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I was wondering why $F=-\frac{dU}{dr}$ would give me a vector quantity when a scalar quantity is differentiated. There are similar pre-existing queries but I think this issue has yet to be properly addressed.

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    $\begingroup$ In which context did you encounter that formula? This could be a bad notation for the gradient. See e.g. en.wikipedia.org/wiki/Gradient $\endgroup$ – AlmostClueless Aug 11 at 6:20
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    $\begingroup$ Probably this is simply an abuse of notation, your source likely means the gradient of U. $\endgroup$ – Noumeno Aug 11 at 9:08
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I was wondering why $F=-\frac{dU}{dr}$ would give me a vector quantity when a scalar quantity is differentiated.

It doesn’t. It gives you only the radial component of the force, $F_r$. Perhaps you are thinking of the negative gradient of the potential energy, $-\vec\nabla U$, which gives the force vector $\vec F$.

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  • $\begingroup$ If $U$ decreases along all the $\hat e_{x,y,z}$ axes, then in your "negative" gradient all the components are positive. Perhaps "negated" would be a more appropriate way to spell out the minus sign. $\endgroup$ – Ruslan Aug 11 at 10:39
  • $\begingroup$ Negative gradient is fine terminology used in lots of places. The fact that the gradient can be positive or negative doesn’t matter, because, as implied, you take the negative of it. $\endgroup$ – Jake Rose Aug 11 at 12:08
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$F=-\frac{dU}{dr}$ is actually $F \,\hat r=-\frac{dU}{dr}\,\hat r$ where $F$ is the radial component of the force.

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The answer to your question is more mathematics than physics. But I will try to answer it in close relation to forces so that it is easier to understand.

Let us consider a Scalar point function such as the Gravitational Potential (U). It is basically some scalar value that is associated to a coordinate point i.e. each possible vector in space has a corresponding scalar value associated with it. This Gravitational Potential is basically the work done in moving a point unit mass from infinity to a distance (r) from the center of the reference body with 0 Kinetic Energy i.e. very slowly (let us consider Earth here). Now that since every point in space that surrounds the earth has a specific value of potential associated with it, and considering that we are (lets assume X-Y-Z coordinate system....you can consider spherical coordinate system too, but lets go with the classic one as a start) we are in the XYZ Coordinate system somewhere in the space surrounding the Earth and let us also assume that we know the exact value of U at any given point in space, what is a quantity that we can measure in the given situation? It is the change in U values in X,Y and Z direction if we take a tiny steps dx,dy and dz from the point being considered.Now since these new values are specific for a given point and are dependent on the direction of the step we take, we get the output as a vector field. In the case of Gravitation we have:

enter image description here

And, this vector field now denotes the Gravitational Force per unit mass at a given point in space.Also, one can thus have value of Force acting on a unit mass at any point p in space in the direction of a vector (say) v as:

enter image description here

where F is Gravitational force per unit mass at point p.

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    $\begingroup$ Please don't post equations as images. Physics.SE uses MathJax to typeset the equations nicely. See this page for a MathJax tutorial and quick reference. $\endgroup$ – Ruslan Aug 11 at 10:44
  • $\begingroup$ Thanks. I will remember this hence forth. $\endgroup$ – Meghaj.B Aug 13 at 8:05
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$F=-\frac{dU}{dr}$ is just a derived formula from $$dU=-\vec F\cdot d\vec r$$

Note that here $\vec F\cdot d\vec r$ represents the dot product between $\vec F$ and $d\vec r$.

So this represents $$dU=-F\times dr\times \cos (\theta)$$

If you convert it to your form, you get

$$F \cos(\theta)=-\frac{dU}{dr}$$

So here you are only dealing with the magnitude of the component of $\vec F$ along $d\vec r$.

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  • $\begingroup$ Formatting hints for math in MathJax: to enter $\times$, use \times, to enter $\cos(\theta)$ (with correctly upright $\cos$) use \cos(\theta), to make displayed equation like $$a=b,$$ use double dollars instead of single: $$a=b$$. For dot product symbol $\cdot$ use \cdot. $\endgroup$ – Ruslan Aug 11 at 6:35
  • $\begingroup$ @Ruslan Thank you so much. I have used these to edit it. Can you tell me if it's fine now? $\endgroup$ – Vamsi Krishna Aug 11 at 6:39
  • $\begingroup$ Yeah, it's much more readable now. Not that I'd agree with the point being made here (or in any of the other answers on this page) though, since the question itself is not completely unambiguous. $\endgroup$ – Ruslan Aug 11 at 10:47
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A scalar and a vector quantity are referred in this case to the $\text{SO}(3)$ group, that is the group of special ortogonal matrices in tridimensional eucliden metric: translated, the group of rotations.

$U$ is a scalar and that means that for a generic $\text{SO}(3)$ transformation sending the coordinates in $\boldsymbol{x}\to\tilde{\boldsymbol{x}}$, you will have $U(\boldsymbol{x})\to\tilde{U}(\tilde{\boldsymbol{x}})$ such that $U(\boldsymbol{x})=\tilde{U}(\tilde{\boldsymbol{x}})$

$\nabla\doteq\{\partial_i\}$ is a covariant vector and that means that transforms like $\tilde{\partial}_i=\tilde{\partial}_i x^j\partial_j$ (hoping the notation is self-explaining; if not write in the comment and I'll try to be more clear); notice the term $\tilde{\partial}_i x^j$ because that is a suggestion of the vectorial nature of the gradient (this just holds in a flat space like the euclidean one, otherwise the gradient is not a vector anymore!)

So when you consider the $-\nabla U(\boldsymbol{x})$ quantity, following a transformation that goes in $-\tilde{\nabla}\tilde{U}(\tilde{\boldsymbol{x}})$ , but for what I said before the relation to previous quantity is $\tilde{\partial}_i\tilde{U}(\tilde{\boldsymbol{x}})=\tilde{\partial}_i x^j \partial_j U(\boldsymbol{x})$ and having this the same form of the $\nabla$ transformation that's a vector too.

If you consider a spherical coordinate system the $r$ component of $-\nabla U(\boldsymbol{x})$ will be the component of a vector

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    $\begingroup$ Some would say, that this was overkill. $\endgroup$ – Jake Rose Aug 11 at 12:08
  • $\begingroup$ yes maybe a bit off; consider that I'm studying these topics for the first time in my life and all the time before I was wondering why a vector was a vector and not just a set of values. The question was about the scalar and vectorial nature of the quantities so I thought that if I had the same doubt, having said that $-\partial U/\partial r\hat{\boldsymbol{r}}$ is a vector would not be of much help to me; at the same time I recognize that would not be simple to read if I never did something like this. I just wanted to give an "inspiration" not expecting to be able to give a full answer $\endgroup$ – Rob Aug 11 at 12:20
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If you have scalars related to the change of force acting on different points on the wire you can get different scalars for different points on the wire and the cange of these scalars as you move from point to point represents a vector. Let say at the center of the wire you measure 80 of potential and at two points 5 cm on the left and right of that central point 78. So You get to vectors in opposite directions as measures of change of the potential scalar. One is (80-78)/5 that points to the left and the other one (80-78)/5 that points to the right... So two distinct vectors with same magnitude but different direction....

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