Physics in troubles: the real equation of force is F = ma and not F = dp/dt

[Genady]

Just now, martillo said:

Momentum p is defined as P = mv. By definition.

I never see momentum defined other way.

You are wrong. This is not a general definition.

A general definition is based on Lagrangian (remember Variational Calculus?). It becomes mv in this specific case.

Did you ever see formula for momentum of photon, for example? It is not mv, because for photon m=0.

[martillo]

2 minutes ago, Genady said:

You are wrong. This is not a general definition.

A general definition is based on Lagrangian (remember Variational Calculus?). It becomes mv in this specific case.

Did you ever see formula for momentum of photon, for example? It is not mv, because for photon m=0.

For the case of systems with mass, p is defined as p = mv whatever m being constant or variable.

You are right that in the case of photons without mass but with energy, the definition is in terms of energy. I must point out that in Relativity there is a total correspondence between energy and mass by the equation E = mc² and so it is not clear for me how photons can have energy and not mass but this subject goes beyond the scope of this thread. May be in other thread...

As I said other place I'm not able to discuss in terms of Variational Calculus, I have no expertise at all in this area.

 

[Genady]

1 minute ago, martillo said:

For the case of systems with mass, p is defined as p = mv whatever m being constant or variable.

You are right that in the case of photons without mass but with energy, the definition is in terms of energy. I must point out that in Relativity there is a total correspondence between energy and mass by the equation E = mc² and so it is not clear for me how photons can have energy and not mass but this subject goes beyond the scope of this thread. May be in other thread...

As I said other place I'm not able to discuss in terms of Variational Calculus, I have no expertise at all in this area.

 

Well, you are wrong about this and you are wrong about E = mc² .  

[martillo]

1 minute ago, Genady said:

Well, you are wrong about this and you are wrong about E = mc² .  

Please explain how is p defined for things with variable mass then? May be you could explain E = mc² later if it would matter...

[Genady]

2 minutes ago, martillo said:

Please explain how is p defined for things with variable mass then? May be you could explain E = mc² later if it would matter...

I can't, because

 

15 minutes ago, martillo said:

Variational Calculus, I have no expertise at all in this area

but that is where p is defined for general systems.

[martillo]

2 minutes ago, Genady said:

I can't, because

 

but that is where p is defined for general systems.

Well, can you tell me if for systems with mass m it is valid the equation p = mv although the momentum p is not defined this way?

[Genady]

1 minute ago, martillo said:

Well, can you tell me if for systems with mass m it is valid the equation p = mv although the momentum p is not defined this way?

Sorry, I don't understand your question. As I said, momentum is mass times velocity only for a system with constant mass.

What is p in your question? Momentum or something else?

[martillo]

Just now, Genady said:

Sorry, I don't understand your question. As I said, momentum is mass times velocity only for a system with constant mass.

Well, the question is: which would be the equation for p when mass varies.

[Genady]

Just now, martillo said:

Well, the question is: which would be the equation for p when mass varies.

It depends on the specifics of the system. No general formula. You need to construct a system Lagrangian and then apply some formulas to this Lagrangian.

[martillo]

1 minute ago, Genady said:

It depends on the specifics of the system. No general formula. You need to construct a system Lagrangian and then apply some formulas to this Lagrangian.

The only thing I think I know is that Lagrangian is about paths with minimum distances and so all would depend on the system of coordinates and its definition of distance, am I wrong?

[Genady]

1 minute ago, martillo said:

The only thing I think I know is that Lagrangian is about paths with minimum distances and so all would depend on the system of coordinates and its definition of distance, am I wrong?

Yes, you are.

[martillo]

8 minutes ago, Genady said:

Yes, you are.

Well, I think that in Classical Physics p is defined as p = mv and it is valid the equation dp/dt = mdv/dt + v dm/dt. This things valid for variable m because if not dm/dt would have not make sense. The Thrust equation for the rockets would have no sense.  I must consider then that they are valid within Classical Physics only. Can I say that?

Edited February 22, 2023 by martillo

[martillo]

The definition of momentum p in my Frank J. Blatt's book of Physics (Spanish version) is p = mv.

Note: p is called "cantidad de movimiento" in Spanish and "linear momentum" or just "momentum" in English as mentioned in the foot note of the page (photo below).

Here the photos of the definition section and the foot note of the page:

N del T in the foot note means note of the traductor.

The definition doesn't mention if the mass must be constant or not. It is said that the definition is a direct translation of the definition made by the proper Newton.

I think Newton didn't considered the mass as variable so may be the original definition is valid for constant mass.

In the link I provided for the page about rockets (http://www.braeunig.us/space/propuls.htm) the same definition p = mv is considered and it is worked with the variable mass of the rockets.

The question of if the definition of the definition of P = mv applies to variable masses can be solved considering that if it is valid, then is valid the equation:

dp/dt = mdv/dt + vdm/dt

This equation is which I considered in my derivation of the Thrust Equation for the rockets in the description of the problem about the rockets we are discussing which is presented at the OP of the thread.

I have made a search with google and found that the definition p = mv is valid in Classical Physics while in Relativity Theory the definition for momentum is p = γmv.

The classic definition would apply then as the particular case of slow velocities (γ approximated to 1).

I conclude then that all my treatment about rockets (including the provided link) is valid within Classical Physics but not within Relativity Theory.

Edited February 22, 2023 by martillo

[studiot]

1 hour ago, martillo said:

I conclude then that all my treatment about rockets (including the provided link) is valid within Classical Physics but not within Relativity Theory.

There is indeed additional theory to apply relativity to rockets.

But I would suggest that you deferred this until you have sorted out non relativistic theory.

I say non relativistic because relativity is now counted as part of classical physics.

 

When doing this sort of thing it is useful to make a list of symbols and what they represent, to avoid using the same symbol for different things in different places.

So first of all what do you mean by F ?
You have one overall system and two subsystems viz the exhaust, the rocket itself and the combination of rocket and exhaust.
That gives the possibility of 9 forces acting to choose from.
Some of these may be zero or equal to other forces.

Each of these 3 (sub)systems will have their own mass, acceleration, velocity and momenta as a result.

 

Now consider the case where the rocket itself has zero acceleration.
What do you think this means ?

[martillo]

17 minutes ago, studiot said:

There is indeed additional theory to apply relativity to rockets.

But I would suggest that you deferred this until you have sorted out non relativistic theory.

I say non relativistic because relativity is now counted as part of classical physics.

Is the inverse: classical physics is a part of relativity theory when only slow velocities (in relation to c velocity) are present.

17 minutes ago, studiot said:

When doing this sort of thing it is useful to make a list of symbols and what they represent, to avoid using the same symbol for different things in different places.

So first of all what do you mean by F ?
You have one overall system and two subsystems viz the exhaust, the rocket itself and the combination of rocket and exhaust.
That gives the possibility of 9 forces acting to choose from.
Some of these may be zero or equal to other forces.

Each of these 3 (sub)systems will have their own mass, acceleration, velocity and momenta as a result.

 

Now consider the case where the rocket itself has zero acceleration.
What do you think this means ?

What are you thinking on? If you think to calculate now the thrust equation and the thrust force from scratch I think I will not follow you.

We all already had a lot of discussions in the thread. Please summarize in what you disagree with me.

Edited February 22, 2023 by martillo

[Genady]

3 hours ago, martillo said:

The question of if the definition of the definition of P = mv applies to variable masses can be solved considering that if it is valid, then is valid the equation:

dp/dt = mdv/dt + vdm/dt

I get a different result. I assume that P = mv applies to variable masses, but I apply dP/dt carefully, keeping track of what P refers to.

Before the change of mass, it refers to a body with mass m and velocity v. At this time,

P₁ = mv

After losing a mass dm, it does not refer to a body with mass (m-dm) but rather to two bodies: one with mass (m-dm) and velocity (v+dv), and another one with mass dm and some velocity u. At this time,

P₂ = (m-dm)(v+dv) + udm

So, in time dt the momentum change is

dP = P₂ - P₁ = (m-dm)(v+dv) + udm - mv = (mv - vdm + mdv - dmdv + udm) - mv = mdv + (u - v - dv)dm = 

mdv + wdm - dvdm

where w = u - v is a relative velocity between the two bodies.

Now, dividing by dt I get the result,

dP/dt = mdv/dt +wdm/dt - dvdm/dt

The last term, dvdm/dt, is infinitesimally small, and the final result is,

dP/dt = mdv/dt + wdm/dt

rather than

dP/dt = mdv/dt + vdm/dt

[swansont]

9 hours ago, martillo said:

You said this. So you agree then, that it is considered in the link that the force on the rocket is F = ma. 

You need to work on paying attention to detail. Nowhere in the part you quoted did I use the equation F = ma

I agreed that there is a term in there that is m dv/dt, which is a force. They derived this by looking at the momentum of the rocket of mass m has as the result of ejecting a mass ∆m. They did not arrive at this by applying F = ma

8 hours ago, martillo said:

For the case of systems with mass, p is defined as p = mv whatever m being constant or variable.

In Newtonian physics, when considering conservation of momentum, mass can't change. Once you have defined the system (mass M) you can't change it. If a mass m2 is ejected, for example, leaving mass m1, you have to consider both masses, m1 and m2 = (m1 + m2 = M)

[martillo]

19 minutes ago, swansont said:

You need to work on paying attention to detail. Nowhere in the part you quoted did I use the equation F = ma

I agreed that there is a term in there that is m dv/dt, which is a force. They derived this by looking at the momentum of the rocket of mass m has as the result of ejecting a mass ∆m. They did not arrive at this by applying F = ma

Let us forget now how they arrived to the Thrust Force, I ask you now: you agree or disagree that the applied force on the rocket is F = ma = -v_edm/dt?

28 minutes ago, Genady said:

I get a different result. I assume that P = mv applies to variable masses, but I apply dP/dt carefully, keeping track of what P refers to.

Before the change of mass, it refers to a body with mass m and velocity v. At this time,

P₁ = mv

After losing a mass dm, it does not refer to a body with mass (m-dm) but rather to two bodies: one with mass (m-dm) and velocity (v+dv), and another one with mass dm and some velocity u. At this time,

P₂ = (m-dm)(v+dv) + udm

So, in time dt the momentum change is

dP = P₂ - P₁ = (m-dm)(v+dv) + udm - mv = (mv - vdm + mdv - dmdv + udm) - mv = mdv + (u - v - dv)dm = 

mdv + wdm - dvdm

where w = u - v is a relative velocity between the two bodies.

Now, dividing by dt I get the result,

dP/dt = mdv/dt +wdm/dt - dvdm/dt

The last term, dvdm/dt, is infinitesimally small, and the final result is,

dP/dt = mdv/dt + wdm/dt

rather than

dP/dt = mdv/dt + vdm/dt

Well, one step further. The question is, do you arrive to the Thrust Force from this your Thrust Equation?

[Genady]

26 minutes ago, martillo said:

Well, one step further. The question is, do you arrive to the Thrust Force from this your Thrust Equation?

Yes, I do.

The expression, dP/dt = mdv/dt + wdm/dt, represents the total force on the combined system of the two bodies.

One component of this total force is mdv/dt, the force on the rocket.

[swansont]

31 minutes ago, martillo said:

Let us forget now how they arrived to the Thrust Force, I ask you now: you agree or disagree that the applied force on the rocket is F = ma = -v_edm/dt?

Yes, that’s the result from equation 1.5

 

[joigus]

2 hours ago, martillo said:

The definition of momentum p in my Frank J. Blatt's book of Physics (Spanish version) is p = mv.

Note: p is called "cantidad de movimiento" in Spanish and "linear momentum" or just "momentum" in English as mentioned in the foot note of the page (photo below).

Here the photos of the definition section and the foot note of the page:

 

2 hours ago, martillo said:

I think Newton didn't considered the mass as variable so may be the original definition is valid for constant mass.

Not necessarily.

 

8 hours ago, martillo said:

Please explain how is p defined for things with variable mass then? May be you could explain E = mc² later if it would matter...

Here's a sketch of how it would be done.

1) We don't care about the exhaust at all. Our system is the rocket; but it's an open system. To simplify, let's make it 1-degree of freedom, with coordinate of rocket = \( x \).

\[ L=\frac{1}{2}m\dot{x}^{2}-V\left(x\right) \]

Momentum of rocket is partial derivative of \( L \) with respect to generalised velocity. Just following our noses, and blindly believing in variational calculus and our guessed-at Lagrangian being good:

\[ p_{x}=\frac{\partial L}{\partial\dot{x}}=m\dot{x} \]

Euler-Lagrange equation:

\[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow\frac{d}{dt}\left(m\dot{x}\right)=-\frac{dV}{dx}\Rightarrow m\ddot{x}+\dot{m}\dot{x}=F_{\textrm{ext}}=-V'\left(x\right) \]

which is, of course, the rocket equation.

2) We care about the exhaust. 

Take all this with a pinch of salt, because I'm not 100% sure this is correct. It produces consistent equation for the rocket though. It's possible that the law of change of \( m \) with time had better be treated as a time-dependent constraint, rather than a coordinate.

The exhaust is complicated to describe. But, as we don't care about it very much, we will only describe it with one generalised coordinate, \( X \) wich accounts for the CoM of the whole thing at time \( t \).

Very important: The system configuration needs specifying three coordinates: \( x \), \( X \) and \( m\left( t \right) \).

We will have 3 Euler-Lagrange equations. The Lagrangian should be,

\[ L=\frac{1}{2}\left(M-m\right)\dot{X}^{2}+\frac{1}{2}m\dot{x}^{2}-V_{r}\left(x\right)-V_{e}\left(X\right) \]

and \( V_{r}\left(x\right) \), \( V_{e}\left(X\right) \) the respective potential energies of rocket and exhaust CoM. We will also treat \( m \) as a generalised coordinate. The Euler-Lagrange equations are,

\[ \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0\Rightarrow m\ddot{x}+\dot{m}\dot{x}=-V_{r}'\left(x\right) \]

\[ \frac{d}{dt}\frac{\partial L}{\partial\dot{X}}-\frac{\partial L}{\partial X}=0\Rightarrow\frac{d}{dt}\left[\left(M-m\right)\dot{X}\right]=-V_{e}'\left(X\right) \]

\[ \frac{d}{dt}\frac{\partial L}{\partial\dot{m}}-\frac{\partial L}{\partial m}=0\Rightarrow V_{r}\left(x\right)=V_{e}\left(X\right) \]

And those are your evolution equations, at least within this very simple model, and as you can see you get a generalised expression for the momentum of the rocket that accounts for the mass it's losing.

My conclusion thus is that @Genady --and everybody else-- is right.

As I told you, your trusty momentum can be generalised. This version is called 'canonical momentum associated to coordinate \( x \).' In the case of a particle in a magnetic field, it has a 'tail' that contains the vector potential. It doesn't have to conform to any previous ideas that you have about what it is. The only thing that spoils your Lagrangian methods is dissipation in those cases where you can't follow where the energy has gone. If you don't care about details, and are happy to describe only the CoM of the bulk of exhaust, you can get a good understanding of what's going on.

Forgot to mention: \( M \) is a certain total initial mass of rocket when it's fully charged of fuel.

[martillo]

27 minutes ago, Genady said:

Yes, I do.

The expression, dP/dt = mdv/dt + wdm/dt, represents the total force on the combined system of the two bodies.

One component of this total force is mdv/dt, the force on the rocket.

 

24 minutes ago, swansont said:

Yes, that’s the result from equation 1.5

 

So you both agree that the Thrust Force is F = ma = -v_edm/dt.

Well, in the rocket the mass m varies then there is a problem because you both agree that even with a varying mass m is valid: F = ma and not F = dp/dt.

I mean F = ma on the rocket and not F = dp/dt.

dp/dt ≠ a when mass varies.

 

17 minutes ago, joigus said:

Not necessarily.

Not necessarily, I agree.

Edited February 22, 2023 by martillo

[Genady]

25 minutes ago, martillo said:

mean F = ma on the rocket and not F = dp/dt.

dp/dt ≠ a when mass varies.

This is incorrect. Both F = ma and F = dP/dt are correct. Your mistake is that you apply them incorrectly.

In F = ma you consider force on the rocket only, while in F = dP/dt, F and P are the force and the momentum of the entire system. So, the F in the first equation is not the same as the F in the second.

Of course, dP/dt should be ≠ a, because dP/dt is rate of change of the momentum of the entire system, while a is an acceleration of its part only, the rocket. The whole is not equal to its part.  

47 minutes ago, joigus said:

 

Not necessarily.

 

Here's a sketch of how it would be done.

1) We don't care about the exhaust at all. Our system is the rocket; but it's an open system. To simplify, let's make it 1-degree of freedom, with coordinate of rocket = x .

 

L=12mx˙2−V(x)

 

Momentum of rocket is partial derivative of L with respect to generalised velocity. Just following our noses, and blindly believing in variational calculus and our guessed-at Lagrangian being good:

 

px=∂L∂x˙=mx˙

 

Euler-Lagrange equation:

 

ddt∂L∂x˙−∂L∂x=0⇒ddt(mx˙)=−dVdx⇒mx¨+m˙x˙=Fext=−V′(x)

 

which is, of course, the rocket equation.

2) We care about the exhaust. 

Take all this with a pinch of salt, because I'm not 100% sure this is correct. It produces consistent equation for the rocket though. It's possible that the law of change of m with time had better be treated as a time-dependent constraint, rather than a coordinate.

The exhaust is complicated to describe. But, as we don't care about it very much, we will only describe it with one generalised coordinate, X wich accounts for the CoM of the whole thing at time t .

Very important: The system configuration needs specifying three coordinates: x , X and m(t) .

We will have 3 Euler-Lagrange equations. The Lagrangian should be,

 

L=12(M−m)X˙2+12mx˙2−Vr(x)−Ve(X)

 

and Vr(x) , Ve(X) the respective potential energies of rocket and exhaust CoM. We will also treat m as a generalised coordinate. The Euler-Lagrange equations are,

 

ddt∂L∂x˙−∂L∂x=0⇒mx¨+m˙x˙=−V′r(x)

 

 

ddt∂L∂X˙−∂L∂X=0⇒ddt[(M−m)X˙]=−V′e(X)

 

 

ddt∂L∂m˙−∂L∂m=0⇒Vr(x)=Ve(X)

 

And those are your evolution equations, at least within this very simple model, and as you can see you get a generalised expression for the momentum of the rocket that accounts for the mass it's losing.

My conclusion thus is that @Genady --and everybody else-- is right.

As I told you, your trusty momentum can be generalised. This version is called 'canonical momentum associated to coordinate x .' In the case of a particle in a magnetic field, it has a 'tail' that contains the vector potential. It doesn't have to conform to any previous ideas that you have about what it is. The only thing that spoils your Lagrangian methods is dissipation in those cases where you can't follow where the energy has gone. If you don't care about details, and are happy to describe only the CoM of the bulk of exhaust, you can get a good understanding of what's going on.

Forgot to mention: M is a certain total initial mass of rocket when it's fully charged of fuel.

This is a nice back-of-envelope derivation! Thank you.

[martillo]

18 minutes ago, Genady said:

This is incorrect. Both F = ma and F = dP/dt are correct. Your mistake is that you apply them incorrectly.

In F = ma you consider force on the rocket only, while in F = dP/dt, F and P are the force and the momentum of the entire system. So, the F in the first equation is not the same as the F in the second.

Of course, dP/dt should be ≠ a, because dP/dt is rate of change of the momentum of the entire system, while a is an acceleration of its part only, the rocket. The whole is not equal to its part.  

No, I'm considering just the rocket now. The force on the rocket and the momentum of the rocket. Let us call F_r the force on the rocket and p_r the momentum of the rocket. Then the force on the rocket is the thrust force: F_r = -v_edm/dt and it must verify one of the two possibilities:

1) F_r = ma_r

2)F_r = dp_r/dt

For me and in the provided link (http://www.braeunig.us/space/propuls.htm) the first applies and agrees with F_r = mdv/dt.

For you, as your definition of force is F = dp/dt, F_r must verify: F_r = dp_r/dt.

Now, dp_r/dt ≠ a_r for the variable mass m of the rocket so only one can be true.

The rockets' Thrust Equation (mdv/dt = -v_edm/dt) in which all of us agree shows that our first possibility is the right one. Your second possibility is wrong then.

The rockets' dynamics shows that F = ma is valid and not F = dp/dt as the title of the thread says. The claim of the thread is right.

Edited February 22, 2023 by martillo

[Genady]

5 minutes ago, martillo said:

No, I'm considering just the rocket now. The force on the rocket and the momentum of the rocket. Let us call F_r the force on the rocket and p_r the momentum of the rocket. Then the force on the rocket is the thrust force: F_r = -v_edm/dt and it must verify one of the two possibilities:

1) F_r = ma_r

2)F_r = dp_r/dt

For me and in the provided link (http://www.braeunig.us/space/propuls.htm) the first applies and agrees with F_r = mdv/dt.

For you, as your definition of force is F = dp/dt, F_r must verify: F_r = dp_r/dt.

Now, dp_r/dt ≠ a_r for the variable mass m of the rocket so only one can be true.

The rockets' Thrust Equation (mdv/dt = -v_edm/dt) in which all of us agree shows that our first possibility is the right one. Your second possibility is wrong then.

The rockets' dynamics shows that F = ma is valid and not F = dp/dt as the title of the thread says. The claim of the thread is right.

You say that you are considering only momentum of a rocket, but when you write P = mv, you work with the momentum of the entire system, and it does not matter what you say about it.

You can insist on making the same mistake, and it will not lead you anywhere. But if it makes you happy, enjoy!