- #1

- 8

- 0

Thanks in advance

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Kieran
- Start date

- #1

- 8

- 0

Thanks in advance

- #2

BvU

Science Advisor

Homework Helper

- 14,268

- 3,631

- #3

- #4

BvU

Science Advisor

Homework Helper

- 14,268

- 3,631

[edit] adding: can't you use something like $${\sigma_{T^2}\over T^2 } = 2 {\sigma_T \over T} \ \ \ \rm ?$$

- #5

- 8

- 0

I think so, this is what I'm having difficulty with. There must be error in <dEk> because there is error in <T>?_{k}> 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 $${\sigma_{T^2}\over T^2 } = 2 {\sigma_T \over T} \ \ \ \rm ?$$

- #6

- 8

- 0

I know of this formula but I wasn't sure if it would work with the error being in the average of T...

- #7

BvU

Science Advisor

Homework Helper

- 14,268

- 3,631

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 ?

- #8

- 8

- 0

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..

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 ?

- #9

- 31,055

- 7,737

https://en.m.wikipedia.org/wiki/Propagation_of_uncertainty

- #10

- 8

- 0

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

https://en.m.wikipedia.org/wiki/Propagation_of_uncertainty

d<T^2>/<T^2> = 2d<T>/<T>

- #11

- 31,055

- 7,737

It probably depends on the mean, so you might try mean 100, variance 9 or something similar.

- #12

- 8

- 0

Ah that sounds like a good plan, thanks for your advice

It probably depends on the mean, so you might try mean 100, variance 9 or something similar.

- #13

- 31,055

- 7,737

You are welcome. Sometimes a few simulations are nearly as good as a proof.

- #14

BvU

Science Advisor

Homework Helper

- 14,268

- 3,631

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}##.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..

Fill us in on the result of Dale's suggestion ! My bet is on the simplest guess: double the relative error in T to get the relative error in T

- #15

FactChecker

Science Advisor

Gold Member

- 6,177

- 2,387

I think using (x+ε)^{2} = x^{2}+2xε+ε^{2} would give the answer that the error would be 2[itex]\bar{x}[/itex]ε + ε^{2}

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 ε^{2}, 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.)

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 ε

Last edited:

Share: