# How does $F = \frac{ \Delta (mv)}{ \Delta t}$ equal $( m \frac { \Delta v}{ \Delta t} ) + ( v \frac { \Delta m}{ \Delta t} )$?

That's how it's framed in my Physics school-book.

The question (or rather, the explanation) is that of the thrust of rockets and how the impulse is equal (with opposite signs) on the thrust-gases and the rocket itself.

$( m \frac { \Delta v}{ \Delta t} ) = -(v \frac {\Delta m}{ \Delta t} ) = F_i$

I suppose it's a problem with how I see the transfer of impulse and exactly which part of the equation relates to which part of the physical world (gases, rocket). So we can start from there.

Title equation:

$F = \frac{ \Delta (mv)}{ \Delta t} = \left ( m \frac { \Delta v}{ \Delta t} \right) + \left ( v \frac { \Delta m}{ \Delta t} \right)$

Grade: The equivalent of G-10 in the US.

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Hi M.Na'el, and welcome to Physics Stack Exchange! I took the "homework" tag off because it looks like you're really asking a conceptual question, not a homework question - but that's good! I rather like this question. –  David Z Apr 26 '12 at 17:59
@DavidZ, Maybe, but I just wanted to be sure I don't get a final answer as my book gave. I wanted thorough steps hence the tag. In any case, you're the experts around here :) –  Noein Apr 28 '12 at 10:48
Your curiosity is much appreciated :-) No need to worry, though; we always give explanations, not plain answers, regardless of what tags may be (or not be) on the question. –  David Z Apr 28 '12 at 11:04

Here is a visualization:

Momentum is mass times velocity, so draw it as the area of a rectangle:

If we change the mass and velocity a little, we change the momentum:

The total change in the momentum is the sum of green, blue, and purple rectangles. Their sizes are just length times width, so overall we have

$\Delta p = m\Delta v + v\Delta m + \Delta v \Delta m$

This looks like the answer you're seeking except for the extra term at the end.

Suppose we cut $\Delta m$ and $\Delta v$ down to one tenth their current size. Then the first two terms become one tenth as large, but $\Delta m \Delta v$ becomes one hundredth as large. The purple box shrinks away much faster than the blue and green ones. Therefore, for very small changes, we can ignore the purple box and write

$\Delta p = \Delta(mv) = m\Delta v + v\Delta m$

we usually indicate this limiting procedure by changing the $\Delta$ to $\mathrm{d}$, so

$\mathrm{d} p = \mathrm{d}(mv) = m\mathrm{d} v + v\mathrm{d} m$

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nice diagrams, great answer. –  tmac Apr 26 '12 at 21:54

It's only true when the changes $\Delta t$, $\Delta v$, $\Delta m$ are small and then it is known as the Leibniz rule, the rule for the derivative of a product, which Leibniz (but also Newton) discovered when they invented the calculus 3 centuries ago.

Just look at this proof: $$\frac{\Delta (mv)}{\Delta t} = \frac{(mv)_{new} - (mv)_{old}}{\Delta t} =\frac{(m_{old}+\Delta m)(v_{old}+\Delta v)-m_{old}v_{old}}{\Delta t} = \dots$$ Here I just used $x_{new}=x_{old}+\Delta x$ which holds for $X=m,v,t,mv$ or anything else. The new value is the old value plus the increment.

But now, expand the parentheses' products via the distribution law. You get; $$=\dots \frac{m_{old}\Delta v+ \Delta m v_{old}+\Delta m\,\Delta v}{\Delta t}$$ because the term $m_{old}v_{old}$ canceled (it was subtracted). Now, if each $\Delta X$ is significantly smaller than $X$, e.g. 100 times, then $\Delta m \,\Delta v$ is 10,000 times smaller and can be totally neglected. So you're only left with the first two terms in the numerator and they give you exactly the two terms you wanted to find.

So the increase of $mv$ is obtained either from an increase of $m$ or from an increase of $v$. The equation just encodes this simple observation quantitatively.

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So I should get it that the $\Delta t$ value is an infinitesimal and constantly changing along the time-line (but not in value) so that the change in $m$ and $v$ is also very small and can be neglected for a larger picture? I'm sorry but I only know the basics of Calculus from wikipedia; nothing more... –  Noein Apr 26 '12 at 12:41
Yup, your comment sounds totally fine and is a way to learn Calculus in Newton's way. But the $\Delta$ symbols are meant for people who don't have to know Calculus and derivatives. You may still want to imagine that all these things are finite, just small, and the very small terms are neglected. But what they mean is really $dt$, $dm$, $dv$ etc. in the calculus-infinitesimal sense while these people also implicitly say "don't ask me about calculus, it is not my goal to explain it now, instead, I want to explain some physics you would quite properly formulate only if you knew calculus". –  Luboš Motl Apr 26 '12 at 16:47
Historically, the story I heard was the Leibniz actually had it wrong, saying $\mathrm{d}(ab) = \mathrm{d}a \, \mathrm{d}b$, and he and Newton had a scrap about it. I can't find a source on this now, but this shows Leibniz definitely had a hard time with it: math.usma.edu/people/rickey/hm/CalcNotes/ProductRule.pdf –  Mark Eichenlaub Apr 26 '12 at 20:50
@LubosM, I think it's now a problem of $\Delta$ and $d$... Can you provide any info-links on this point? I've only encountered cap-Delta in my studies and I thought $d$ was just a synonym used in wikipedia... –  Noein Apr 28 '12 at 11:08

The idea about this school book derivation is that you can change moment by changing velocity (common case) or by changing mass.

Since you can change the momentum of the system (rocket plus gasses) only by the external force, and in case of the rocket there is no external force (neglect gravity for a moment), so the question is, why is rocket getting faster and faster?

$$F_\text{ext} = \frac{\text{d}p}{\text{d}t} = m \frac{\text{d}v}{\text{d}t} + v \frac{\text{d}m}{\text{d}t} = 0,$$

$$m \frac{\text{d}v}{\text{d}t} = - v \frac{\text{d}m}{\text{d}t}.$$

Important contribution comes from the fact that as rocket pushes gasses away it effectively decreases its mass (right side of the equation). If momentum is conserved, this means that velocity of the rocket increases (left side of equation).

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That's the way I saw at first but I still don't see how the non-changing $m$ and $v$ get into the same equation. The only explanation I had (which the book said nothing about) was that the equation measured momentum-change in very small time intervals hence why, for a brief moment, either $m$ or $v$ are constant-ized to measure the other. Doesn't this sound like the uncertainty principle? –  Noein Apr 26 '12 at 12:39
The idea is that this process happens in very short moment and changes are very small. You could imagine that you have $\bar{m}$ on the left side and $\bar{v}$ on the right. The very next moment you write the same equation but with a new mass and new velocity, so this is step-by-step process, analogous to integrating. –  Pygmalion Apr 26 '12 at 13:45