Understanding Force: Velocity and Mass Expulsion

[phymatter]

i can write the force on a body as F=v dm/dt , i am a bit confused with what this v is , i mean is it the velocity of body experiencing force or the mass expelled?

 

[HallsofIvy]

i can write the force on a body as F=v dm/dt , i am a bit confused with what this v is , i mean is it the velocity of body experiencing force or the mass expelled?

Who told you you could write the force on a body like that? More common is F= m dv/dt or, if mass is not constant, F= d(mv)dt. Where did F= v dm/dt come from? I think any situation where that was true, that velocity was constant while mass was changing, would be a very strange situation. Without knowing what situation gave rise to that equation, I don't see how anyone can answer your question.

 

[phymatter]

Without knowing what situation gave rise to that equation, I don't see how anyone can answer your question.

As you did point ,
the only condition in this equation is that the velocity of the object is constant ,

I think any situation where that was true, that velocity was constant while mass was changing, would be a very strange situation.

i can think of a situation where an object goes along +ve x direction gradually dropping sand filled in it on the track , now since the momentum needs to be conserved along x direction , the object should experience a force along +ve x which would cause change i its velocity .

well my question remains the same ,

i can write the force on a body as F=v dm/dt , i am a bit confused with what this v is , i mean is it the velocity of body experiencing force or the mass expelled?

 

[arildno]

phymatter:

Remember that forces act upon material particles, NOT upon arbitrarily, though well-defined, SYSTEMS of such particles!

Now, for non-relativistic speeds, material particles obey the law of CONSERVATION OF MASS.

This does NOT mean we cannot define SYSTEMS of particles where the system mass will vary in time, by reason of unequal entry and departure of material particles in the system.

That is, we can have systems that experience mass flux, and also momentum flux.

In order to develop the correct equations for such systems (typically they are called "geometric systems, as in "whatever particles inhabit this or that spatial region"), I have made a thread here:
https://www.physicsforums.com/showthread.php?t=72176

 

[MaxwellsDemon]

v would be the velocity of the object in question, not the expelled matter...this is because the force is acting on the body doing the expelling and the body is what you are considering traveling at a constant v. As a physical example of your question, I think of a rocket ship moving under the influence of a gravitational field. I imagine a situation where the ship has to burn fuel and expel propellent to keep moving at a constant speed.

 

[Dale]

i can write the force on a body as F=v dm/dt , i am a bit confused with what this v is , i mean is it the velocity of body experiencing force or the mass expelled?

It is always the m and the v of the body experiencing the force.

In general, by the chain rule, we have

F = dp/dt = d(mv)/dt = m dv/dt + v dm/dt

So the first term only drops out if dv/dt = 0, likewise the second term drops out if dm/dt = 0 which is the usual case.

 

[arildno]

It is always the m and the v of the body experiencing the force.

In general, by the chain rule, we have

F = dp/dt = d(mv)/dt = m dv/dt + v dm/dt

So the first term only drops out if dv/dt = 0, likewise the second term drops out if dm/dt = 0 which is the usual case.

This is incorrect, and does not give the appropriate expression for the momentum flux.

 

[A.T.]

i can think of a situation where an object goes along +ve x direction gradually dropping sand filled in it on the track , now since the momentum needs to be conserved along x direction , the object should experience a force along +ve x which would cause change i its velocity .

If the sand is just dropped, it initially travels at the same speed as the car, so the total momentum of object + sand is conserved. There is no force on the object.

The expelled mass becomes interesting in the case of a rocket hovering at constant height. The rocket gets lighter by expelling fuel, so it needs less thrust force to hover. The thrust force depends on the mass and velocity of the expelled fuel.

 

[Dale]

This is incorrect, and does not give the appropriate expression for the momentum flux.

Could you elaborate?

 

[arildno]

Could you elaborate?

See my thread on this, https://www.physicsforums.com/showthread.php?t=72176, in particular post 3:
https://www.physicsforums.com/showpost.php?p=537060&postcount=3

Note, for example, that your expression is not Galilean invariant, since the velocity is not a Galilean invariant quantity.

 

[Bob S]

The relativistic expression for momentum of a particle of rest mass m₀ is

p = βγ m₀c

So F = dp/dt = c d(βγ m₀) /dt = c m₀ d(βγ)/dt

Bob S

 

[arildno]

The relativistic expression for momentum of a particle of rest mass m₀ is

p = βγ m₀c

So F = dp/dt = c d(βγ m₀) /dt = c m₀ d(βγ)/dt

Bob S

Indeed. And that holds for a MATERIAL system, which is the only sensible type of system when working with relativistic speeds.

However, at non-relativistic speeds, where we can regard space&time as absolute dimensions independent of observer velocities, designating a physical system as, for example: "The streaming fluid contained by this U-tube" is perfectly well-defined over time, but it won't consist of the same particles over time.
Hence, that physical system is NOT a material system, but a geometric system.

 

[Dale]

That's fine, I don't think the OP is looking at that level of detail.

 

[arildno]

Okay, I'll give a puzzle that highlights and problematizes the naive idea that with "variable mass" and constant velocity V, the force acting upon a system is given as dm/dt*V:


Suppose you have a fluid that flows with a uniform velocity V, say in the horizontal direction.

Our system will initially at t=0 be the unit square, but with a twist:
We let the right-hand vertical side move with constant horizontal velocity u, the other sides stationary, i.e, we have an expanding rectangle as our system.

Now, with "r" being the density, the total momentum p(t) within this system is:
\vec{p}(t)=r*1*(1+ut)*V\vec{i}
Thus, we have: \frac{d\vec{p}}{dt}=ruV\vec{i}\neq\vec{0}

But, obviously, none of the material fluid elements either inside or outside our expanding rectangle can experience any net force, precisely because they have constant velocity!

Something is evidently missing here, and my thread explores just those issues.

 

[A.T.]

Okay, I'll give a puzzle that highlights and problematizes the naive idea that with "variable mass" and constant velocity V, the force acting upon a system is given as dm/dt*V:

Your example is the same as the box-car dropping sand (see post #8). In both cases the mass removed from or added to the system, is at rest to the system. Obviously no force is resulting from that.

In the formula F = dm/dt*V the 'm' denotes the amount of mass that changed it's velocity by V.

Example:

Space is filled uniformly with massive particles a rest to each other. A bucket moves trough that space with a constant speed V relative to the particles, and collects dm particles in the time dt. The force needed to move that bucket with a constant speed is given by F = dm/dt*V

 

[arildno]

Your example is the same as the box-car dropping sand (see post #8). In both cases the mass removed from or added to the system, is at rest to the system. Obviously no force is resulting from that.

Here, you suddenly switch to regarding velocities relative to the rest frame of the system, rather than regarding it from an ARBITRARY reference frame.

In the formula F = dm/dt*V the 'm' denotes the amount of mass that changed it's velocity by V.

Here, you suddenly shrink yourself into merely considering a portion of the whole system, i.e, rather than the whole system.

Example:

Space is filled uniformly with massive particles a rest to each other. A bucket moves trough that space with a constant speed V relative to the particles, and collects dm particles in the time dt. The force needed to move that bucket with a constant speed is given by F = dm/dt*V

Sure enough.

It is, however, better to have a fully general and rigorous theory of this, rather than the melange of ad hoc rules you propose.

 

[arildno]

In order to understand WHY we get the correct result by always considering the system's rest frame, we'll work with a simplified model of the proper equation with an express term for the momentum flux:
\frac{d\vec{p}}{dt}+\dot{m}\vec{u}=\vec{F} (1)
where \dot{m} is the outward mass flux, \vec{u} the velocity of the leaving mass.
Hence, if the system's mass is m(t), we have \frac{dm}{dt}=-\dot{m}, with \vec{v} the system's velocity.

Thus, (1) is readily re-written as:
\vec{F}=m\vec{a}+\dot{m}(\vec{u}-\vec{v})

This follows, therefore, trivially by my formalism; which equally well handles complex flow patterns and varying control volumes, in contrast to a bundle of ad hoc rules.

The salient aspect is that whereas the given ad hoc rules PRIVILEGE the system rest frame as a presupposition, my formalism does not; rather, it DERIVES the system's rest frame as being the simplest to work with.

Furthermore, it brings out that control volumes are artificial (although often helpful!), in the sense of being structures that are not acted upon by any forces; only real particles are, not the spatial region we choose to define as our physical system within which those particles happen to be at times.

This is highlighted by the momentum flux term:
Namely, that in a geometric system, the set of particles we are considering the momentum of as a function of time is a CHANGING set of particles, rather than a CONSTANT set of the same particles.

Old ones leave and new ones arrive in the system, and there is no reason at all why the new ones should have the same "force/momentum history" as the old ones, since they are totally different entities.

Therefore, the "force law" of such a system is tweaked, relative to the force law for the material system (i.e, Newton's 2.law of motion)

 

[Dale]

This is incorrect, and does not give the appropriate expression for the momentum flux.

Sorry to go back to this. But I know that I applied the chain rule correctly, and I know that F=dp/dt, so the only thing that I could see you objecting to would be saying p=mv. Do you have a better expression for momentum?

 

[arildno]

Sorry to go back to this. But I know that I applied the chain rule correctly, and I know that F=dp/dt, so the only thing that I could see you objecting to would be saying p=mv. Do you have a better expression for momentum?

Well, it is the correct expression for the rate of change of the momentum of a material system, i.e, consisting of the same stuff/particles over time.

However, a physical system is something WE interpose on reality, and two approaches, conceptually wholly distinct, readily suggest themselves:

1. We pick a set of particles/stuff (say, a particular bunch of molecules) and observe/predict their motions/behaviour over time, for example predicting what regions of space that set will occupy at various times.
This is what we call a "material system"

2. We focus on a particular REGION of space, and endeavour to describe the motions/behaviour of whatever particles might inhabit it, at various times, i.e, our set of particles (contained in our specified region) might well change over time.
This is a geometric system.

1. corresponds to what is known as "Lagrangian" formulations, 2. to "Eulerian" descriptions.

Is it really that difficult to understand the difference between these two perspectives?
Either you follow the stuff wherever it might go (or transmute into), or you observe what happens within a set region, whatever stuff inhabits it.

Newtons law, F=dp/dt, is formulated with respect to the behaviour of a particular piece of MATTER, i.e, NOT as such, with reference to what happens within some spatial region.

Thus, if you want to switch from perspective 1. to perspective 2., you need to take into account that you no longer have a single material system that you observe; but rather, a continuous sequence of material systems, the sequence being defined by the assigned spatial region they will inhabit at different times.

But then, if you are to set up a law for the forces acting upon the particles inhabiting that region (the forces do NOT act upon the region itself, but upon the material particles within it), your balance sheet must include, beside the rate of change of momentum WITHIN your spatial region also the flux of momentum out of the region.

This is so, because forces only act upon the particles themselves, not upon any region we have chosen to limit ourselves to look upon.

 

[arildno]

I re-post hereby post 3 in my mentioned thread, and that details the balancing act of momenta in order to gain the proper formulation of Newton's 2.law of motion valid for non-material systems:

2b). THE COINCIDENT MATERIAL SYSTEM
Our trick will be to consider the dynamics of a material system instantaneously coincident with our geometric system, and through that, determine what laws a geometric system must obey.

The fancy way of doing this is by deriving what is known as Reynolds' transport theorem, we'll choose a more visual approach.

First, let us describe our geometric system:
Let S be a region in space, and let there exist material particles enclosed by S and outside of S.
S can be moving through space or undergo deformations, or it might be fixed and stationary.

However, we require that material particles can both enter and leave S, that is :
S is an arbitrarily chosen GEOMETRIC region, where in particular the boundary of S is not determined by whatever material particles happen to be there.

For example, our region S might be a specified section of a tube, with an inlet and an outlet.

Let us consider S and its surroundings at the two times t and t+\delta{t}
Now, material particles can be subdivided into 4 base types:
1) Those particles which were enclosed in S at time t and which is also enclosed in S at time t+\delta{t}
We assign to that subset of particles a momentum \vec{p}_{E}(t) at time "t", and momentum \vec{p}_{E}(t+\delta{t}) at time t+\delta{t}
2)Those particles which were enclosed in S at time t and but which are NOT enclosed in S at time t+\delta{t}
We assign to that subset of particles a momentum \vec{p}_{-}(t) at time "t", and momentum \vec{p}_{-}(t+\delta{t}) at time t+\delta{t}
3)Those particles which were NOT enclosed in S at time t and but which ARE enclosed in S at time t+\delta{t}
We assign to that subset of particles a momentum \vec{p}_{+}(t) at time "t", and momentum \vec{p}_{+}(t+\delta{t}) at time t+\delta{t}
4) Those particles which is outside of S at both times; these are ignored in the following.

The MATERIAL system which contains particles of type 1),2),3) is called M

The S-momentum:
At every time, there is an amount of momentum enclosed in S called \vec{p}_{S}, and from the description above, we have:
\vec{p}_{S}(t)=\vec{p}_{E}(t)+\vec{p}_{-}(t) (1)
\vec{p}_{S}(t+\delta{t})=\vec{p}_{E}(t+\delta{t})+\vec{p}_{+}(t+\delta{t}) (2)

Note, in particular, that the set of particles inhabiting S at time t+dt is DIFFERENT from the set of particles inhabiting S at time t!

The M-momentum
The amount of momentum in M is called \vec{p}_{M}, and from the description above, we have:
\vec{p}_{M}(t)=\vec{p}_{E}(t)+\vec{p}_{-}(t)+\vec{p}_{+}(t) (3)
\vec{p}_{M}(t+\delta{t})=\vec{p}_{E}(t+\delta{t})+\vec{p}_{-}(t+\delta{t})+\vec{p}_{+}(t+\delta{t})(4)

Thus, the particles themselves are the same at both instances, whether or not they reside in S

Now, M is a material system, and thus, we know the rate of change of its momentum is related to the external forces acting upon it in \vec{F}=\frac{d\vec{p}_{M}}{dt}

What we're interested in, is the appropriate relation which holds for the rate of change of momentum enclosed in S, that is, \frac{d\vec{p}_{S}}{dt}
But, by subtracting eq. (1) from (2), adding zero on the right-hand side in an intelligent manner, dividing with \delta{t} and using (3)+(4), we gain:
\frac{\vec{p}_{S}(t+\delta{t})-\vec{p}_{S}(t)}{\delta{t}}=\frac{\vec{p}_{M}(t+\delta{t})-\vec{p}_{M}(t)}{\delta{t}}-\frac{\vec{p}_{-}(t+\delta{t})-\vec{p}_{+}(t)}{\delta{t}}

Now, the left hand side will go to \frac{d\vec{p}_{S}}{dt} as \delta{t}\to0, whereas the first term on the right-hand-side will go to \frac{d\vec{p}_{M}}{dt}, and hence, by invoking Newton 2-law, \vec{F}

We'll focus on the second term on the right hand-side:
Simplify particles of type 2) as a "single" continously leaving particle, whose momentum may be written as:
\vec{p}_{-}(t+\delta{t})=\delta{m}_{-}\vec{v}^{(-)}
where \delta{m}_{-} is its mass, and \vec{v}^{(-)} is the velocity it has right AFTER it has left S.
Make the analogous rewriting for the particle of type 3):
\vec{p}_{+}(t)=\delta{m}_{+}\vec{v}^{(+)}
where \delta{m}_{+} is its mass, and \vec{v}^{(+)} is the velocity it has right BEFORE it has entered S.
Thus, by setting \frac{\delta{m}_{-}}{\delta{t}}=\dot{m}_{-},\frac{\delta{m}_{+}}{\delta{t}}=\dot{m}_{+}
we gain, in the limit when rearranging:
\frac{d\vec{p}_{S}}{dt}+\dot{m}_{-}\vec{v}^{(-)}-\dot{m}_{+}\vec{v}^{(+)}=\vec{F}(5)
This is the law we were after!
.

I am not interested in answering any other posts, unless they directly concern the maths and pre-suppositions in that post.