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

  • #1
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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
A.T.
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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
tnich
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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
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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.

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.

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

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.

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
vanhees71
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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).
 
  • #6
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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
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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
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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
stevendaryl
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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).
 
  • #10
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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
vanhees71
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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 lets 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
vanhees71
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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
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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
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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
vanhees71
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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
stevendaryl
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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
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You left out that the expelled mass, ##m## has momentum, also.
Of course I did. The expelled mass is not part of the rocket.

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.

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
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But I answered that (or at least I tried to).
Than you didn't understand my question.

You cannot blindly use ##\vec{F}=\dot{\vec{p}}## in this example
Equations should never be blindly used - not only in this example.

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
stevendaryl
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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
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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
vanhees71
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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
stevendaryl
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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
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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.

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
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Thanks for all your contributions. A question of mine remained unanswered, though:
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##?
What's so subtle about that?
 
  • #25
vanhees71
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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.
 

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