Timeline for Summing graphs in the partition function (statistical physics)
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 17, 2023 at 22:53 | comment | added | Yvan Velenik | The sum is not over $m_l$, but over $\{m_l\}$, that is, the collection of all numbers $m_l$, for all possible values of $l$. So, each term of the sum specifies the values of $m_1, m_2, m_3$, etc. The product over $l$ is then just a short way of writing $f(m_1)f(m_2)f(m_3)\cdots$. If this helps you, you can write the product as being over $k$ (and replace all $l$ in the expression following the product by $k$, of course). | |
| Mar 17, 2023 at 20:55 | comment | added | Bedge | I have accepted the answer, but thought of one other thing to clarify: In the sum of the product $\sum_{m_l}\prod_{l}(a^{lm_l}\frac{1}{m_l!}b_l^{m_l})$, does fixing the value in the sum, e.g. $m_3$, fix the value of $l$ (to be 3 in the example), or does the product run separately? | |
| Mar 17, 2023 at 20:44 | vote | accept | Bedge | ||
| Mar 17, 2023 at 20:43 | comment | added | Yvan Velenik | I have added the last explanation to the answer, since it's an important point. | |
| Mar 17, 2023 at 20:43 | history | edited | Yvan Velenik | CC BY-SA 4.0 |
Added an explanation
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| Mar 17, 2023 at 20:40 | comment | added | Yvan Velenik | Since I wrote explicitly the constraint as an indicator function, the sum over $\{m_l\}$ is never restricted (of course the summand vanishes when the constraint is not satisfied, because of the indicator). | |
| Mar 17, 2023 at 20:39 | comment | added | Bedge | In that case, is this understanding correct: On the LHS of 4th equality (i.e. line 3) the sum over $\{m_l\}$ is no longer subject to the constraint? | |
| Mar 17, 2023 at 20:37 | comment | added | Yvan Velenik | Let me explain on a particular case: $\sum_{m_1,m_2} f(m_1)f(m_2) = \bigl(\sum_{m_1}f(m_1)\bigr)\bigl(\sum_{m_2}f(m_2)\bigr) = \prod_{l=1}^2 \sum_{m_l}f(m_l)$. Does this help? | |
| Mar 17, 2023 at 20:34 | comment | added | Bedge | Thanks for you answer. I'm still not quite clear how the 4th equality works - I don't understand why the sum is now over all $m_l$? | |
| Mar 17, 2023 at 20:29 | history | edited | Yvan Velenik | CC BY-SA 4.0 |
[Edit removed during grace period]
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| Mar 17, 2023 at 20:18 | history | edited | Yvan Velenik | CC BY-SA 4.0 |
Added parentheses for clarity...
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| Mar 17, 2023 at 20:12 | history | edited | Yvan Velenik | CC BY-SA 4.0 |
Added details.
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| Mar 17, 2023 at 20:11 | comment | added | Yvan Velenik | Tell me if one step is still unclear and I'll try to add an explanation. | |
| Mar 17, 2023 at 20:07 | history | answered | Yvan Velenik | CC BY-SA 4.0 |