Calculating the error in <x^2> from the error in <x> (Molecular Dynamics)

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Kieran
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Hi, does anyone know of an easy way to calculate the error in <x^2> from the error in <x>? I am running a molecular dynamics simulation and trying to work out the error in the fluctuation of kinetic energy <dEk> = <3/2NT^2> - <3/2NT>^2 from the error in <T>.

Thanks in advance
 
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BvU said:
You say you have the error in <T> ? Is'n that the same as ##<\sigma^2> = <(T-<T>)^2> = <T^2> - (<T>)^2 \ \ ## ?
Yes, I have the standard error in the mean of T but I also need the error in the mean of T^2
 
So you don't just want the spread <dEk> but also the error in this spread ? Doesn't that depend on the size of the system ?

[edit] adding: can't you use something like $${\sigma_{T^2}\over T^2 } = 2 {\sigma_T \over T} \ \ \ \rm ?$$
 
BvU said:
So you don't just want the spread <dEk> but also the error in this spread ? Doesn't that depend on the size of the system ?

[edit] adding: can't you use something like $${\sigma_{T^2}\over T^2 } = 2 {\sigma_T \over T} \ \ \ \rm ?$$
I think so, this is what I'm having difficulty with. There must be error in <dEk> because there is error in <T>?
 
[edit] adding: can't you use something like $${\sigma_{T^2}\over T^2 } = 2 {\sigma_T \over T} \ \ \ \rm ?$$[/QUOTE]
I know of this formula but I wasn't sure if it would work with the error being in the average of T...
 
I suppose you are doing something very sophisticated ?
The Maxwell Boltzmann distribution has well-defined characteristics with 'exact' ##\sigma##. It's only when you generate samples, that the error in such ##\sigma## (determined from the sample) comes into the picture ?
 
BvU said:
I suppose you are doing something very sophisticated ?
The Maxwell Boltzmann distribution has well-defined characteristics with 'exact' ##\sigma##. It's only when you generate samples, that the error in such ##\sigma## (determined from the sample) comes into the picture ?
Well I'm using a molecular dynamics program to determine a set of T values over time, so I get some fluctuation about the average T and I've worked out the error in the average of T. The problem is that I now need the error in <dEk> which I think needs the error in T but I'm not completely sure.. :confused:
 
Dale said:
I think you want to use the propagation of errors formula.

https://en.m.wikipedia.org/wiki/Propagation_of_uncertainty
I've looked at this carefully but I am still unsure and can't seem to find any information of whether these formulas work for errors in the mean of quantities. For example, I'm not sure that

d<T^2>/<T^2> = 2d<T>/<T>
 
Hmm, I can't answer for certain, but when I have been in similar situations I would just generate 10000 normally distributed numbers and see if the formula works.

It probably depends on the mean, so you might try mean 100, variance 9 or something similar.
 
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Dale said:
Hmm, I can't answer for certain, but when I have been in similar situations I would just generate 10000 normally distributed numbers and see if the formula works.

It probably depends on the mean, so you might try mean 100, variance 9 or something similar.
Ah that sounds like a good plan, thanks for your advice
 
Kieran said:
Well I'm using a molecular dynamics program to determine a set of T values over time, so I get some fluctuation about the average T and I've worked out the error in the average of T. The problem is that I now need the error in <dEk> which I think needs the error in T but I'm not completely sure.. :confused:
Looks a bit like sample mean sigma ##\sigma_m## is ##\displaystyle \sigma_{\rm\ population}\over \sqrt N## and since you only have the sample it's ##\displaystyle \sigma_m## is ##\displaystyle\sigma_{\rm\ sample}\over \sqrt {n-1}##.

Fill us in on the result of Dale's suggestion ! :smile: My bet is on the simplest guess: double the relative error in T to get the relative error in T2
 
I think using (x+ε)2 = x2+2xε+ε2 would give the answer that the error would be 2[itex]\bar{x}[/itex]ε + ε2
But I do agree that a simulation would be the best way to confirm a result one way or another.

EDIT (CORRECTION): Since the error, ε, will be randomly positive or negative, I think the 2xε term will have an expected value of 0. So that leaves just ε2, which I think many others have already said. (But 2xε the term can increase the variance of the error greatly, depending on the magnitude of x.)
 
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