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Separation Vector 📂Mathematical Physics

Separation Vector

Definition1

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The vector from the source point to the observation point is called the separation vector.

$$ \bcR = \mathbf{r} - \mathbf{r}^{\prime} $$

Description

  • Source vector $\mathbf{r}^{\prime}$: The place where there is a charge or current. That is, it represents the coordinates of the origin of the electromagnetic field.
  • Position vector $\mathbf{r}$: Represents the coordinates of where the electric field $\mathbf{E}$ or magnetic field $\mathbf{B}$ is measured.
  • Separation vector $\bcR$: The difference between the position vector and the source vector (origin vector).

There is no standard notation for the separation vector, and it differs widely. Some people do not designate a symbol and just write $\mathbf{r} - \mathbf{r}^{\prime}$. At the shrimp sushi restaurant, like in Griffiths’ electrodynamics, it’s denoted by the cursive $r$(Kaufmann font) $\bcR$. Other characters used include the Greek letter eta $\eta$. The magnitude and unit vector of the separation vector are as follows.

$$ \left| \bcR \right| = \cR = \left| \mathbf{r} - \mathbf{r}^{\prime} \right| $$

$$ \crH = \dfrac{\bcR}{\cR} = \dfrac{\mathbf{r} - \mathbf{r}^{\prime}}{ \left| \mathbf{r} - \mathbf{r}^{\prime} \right|} $$

In the Cartesian coordinate system, it looks as follows.

$$ \bcR = (x-x^{\prime})\hat {\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}} $$ $$ \cR = \sqrt{ (x-x^{\prime})^2 + (y-y^{\prime})^2 + (z-z^{\prime})^2 } $$ $$ \crH = \dfrac{ (x-x^{\prime})\hat {\mathbf{x}} + (y-y^{\prime})\hat{\mathbf{y}} + (z-z^{\prime})\hat{\mathbf{z}}}{\sqrt{ (x-x^{\prime})^2 + (y-y^{\prime})^2 + (z-z^{\prime})^2 }} $$

Example

Find the separation vector $\bcR$ from the source point (2,8,7) to the observation point (4,6,8). Also, calculate its magnitude and unit vector.


$$ \bcR=(4,6,8)-(2,8,7)=(2,-2,1)=2\hat{\mathbf{x}} -2\hat{\mathbf{y}}+\hat{\mathbf{z}} $$

$$ \cR=\sqrt{ 2^2+ (-2)^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3 $$

$$ \crH=\left( \dfrac{2}{3}, - \dfrac{2}{3},\dfrac{1}{3} \right) = \dfrac{2}{3}\hat{\mathbf{x}} -\dfrac{2}{3}\hat{\mathbf{y}}+\dfrac{1}{3}\hat{\mathbf{z}} $$


  1. David J. Griffiths, Introduction to Electrodynamics (Translated by Jin-Seung Kim)(4th Edition). 2014, p9-10 ↩︎