Second law of Newton for variable mass systems

Question

Frequently I see the expression $$F = \frac{dp}{dt} = \frac{d}{dt}(mv) = \frac{dm}{dt}v + ma,$$ which can be applied to variable mass systems.

But I'm wondering if this derivation is correct, because the second law of Newton is formulated for point masses.

Furthermore if i change the inertial frame of reference, only $v$ on the right side of the formula $F = \frac{dm}{dt}v+ma$ will change, meaning that $F$ would be dependent of the frame of reference, which (according to me) can't be true.

I realize there exists a formula for varying mass systems, that looks quite familiar to this one, but isn't exactly the same, because the $v$ on the right side is there the relative velocity of mass expulsed/accreted. The derivation of that formula is also rather different from this one.

So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake.

Answer 1

But I'm wondering if this derivation is correct, because the second law of Newton is formulated for point masses.

So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake.

Newton's second law is not meant only for point masses, but for any massive body which does not lose or gain massive parts.

The formula $$ \mathbf F_{ext} = \frac{d\mathbf p}{dt},~~~\tag{1} $$ expressing 2nd law mathematically, thus $is$ correct, provided $\mathbf F_{ext}$ is external force on the system and $\mathbf p$ is momentum of the system, where system is a body which does not lose or gain massive parts. Thus mass of the body is assumed constant. But then we have $d\mathbf p/dt = m\mathbf a$, and we can rewrite the equation as $$ \mathbf F_{ext} = m\mathbf a.~~~\tag{2} $$ It turns out, surprisingly, that the former equation is not easily adaptable to the case of variable mass systems, but the latter equation can be generalized to the case of variable mass easily, by adding new force term on the left-hand side.

But there is a persistent idea in literature and many minds, that (1) is actually the more general equation, and that it applies to variable mass systems.

An example of such a variable mass system is e.g. all matter considered to be part of a rocket (thus including the fuel and oxidizer inside, but without the expelled gases far away from the engine nozzle). Obviously, such system has variable mass. More generally, we can consider all matter inside a well-delimited, possibly moving control volume, as the subject system for which we seek the equation of motion.

The above idea that (1) is more general and is applicable to variable mass systems keeps around in part because of a) Newton writing 2nd law this way; b) some mechanics textbooks suggesting so and also c) (probably) because people get the impression that since $\mathbf F_{ext} = d\mathbf p/dt$ is the more accurate equation in special relativity, where "mass increases with speed", while the other equation $\mathbf F_{ext} = m\mathbf a$ is less correct, this "ability to account for variable mass" should be true also in Newtonian mechanics.

But a) Newton meant bodies that do not lose/gain mass; b) those textbooks are either misleading/confusing, or even wrong on this; c) notice that even in special relativity, (1) is valid only for systems of constant rest mass. If rest mass changes, like for a rocket that loses rest mass, (1) is not correct in special relativity. Similarly, in non-relativistic mechanics, (1) is not correct when the system loses or gains massive parts.

In the interesting case where the system of interest (inside the control volume) does lose or gain massive parts (like a rocket), the equation (1) is no longer valid. An easy way to see this is that this equation is not Galilei-invariant, and thus cannot be true in all inertial frames; when we write it in different frame nevertheless, the external force does not change, but the right-hand side does, so the equation cannot hold in all frames, and thus cannot be general equation of motion for such system.

The second equation (2) is not correct in this case either, as the system with variable mass can have non-zero acceleration even in absence of external forces (let's not count the force of the exhaust gases on the rocket nozzle as part of the external force for now, as it is due to mass which is very close to the system (touching it), and which may be considered part of the system, when we put the boundary of the control volume far enough from the surface of the rocket nozzle).

However, a different and correct equation for $\mathbf p$ (a variable-mass system's momentum) may be derived from Newton's 2nd law when applied to each particle in the control volume¹. This can be written as

$$ \mathbf F_{ext} + \mathbf F_{parts~outside} = \frac{d\mathbf p}{dt} + \frac{d\mathbf p_{lost}}{dt}~~~(4). $$

¹⁾ <sub>This can be done because each particle $i$ obeys the equation $\mathbf F = m_i\mathbf a_i$, as it does not lose or gain parts. One way to derive the equation above goes like this.
Let us use a convention where $\mathbf F_{a,-b}$ means force due to particle (or set of particles) $a$ acting on the particle (or set of particles) $b$.
Let $V(t)$ be the region of space that is our control volume, which defines "the system". It is easy to see that at time $t$ $$ \sum_{i\in V(t)} \mathbf{F}_{ext,-i} + \sum_{i \in V(t)} \mathbf F_{parts~outside,-i} = \sum_{i\in V(t)} m_i \mathbf a_i.~~~(a) $$ We would like to express this without summing over index $i$, using only quantities referring to the system inside and outside as a whole.
Let $\mathbf p$ be momentum inside the control volume $V$. This changes in time for two reasons:
- the particles that stay inside change their momentum in time;
- some particles leave or come in to the control volume.

So we can write $$ \frac{d\mathbf p}{dt} = \sum_{i\in V} m_i \mathbf a_i - \frac{d\mathbf p_{lost}}{dt}~~~(b) $$
where $d\mathbf p_{lost}$ is momentum lost from the control volume due to particles leaving the control volume, per time $dt$.
Comparing $(a)$ and $(b)$, we can see that summation of terms $m_i\mathbf a_i$ over $i$ can be replaced by an expression involving $d\mathbf p/dt$ and $d\mathbf p_{lost}/dt$.</sub>

_{Introducing $\mathbf F_{ext}=\sum_{i\in V} \mathbf F_{ext,-i}$ and $\mathbf F_{parts~outside}=\sum_{i\in V} \mathbf F_{parts~outside,-i}$, we can express the above equation as $$ \mathbf F_{ext} + \mathbf F_{parts~outside} = \frac{d\mathbf p}{dt} + \frac{d\mathbf p_{lost}}{dt}, $$ which is the sought equation (4).}

Thus we can see that net force on the particles in the system (those inside the control volume) does not equal rate of change of momentum of the system; but there is an additional term, rate of change of momentum lost from the control volume by particles crossing the boundary. This is why the idea that (1) applies to variable mass systems is untenable. Howsoever we re-interpret the term $\mathbf F_{ext}$ to include the forces due to outside particles acting back on the system, we won't get from (1) the correct equation involving $d\mathbf p/dt$. We have to use (4) instead, to take into account the change of momentum not due to forces, but due to gain/loss of particles.

When we apply this to a rocket (with control volume that follows the rocket shell, completely containing it), we can see that momentum of the rocket $\mathbf p$ changes due to 1) external force, 2) force of the exhaust gases acting back on the system in the control volume (rocket+some exhaust gas inside the control volume), and 3) some momentum going out of the control volume (due to exhaust gas leaving the system).

This expression of the equation of motion in terms of momenta is somewhat foreign to the usual treatment of a rocket in mechanics, because for the purpose of rocket mechanics in gravity field, we are interested in its velocity, rather than momentum.

In the simplest case where the lost particles all leave in direction same or opposite to the body velocity $\mathbf v$ (idealized rocket), the equation of motion can be further simplified, as is common in textbooks, in terms of velocities. Let the boundary of the comoving control volume be far from the rocket, so that force $\mathbf{F}_{parts~outside}$ acting back on the system in the control volume is negligible (the exhaust gas is rarified and has low pressure on the control volume boundary). Velocities of gas particles crossing the boundary of the control volume are close to average value $\mathbf v + \mathbf{c}$, where $\mathbf c$ is the average final velocity of the particles when very far away from the nozzle, in the frame of the rocket. Then the momentum that gets lost per unit time is

$$ \frac{d\mathbf{p}_{lost}}{dt} = - \frac{dm}{dt}(\mathbf c+\mathbf v).~~~(5) $$ Using these findings, the equation of motion (4) simplifies into

$$ \mathbf{F}_{ext} + \frac{dm}{dt} \mathbf c = m\frac{d\mathbf{v}}{dt}.~~~(6) $$ This equation has the same form as (2), only with a different net force, where we have to add the term $\frac{dm}{dt} \mathbf c$ to the external force. It is tempting to interpret this term as value of the force of the gas acting on the nozzle; however, such interpretation is hard to justify based just on the above derivation. Notice that this term comes from the term giving lost momentum per unit time, not from the term giving the force of the outside gas on the system in the control volume. We would have to do another derivation with a different control volume whose boundary would copy the rocket surface, including the nozzle surface and the pipes surface right to the place where the fuel is at rest wrt to the rocket shell. Then, comparing to the above, we would find out that force acting on the nozzle and pipes above it is indeed $\frac{dm}{dt} \mathbf c$, where $\mathbf c$ is the final velocity of the exhaust gas, far away from the nozzle.

It is also interesting to note that $\mathbf v$ in this equation is not velocity of the center of mass of the rocket, but velocity of its solid shell. It satisfies the relation $\mathbf p = m \mathbf v$ where $\mathbf p$ and $m$ are momentum and mass inside the (latter, close to the rocket surface) control volume. But because most of the mass gets lost, and center of mass moves closer to the tip of the rocket, this center moves somewhat faster than the solid rocket shell.

Answer 2

You'll probably find the wikipedia article useful: http://en.wikipedia.org/wiki/Variable-mass_system

It is formulated so that $F+v_{rel} \frac{dm}{dt} = ma$, where $v_{rel}$ is the relative velocity of the mass being ejected to the center of mass of the body. This takes care of your question about reference frames, because $v$ will be the same in all frames. The term gets moved to the left side of the equation because $-v$ describes the velocity of the center of mass relative to the ejected matter.

Answer 3

You can only apply Newton's second law to closed systems. But, since you are applying second law to a open system, you are getting contradictory results. The correct procedure for solving variable mass system, is by calculating the change in momentum and then equating it to

    Force = (change in momentum)/small time interval in which change occurred.

Here is a article on it that you can find useful, apart from the wikipedia article. See, this website. http://www.thestudentroom.co.uk/wiki/Revision:Motion_With_Variable_Mass

Answer 4

First, we should remove some misconceptions that are actually getting in the way of the discussion here. Newton's laws were not formulated for point masses! The very first two definitions in the Principia frame bodies as continua, irrespective of any distinction between composite versus elementary, and do so with the implied understanding that the same laws apply to a whole body as apply to each of its parts. It is for this reason that you also have the third law - as that is the key enabling condition that allows you to scale the first and second law up from a body's parts to the body as a whole.

Definition 1 states that the quantity of matter comprising a body (its "mass") is - what we would say in contemporary language - the integral of its density over its volume. In so stating, it asserts that mass is additive and (tacitly) that it is positive.

By "additivity" I will also be referring to the stronger sense where two bodies can occupy overlapping regions, such as with the mixing of fluids or the lodging of a soda straw into a tree trunk by a tornado.

Definition 2 associates each part of the body with a velocity, and states - as we would today say - that the quantity of motion of the body (its "momentum") is the integral of the product of the density and velocity over the volume of the body. This means that the momentum is also additive, and it also gives you a definition for the average velocity of a body and - by these means - the average position of a body, up to a constant of time integration. It tacitly states (by the context of subsequent discussion) that this constant (that is: the constant of integration of the time integral of the "quantity of motion") is also additive; specifically that mass-moment-at-a-fixed-time is additive.

This part-whole and upward-scalability approach, where no consideration is paid to "elementariness", in fact, might be seen as a precursor to the approach Wilson later adopted in Quantum Field Theory.

There is nothing said in Newton's formulation of his laws about point masses in any of this. The only assumption (tacitly) made, in carrying on to formulate the laws, is that a body has consistent contents - that its mass be constant. The assumption incurs no real loss of generality, since a body of variable mass can be treated as a component of a larger body of constant mass, where the choice of where to draw the dividing line between what comprises that body, versus what comprises the rest of the larger body it is contained in, is allowed to vary with time.

It might do well to review the reply I gave here Rocket Equation where I spell out the derivation in general terms. That's exactly the approach adopted there.

There's nothing wrong with continuing to assert $𝐅 = d𝐩/dt$, for a body with mass $m$ and momentum $𝐩$, as long as you understand that when the mass is variable, then $𝐅$ is no longer frame-independent. Under Galilean a boost by $𝐮$, the coordinates, coordinate differentials, velocity, partial derivative operators and moving time derivative transform as $$(𝐫,t) → (𝐫 - 𝐮t, t), \quad (d𝐫,dt) → (d𝐫 - 𝐮dt, dt), \quad 𝐯 → 𝐯 - 𝐮,\\ \left(∇,\frac{∂}{∂t}\right) → \left(∇, \frac{∂}{∂t} + 𝐮·∇\right), \quad \frac{d}{dt} → \frac{d}{dt},$$ where $$𝐫 = (x, y, z), \quad ∇ = \left(\frac{∂}{∂x}, \frac{∂}{∂y}, \frac{∂}{∂z}\right), \quad \frac{d}{dt} = \frac{∂}{∂t} + 𝐯·∇. $$

This is what leads to the transport laws for continuua: $$m_Ω = \int_Ω ρ dV\quad⇒\quad\frac{dm_Ω}{dt} = \frac{d}{dt} \int_Ω ρ dV = \int_Ω \left(\frac{∂ρ}{∂t} + ∇·(𝐯ρ)\right) dV,$$ where the region $Ω$ comprising the body can be time-varying, where $ρ$ is the density within the region, $𝐯$ the associated velocity and $dV = dx∧dy∧dz$ is the volume 3-form. If the region follows the body's contents, then the mass is constant and that leads to the transport equation: $$\frac{∂ρ}{∂t} + ∇·(ρ𝐯) = 0.$$

Otherwise if the contents are not constant, then you're treating this as part of a larger body where the boundary of $Ω$ is allowed to have flow in and out. And that leads to the transport law for momentum: $$\frac{∂(ρ𝐯)}{∂t} + ∇·(ρ𝐯𝐯) = 𝞅,$$ where tensor-dyad notation is used, and the decomposition $𝞅 = ρ𝐟$ for (the additive) force density $𝞅$ into the product of mass density $ρ$ and force per unit mass $𝐟$ is frequently employed. As stated, it is not additive. Therefore, solely on consistency grounds, there needs to be an additional term $𝐏$ alongside the dyad $ρ𝐯𝐯$ that absorbs the non-additivity, thus leading to the modified equation $$\frac{∂(ρ𝐯)}{∂t} + ∇·(ρ𝐯𝐯 + 𝐏) = 𝞅.$$ The dyad $𝐏$ is then identified as the body's stress tensor. Thus, for two bodies with respective densities, velocities and stress tensors $\left(ρ_0, 𝐯_0, 𝐏_0\right)$ and $\left(ρ_1, 𝐯_1, 𝐏_1\right)$ occupying respective regions $Ω_0$ and $Ω_1$ - that may overlap - you can treat them as being part of one body within a region $Ω ⊇ Ω_0 ∪ Ω_1$, by just requiring $\text{supp }ρ_0 ⊆ Ω_0$ and $\text{supp }ρ_1 ⊆ Ω_1$ and writing $$ \int_{Ω_0} ρ_0 dV + \int_{Ω_1} ρ_1 dV = \int_Ω ρ dV,\\ \int_{Ω_0} ρ_0 𝐯_0 dV + \int_{Ω_1} ρ_1 𝐯_1 dV = \int_Ω ρ 𝐯 dV,\\ \frac{∂ρ_0}{∂t} + ∇·(ρ_0𝐯_0) + \frac{∂ρ_1}{∂t} + ∇·(ρ_1𝐯_1) = \frac{∂ρ}{∂t} + ∇·(ρ𝐯),\\ \frac{∂\left(ρ_0𝐯_0\right)}{∂t} + ∇·\left(ρ𝐯_0𝐯_0 + 𝐏_0\right) + \frac{∂\left(ρ_1𝐯_1\right)}{∂t} + ∇·\left(ρ𝐯_1𝐯_1 + 𝐏_1\right) = \frac{∂\left(ρ𝐯\right)}{∂t} + ∇·\left(ρ𝐯𝐯 + 𝐏\right),\\ 𝞅_0 + 𝞅_1 = 𝞅, $$ then the additivity entails the following part-whole decompositions: $$ ρ_0 + ρ_1 = ρ ⇒ ρ = ρ_0 + ρ_1,\\ ρ_0𝐯_0 + ρ_1𝐯_1 = ρ𝐯 ⇒ 𝐯 = \frac{ρ_0𝐯_0 + ρ_1𝐯_1}{ρ_0 + ρ_1},\\ ρ_0𝐯_0 + 𝐏_0 + ρ_1𝐯_1 + 𝐏_1 = ρ𝐯 + 𝐏 ⇒ 𝐏 = 𝐏_0 + 𝐏_1 + \frac{ρ_0ρ_1}{ρ_0 + ρ_1}(𝐯_0 - 𝐯_1)(𝐯_0 - 𝐯_1),\\ 𝞅_0 + 𝞅_1 = 𝞅 ⇒ 𝞅 = 𝞅_0 + 𝞅_1. $$

If, on the other hand, you do not follow a body's contents, by adopting a velocity field whose flow does not go with the mass density, then continuity law for the mass will no longer be homogeneous, but will take the form $$\frac{∂ρ}{∂t} + ∇·(ρ𝐯) = μ.$$ That's equivalent to dividing a body into parts where the enclosing boundary may vary with time - in a way that cuts progressively across the body. Under transform, we have: $$\begin{align} \frac{∂ρ}{∂t} + ∇·(ρ𝐯) &→ \left(\frac{∂}{∂t} + 𝐮·∇\right)ρ + ∇·(ρ(𝐯 - 𝐮))\\ &= \frac{∂ρ}{∂t} + ∇·(ρ𝐯),\\ \frac{∂(ρ𝐯)}{∂t} + ∇·(ρ𝐯𝐯 + 𝐏) &→\left(\frac{∂}{∂t} + 𝐮·∇\right)(ρ(𝐯 - 𝐮)) + ∇·(ρ(𝐯 - 𝐮)(𝐯 - 𝐮) + 𝐏)\\ &= \frac{∂(ρ𝐯)}{∂t} + ∇·(ρ𝐯𝐯 + 𝐏) - \left(\frac{∂ρ}{∂t} + ∇·(ρ𝐯) \right)𝐮, \end{align}$$ thus $$μ → μ, \quad 𝞅 → 𝞅 - μ𝐮.$$

To make a boost-invariant out of $𝞅$ will then require associating the mass flux $μ$ with a momentum flux $𝝿$ and an associated velocity $𝐯_μ = 𝝿/μ$. This is the rate at which the mass and momentum are passing through the boundary of the region $Ω$ associated with the body, due to the fact that the velocity field $𝐯$ is not following the body's mass flow. Then we can define a boost-invariant version $𝞅_0$ of $𝞅$ by $𝞅_0 = 𝞅 - 𝝿$, with the transforms: $$𝐯_μ → 𝐯_μ - 𝐮, \quad 𝝿 → 𝝿 - μ𝐮, \quad 𝞅_0 → 𝞅_0.$$

Similarly, for the body, itself, we have the transform: $$\frac{dm}{dt} → \frac{dm}{dt}, \quad \frac{d𝐩}{dt} → \frac{d𝐩}{dt} - \frac{dm}{dt}𝐮\quad→\quad𝐅 → 𝐅 - \frac{dm}{dt}𝐮.$$ So, associating a momentum flux $𝝥$ and velocity $𝐯_{\dot{m}}$ with the mass flux $dm/dt$, with $$𝝥 = \frac{dm}{dt}𝐯_{\dot{m}}, \quad 𝐯_{\dot{m}} → 𝐯_{\dot{m}} - 𝐮, \quad 𝝥 → 𝝥 - 𝐮\frac{dm}{dt},$$ we can define the boost-invariant force by: $$𝐅_0 = 𝐅 - 𝝥.$$ The force law will then read: $$\frac{d𝐩}{dt} = 𝐅 = 𝐅_0 + 𝝥 = 𝐅_0 + \frac{dm}{dt}𝐯_{\dot{m}}\quad→\quad\frac{d𝐩}{dt} - 𝐯_{\dot{m}}\frac{dm}{dt} = 𝐅_0.$$

Answer 5

1. Whether a mass can be considered a point or not depends on the scale at which it is studied. The earth can be considered to be a point mass when we are studying its motion around the sun but not so when we are studying its own rotation. Newton's laws are applied to systems of many particles.

2. Newton's second law says that the rate of change of momentum of a system is proportional to the applied force. We choose units in such a manner that the constant of proportionality is 1. With this definition, the equation $md\vec{v}/dt + \vec{v} dm/dt = \vec{F}$ makes sense with $\vec{v}$ being the velocity of the body in the same frame of reference in which other vectors are measured.