Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am reading a paper and saw the author wrote something like,

Because of no-slip wall assumption, so velocity vector $\vec{v}$ is $0$ on the wall, and also the variation of this velocity on the wall is $\delta \vec{v} = 0$. And he also said $\delta \vec{v}_t = 0$.

But it seems his subtext is $\delta v_n$ is not $0$. (btw, n is normal, and t is tangent)

Hmm, this seems quite obvious to the author, but not to me.

Could someone elaborate on it? I'd like to know the physical meaning of this, so to better understand it. Why it is so obvious?

Thanks a lot.

share|improve this question
    
I do not know details, but my guess would be: Everywhere along the wall velocity is zero. So if you move in the tangential direction (along the wall), the difference $\delta\vec{v}_t$ is zero. However, if you are moving away from the wall, velocity is no longer zero, so $\delta v_n$ is not zero. –  Pygmalion May 25 '12 at 20:13
1  
variation for what purpose? Is it the change in time, or a variation for a functional? Can you link the paper? –  Ron Maimon May 26 '12 at 2:35
    
Pygmalion's explanation is clear to me. Thanks. @Ron Maimon: I mean variation in space. –  Daniel May 27 '12 at 6:26
    
@Pygmalion Maybe you can post it as an answer, so Daniel can accept it? –  Bernhard Jun 3 '12 at 14:50
    
@Bernhard Thank you for the suggestion. I was not even aware that my comment answered Daniel's question. –  Pygmalion Jun 3 '12 at 15:19

1 Answer 1

up vote 1 down vote accepted

Everywhere along the wall velocity is zero. So if you move in the tangential direction (along the wall), the difference $\delta\vec{v}_t$ is zero. However, if you are moving away from the wall, velocity is no longer zero, so $\delta v_n$ is not zero.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.