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For a conservative force $\vec{F}=-\vec{\nabla } U \implies \mathrm dW= -\vec{\nabla} U \cdot \mathrm d\vec{s} $

Where $\mathrm d\vec{s}$ is the infinitesimal displacement.

For a differentiable function $f$ the directional derivative in the direction of a vector $\vec{a}$ is $$\frac{\partial f}{\partial \vec{a}}= \vec{\nabla} f \cdot \vec{a}\;.$$

So is it possible to say the following?

$$\mathrm dW= -\vec{\nabla} U \cdot \mathrm d\vec{s}= -\frac{\partial U}{\partial \vec{s}}$$

I don't think that this i right because, in one dimension, $\vec{F}= -\frac{\mathrm dU}{\mathrm dx}\;.$ The spatial derivative of the potential energy is not the infinitesimal work but the force. How's that possible? What's the directional derivative of the potential energy, in the direction of $\vec{\mathrm ds}\,,$ in the case of potential energy in three dimensions? And what is its physical meaning?

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    $\begingroup$ Yes you can write that. Your source of confusion is merely one of notation. If you denoted the directional derivative as $D_\vec{s}U$ instead of $\partial U/\partial\vec{s}$ does it still cause you confusion? $\endgroup$ – lemon Apr 4 '16 at 8:09
  • $\begingroup$ @lemon Thanks for the answer! Besides notation what bothers me is that in the one dimensional case the derivative of potential energy is (minus) the force while in three dimensions the (directional) derivative of potential energy is (minus) infinitesimal work. How is that possible? $\endgroup$ – Sørën Apr 4 '16 at 8:59
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    $\begingroup$ In the 1D case, $-\nabla U\cdot d\vec{s}$ becomes $-(\partial U/\partial x)ds$. I don't see why you think the displacement term disappears... $\endgroup$ – lemon Apr 4 '16 at 9:46
  • $\begingroup$ @lemon Ok thanks I see my mistake now. If I may ask, I don't think that the expression $\vec{F}=-\frac{dU}{dx}$ is correct, since here we have a vector equal a scalar quantity, which is not possible. So is it more correct to write $\mid\vec{F} \mid=-\frac{dU}{dx} $ and $\vec{F}=-\frac{dU}{dx} \hat{u_x}$ (where $\hat{u_x}$ is the versor in the direction of $x$ axis)? Then rewriting what you mentioned in last comment (in 1d case) $dW=-\frac{\partial U}{\partial x} \hat{u_x} \cdot d\vec{s}=-\frac{\partial U}{\partial x} dx=-dU$. Can this be correct? $\endgroup$ – Sørën Apr 4 '16 at 10:05
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    $\begingroup$ In one dimension the force isn't a vector, so it's just $F=-dU/dx$... $\endgroup$ – lemon Apr 4 '16 at 10:26
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Your first line is fine, everything else is wrong (except the one time you repeated something from the first line). How wrong? You don't even have the right units, so pretty much as wrong as you can be. It's like if I asked for the surface area of a house and you said 5m or 8s or 20N or 80K.

For a differentiable function $f$ the directional derivative in the direction, $\hat a,$ of a nonzero vector $\vec{a}$ is $$\frac{\partial f}{\partial a}= \hat a\cdot\vec{\nabla} f,$$ where $\hat a =\vec a/\|\vec a\|.$

So since $\mathrm dW= -\vec{\nabla} U \cdot \mathrm d\vec s$ you get: $$\mathrm dW= -\left(\frac{\partial U}{\partial x}\hat x+\frac{\partial U}{\partial y}\hat y+\frac{\partial U}{\partial z}\hat z\right)\cdot\left(\mathrm dx\hat x+\mathrm dy\hat y+\mathrm dz\hat z\right),$$ or equivalently

$$ \mathrm dW= -\frac{\partial U}{\partial x}\mathrm dx-\frac{\partial U}{\partial y}\mathrm dy-\frac{\partial U}{\partial z}\mathrm dz.$$

Notice that now the units are correct. Any time your units are wrong it's a sign that you made at least one mistake.

What's the directional derivative of the potential energy, in the direction of $\vec{\mathrm ds}\,,$ in the case of potential energy in three dimensions?

It's the (negative of the) component of the force in the direction of $\mathrm d \vec s$. And note that the direction is a dimensionless unit vector equal to $\mathrm d\vec s/\|\mathrm d\vec s\|$.

For perspective, notice that each of the terms $\mathrm dx,$ $\mathrm dy,$ and $\mathrm dz,$ are proportional to $\mathrm ds$ (the longer your segment, the longer each of its three components) so when you divide by the length of your segment $\mathrm ds=\|\mathrm d \vec s\|$ instead of getting how much $U$ changed, you get the rate it changes per length in that particular direction. That's what a directional derivative such as $\partial U/\partial s$ is.

And what is its physical meaning?

The physical meaning of a component of the force along a direction is the component of a force along a direction. Sure, multiplying by the length of the displacement when the direction is the direction of the displacement give you the amount of work done in that displacement. But really you just need to know what potential energy is, and what a direction derivative is, how unit vectors are related to directions which are different than vectors with units and magnitudes. And what a directional deriavtive is.

A directional deriavtive is a scalar that you tell you rate (per length of movement in the domain) that a function undergoes when you evaluate it at different places in the domain that are displaced along a particular direction.

You could imagine a line in the domain and then evaluate the function along that line and then take its slope at that point.

Or if you have a restricted domain you might need to take a curve that goes through the point whose curve has that direction as a tangent and then find the slope of the line whose projection onto the domain is tangent to the curve at that point and which is tangent to the function at that point.

It's just about rates of change in various directions. It's as fundamental as that.

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