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Math Help - Elemental displacement question

  1. #1
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    Elemental displacement question

    I was reading though my lecture notes and elemental displacement is given as \delta{\underline{\textbf{r}}} = \underline{\textbf{r}}_b - \underline{\textbf{r}}_a for two points \underline{\textbf{r}}_a = x_1\underline{\textbf{i}} + x_2\underline{\textbf{j}} + x_3\underline{\textbf{z}} and \underline{\textbf{r}}_b = (x_1 + \delta{x_1})\underline{\textbf{i}} + (x_2 + \delta{x_2})\underline{\textbf{j}} + (x_3 + \delta{x_3})\underline{\textbf{z}}

    but then elemental distance is given as \delta{\underline{\textbf{s}}^2} = \delta{\underline{\textbf{r}}}.\delta{\underline{\  textbf{r}}} and i dont understand how this could be true as the square of displacement is not the distance... maybe theres something im missing something
    Last edited by renlok; October 31st 2010 at 01:19 PM.
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  2. #2
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    But what are \delta{\underline{\textbf{r}}_b} and \delta{\underline{\textbf{r}}_a}?
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  3. #3
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    oh woops its supposed to be \delta{\underline{\textbf{s}}^2} = \delta{\underline{\textbf{r}}}.\delta{\underline{\  textbf{r}}} ive edited it now
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    Since \delta{\underline{\textbf{r}}} = (\delta x_1,\delta x_2, \delta x_3), the dot product \delta{\underline{\textbf{r}}}\cdot \delta{\underline{\textbf{r}}} = (\delta x_1)^2+(\delta x_2)^2+(\delta x_3)^2.
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  5. #5
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    Quote Originally Posted by renlok View Post
    I was reading though my lecture notes and elemental displacement is given as \delta{\underline{\textbf{r}}} = \underline{\textbf{r}}_b - \underline{\textbf{r}}_a for two points \underline{\textbf{r}}_a = x_1\underline{\textbf{i}} + x_2\underline{\textbf{j}} + x_3\underline{\textbf{z}}
    Your basis vectors should be \underline{\textbf{i}}, \underline{\textbf{j}}, and \underline{\textbf{k}}, NOT \underline{\textbf{z}}

    and \underline{\textbf{r}}_b = (x_1 + \delta{x_1})\underline{\textbf{i}} + (x_2 + \delta{x_2})\underline{\textbf{j}} + (x_3 + \delta{x_3})\underline{\textbf{z}}

    but then elemental distance is given as \delta{\underline{\textbf{s}}^2} = \delta{\underline{\textbf{r}}}.\delta{\underline{\  textbf{r}}} and i dont understand how this could be true as the square of displacement is not the distance... maybe theres something im missing something
    That isn't distance, it is "distance squared".
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