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Determining the torsion energy profile of butadiene by molecular dynamics
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Nissim Levy
Determining the torsion energy profile of butadiene by molecular dynamics
I really only want to make a comment, but it is too long...
Your equation $n_i = \exp^{-E_i/RT}/N$ does not make sense to me.
I agree with
\begin{equation}
P_i = \frac{\exp ^{-E_i/RT}}{Z}
\end{equation}
where $Z = \sum \exp ^{-E_i/RT}$ is the configurational partition function.
I can also accept that using a histogram, you could say $P_i = n_i/N$ where $n_i$ is number of samples in that bin, and N is total samples.
I do not see however how
\begin{equation}
n_i N = \exp ^{-E_i/RT}
\end{equation}
This does not make sense to me. Perhaps you could swap $P_i$ for $n_i/N$ so that you have
\begin{equation}
P_i =\frac{n_i}{N} = \frac{\exp ^{-E_i/RT}}{Z}
\end{equation}
which could lead to
\begin{equation}
n_i = \frac{N}{Z}\exp ^{-E_i/RT}
\end{equation}
Which could be rearranged to get
\begin{equation}
E_i =-RT \ln \left(\frac{n_i }{N}Z\right)
\end{equation}
If you are comparing to a reference energy then you could get
I really only want to make a comment, but it is too long...
Your equation $n_i = \exp^{-E_i/RT}/N$ does not make sense to me.
I agree with \begin{equation}P_i = \frac{\exp ^{-E_i/RT}}{Z}\end{equation}
where $Z = \sum \exp ^{-E_i/RT}$ is the configurational partition function.
I can also accept that using a histogram, you could say $P_i = n_i/N$ where $n_i$ is number of samples in that bin, and N is total samples.
I do not see however how
\begin{equation}n_i N = \exp ^{-E_i/RT}\end{equation}
This does not make sense to me. Perhaps you could swap $P_i$ for $n_i/N$ so that you have\begin{equation}P_i =\frac{n_i}{N} = \frac{\exp ^{-E_i/RT}}{Z}\end{equation}
which could lead to\begin{equation}n_i = \frac{N}{Z}\exp ^{-E_i/RT}\end{equation}
Which could be rearranged to get\begin{equation}E_i =-RT \ln \left(\frac{n_i }{N}Z\right) \end{equation}
If you are comparing to a reference energy then you could get
You are right, I made a leap at that part, thanks for pointing it out. However, it does not appear to cause the mismatch. Since N and Z and in the logarithm, they will subtract/add a constant to the energies. If I calculate the relative energies afterwards, the result is the same.More
@Raphaël The probability of one conformer to another would not need the partition function, but that is a special case. Overall, just by making things relative, you still have a Z, just a simpler Z made up of a sum of $\Delta E$s rather than a sum of $E_i$s. i.e, if $n_i = N \frac{\exp^{-E_i/RT}}{Z}$ then you could say $\frac{n_i}{n_j}=\exp^{-\Delta E_{\rm ij}/RT}$, but that cancellation of both N and Z is because it is a ratio of 2 specific configurations.More
I really only want to make a comment, but it is too long...
Your equation $n_i = \exp^{-E_i/RT}/N$ does not make sense to me.
I agree with \begin{equation} P_i = \frac{\exp ^{-E_i/RT}}{Z} \end{equation}
where $Z = \sum \exp ^{-E_i/RT}$ is the configurational partition function.
I can also accept that using a histogram, you could say $P_i = n_i/N$ where $n_i$ is number of samples in that bin, and N is total samples.
I do not see however how
\begin{equation} n_i N = \exp ^{-E_i/RT} \end{equation}
This does not make sense to me. Perhaps you could swap $P_i$ for $n_i/N$ so that you have \begin{equation} P_i =\frac{n_i}{N} = \frac{\exp ^{-E_i/RT}}{Z} \end{equation}
which could lead to \begin{equation} n_i = \frac{N}{Z}\exp ^{-E_i/RT} \end{equation}
Which could be rearranged to get \begin{equation} E_i =-RT \ln \left(\frac{n_i }{N}Z\right) \end{equation}
If you are comparing to a reference energy then you could get
\begin{eqnarray} E_i - E_{\rm ref} = \Delta E_{\rm i,ref} &=&-RT \ln \left(\frac{n_i }{N}Z\right) +RT \ln \left(\frac{n_{\rm ref} }{N}Z\right) \\ &=& -RT\ln \left(\frac{n_i }{n_{\rm ref}}\right) \end{eqnarray}
It would be interesting to see if this works? I don't have the raw data, so I cannot try it (I also don't have the time, only the interest).
I really only want to make a comment, but it is too long...
Your equation $n_i = \exp^{-E_i/RT}/N$ does not make sense to me.
I agree with \begin{equation}P_i = \frac{\exp ^{-E_i/RT}}{Z}\end{equation}
where $Z = \sum \exp ^{-E_i/RT}$ is the configurational partition function.
I can also accept that using a histogram, you could say $P_i = n_i/N$ where $n_i$ is number of samples in that bin, and N is total samples.
I do not see however how
\begin{equation}n_i N = \exp ^{-E_i/RT}\end{equation}
This does not make sense to me. Perhaps you could swap $P_i$ for $n_i/N$ so that you have\begin{equation}P_i =\frac{n_i}{N} = \frac{\exp ^{-E_i/RT}}{Z}\end{equation}
which could lead to\begin{equation}n_i = \frac{N}{Z}\exp ^{-E_i/RT}\end{equation}
Which could be rearranged to get\begin{equation}E_i =-RT \ln \left(\frac{n_i }{N}Z\right) \end{equation}
If you are comparing to a reference energy then you could get
\begin{eqnarray}E_i - E_{\rm ref} = \Delta E_{\rm i,ref} &=&-RT \ln \left(\frac{n_i }{N}Z\right) +RT \ln \left(\frac{n_{\rm ref} }{N}Z\right) \\&=& -RT\ln \left(\frac{n_i }{n_{\rm ref}}\right)\end{eqnarray}
It would be interesting to see if this works? I don't have the raw data, so I cannot try it (I also don't have the time, only the interest).
More
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