Is the acceleration vector half of the gradient of velocity squared?

Question

Consider the differentiation of speed squared with respect to time: $$\frac{d(v^2)}{dt}=\frac{d(\mathbf v\cdot\mathbf v)}{dt}$$ $$=2\mathbf v\cdot\frac{d\mathbf v}{dt}$$ $$=2\mathbf v\cdot\mathbf a$$ $$=2\mathbf a\cdot\mathbf v$$ Which gives

$$d(v^2)=2\mathbf a\cdot\mathbf vdt$$ $$=>d(v^2)=2\mathbf a\cdot d\mathbf r$$ This is similar to the differential relation of potential energy U(r) with conservative force $\mathbf F$ $$dU(r)=-\mathbf F\cdot d\mathbf r$$ Since this relation can be expressed as $$\mathbf F=-\nabla U(r)$$ So does it imply $$\mathbf a=\frac{1}{2}\nabla v^2$$ I checked this relation for one and two dimensional motion and it does seem to give correct acceleration. So is it correct?

Answer 1

The main difference between $\mathbf{F} = -\nabla U(\mathbf{r})$, which is true for a conservative force derived from a potential $U(\mathbf{r})$, and the proposed equation $\mathbf{a} = \dfrac{1}{2}\nabla(v^2)$, which is not true, derives from a subtle difference in how we should understand the functional dependence of $\mathbf{F}$ and $\mathbf{a}$ on $\mathbf{r}$ in each case.

The identification $dU = -\mathbf{F}\cdot d\mathbf{r}$ is true for any spatial displacement in the position $\mathbf{r}$ in any trajectory---after all, it is essentially the defining relation between force and potential energy in a conservative system. The step where you replaced $\mathbf{v}dt$ with $d\mathbf{r}$, on the other hand, is only valid once a trajectory has been fixed. This automatically means that any functional relation between $\mathbf{a}$ and $\mathbf{r}$ obtained after that step is only valid for one given trajectory, and is therefore no longer true if you treat $\mathbf{r}$ as an independent variable (as opposed to a collection of implicit functions of time, $\mathbf{r} = \mathbf{r}(t)$).

Just to illustrate how important this distinction is, let us assume for a minute that the answer to your question is affirmative, so that we can essentially treat $v^2$ as an independent function of $\mathbf{r}$ from which the acceleration is derived through $\mathbf{a} = \frac{1}{2}\nabla (v^2)$. By making use of $\mathbf{F} = m\mathbf{a} = -\nabla U$, we would get to $$\nabla\left(\dfrac{1}{2}mv^2 + U(\mathbf{r})\right) = 0.$$

Now, at a superficial level, this may just look like the good old law for conservation of energy, $\dfrac{1}{2}mv^2 + U(\mathbf{r}) = \text{constant}$. But if this were true as a vector field in $\mathbb{R}^n$ (if space is $n$-dimensional), it would mean something much more restrictive than conservation of energy: it would actually imply that

$$\dfrac{1}{2}mv_1^2 + U(\mathbf{r}_1) =\text{constant}=\dfrac{1}{2}mv_2^2 + U(\mathbf{r}_2)$$

even for points $\mathbf{r}_1$ and $\mathbf{r}_2$ which are not connected by a solution to the equations of motion. In other words, not only would energy be conserved, but it would actually attain the same value (which we might as well set to zero by shifting the potential by a constant) among all solutions to the equations of motion. This is clearly not true in general, even for conservative systems. The version of the equation that is true is very similar to the previous one, but instead of having the gradient, we just have $$\dfrac{d}{dt}\left(\dfrac{1}{2}mv^2 + U(\mathbf{r})\right) = 0$$ with $\mathbf{v}$ and $\mathbf{r}$ being treated as implicit functions of $t$.

Answer 2

When deriving the work-energy theorem, the steps are similar. From the Newton's second law $\mathbf F = m\mathbf a$. Making the dot product with the displacement $\mathbf {dr}$: $$dw = \mathbf F \cdot \mathbf {dr} = m \frac{\mathbf{dv}}{dt} \cdot \mathbf{dr} = m \frac{\mathbf{dr}}{dt}\cdot \mathbf{dv} = m\mathbf v\cdot \mathbf{dv} = d(\frac{1}{2}mv^2)$$

If $\mathbf F$ is minus the gradient of a scalar potential: $$-\nabla V \cdot \mathbf {dr} = d(\frac{1}{2}mv^2)$$ From this we can not derive that: $$d(\frac{1}{2}mv^2) = \nabla (\frac{1}{2}mv^2)\cdot \mathbf{dr} $$ because it is to say that the kinetic energy is a function of the position, when it could be written: $$d(\frac{1}{2}mv^2) = \frac {\partial(\frac{1}{2}mv^2)}{\partial x}dx + \frac {\partial(\frac{1}{2}mv^2)}{\partial y}dy + \frac {\partial (\frac{1}{2}mv^2)}{\partial z}dz $$ But that is not the case.

Answer 3

Consider constant a, so $v=a t$, $v^2=a^2 t^2$, $\frac{dv}{dt}=2a^2t$. Your error $v^2$ has no information of direction, so your following vectors do not follow.