# 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

• 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. Commented Feb 23, 2012 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. Commented Feb 23, 2012 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. Commented Feb 24, 2012 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).

• 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 $\mathbf{g}\left(\mathbf{r}\right)=-\mathbf{\nabla\phi}$? Commented Feb 23, 2012 at 15:43
• I made the notation easier to comprehend by using < > for vectors, eg. r = <x, y, z> Commented Feb 23, 2012 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. Commented Feb 23, 2012 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. Commented Feb 23, 2012 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. Commented Feb 23, 2012 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.