The rocket equation, one more time

[A.T.]

In all the situations that I have encountered so far, it didn't matter whether I wrote $\Delta$ or $d$.

In those situations $\Delta$ was apparently defined conveniently, to match the sign of the derivative.

For $\Delta y$ you always have a choice of which $y$ value is subtracted from which other $y$ value, and you can even force a positive sign by defining $\Delta y$ as the absolute difference.

For $\frac{dy}{dx}$ the sign is determined by how $y$ changes with increasing $x$.

 

[Rick16]

There is one more complication. I learned the foundations of classical mechanics with John Taylor's Classical Mechanics, and I have now traced the origin of my misunderstanding about deltas and differentials back to this book, specifically to footnote 4 on page 58, where Taylor writes:

As physicists we know that dv/dt is the limit of Δv/Δt, as both Δv and Δt become small, and I shall take the view that dv is just shorthand for Δv (and likewise dt for Δt), with the understanding that it has been taken small enough that the quotient dv/dt is within my desired accuracy of the true derivative.

Taylor applies this footnote to a specific problem. However, when developing the rocket equation on page 85, he adds a similar footnote:

Concerning the use of the small quantities like dt and dm, I recommend again the view that they are small but nonzero increments, with dt chosen sufficiently small that dm divided by dt is (within whatever we have chosen as our desired accuracy) equal to the derivative dm/dt.

I just learned in this thread that I have to distinguish between expressions like Δx/Δt, which can be either positive or negative, and expressions like dx/dt, which is positive by definition. Taylor blurs this distinction and I wonder if I should simply forget this view that "dv is just shorthand for Δv"?

 

[PeroK]

I just learned in this thread that I have to distinguish between expressions like $\frac {\Delta x} {\Delta t}$, which can be either positive or negative, and expressions like $\frac {dx}{dt}$, which is positive by definition. Taylor blurs this distinction and I wonder if I should simply forget this view that "$dv$ is just shorthand for $\Delta v$"?

It's simpler than that. Strictly speaking, $\Delta X$ has a well-defined meaning as the difference in quantity $X$, with the convention that $\Delta X = X_{final} - X_{initial}$. If you want to take $dX$ as the limit of $\Delta X$, then you can't have it any other way.

Sometimes, however, you want to work with the quantity $X_{initial} - X_{final}$. This is, of course, $-\Delta X$. If you call this $\Delta X$, then you need to keep in mind that this is non-standard usage. And, $dX$ is the limit of $\Delta X$, then it can't be the limit of $-\Delta X$.

It's like using binary, where 11 is the number 3 and then deciding that's it's actually the number eleven. If you are using binary, then you cannot suddenly start interpreting binary numbers as decimal numbers.

In a nutshell, non-standard usage of symbols is not wrong, as long as a) you are clear about the non-standard usage; and, b) you remain consistent and don't suddenly (half way through a problem) reinterpret the non-standard usage as though it was standard.

 

[PeroK]

Note that under all circumstances: $\frac{dX}{dt}$ is the limit of $\dfrac{X_{final} - X_{initial}}{t_{final} - t_{initial}}$.

You can't have it any other way, no matter how you define anything else in your problem.

 

[Rick16]

Sometimes, however, you want to work with the quantity Xinitial−Xfinal.

When I wrote $m-\Delta m$ for the change of the mass of the rocket, I did not see this as $m -(m_{initial}-m_{final})$. I rather considered $\Delta m$ the mass of the exhaust gases, which is positive, and I simply subtracted this positive mass from the total initial mass of the rocket. I guess that in doing so I mixed two concepts that belong to two different subsystems, i.e. I took $\Delta m$ from the gas subsystem and used it in the rocket subsystem. The result was the same, but this view prevented me from properly interpreting the meaning of $\Delta m$ for the rocket.

 

[jbriggs444]

There is one more complication. I learned the foundations of classical mechanics with John Taylor's Classical Mechanics, and I have now traced the origin of my misunderstanding about deltas and differentials back to this book, specifically to footnote 4 on page 58, where Taylor writes:

As physicists we know that dv/dt is the limit of Δv/Δt, as both Δv and Δt become small, and I shall take the view that dv is just shorthand for Δv (and likewise dt for Δt), with the understanding that it has been taken small enough that the quotient dv/dt is within my desired accuracy of the true derivative.

Taylor applies this footnote to a specific problem. However, when developing the rocket equation on page 85, he adds a similar footnote:

Concerning the use of the small quantities like dt and dm, I recommend again the view that they are small but nonzero increments, with dt chosen sufficiently small that dm divided by dt is (within whatever we have chosen as our desired accuracy) equal to the derivative dm/dt.

I just learned in this thread that I have to distinguish between expressions like Δx/Δt, which can be either positive or negative, and expressions like dx/dt, which is positive by definition.

dx/dt is not positive by definition. It can be either positive or negative.

If $x$ is a differentiable and strictly monotone increasing function of $t$ then $\frac{dx}{dt}$ is indeed positive everywhere.

If $x$ is a differentiable strictly monotone decreasing function of $t$ then $\frac{dx}{dt}$ is negative everywhere.

If $x$ is an arbitrary differentiable function of $t$ then $\frac{dx}{dt}$ can be negative for some values of $t$, positive for other values of $t$ and even zero for some values of $t$. An example would be $x = \sin t$.

It is true that there is no ambiguity about the sign of $\frac{dx}{dt}$. If we phrase the limit appropriately then the sign conventions for $\Delta x$ and $\Delta t$ do not even enter in. For instance, we could take the limit as $t_1 \to t_0$ of $\frac{x_1 - x_0}{t_1-t_0}$. No $\Delta$ anywhere.

 

[Rick16]

I see. I have to be clear about what $\Delta x$ actually represents, whether it represents $x_{final}-x_{initial}$ or $x_{initial}-x_{final}$. Everything else should then fall into place automatically. I had a rather vague idea of $\Delta m$ as simply the mass of the exhaust gases that were expelled within a certain time interval. I did not see it as a difference at all, just some mass. That was apparently my major mistake.

 

[PeroK]

When I wrote $m-\Delta m$ for the change of the mass of the rocket, I did not see this as $m -(m_{initial}-m_{final})$. I rather considered $\Delta m$ the mass of the exhaust gases, which is positive, and I simply subtracted this positive mass from the total initial mass of the rocket. I guess that in doing so I mixed two concepts that belong to two different subsystems, i.e. I took $\Delta m$ from the gas subsystem and used it in the rocket subsystem. The result was the same, but this view prevented me from properly interpreting the meaning of $\Delta m$ for the rocket.

Strictly speaking, the mass of the exhaust gases is not itself a "delta". Instead, with $m$ as the mass of the rocket:
$$m_{final} = m - m_{gas}$$And:
$$\Delta m = m_{final} - m_{initial} = (m - m_{gas}) - m = -m_{gas}$$This shows the importance, when things go wrong (as they often do), of being able to understand and deconstruct what you're doing.

 

[Rick16]

Strictly speaking, the mass of the exhaust gases is not itself a "delta". Instead, with $m$ as the mass of the rocket:
$$m_{final} = m - m_{gas}$$And:
$$\Delta m = m_{final} - m_{initial} = (m - m_{gas}) - m = -m_{gas}$$This shows the importance, when things go wrong (as they often do), of being able to understand and deconstruct what you're doing.

Thank you. This is a very instructive way to look at it.​

 

[Rick16]

I cannot believe how simple this turned out to be. A long time ago, I had decided that Δm was the mass of the expelled gases and I stubbornly stuck with this misconception. That's why I always wanted to write m - Δm. Writing m + Δm seemed nonsensical to me, because in my mind this meant adding the mass of the expelled gases to the mass of the rocket.

This is one of those moments where I don't know whether I should feel euphoric because I finally understood something that had plagued me for a long time, or whether I should feel stupid because it took me so long to understand such a simple thing.

 

[Rick16]

I just came across a post in another thread, “Some thoughts about self-education“, where gleem wrote the following:

“Sometimes, after many responses, the person may give a courteous thank you, but leave us wondering if our help was understood.

One thing that I thought might be of value is for one of the advisors to summarize the solution or post the" best" solution at the end of the discussion. This might be helpful to others who later peruse the Forum, find the problem of interest, but do not want to spend too much time working through the multitude and sometimes circuitous suggestions and approaches.”

I thought I would follow this advice and summarize the solution myself. As the OP, I know exactly where my mistake was and how it was solved in this thread.

The situation in question is a rocket flying in outer space with momentum mv at time t. The rocket’s engines are running, it expels gases and its velocity increases. Here is my original reasoning:

In a time $dt$ the rocket’s velocity increases, so I express this as $v+dv$. Also, in a time $dt$ the rocket’s mass decreases, so I express this as $m-dm$. Therefore, the rocket’s momentum at time $t+dt$ should be $(m-dm)(v+dv)$. This reasoning seemed perfectly logical to me, but it is wrong.

My mistake was to see $dm$ and $dv$ only as small quantities, but they are not just any small quantities, they are differentials, which are defined in the differential quotient. Specifically, $\frac {dm}{dt}=\lim_{\Delta t \rightarrow 0} {\frac {\Delta m} {\Delta t}}=\lim_{\Delta t \rightarrow 0} {\frac {m(t+\Delta t)-m(t)} {\Delta t}}$. According to this definition, $\Delta m=m(t+\Delta t)-m(t)=m_{final}-m_{initial}$. In the case of the rocket, $m_{final}<m_{initial} \Rightarrow \Delta m <0$ and consequently $dm<0$. Therefore I must write $m+dm$ for the change in the mass of the rocket, because writing $m-dm$ would mean to subtract a negative number, which would increase the rocket’s mass instead of decreasing it.

If I use differences instead of differentials, I am free to write $m-\Delta m$ for the rocket's mass at time $t+\Delta t$, but when I do this, I have to consider that $\Delta m$ must be positive in order for the rocket’s mass to decrease. What does it mean for $\Delta m$ to be positive in the case of the rocket? It means that I must define it as $\Delta m = m_{initial}-m_{final}$. When I then switch to differentials, I cannot simply make the substitution $\Delta m \rightarrow dm$, because $\Delta m = m_{final} - m_{initial}$ in the definition of the differential quotient. Using $\Delta m_+$ for positively defined $\Delta m$ and $\Delta m_{DQ}$ for $\Delta m$ according to the definition of the differential quotient, I get: $\Delta m_+=m_{initial}-m_{final}=-(m_{final} - m_{initial})=-\Delta m_{DQ}$. It follows that in this case, when switching from differences to differentials, I must make the substitution $\Delta m \rightarrow -dm$, which again leads to $m+dm$.

For the velocity, the situation is more straightforward: $\Delta v = v_{final}-v_{intial}>0$ and therefore writing $v+\Delta v$ or $v+dv$ expresses an increase in the rocket’s velocity. I could write $v-\Delta v$, but then $\Delta v$ would have to be negative and I would end up with the substitution $\Delta v \rightarrow -dv$, which would again lead to $v+dv$. No matter how I define my differences, I always end up with $(m+dm)(v+dv)$.

All this is probably obvious for many people, but it was not obvious for me and I have never seen it presented in such detail anywhere.

 

[Dale]

Thanks, that is, a helpful way to follow up!