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Line integrals distance elements

  1. Aug 21, 2011 #1
    in line integrals we always need a vector element of distance. I cant understand the difference between ds and dr. is ds for all kinds of paths (even curly ones) and dr only for straight lines, or theyre the same? Im confused, or maybe dr is just the magnitude of ds, and the vector here is the ds,which one is true?
     
  2. jcsd
  3. Aug 21, 2011 #2

    BruceW

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    If [itex]\vec{r} \ [/itex] represents the possible points on some path, then [itex]d \vec{r}[/itex] and [itex]d \vec{s} \ [/itex] mean the same thing. And also dr and ds would mean the same thing.
    So they are both used for curly lines. I think some professors will write one, and others write the other. Its just a different notation.
     
  4. Aug 21, 2011 #3

    BruceW

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    For example, ds is defined as:
    [tex] ds = \sqrt{ d \vec{r} \cdot d \vec{r} } [/tex]
    and so ds = dr. I think ds is also used to mean paths through spaces other than normal space (i.e. spacetime or momentum-space). So ds just means a general path integral, but dr is specifically through normal space.
    So if your teacher says "ds as a path through space" then it does mean the same as dr.
     
  5. Aug 21, 2011 #4
    Good morning, y.moghdamnia, welcome to physics forums.

    dr and ds are not the same thing at all. We actually want ds, but use dr as the next best thing.

    Since you are approaching line integrals through vectors here is a vector explanation of what is going on.

    With reference to the attached diagram.

    ds is an element of any suitable curve C.
    Note ds is curved and measured along C.

    In order to specifiy the curve we consider a centre O and a vector r from O to any point (A) on the curve.

    Now let us move along the curve to another point, B.

    the vector r changes to another vector r+dr
    The distance along the curve is ds. Note it is curved.

    In order to recover dr we take the vector difference (r+dr - r) = dr
    Note that this is like all vectors, a straight line. Further it is tangent to the curve at A.
    Further note that this vector difference is given by the closure of triangle AOB as in the diagram.

    Now we are doing some (simple) vector calculus, which follows the same pattern as elementary scalar calculus you are already familiar with.

    We let dr approach smaller and smaller values (zero) and take the limit, where we find that
    dr and ds concide. But only at A.

    does this help?
     

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