- #1

- 15

- 0

for eg. < [tex]\frac{df}{dt}[/tex] > = [tex]\frac{d<f>}{dt}[/tex]

Here < > is temporal averaging. If the differentiation is w.r.t. a spatial coordinate it makes sense, but could someone help me with the above equation?

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.

- Thread starter plasmoid
- Start date

- #1

- 15

- 0

for eg. < [tex]\frac{df}{dt}[/tex] > = [tex]\frac{d<f>}{dt}[/tex]

Here < > is temporal averaging. If the differentiation is w.r.t. a spatial coordinate it makes sense, but could someone help me with the above equation?

- #2

Stephen Tashi

Science Advisor

- 7,707

- 1,515

Is it as simple as moving differentiation past the integral sign? Physicists do that all the time.

[tex] \int \frac{d}{dt}(f(t)) [/tex]

[tex] = \frac{d}{dt} \int f(t) dt [/tex]

- #3

- 15

- 0

Is it as simple as moving differentiation past the integral sign? Physicists do that all the time.

[tex] \int \frac{d}{dt}(f(t)) [/tex]

[tex] = \frac{d}{dt} \int f(t) dt [/tex]

That's it actually. But all the math texts I saw move differentiation past the integral sign only when the two involve

Edit - I just cant get the latex to work :( - you're missing a dt after f(t) in the L.H.S. of your equation.

Last edited:

- #4

lurflurf

Homework Helper

- 2,440

- 138

- #5

Stephen Tashi

Science Advisor

- 7,707

- 1,515

That's it actually. But all the math texts I saw move differentiation past the integral sign only when the two involveindependentvariables.

That's a good point. What I wrote suggests an idea, but it doesn't really make sense to write

[tex] \frac{d}{dt} \int f(t) dt [/tex]

since the integralion shouldn't leave a function that still depends on t. So you should double check what you said about <f> being a temporal average. Perhaps it is a spatial average.

Is there a rigorous justification for doing it when both the differentiation and integration involve thesamevariable?

Yes, there are conditions that say when one may do this. I can look them up. They aren't something that most people remember since functions describing physical phenomena are usually "well behaved".

Edit - I just cant get the latex to work :( - you're missing a dt after f(t) in the L.H.S. of your equation.

LaTex on the forum looks crazy because of a problem with the pages showing a stale cache of LaTex expressions. (There are threads discussing this.) What you must do is to "preview" your page and then uses the browsers refresh or reload function after that.

- #6

lurflurf

Homework Helper

- 2,440

- 138

It would help if the definition given were given here. The usual temporal average does not remove temporal dependence. Often an overbar is used instead of a bracket to avoid confusion. For example

[tex]\int_{-\infty}^{\infty} f(x,s)w(s-t) ds[/tex]

or

[tex]\frac{1}{2h}\int_{t-h}^{t+h} f(x,s) ds[/tex]

[tex]\int_{-\infty}^{\infty} f(x,s)w(s-t) ds[/tex]

or

[tex]\frac{1}{2h}\int_{t-h}^{t+h} f(x,s) ds[/tex]

Last edited:

- #7

- 15

- 0

<f(t,x)> = [tex]\stackrel{lim}{T\rightarrow\infty}[/tex] [tex]\frac{1}{T}[/tex] [tex]\int_{t_{0}}^{t_{0}+T} f(t,x) dt [/tex]

And it's all making even less sense to me now. With this definition, <f(t,x)> is independent of t (as it should if the "average" is to make sense.).

So is

[tex]\frac{\partial<f>}{\partial t}[/tex] = < [tex]\frac{\partial f}{\partial t} [/tex] > ?

Shouldn't the L.H.S. be zero? I'm begiining to think there's something wrong in the problem I'm working on ....

- #8

lurflurf

Homework Helper

- 2,440

- 138

It would be more usuual and less nonsensical if it were

[tex]\frac{1}{T}\int_{t}^{t+T} f(s,x) ds [/tex]

Share: