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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.

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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
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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

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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.

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