defined by its macroscopic properties, which in this case is the total magnetic dipole moment.

PREREQUISITES
  • Understanding of two-state paramagnetism
  • Familiarity with combinatorial mathematics
  • Knowledge of microstates and macrostates in statistical mechanics
  • Ability to apply the formula ##\Omega(N,n)=\frac{N!}{n!(N-n)!}##
NEXT STEPS
  • Study the concept of microstates and macrostates in statistical mechanics
  • Learn about the implications of magnetic dipole moments in thermodynamics
  • Explore combinatorial methods for calculating arrangements and states
  • Investigate the applications of two-state systems in real-world scenarios
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This discussion is beneficial for physics students, researchers in thermodynamics, and anyone interested in the statistical mechanics of magnetic systems.

learningastronomy
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Homework Statement
A two-state paramagnet has 40 magnetic dipoles. Determine the amount of microstates and macrostates?
Relevant Equations
##\Omega(N,n)=\frac{N!}{n!(N-n)!}## or ##\omega(N_{\uparrow})=\frac{N!}{N_{\uparrow}!N_{\downarrow}!}##
Since this is a two-state paramagnet where N = 40, therefore the microstate is ##40^2##? But I am not sure how to proceed to count the number of macrostates? Because from what I understand of what a macrostate is, shouldn't there a specific outcome to be stated in order to determine how many macrostate there are? For instance suppose we have 40 coins then the number of microstate (all different states of heads and tails) will be ##40^2## but in order to compute macrostate then you will have to explicitly provide the state it is in, for example determine the amount of macrostate of the 40 coins that only have two heads then you can use $$\Omega(N,n)=\frac{N!}{n!(N-n)!}$$ $$\therefore\Omega(40,2)=\frac{40!}{2!(40-2)!}=780?$$ And since the question didn't explicitly provide the macrostate then will it just be the general case of $$\Omega(40,n)=\frac{40!}{(40-n)!}?$$

Is my reasoning correct?
 
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learningastronomy said:
Is my reasoning correct?
I don't think so. Not my area but this might help.

You have 3 coins in a line. How many head-tail patterns are possible? (It's not 3².) What about 40 coins (it's not 40²)?

As for macrostates, I would assume the macrostates are simply (continuing the 40-coin example):
0 heads, 40 tails
1 head, 39 tails
2 heads, 38 tails
etc.
 
there are 41 macro states total. A macro state is different from each other by its macroscopic properties in this case it would be the total magnetic dipole moment.
learningastronomy said:
Homework Statement:: A two-state paramagnet has 40 magnetic dipoles. Determine the amount of microstates and macrostates?
Relevant Equations:: ##\Omega(N,n)=\frac{N!}{n!(N-n)!}## or ##\omega(N_{\uparrow})=\frac{N!}{N_{\uparrow}!N_{\downarrow}!}##

Since this is a two-state paramagnet where N = 40, therefore the microstate is ##40^2##? But I am not sure how to proceed to count the number of macrostates? Because from what I understand of what a macrostate is, shouldn't there a specific outcome to be stated in order to determine how many macrostate there are? For instance suppose we have 40 coins then the number of microstate (all different states of heads and tails) will be ##40^2## but in order to compute macrostate then you will have to explicitly provide the state it is in, for example determine the amount of macrostate of the 40 coins that only have two heads then you can use $$\Omega(N,n)=\frac{N!}{n!(N-n)!}$$ $$\therefore\Omega(40,2)=\frac{40!}{2!(40-2)!}=780?$$ And since the question didn't explicitly provide the macrostate then will it just be the general case of $$\Omega(40,n)=\frac{40!}{(40-n)!}?$$

Is my reasoning correct?
 
Steve4Physics said:
I don't think so. Not my area but this might help.

You have 3 coins in a line. How many head-tail patterns are possible? (It's not 3².) What about 40 coins (it's not 40²)?

As for macrostates, I would assume the macrostates are simply (continuing the 40-coin example):
0 heads, 40 tails
1 head, 39 tails
2 heads, 38 tails
etc.

Ops, I meant to say ##2^{40}## not ##40^2##, thanks for catching that.
 
guv said:
there are 41 macro states total. A macro state is different from each other by its macroscopic properties in this case it would be the total magnetic dipole moment.
Hmm can you please elaborate why it is 41 macro states?
 
learningastronomy said:
Hmm can you please elaborate why it is 41 macro states?
@guv is saying that a macro state is defined by its total dipole moment. How many different possibilities are there for that total?
 
haruspex said:
@guv is saying that a macro state is defined by its total dipole moment. How many different possibilities are there for that total?
Oh I see, I may need to revisit the definition of macrostate then because I was thinking of a different interpretation of it. Also, the total different possibilities will be 40+1 therefore 41.
 
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learningastronomy said:
Oh I see, I may need to revisit the definition of macrostate then because I was thinking of a different interpretation of it. Also, the total different possibilities will be 40+1 therefore 41.
Macrostates are whatever matters to you in a particular context. For the result of a tennis match, we might only care who won, or we might care what the individual set scores were, etc.
 
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