Separation Vector: Showing $\nabla(\frac{1}{||\vec{r}||})$

  • Thread starter Yeldar
  • Start date
  • Tags
    Vector
In summary, the conversation discusses how to calculate the gradient of a function using the separation vector in spherical coordinates. The speaker attempts to solve the problem using the definition of the gradient but struggles. However, they eventually find a different approach that simplifies the problem without having to use spherical coordinates.
  • #1
Yeldar
6
0
Separation Vector

Let [itex]\vec{r}[/itex] be the separation vector from a fixed point [itex](\acute{x},\acute{y},\acute{z})[/itex] to the source point [itex](x,y,z)[/itex].

Show that:

[tex]\nabla(\frac{1}{||\vec{r}||}) = \frac {-\hat{r}} {||\vec{r}||^2} [/tex]

Now, I've attempted this comeing from the approach that [itex]||\vec{r}|| = (\vec{r} \cdot \vec{r})^\frac {1} {2} [/itex] but it dosent seem to get me anywhere, am I missing something blatently obvious?

Thanks.
 
Last edited:
Physics news on Phys.org
  • #2
Go back to the definition of the gradient in spherical coordinates.
 
  • #3
Wouldnt that just complicate things further?


In Spherical Coordinates:

[tex]\displaystyle{ \nabla = \hat{r} \frac {\partial{}{}} {\partial{}{r}} + \frac {1}{r} \hat{\phi}\frac {\partial{}{}} {\partial{}{\phi}} + \frac {1}{r sin \phi} \hat{\theta}\frac {\partial{}{}} {\partial{}{\theta}} }[/tex]



I just don't see how that could simplify things?
 
  • #4
Okay, nevermind on this...

Went with a totally different appraoch and things worked out nicely without having to go into spherical coordinates.


Thanks again.
 

1. What is the Separation Vector and why is it important in science?

The Separation Vector, denoted as $\vec{r}$, is a vector that represents the distance and direction between two points or objects. It is important in science because it allows us to calculate and understand the relationship between different points or objects in a system, such as the force or potential energy between them.

2. How is the Separation Vector related to the gradient of the inverse distance function?

The gradient of the inverse distance function, denoted as $\nabla(\frac{1}{||\vec{r}||})$, is a mathematical representation of the direction and magnitude of the change in the inverse distance function at a specific point. This is directly related to the Separation Vector, as it is the vector that points in the direction of the maximum change in the inverse distance function and has a magnitude equal to the rate of change at that point.

3. Can you explain the physical significance of the Separation Vector in real-world applications?

In real-world applications, the Separation Vector is used to understand the forces and interactions between objects or particles. For example, in physics, it is used to calculate the gravitational force between two objects, while in chemistry, it is used to understand the bonding between atoms in a molecule. In engineering, it is used to analyze the stress and strain on a structure.

4. How is the Separation Vector calculated and represented mathematically?

The Separation Vector is calculated by subtracting the position vector of one point from the position vector of another point. In math notation, it can be represented as $\vec{r} = \vec{r}_2 - \vec{r}_1$, where $\vec{r}_2$ and $\vec{r}_1$ are the position vectors of the two points. To calculate the gradient of the inverse distance function, we use the formula $\nabla(\frac{1}{||\vec{r}||}) = \frac{\vec{r}}{||\vec{r}||^3}$.

5. What are some common misconceptions about the Separation Vector?

One common misconception about the Separation Vector is that it represents only the distance between two points. In reality, it also includes the direction of the vector. Another misconception is that the Separation Vector is always a straight line, when in fact it can have any direction and magnitude depending on the positions of the two points. It is important to understand the full meaning and implications of this vector in order to accurately use it in scientific calculations and applications.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
891
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
204
  • Introductory Physics Homework Help
Replies
12
Views
202
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
720
  • Introductory Physics Homework Help
Replies
4
Views
544
  • Introductory Physics Homework Help
Replies
2
Views
259
Back
Top