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

[haushofer]

Once again, I would like to see a single example where considering $\frac{dM}{dt} V$ a force helps to understand a problem.

Maybe if you consider the geodesic equation in which mass is not constant anymore? I didn't do the calculation, but don't you then get something like

<br /> \ddot{x}^{\rho} + \Gamma^{\rho}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} = \frac{\dot{m}}{m}\dot{x}^{\rho}<br />

? The right hand side can then be considered as a force acting on the particle, pulling it out of geodesic motion.

 

[DrStupid]

It's an axiom, or assumption, not a definition.

It's an axiomatic definition.

But there is a physical difference between physical forces and the nonphysical quantity $\frac{dM}{dt} V$.

I am talking about forces as defined by Newton and not about "physical forces" (whatever that means).

Given that mass is conserved (and more specifically, obeys a continuity equation), you can always solve problems by considering forces on constant-mass objects. So it's certainly not a limitation on what can be calculated. If you want to compute the motion of a rocket, you consider the rocket plus the fuel together.

With limitation of the laws of motion to closed systems you need to divide the fuel into infinite pieces in order to apply forces. Without this limitation you just need two systems: the rocket and the exhaust. What is the benefit of the first option over the second?

 

[stevendaryl]

It's an axiomatic definition.

That's twisting the notion of "definition" to the point of meaninglessness.

With limitation of the laws of motion to closed systems you need to divide the fuel into infinite pieces in order to apply forces.

It seems straightforward to me: You consider two times: $t$ and $t+\delta t$. Let $M_{fuel}$ be the amount of fuel expended during time $\delta t$, and let $V$ be the velocity of the rocket at time $t$, and let $\delta V$ be the change in the velocity of the rocket, and let $U$ be the relative velocity of the fuel after it is expelled. Then conservation of momentum gives you:

$(M + M_{fuel}) V = M (V + \delta V) + M_{fuel} (V + U)$

Ignoring higher-order terms gives:

$M \delta V = - M_{fuel} U$

Since the mass of the rocket decreases as fuel is expended, we can write: $\frac{dM}{dt} = - M_{fuel}/\delta t$. So dividing through by $\delta t$ gives:

$M \frac{dV}{dt} = \frac{dM}{dt} U$

The quantity $\frac{dM}{dt} V$ just doesn't come into play, at all. This is just conservation of momentum. The force on the rocket is just

$M \frac{dV}{dt} = \frac{dM}{dt} U$

If you have an external force such as gravity, then the equations of motion are:

$M \frac{dV}{dt} = \frac{dM}{dt} U - M g$

So:

$F_{rocket} = \frac{dM}{dt} U - M g$

The total force on the rocket is the gravitational force, $-Mg$ plus the contact force of the fuel against the rocket, $\frac{dM}{dt} U$.

If you want to call $\frac{dM}{dt} V$ a force, where does it come into play? It doesn't. You can certainly add and subtract it from both sides:

$M \frac{dV}{dt} + \frac{dM}{dt} V = \frac{dM}{dt} U - M g + \frac{dM}{dt} V$

Then you can say that the total force on the rocket has three parts:

1. $\frac{dM}{dt} U$: the contact force of the fuel pressing against the rocket
2. $-Mg$: the force of gravity
3. $\frac{dM}{dt} V$: the force due to changing mass

But term number three serves no purpose at all. It's just a term that "goes along for the ride" on both sides of the equal sign.

 

[vanhees71]

Maybe if you consider the geodesic equation in which mass is not constant anymore? I didn't do the calculation, but don't you then get something like

<br /> \ddot{x}^{\rho} + \Gamma^{\rho}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} = \frac{\dot{m}}{m}\dot{x}^{\rho}<br />

? The right hand side can then be considered as a force acting on the particle, pulling it out of geodesic motion.

How do you come to this equation? In GR, the EoM reads
$$\mathrm{D}_t p^{\mu}=\dot{p}^{\mu} + \Gamma^{\mu}_{\rho \sigma} p^{\rho} p^{\sigma} = K^{\mu},$$
where $p^{\mu}$ is the canonical (sic!) momentum and $K^{\mu}$ the four-force of all non-gravitational interactions acting on the particle.

 

[DrStupid]

It seems straightforward to me: [...]

Not for me. Without the limitation to closed systems I can directly apply the second law to rocket and exhaust:

F_R = m_R \cdot \dot v_R + \dot m_R \cdot v_R

F_E = m_E \cdot \dot v_E + \dot m_E \cdot v_E

The already exhausted reaction mass can be assumed to be zero (as if the rocked just started) because it does not interact with the rocket anymore:

m_E = 0

The relative exhaust velocity is

u = v_E - v_R

Conservation of mass results in

\dot m_E = - \dot m_R

And the third law says

F_R = - F_E

Everything together results in the differential rocket equation

m_R \cdot \dot v_R = \dot m_R \cdot u

without considering differential time steps, ignoring of higher-order terms or hand waving force equations. Additional forces can easily be included in the last condition. That means for gravity

F_R = m_R \cdot g - F_E

which results in

m_R \cdot \dot v_R = g + \dot m_R \cdot u

That is straightforward for me.

 

[haushofer]

How do you come to this equation? In GR, the EoM reads
$$\mathrm{D}_t p^{\mu}=\dot{p}^{\mu} + \Gamma^{\mu}_{\rho \sigma} p^{\rho} p^{\sigma} = K^{\mu},$$
where $p^{\mu}$ is the canonical (sic!) momentum and $K^{\mu}$ the four-force of all non-gravitational interactions acting on the particle.

It was a sophisticated guess if one makes the mass time dependent in the geodesic equation, but as I said, it could be wrong.

 

[stevendaryl]

Not for me. Without the limitation to closed systems I can directly apply the second law to rocket and exhaust:

F_R = m_R \cdot \dot v_R + \dot m_R \cdot v_R

F_E = m_E \cdot \dot v_E + \dot m_E \cdot v_E

I consider your derivation a little bogus. When you treat $M_E$ as zero, it means that you aren't considering the mass of the exhaust, but the incremental change in the mass of the exhaust. So you're actually doing the differential case that I was doing, but hiding that fact.

The mass of the exhaust isn't zero, except when the rocket first launches. But if you are really talking about a nonzero mass of the exhaust, then there is no single velocity to use in computing the term $V_E \frac{dM_E}{dt}$. Different parts of the velocity are moving at different speeds. The only times when it is meaningful to talk about an object having a single speed is when the object is rigid (so all the parts are moving at the same speed) or when the object is confined to a small region of space, so that all the pieces can be treated as approximately having the same velocity. But in those cases, you have well-defined objects with well-defined masses, so there is no point.

If you want to talk about a more complex case where you have extended objects that are not rigid, then to me, the way to do the analysis is like this:

$\frac{dP}{dt} = F_{ext} +$ the flux of momentum through the surface defining the object

where $F_{ext}$ is the external force acting on the object. The second term is not $V \frac{dM}{dt}$. It's a surface integral; something like this:

Flux = $\int_\mathcal{S} \rho \overrightarrow{V} (\overrightarrow{V} \cdot \overrightarrow{d \mathcal{S}})$

where $\rho$ and $V$ are the local mass density and velocity field, and $\mathcal{S}$ is the surface defining the object.

(I think that's right for a time-independent surface. If the definition of the surface changes with time, I'm not sure how to generalize it, right off the bat.)

My point is that there are almost no cases where $\frac{dM}{dt} V$ comes into play. If the situation is too complicated to analyze using fixed-mass objects, then that term is too primitive to be useful, anyway.

 

[vanhees71]

I don't get what you are after here. Take your example of the rocket. You clearly need the correct momentum balance equation. For the rocket with mass $M(t)$ you have (I just put the most simple 1D equation, for a rocket going straight up near the Earth, where gravity can be described by $g=\text{const}$). Then you have
$$\mathrm{d} p =\mathrm{d} t (M \dot{v} + \dot{M}(v-u))=-M g \mathrm{d} t.$$
Here, $u$ is the velocity of the center of mass of exhaust. The velocity $v-u=v_{\text{rel}}$ is usually taken as constant (it's the velocity of the exhaust in the rest frame of the rocket). Then the equation gets
$$M \dot{v}=-\dot{M} v_{\text{rel}}-M g.$$
Dividing by $M$ and integrating over $t' \in [0,t]$ gives
$$v(t)=-v_{\text{rel}} \ln \left (\frac{M(t)}{M_0} \right) - g t + v_0.$$

 

[DrStupid]

The mass of the exhaust isn't zero, except when the rocket first launches.

It makes no difference if the rocket first launches or not because the already ejected reaction mass doesn't interact with the rocket anymore. Therefore I can assume the rocket to be first launched in every point of the trajectory. I already explained that in my derivation above.

If you don't like this simplification you can include the exhaust. But that doesn't change anything. In absence of external forces all parts of the exhaust are moving with constant velocity. Therefore its total momentum changes by the mass transfer only, according to

\dot p_T = \dot m_T \cdot v_T = F_T

where $v_T$ is the velocity of the reaction mass crossing the system boundary between rocket and exhaust. This interaction doesn't change with inclusion of external forces as long as they do not affect the internal processes in the rocket engine.

My point is that there are almost no cases where $\frac{dM}{dt} V$ comes into play.

That would mean that there are almost no open systems. I do not agree.