Partial derivative in Newtons Second law

Question

Newton's second law states Force is the time derivative of momentum. But is it a total derivative or partial derivative? What is the reason behind it?

Answer 1

As a rule of thumb, physically meaningful quantities use total derivatives. Partial derivatives are mostly mathematical tools that help us express total derivatives, since the implicit functional dependencies that even allow us to speak of a difference between partial derivatives and total ones aren't inherent features of the physical system. Rather, they depend on our choice of how we think it's most convenient to do the math. For instance, consider the function $f(t)=3t+t^2$. We can write this as a function of two variables with implicit dependencies in different ways. For instance, we could introduce the variable $u=t^2$ and express $f$ as $f(t,u)=3t+u^2$. Or we could introduce $v=3+t$ and express $f$ as $f(t,v)=tv$. The functions are really the same, but their partial derivatives are different, while their total derivatives are exactly the same. Physically meaningful things like momentum do not depend on our arbitrary choice of how to express quantities like force, so they shouldn't actually depend on partial derivatives, but total ones. Of course, total derivatives can be expressed through partial derivatives, but again, that's just math, not physics.

So in this case, since Newton's laws are physical laws, not mathematical ones, they should work with total derivatives.

Answer 2

It is total derivative of momentum $p=mv$ w.r.t time: i.e. $F=d(mv)/dt=mdv/dt=ma$

Answer 3

The $\mathbf p$ which appears in Newton's second law is a function of one variable, so the derivative which appears is just the derivative $\mathbf p'(t)$. There is no notion of partial derivatives because there is only one variable, and there is no such thing as the "total derivative" of a function all by itself.

This is how it works: If you have a function of one variable, then you can take its derivative in the usual way. $$f(x) = x^2+3 \implies f'(x) = 2x$$

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If you have a function of two variables, then you can take its derivative with respect to one of its entries while holding the other constant.
$$g(x,y)=x^2+y^3+2xy \implies \begin{cases}(\partial_1g)(x,y)=2x+2y \\(\partial_2g)(x,y) = 3y^2+2x\end{cases}$$

Here, $\partial_1g$ is the derivative of $g$ with respect to its first slot while holding the second one constant, and similarly for $\partial_2 g$. These are the partial derivatives of $g$. In an abuse of notation, we often understand $x$ to be the thing we plug into the first slot and $y$ the thing we plug into the second, and write $\partial_1g$ as $\frac{\partial g}{\partial x}$, but I'm specifically not doing that here.

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Now consider what happens when you have a function of multiple variables (like $g$), but then we plug two different functions $a$ and $b$ of a variable (which we'll call $t$) into each of $g$'s two slots. The result will be a composite function:

$$h(t) = g\bigg(a(t),b(t)\bigg)= [a(t)]^2 + [b(t)]^3 + 2 a(t) b(t)$$

This function can be differentiated with respect to $t$. If we do so, the chain rule tells us that

$$h'(t) = a'(t) \cdot (\partial_1 g)\big(a(t),b(t)\big) + b'(t) \cdot (\partial_2g)\big(a(t),b(t)\big)$$

This is what people call the "total derivative of $g$." My argument is that this name makes no sense, because it is not a property of $g$ - it is a property of $g$ and the two functions $a$ and $b$.

Here's a more concrete example. The temperature in the room you're currently sitting in is a function of position $(x,y,z)$. Pick a point - maybe the point just in front of your nose - and think about the rate of change of temperature with respect to the $x$ coordinate. That makes sense - you can move a little bit in the $x$ direction while keeping $y$ and $z$ constant, see how the temperature changes, and calculate the derivative. You can do the same for $y$ and $z$ to obtain the partial derivatives of the temperature with respect to position.

If I come into the room and ask you for the total derivative of the temperature at that point, I'd be making no sense. What I could do is take a thermometer and move it past that point, and then ask you the rate of change of the temperature with respect to time. That's well-defined, but hopefully you can see that it depends not only on the temperature field $T(x,y,z)$, but also on the details of how I'm moving the thermometer.

Answer 4

Newton second law

$$\frac{d}{dt}(m\,\vec v)=m\,\frac{d\vec v}{dt}=\vec{F}\tag 1$$

the velocity $~\vec v$ is in general depending on the generalized coordinates $~\vec q~$ the velocity of the generalized coordinates $~\vec{\dot{q}}~$ and the time $~t$

$\Rightarrow$

the total derivative is

$$d\vec v=\frac{\partial \vec v}{\partial \vec q}\,d\vec q+ \frac{\partial \vec v}{\partial \vec{\dot{q}}}\,d\vec{\dot{q}}+\frac{\partial \vec v}{\partial t}\,dt$$

Eq. (1)

$$m\,\left(\frac{\partial \vec v}{\partial \vec q}\,\frac{d\vec q}{dt}+ \frac{\partial \vec v}{\partial \vec{\dot{q}}}\,\frac{d\vec{\dot{q}}}{dt}+\frac{\partial \vec v}{\partial t}\right)=\vec{F}$$

but if you obtain only the partial derivative , you get:

$$m\,\frac{\partial \vec v}{\partial t}=\vec{F}$$

this is not correct