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I know the definitions of vector and scalar field but I don't know what is meant by them. Suppose a scalar field is given by $\phi(x,y,z) = 3xyz$. Then what does it mean? What is the relationship between this function and the scalar field?

I am having the same difficulty in understanding vector fields.

Can anyone show me how I can express the gravitational field by a function like I wrote? I think tgat will help me very much.

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  • $\begingroup$ This function is the scalar field. It take the position in space $(x,y,z)$ and returns a number (a scalar) $3xyz$. A vector field takes a position in space in returns a vector, which should be thought of as pointing from that position. The gravitational force created by a body is a vector field, while the potential is a scalar field. $\endgroup$ – tsufli May 18 '18 at 14:34
  • $\begingroup$ @tsufli If you add some example expression for a vector field that could easily be an answer. $\endgroup$ – Javier May 18 '18 at 15:00
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This function is the scalar field. It take the position in space (x,y,z) and returns a number (a scalar) 3xyz. A vector field takes a position in space in returns a vector, which should be thought of as pointing from that position. The gravitational force created by a body is a vector field, while the potential is a scalar field.

For example, the gravitational force field of a point particle in position (0,0,0) of mass M is: $\vec F(x,y,z)=-\frac {GM}{r^{2}}\hat r=-\frac {GM}{(x^{2}+y^{2}+z^{2})^{3/2}}(x,y,z)$

While the potential is $\phi(x,y,z)=-\frac {GM}{r}=-\frac {GM}{(x^{2}+y^{2}+z^{2})^{1/2}}$

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  • $\begingroup$ Can't I write r hat as $xi+yj+zk$ and thus calculate the curl? $\endgroup$ – Theoretical May 18 '18 at 17:09
  • $\begingroup$ Why isn't there another m in your function for force field? $\endgroup$ – Theoretical May 18 '18 at 18:17
  • $\begingroup$ You can write $(x,y,z)=xi+yj+zk$, but $\hat r$ is this divided by r . There is no additional m in the force field - the force field is defined to be the force a mass m feels divided by m. (This is a logical thing to do, because it gives the information about the force generated by the mass, without regard to "what test particle is feeling it") $\endgroup$ – tsufli May 18 '18 at 19:06
  • $\begingroup$ I tried to calculate the curl but I found a non zero value which I think shouldn't happen. $\endgroup$ – Theoretical May 19 '18 at 9:49

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