##F=\dot{p}=\dot{m}v+m\dot{v}## and Galilean invariance

In summary: Since ##\Delta M = -m##, we can substitute in the original equation and get:##F \Delta T \approx M \Delta V + m U##
  • #1
greypilgrim
546
38
Hi.

In Newtonian physics, total mass is conserved, but open systems can obviously gain or lose mass, such as a rocket. But how can the term ##\dot{m}v## be Galilean invariant?
 
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  • #2
greypilgrim said:
In Newtonian physics, total mass is conserved, but open systems can obviously gain or lose mass, such as a rocket. But how can the term ##\dot{m}v## be Galilean invariant?
Why should it be conserved for an open system?
 
  • #3
I don't think it is. Galilean invariance says that Newton's laws are invariant under a set of assumptions:
1) An absolute reference for space and time irrespective of motion of the observer
2) The velocity of an object in a frame moving at velocity ##V## relative to the absolute frame is
##\dot X' = \dot X - V##
where ##\dot X## is the velocity of the object in the absolute frame
This implies that ##\ddot X' = \ddot X##
3) Mass is constant and invariant

If mass changes with time, then force is not invariant:

##F = \dot p = \dot m \dot X + m \ddot X##
##F' = \dot p' = \dot m \dot X' + m \ddot X'\\
= \dot m (\dot X - V) + m \ddot X\\
=F - \dot m V ##
 
  • #4
A.T. said:
Why should it be conserved for an open system?
What do you mean? I did write that mass doesn't have to be conserved in open systems.

tnich said:
Galilean invariance says that Newton's laws are invariant under a set of assumptions:
I have never seen those assumptions and all three of them seem odd to me.

tnich said:
1) An absolute reference for space and time irrespective of motion of the observer
Why would Galilean/Newtonian physics require an absolute reference frame?

tnich said:
2) The velocity of an object in a frame moving at velocity VV relative to the absolute frame is [...]
Those transformations for velocity and acceleration are derivatives of the Galilean transformation of the coordinates, not assumptions.

tnich said:
3) Mass is constant and invariant
Never seen this as an assumption. As far as I remember, the conservation of total mass in Newtonian physics can be derived from another conservation law (I think momentum).
 
  • #5
Only closed systems can be Galileo invariant. E.g., the motion of a planet around the sun at a fixed point is an approximation for a very heavy (compared to the planet) sun, where you can neglect its motion due to the interaction with the planet. Only if you consider the system consisting of the Sun and the planet together, everything is Galileo invariant, and all 10 conservation laws (conservation of energy, momentum, angular momentum, center-of-mass velocity) hold true. Why mass is conserved in Newtonian is a very subtle issue, which as far as I know can only be explained from a careful analysis of the representation theory of the Galileo group in quantum mechanics (see Ballentine, Quantum Mechanics, for a very good reference about this).

In your example of a system with non-constant mass, you should consider a concrete example, like a rocket. The rocket alone, ejecting it's fuel to accerlate forward, is of course an open system, and thus Galileo invariance doesn't hold in this description. If you take into account both the rocket and the ejected fuel, Galileo invariance holds true and again the corresponding 10 conservation laws are fulfilled (+conservation of mass).
 
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  • #6
vanhees71 said:
Why mass is conserved in Newtonian is a very subtle issue
Isn't it a rather trivial implication from conservation of total linear momentum ##P## (symmetry under translations) and linear motion of the center of mass ##V## (symmetry under boosts) and ##P=MV##?
 
  • #7
greypilgrim said:
Isn't it a rather trivial implication from conservation of total linear momentum ##P## (symmetry under translations) and linear motion of the center of mass ##V## (symmetry under boosts) and ##P=MV##?

Only if the uniform motion of the center of mass is not derived from conservation of momentum and mass.
 
  • #8
DrStupid said:
Only if the uniform motion of the center of mass is not derived from conservation of momentum and mass.
It's the conserved quantity corresponding to symmetry under boosts.
 
  • #9
I'm not sure if this going off on a tangent, or not, but it is common to see people write that in cases of variable-mass systems (such as a rocket expending fuel), you should use ##F = M \frac{dV}{dt} + \frac{dM}{dt} v## instead of ##F = M \frac{dV}{dt}##. I actually don't think that's a good way to think about variable mass systems.

Assume that initially you have a rocket of mass ##M## traveling at velocity ##V## and during a period ##\Delta T## it is acting on by a force ##F## and throws off a smaller mass ##m## at a relative velocity of ##U##.

Then Newton's laws applied to the composite system gives:

##F \Delta T = \Delta P = (M-m) (V+\Delta V) + m (V+U) - MV = M \Delta V + m U - m \Delta V##

Assuming that ##m## and ##\Delta V## are both small, so that their product is negligible, this gives

##F \Delta T \approx M \Delta V + m U##

Since ##\Delta M = -m##, we can write this as:

##F \Delta T \approx M \Delta V - \Delta M U##

Dividing through by ##\Delta T## gives (in the limit as ##\Delta T \Rightarrow 0##

##F = M \frac{dV}{dt} - \frac{dM}{dt} U##

So the rule ##F = M \frac{dV}{dt} + \frac{dM}{dt} V## is only valid in the case ##U = -V##. Remember, ##U## is the relative velocity between the rocket ##M## and the fuel ##m##. So ##U = -V## is the case where the fuel has zero velocity after being expended (the total velocity of ##m## is ##U+V##). That's not a realistic assumption about rockets. Certainly when a rocket first launches, ##V \approx 0## but ##U < 0## (the fuel is propelled backward).
 
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  • #10
stevendaryl said:
Then Newton's laws applied to the composite system gives:
##F \Delta T = \Delta P = (M-m) (V+\Delta V) + m (V+U) - MV = M \Delta V + m U - m \Delta V##

How do you get this equation? Before the ejection of the reaction mass the rocket has the momentum ##M \cdot V## and if ##\Delta V## is the change of the velocity of the rocket, than it has the momentum ##\left( {M - m} \right) \cdot \left( {V + \Delta V} \right)## after the ejection. The resulting change of momentum is

[itex]F \cdot \Delta t = \Delta p = \left( {M - m} \right) \cdot \left( {V + \Delta V} \right) - M \cdot V = M \cdot \Delta V - m \cdot V - m \cdot \Delta V \approx M \cdot \Delta V - m \cdot V[/itex]

which is consistent with ##F = \dot p = M \cdot \frac{{dV}}{{dt}} + \frac{{dM}}{{dt}} \cdot V##. Could you please show me your full calculation, including the derivation of the first equation?
 
  • #11
greypilgrim said:
Isn't it a rather trivial implication from conservation of total linear momentum ##P## (symmetry under translations) and linear motion of the center of mass ##V## (symmetry under boosts) and ##P=MV##?
The point is that you tacitly assume that ##M## is constant. There's no symmetry principle in classical mechanics that let's you derive this, and special relativity shows that mass is not separately conserved although in special relativity you have the same symmetries as in Newtonian mechanics (homogeneity and isotropy of space and boost invariance), however of course realized by a different symmetry group (Poincare group vs. Galileo group).

In quantum mechanics it turns out that mass occurs as a non-trivial central charge of the Galileo group's Lie algebra. This suggests a superselection rule according to which there are no superpositions of states with different masses, i.e., a non-relativistic quantum system has to live in a Hilbert space, where the total mass is fixed to one value.
 
  • #12
DrStupid said:
How do you get this equation? Before the ejection of the reaction mass the rocket has the momentum ##M \cdot V## and if ##\Delta V## is the change of the velocity of the rocket, than it has the momentum ##\left( {M - m} \right) \cdot \left( {V + \Delta V} \right)## after the ejection. The resulting change of momentum is

[itex]F \cdot \Delta t = \Delta p = \left( {M - m} \right) \cdot \left( {V + \Delta V} \right) - M \cdot V = M \cdot \Delta V - m \cdot V - m \cdot \Delta V \approx M \cdot \Delta V - m \cdot V[/itex]

which is consistent with ##F = \dot p = M \cdot \frac{{dV}}{{dt}} + \frac{{dM}}{{dt}} \cdot V##. Could you please show me your full calculation, including the derivation of the first equation?
The save way to derive the rocket equation is to use the conservation of total momentum, which derives from first (symmetry) principles in Newtonian mechanics, i.e., you have
$$\vec{p}_{\text{rocket}} + p_{\text{fuel}}=\text{const}.$$
Now take the time derivative
$$\dot{\vec{p}}_{\text{rocket}}=-\dot{p}_{\text{fuel}}+\vec{F}.$$
Here, ##\vec{F}## is the total external force on the rocket (usually just the gravitational force of the Earth)
Assuming a given "exhaust of fuel" of ##\mu## (mass of fuel per unit time), you get
$$\dot{M} \vec{V}+M \dot{\vec{V}}=-\mu \vec{V}_{\text{fuel}}+\vec{F}.$$
Now we also assume (!) the conservation of total mass, i.e., ##\dot{M}=-\mu##, and then you get
$$M \dot{\vec{V}} = \mu (\vec{V}-\vec{V}_{\text{fuel}})+\vec{F}.$$
Usually one assumes that the velocity of the fuel relative to the rocket ##\vec{U}=\vec{V}_{\text{fuel}}-\vec{V}=\text{const}.##

Then we get
$$M \dot{\vec{V}}+\mu \vec{U}=\vec{F}.$$
Let's consider a simplified start of a rocket, which stays close to Earth and flies straight upward. Make the ##z## axis pointing upward. Then ##\vec{F}=-M g \vec{e}_z##, and ##\vec{U}=-U \vec{e}_z## with ##U>0##. Then we have
$$M \dot{V}_z+\dot{M} U=-M g$$
or
$$\dot{V}_z=-\frac{\dot{M}}{M} U - g.$$
With ##U=\text{const}## we can integrate wrt. ##t## from ##t=0## to ##t##:
$$V_z(t)=-U \ln \left [\frac{M(t)}{M_0} \right]-g t.$$
 
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  • #13
vanhees71 said:
The point is that you tacitly assume that MM is constant.
Where do I assume that? From symmetry under translations it follows that the total momentum ##P## is constant, from symmetry under boosts that the center of mass moves at a constant velocity ##V##. Since ##P=MV##, the total mass ##M## must necessarily be constant as well.
 
  • #14
vanhees71 said:
The save way to derive the rocket equation is to use the conservation of total momentum [...]

Thank you for the derivation. But I already know that and I asked another question.
 
  • #15
But I answered that (or at least I tried to). You cannot blindly use ##\vec{F}=\dot{\vec{p}}## in this example but you must go back to the fundamental law of momentum conservation for a closed system, and the closed system in this case consists of the rocket (including the unexhausted fuel) and the exhausted fuel. Of course @stevendaryl implicitly did exactly this. All I did was to reformulate his derivation in the hope it would clarify the argument.

For more examples, see Sommerfeld, Lectures on theoretical physics, vol. 1 (in the very beginning), where he treats all kinds of problems with "varying mass" like a chain falling from a table (both variants), a rain drop falling is vapour-saturated atmosphere and the like. As always, Sommerfeld is by at least an ##\epsilon## more comprehensive than many other textbooks!
 
  • #16
DrStupid said:
How do you get this equation? Before the ejection of the reaction mass the rocket has the momentum ##M \cdot V## and if ##\Delta V## is the change of the velocity of the rocket, than it has the momentum ##\left( {M - m} \right) \cdot \left( {V + \Delta V} \right)## after the ejection. The resulting change of momentum is

[itex]F \cdot \Delta t = \Delta p = \left( {M - m} \right) \cdot \left( {V + \Delta V} \right) - M \cdot V = M \cdot \Delta V - m \cdot V - m \cdot \Delta V \approx M \cdot \Delta V - m \cdot V[/itex]

which is consistent with ##F = \dot p = M \cdot \frac{{dV}}{{dt}} + \frac{{dM}}{{dt}} \cdot V##. Could you please show me your full calculation, including the derivation of the first equation?

You left out that the expelled mass, ##m## has momentum, also. The way I set things up:

Before: (separating the rocket from its fuel):
  • Rocket has mass ##M - m##
  • Rocket has velocity ##V##
  • Fuel has mass ##m##
  • Fuel has velocity ##V##
  • Total momentum = ##(M-m)V + mV = MV##
After:
  • Rocket has mass ##M-m##
  • Rocket has velocity ##V + \Delta V##
  • Fuel has mass ##m##
  • Fuel has velocity ##V+U## (##U## is the relative velocity between expended fuel and rocket, which is typically in the opposite direction from ##V##)
  • Total momentum = ##(M-m) (V + \Delta V) + m (V+U) = MV + M \Delta V - mV - m \Delta V + mV + mU = MV + M \Delta V + m (U - \Delta V)##
  • So the change in momentum is ##MV + M \Delta V + m (U - \Delta V) - MV = M \Delta V + m (U - \Delta V)##
  • In terms of ##\Delta M = -m##: ##\Delta P = M \Delta V - (\Delta M) U + \Delta M \Delta V \approx M \Delta V - (\Delta M) U ##
It only reduces to ##\Delta P = M \Delta V + (\Delta M)## in the special case ##U = -V##. Which might conceivably be correct in some circumstances, but not all. Probably not for an actual rocket.
 
  • #17
stevendaryl said:
You left out that the expelled mass, ##m## has momentum, also.

Of course I did. The expelled mass is not part of the rocket.

stevendaryl said:
So the change in momentum is ##MV + M \Delta V + m (U - \Delta V) - MV = M \Delta V + m (U - \Delta V)##

That's the change of the total momentum and as you do not include external forces it is always zero.

stevendaryl said:
It only reduces to ##\Delta P = M \Delta V + (\Delta M)## in the special case ##U = -V##.

You are confusing the force ##M \cdot \dot V + \dot M \cdot V## acting on the rocket with the force ##M \cdot \dot V - \dot M \cdot U = 0## acting on the total system of rocket and exhaust. Of course they are equal for U=-V only. This is the special case where the momentum of the rocket doesn't change.
 
  • #18
vanhees71 said:
But I answered that (or at least I tried to).

Than you didn't understand my question.

vanhees71 said:
You cannot blindly use ##\vec{F}=\dot{\vec{p}}## in this example

Equations should never be blindly used - not only in this example.

vanhees71 said:
but you must go back to the fundamental law of momentum conservation for a closed system

With ##\vec{F}=\dot{\vec{p}}## the conservation of momentum is included in the laws of motion.
 
  • #19
DrStupid said:
Of course I did. The expelled mass is not part of the rocket.

Yes, but for ##F=Ma##, the external force goes into changing the momentum of both.
 
  • #20
stevendaryl said:
Yes, but for ##F=Ma##, the external force goes into changing the momentum of both.

You neither included external forces nor did you talk about ##F=Ma##.
 
  • #21
DrStupid said:
Than you didn't understand my question.
Equations should never be blindly used - not only in this example.
With ##\vec{F}=\dot{\vec{p}}## the conservation of momentum is included in the laws of motion.
Of course you can derive the equations of motion from the fundamental conservation laws and vice versa for closed (!) systems, and that's what I did. If this doesn't answer your question, I didn't understand it, and if you want another clarification you have to tell me what's not clear with @stevendaryls and my derivation.
 
  • #22
DrStupid said:
You neither included external forces nor did you talk about ##F=Ma##.

For the composite system rocket + fuel, the mass is unchanging, so we can we certainly use:

##F_{ext} = \frac{dP}{dt}##

where ##P## is the total momentum of rocket + fuel.

For that composite system, we have:

##F_{ext} = M \frac{dV}{dt} - U \frac{dM}{dt}##

which is not equal to ##\frac{d}{dt} MV## except in the special (unrealistic) case ##U = -V##.
 
  • #23
vanhees71 said:
If this doesn't answer your question, I didn't understand it, and if you want another clarification you have to tell me what's not clear with @stevendaryls and my derivation.

stevendaryls already answered my question. Instead of the force acting on the rocket he derived the force acting on the total system of rocket and exhaust. Not being aware of the difference he erroneously concluded that ##F = \dot p## doesn't work because both forces are equal in the special case V=-U only.

stevendaryl said:
For the composite system rocket + fuel, the mass is unchanging, so we can we certainly use:

##F_{ext} = \frac{dP}{dt}##

where ##P## is the total momentum of rocket + fuel.

For that composite system, we have:

##F_{ext} = M \frac{dV}{dt} - U \frac{dM}{dt}##

which is not equal to ##\frac{d}{dt} MV## except in the special (unrealistic) case ##U = -V##.

You are still confusing the force ##F_{ext}## acting on the total system of rocket and exhaust with the force ##\frac{d}{dt} MV## acting on the rocket. Of course they are equal for V=-U only because that’s the special case where the exhaust leaves the rocket without momentum.
 
  • #24
Thanks for all your contributions. A question of mine remained unanswered, though:
vanhees71 said:
Why mass is conserved in Newtonian is a very subtle issue
greypilgrim said:
Isn't it a rather trivial implication from conservation of total linear momentum ##P## (symmetry under translations) and linear motion of the center of mass ##V## (symmetry under boosts) and ##P=MV##?
What's so subtle about that?
 
  • #25
Well, mass in non-relativistic theory is a quite complicated quantity. It comes into the theory when analysing the Galileo symmetry in quantum mechanics, i.e., you look for the unitary ray representations of the corresponding covering group. Then you realize that the proper unitary transformations do not lead to anything physically sensible. Simplified you can say that these proper representations would mean zero-mass particles, which do not make any sense in Newtonian physics nor in its quantized version. So you have to use also a non-trivial central extension of the Galileo Lie algebra, and the only non-trivial central extension is to introduce mass as a non-trivial central charge. Looked on mass from this very abstract point of view of extending classical Galileo symmetry to its quantum version, the fact that mass is a central charge implies that there is a socalled super selection rule, which forbids the superposition of states with different masses, i.e., if you have a system with a given mass no interaction can change this mass, because the unitary time evolution of quantum theory cannot connect Hilbert spaces with systems of different mass and thus mass is conserved in non-relativistic physics.
 
  • #26
That indeed sounds highly nontrivial. Is QM essential to understand mass, even in classical physics?

Anyway, what's wrong with my simple argument using symmetry under translations and under boosts and ##P=MV##?
 
  • #27
If you ask "why" in the natural sciences the answer can only be, "it's because of that or that fundamental principle". In classical mechanics mass is just a parameter for the inertia of matter. Its conservation is just postulated. So this doesn't answer your "why-question" very satisfactorially. The reason, why I answered with quantum-theoretical arguments is that the most fundamental theory we have today is quantum theory, and there is an argument from simpler principles, namely symmetry principles, which leads to the best answers to "why-questions" we can give today.
 
  • #28
DrStupid said:
You are still confusing the force ##F_{ext}## acting on the total system of rocket and exhaust with the force ##\frac{d}{dt} MV## acting on the rocket. Of course they are equal for V=-U only because that’s the special case where the exhaust leaves the rocket without momentum.

I think that the only sense in which ##F = M \frac{dV}{dt} + V \frac{dM}{dt}## is true is if you make it true by definition.

Consider a rocket, with no engine at all --- it's just drifting in space. But it's an old, crumbly rocket that is falling apart. So pieces of it are breaking off. That implies that ##\frac{dM}{dt}## is negative for the rocket. So that means that there is a force acting on the rocket? I think that's bizarre. Yes, you can say that it's true by definition, but it certainly isn't a force in the intuitive sense.
 
  • #29
stevendaryl said:
So that means that there is a force acting on the rocket? I think that's bizarre.
I do not think it bizarre. There is a change in momentum of the rocket because you are redefining what "the rocket" means. There is a transfer of momentum from "the rocket" to "not the rocket".
stevendaryl said:
Yes, you can say that it's true by definition, but it certainly isn't a force in the intuitive sense.
There are many instances in physics where "intuition" leads you wrong. I think it is the reasonable definition, whether or not it makes "intuitive sense" or not. Intuitive sense is subjective.
 
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  • #30
There are many examples, where for a mechanical system the mass changes. There are some very illuminating exercises in Sommerfeld Lectures on Theoretical Physics.

E.g., there are two types of the problem of a chain falling from a table: (a) the chain is curled up and only the part hanging over the edge of the table moves or (b) the chain moves all the time as a whole. Another one is a raindrop falling in an oversaturated atmosphere and thus is gaining more and more water on its way down. Of course the rocket is among the standard problems with systems changing mass.

The fundamental laws are of course not the forces but the conservation laws and symmetries, i.e., here momentum conservation and translation symmetry.
 
  • #31
greypilgrim said:
That indeed sounds highly nontrivial. Is QM essential to understand mass, even in classical physics?

My 2 cents: no, this already plays at the classical level.

The action of a point particle with embedding coordinates ##x^0(\tau),x^i(\tau)## and wordlineparameter ##\tau## reads

[tex]
S = \frac{m}{2} \int \frac{\dot{x}^i \dot{x}_i}{\dot{x}^0} d \tau
[/tex]

This action is invariant under the infinitesimal Galilei transformations

[tex]
\delta_H x^0 = \zeta^0, \ \ \delta_P x^i = \zeta^i, \ \ \delta_J x^i = \lambda^i{}_j x^i, \ \ \delta_G x^i = v^i x^0
[/tex]

From this you can calculate the commutators. In particular, you'll find

[tex]
[\delta_P, \delta_G] x^0 = [\delta_P, \delta_G] x^i = 0 \ \ \rightarrow [\delta_P, \delta_G] = 0
[/tex]

But now comes the subtlety: the action is invariant, but the Lagrangian ##L## is only quasi-invariant under boosts, i.e. it transforms to a total derivative as you can check:
[tex]
\delta_G L = \frac{d}{d\tau}\Bigl( m x_i v^i \Bigr)
[/tex]

This has an important consequence: the Noether charge ##Q_G## of the boosts needs to be adjusted to this such that it remains conserved:

[tex]
Q_G = p_i v^i x^0 - m x^i v_i
[/tex]

If we now calculate the corresponding Poisson-brackets between the Noether charges ##Q_P = p_i \zeta^i## and ##Q_G##, you will find due to this (signs depend on your conventions for these brackets)

[tex]
\{Q_G, Q_P \}_{poisson} \sim m \zeta_i v^i
[/tex]

where ##m## is the mass appearing in the action. This suggests that the Lie algebra of the Galilei group allows for a central extension (i.e. an extra generator ##M## which commutes with all the other generators), playing the role of mass:

[tex]
[G,P] = M
[/tex]

You can also check by the Jacobi-identities that this works out. Algebraically, in the non-relativistic limit ##M## replaces the Casimir operator ##P_{\mu}P^{\mu}## of the Poincare algebra which plays the role of mass. This centrally-extended Galilei algebra is called the Bargmann algebra. Interestingly, one can gauge this algebra and show that the corresponding gauge field of the generator ##M## roughly plays the role of the Newton potential. I like this result, because usually people tend to associate central extensions to quantization, but they already play an important role at the classical level.

By the way, this should also answer the question of the OP: in varying the action to obtain the EOM or symmetries, we treat the mass parameter ##m## as a constant. If you want to promote it to a function depending on ##x^0## or ##x^i##, you should also vary it. I'm not sure if this makes sense, though, but it should be clear that the Galilei-symmetries are not apparent anymore.
 
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  • #32
By the way, the fact that Galilei boosts play a subtle role in quantum mechanics can be seen by transforming the Schrodinger equation under them. You already see there easily that the wave function cannot be a simple scalar under Galilei boosts. The textbook by Ballentine has a very nice chapter on this.
 
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  • #33
And, as a last fun-fact about the Bargmann algebra: the simplest ##N=1## supersymmetric extension of it necessarily break the relation between supertransformations and spacetime translations which is at the core of supersymmetry; the commutator of two supertransformations don't give a spacetime translation anymore in the non-relativistic case, but...the central element ##M##.

Sorry, spend 4 years of my life on this, had to spit it out.
 
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  • #34
Orodruin said:
I do not think it bizarre. There is a change in momentum of the rocket because you are redefining what "the rocket" means. There is a transfer of momentum from "the rocket" to "not the rocket".

Yes, I agree that momentum is changing. But the way you compute that is by using conservation of momentum, not by analyzing forces.

I think it is the reasonable definition, whether or not it makes "intuitive sense" or not.

What makes it a reasonable definition?
 
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  • #35
stevendaryl said:
What makes it a reasonable definition?

To answer my own question, a definition is reasonable if it does one or both of (1) provides insight, or (2) helps to organize information to calculate answers to problems. If it doesn't do either of those, then I don't think it's a reasonable definition.

As I said previously, the canonical case of variable mass is an object such as a rocket that throws mass behind it to propel forward. In this case, if there are no other forces (such as gravity) involved, then the equations of motion will be:

##M \frac{dV}{dt} - \frac{dM}{dt} U = 0##

where ##U## is the velocity of the exhaust relative to the rest of the rocket (for the rocket, both ##\frac{dM}{dt}## and ##U## will be negative, for a coordinate system in which the rocket is accelerating in the positive-x direction). The quantity ##M \frac{dV}{dt} + V \frac{dM}{dt}## doesn't come into play, at all. Now, of course you can come up with an equation for it:

##M \frac{dV}{dt} + V \frac{dM}{dt} = (U+V) \frac{dM}{dt}##

I guess that sort of makes sense: During a time ##\delta t##, you're throwing off an amount of mass ##- \frac{dM}{dt} \delta t## and its velocity is ##U+V##. So you're tossing out an amount of momentum equal to ##-\frac{dM}{dt} (U+V) \delta t##, so the rocket's momentum must decrease by this amount by conservation of momentum. But I don't see how it helps to think of this in terms of forces, rather than conservation of momentum. It seems very awkward.
 

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