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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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  • $\begingroup$ 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. $\endgroup$
    – Pygmalion
    May 25, 2012 at 20:13
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    $\begingroup$ variation for what purpose? Is it the change in time, or a variation for a functional? Can you link the paper? $\endgroup$
    – Ron Maimon
    May 26, 2012 at 2:35
  • $\begingroup$ Pygmalion's explanation is clear to me. Thanks. @Ron Maimon: I mean variation in space. $\endgroup$
    – Daniel
    May 27, 2012 at 6:26
  • $\begingroup$ @Bernhard Thank you for the suggestion. I was not even aware that my comment answered Daniel's question. $\endgroup$
    – Pygmalion
    Jun 3, 2012 at 15:19

1 Answer 1

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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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