# Why no basis vector in Newtonian gravitational vector field?

In my textbook, the gravitational field is given by$$\mathbf{g}\left(\mathbf{r}\right)=-G\frac{M}{\left|\mathbf{r}\right|^{2}}e_{r}$$ which is a vector field. On the same page, it is also given as a three dimensional gradient$$\mathbf{g}=-\mathbf{\nabla\phi}=-\left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right)$$ As this second equation is also a vector field, why doesn't it contain a basis vector of some sort and why isn't $\mathbf{g}$ given as a function of something or other?

Also, how do you actually get from the gravitational potential field $$\phi=\frac{-Gm}{r}$$

to the second equation? I can see that you apply the operator $\nabla$ but how does that give you $\mathbf{g}$?

Thank you

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I don't get what you mean by 'shouldn't g b a function of something or other'. It's a function of $\phi$(potential) isn't it? The second equation is more general, as it can handle any gravitational distribution; whereas the first only applies to single point mass. And the second equation does have basis vectors, it's written in cartesian shorthand, like normal 3D cartesian coordinates. – Manishearth Feb 23 '12 at 16:31
@Manishearth - Just showing my colossal ignorance! I'm a self learner and have only recently learned that $y=y\left(x\right)$ means $y$ is a function of $x$ so I was wondering why there's a $\mathbf{g}\left(\mathbf{r}\right)=$ for the first equation but only a $\mathbf{g}=$ for the second. – Peter4075 Feb 23 '12 at 19:08
Great! Even I'm a self-learner =D. The reason is that in the second equation, it's not $g(r)$ but actually $g(\left(x,y,z\right))$. Which becomes tedious. Actually when you write a quantity in bold its usually implied that its a vector field (or sometimes a plain vector). A vector field on its own need not have basis vectors, but while defining it, it does need to have them, as you pointed out. The rest is just a matter of the book switching bases without letting you know. Usually $(a,b,c)$ is cartesian unless specified. And if the equation has $x$,$y$,$z$, then its nearly always cartesian. – Manishearth Feb 24 '12 at 1:17

We want to compute the gradient of

$$\phi(r) = \phi(<x,y,z>) =\frac{-Gm}{|r|}=\frac{-Gm}{\sqrt{x^{2}+y^{2}+z^{2}}}$$

It is:

$$\left<\frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right> \phi(<x, y, z>) =$$

$$= Gm \left< \frac{x}{(x^2+y^2+z^2)^{3/2}} , \frac{y}{(x^2+y^2+z^2)^{3/2}} , \frac{z}{(x^2+y^2+z^2)^{3/2}} \right> =$$

Using $(x^2+y^2+z^2)^{3/2} = |r|^3$, we get

$$= \frac{Gm}{|r|^3} \left< x, y, z \right> =$$

By definition $r=<x,y,z>=|r| e_r$

$$= \frac{Gm}{|r|^2} e_r =$$

$$= -g(r)$$

Note that $r$ is a vector, $e_r$ is a vector, $g(r)$ is a vectorfield (maps a vector r to a vector), $\phi(r)$ is a scalarfield (maps a vector r to a scalar).

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Thanks. Afraid I'm still a little confused with the maths notation. Can we do this in baby steps? When you say $\phi\left(x,y,z\right)$ I assume you mean $\phi$ is a function of $x,y,z$. But when you say $\nabla\phi=\frac{Gm}{|r|^{3}}\left(x,y,z\right)$ I think you mean $\left(x,y,z\right)$ is a vector, which is equal to the vector $|r|e_{r}$. So $\mathbf{g}=\mathbf{g}\left(\mathbf{r}\right)=-\mathbf{\nabla\phi}$. So why does the author say $\mathbf{g}=-\mathbf{\nabla\phi}$ and not $\mathbf{g}\left(\mathbf{r}\right)=-\mathbf{\nabla\phi}$? – Peter4075 Feb 23 '12 at 15:43
I made the notation easier to comprehend by using < > for vectors, eg. r = <x, y, z> – mtrencseni Feb 23 '12 at 16:00
@Peter4075 When we write a vector as $(a,b,c)$, the basis vectors are implied. No need to write I/j/k inside (technically, it's wrong to do that). Here they've implied cartesian coordinate by using derivatives with respect to x,y,z. I think they were assuming that you knew what $\nabla$ was. – Manishearth Feb 23 '12 at 16:37
@Peter4075 If you kept the I/j/k inside the brackets, it's wrong as then that would imply $\partial_x\hat{i}\text{ scalar multiplied by }\hat{i}$ etcetera. Notation with brackets is equivalent to scalar multiplication of bracket contents with basis vectors. You can't scalar multiply two vectors, so writing I/j/k inside is technically wrong. Though you need not be nitpicky about it. – Manishearth Feb 23 '12 at 16:43
By scalar multiply I mean scalar times vector, not the scalar (dot) product. Even though they behave similarly, we cannot interchange the two just as we cannot interchange dot and cross. – Manishearth Feb 23 '12 at 16:45

To add to @mtrecseni's derivation, the definition of scalar potential of a field $\mathbf{E}$ is a scalar field $\phi$ such that $\mathbf{E}=-\nabla{\phi}$.

We can easily derive this from the fact that the potential is a path integral (a path-independant path integral actually) $$\phi=-\int\limits_{(x_1,y_1,z_1)}^{(x_2,y_2,z_2)}\mathbf{E}\cdot \vec{dl}$$ We can write the path integral as three integrals, over x, y, z: $$\phi=-\int\limits_{x_1}^{x_2}E_x(x,y_1,z_1).dx-\int\limits_{y_1}^{y_2}E_y(x_2,y,z_1).dy-\int\limits_{z_1}^{z_2}E_z(x_2,y_2,z_2).dz$$ Basically we have first taken the particle from $x_1\to x_2$, keeing $y,z$ constant, and so on. $E_x$ is the x-component of $\mathbf{E}$.

Taking $\nabla$,since y and z are constant in the first integral and so on, the equation reduces to $\nabla\phi=-(E_x\hat{i}+E_y\hat{j}+E_z\hat{k})\implies \mathbf{E}=-\nabla\phi$

This derivation works for any conservative field.

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