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

[martillo]

43 minutes ago, studiot said:

I see that hyperphysics does it my way, but with much better pictures.

http://hyperphysics.phy-astr.gsu.edu/hbase/limn2.html#ln21

http://hyperphysics.phy-astr.gsu.edu/hbase/rocket.html#c2

 

I like hyperphysics. It gives very concise, simple but precise and strict presentations of fundamental things in Physics.

I analyzed many things with the help of it.

But if you think it could change my approach that there is needed a profound review in Physics from the beginnings (say Newton laws) and that a new developing must surge from scratch then you are losing your time.

That will come, in one way or another one, the point is if you are going to stay just waiting for it or if you are going to be a protagonist of it. I'm trying something, it looks a good approach although is just a start-point. Don't pretend me to have all resolved. Impossible. I'm presenting this start-point here in this forum to be analyzed with the total rationality I have seen things are treated here. I think we, here, could make a good progress. 

I would ask you now, just to consider what is proposed just for a while to analyze where it could go, just for a while...

Edited February 23, 2023 by martillo

[joigus]

3 hours ago, martillo said:

The Momentum and the Energy MUST be conserved ALWAYS!

No. It (momentum of either rocket, fuel or both) only has to be conserved if there are no external forces on the system. In the more general case with an external force that I presented to you --in the presence of this external force acting on both the rocket and the exhaust-- neither momentum nor (mechanical) energy have to be conserved.

You say "momentum" and "energy" as if they must mean something independent of what system you apply it to, or which level of detail you want to describe it. They don't.

Rocket:

Energy is not conserved

Momentum is not conserved

Exhaust:

Energy is not conserved

Momentum is not conserved 

Rocket + exhaust:

Mechanical energy is not conserved

Mechanical energy + chemical energy stored in fuel is conserved

Momentum is not conserved if rocket + exhaust are in a gravitational field

Momentum is conserved if fuel + rocket are in free space

You see, when you burn fuel, some kind of potential energy that's stored in the molecular bonds gets converted into heat + kinetic energy for both rocket and exhaust. We call this energy Gibbs free energy (of fuel).

So no, mechanical energy is not conserved in this case. If the rocket is stationary in free space at initial time, K.E = P.E. = 0 initially. Then it starts burning fuel and starts moving. Now P.E. is not cero, but K.E. is K.E. of exhaust + K.E. of rocket. So where did this energy come from?

Edited February 23, 2023 by joigus
minor addition

[martillo]

10 minutes ago, joigus said:

No. It only has to be conserved if there are no external forces on the system. In the more general case with an external force that I presented to you --in the presence of this external force acting on both the rocket and the exhaust-- neither momentum nor (mechanical) energy have to be conserved.

You say "momentum" and "energy" as if they must mean something independent of what system you apply it to, or which level of detail you want to describe it. They don't.

Rocket:

Energy is not conserved

Momentum is not conserved

Exhaust:

Energy is not conserved

Momentum is not conserved 

Rocket + exhaust:

Mechanical energy is not conserved

Mechanical energy + chemical energy stored in fuel is conserved

Momentum is not conserved if rocket + exhaust are in a gravitational field

Momentum is conserved if fuel + rocket are in free space

You see, when you burn fuel, some kind of potential energy that's stored in the molecular bonds gets converted into heat + kinetic energy for both rocket and exhaust. We call this energy Gibbs free energy.

So no, mechanical energy is not conserved in this case. If the rocket is stationary in free space at initial time, K.E = P.E. = 0 initially. Then it starts burning fuel and starts moving. Now P.E. is not cero, but K.E. is K.E. of exhaust + K.E. of rocket. So where did this energy come from?

You are right. They must be conserved for closed systems only without external forces or energies being applied to or delivered from it. It is just that it was the case we were analyzing, a closed system.

Edited February 23, 2023 by martillo

[joigus]

3 minutes ago, martillo said:

You are right. They must be conserved for closed systems only without external forces or energies being applied to or delivered from it. It is just that it was the case we were analyzing, a closed system.

What system? There's your vagueness again.

6 minutes ago, martillo said:

It is just that it was the case we were analyzing, a closed system.

Which is included as a particular case in my analysis, of which you said nothing.

[studiot]

11 minutes ago, joigus said:

No. It (momentum of either rocket, fuel or both) only has to be conserved if there are no external forces on the system. In the more general case with an external force that I presented to you --in the presence of this external force acting on both the rocket and the exhaust-- neither momentum nor (mechanical) energy have to be conserved.

He's just not listening bro.

He doesn't even know the correct definition of a closed system.

So many misconceptions of a fundamental nature.

[martillo]

1 minute ago, joigus said:

What system? There's your vagueness again.

That was said for the case presented by Genady of a melting ice moving at some constant velocity v. We were analyzing that how th principle of conservation of momentum applies.

[joigus]

1 minute ago, studiot said:

He doesn't even know the correct definition of a closed system.

 

I know. This is very close to coming full circle, or is already there. Sigh.

[martillo]

1 minute ago, studiot said:

He doesn't even know the correct definition of a closed system.

So many misconceptions of a fundamental nature.

If I would be wrong in something I would prefer to be corrected by you in it, not being just discarded...

[joigus]

Lee Smolin:

"The trouble with physics..."

Martillo:

"...is F=dp/dt"

"is F=ma"

 

Edited February 23, 2023 by joigus
correction

[martillo]

7 minutes ago, joigus said:

Lee Smolin:

"The trouble with physics..."

Martillo:

"is F=ma"

 

As I could see Lee Smolin looking for a unification of Relativity with Quantum Physics. You are right that this not my approach of Physics from scratch at all. You decide...

Edited February 23, 2023 by martillo

[studiot]

1 hour ago, martillo said:

If I would be wrong in something I would prefer to be corrected by you in it, not being just discarded...

Well I was just following Dr Cowan from London University all those decades ago, when I learned this stuff.

Here is his introduction to the subject.
I was just trying to flesh out the explanations further and show where the mathematical tricks occur.

Note very carfully what he says about f = ma   v  f =mdp/dt

and his use of r as distance moved by centre of mass.

Edited February 23, 2023 by studiot

[martillo]

50 minutes ago, studiot said:

Well I was just following Dr Cowan from London University all those decades ago, when I learned this stuff.

Here is his introduction to the subject.
I was just trying to flesh out the explanations further and show where the mathematical tricks occur.

Note very carfully what he says about f = ma   v  f =mdp/dt

and his use of r as distance moved by centre of mass.

Well, for the total system (the rocket and exhaust together) we have dp/dt = 0, right?

Thinking on...

Edited February 23, 2023 by martillo

[studiot]

5 minutes ago, martillo said:

Well, for the total system (the rocket and exhaust together) we have dp/dt = 0, right? Now if we consider rocket and exhaust separately, say p_r and p_e, then  p_r = -p_e and so p_r + p_e = 0. Is not a conserved quantity unless it means zero. I think something is going wrong there. Let we see how you proceed now...

What is dp/dt ?

I am asking what you thing the symbols mean mathematically, not what is momentum or the tim,e rate of change of momentum.

You are confusing the derived function (derivative) with the value of that function at a single value of time in the case.

Sometimes this distinction is unimportant, sometimes it matters very much.

That is the case here.

Not only is dp/dt a function it is a vector valued function.

 

[Genady]

1 hour ago, joigus said:

This is very close to coming full circle, or is already there. Sigh.

As it was two days ago: https://www.scienceforums.net/topic/128859-physics-in-troubles-the-real-equation-of-force-is-f-ma-and-not-f-dpdt/?do=findComment&comment=1230793

 

(there is a link inside this link)

[martillo]

17 minutes ago, studiot said:

What is dp/dt ?

I am asking what you thing the symbols mean mathematically, not what is momentum or the tim,e rate of change of momentum.

You are confusing the derived function (derivative) with the value of that function at a single value of time in the case.

Sometimes this distinction is unimportant, sometimes it matters very much.

That is the case here.

Not only is dp/dt a function it is a vector valued function.

 

May be I didn't express it correctly. Let me try again:

If we consider the total system (the rocket and exhaust together) we have not only dp/dt = 0 but also p = 0. If we consider now rocket and exhaust separately, say pr and pe, then pr = -pe and so pr + pe = 0. Is not a conserved quantity unless it means zero. I think something is going wrong there...

Let we see how it proceeds...

Edited February 23, 2023 by martillo

[sethoflagos]

On 2/22/2023 at 3:45 PM, martillo said:

You know, you can't escape from the fact that what works for the rockets is the definition of force as F = ma and not F = dp/dt. It is what works in practice...

F_total = dp/dt = mdv/dt + vdm/dt = F₁ - F₂ = 0  For total system

Consider momentum change of body A (rocket +unused fuel)

F_A = dp_A/dt = m_Adv/dt + vdm_A/dt = F₁ - F₃ = X (non-zero!)

Consider momentum change of body B (exhaust) 

F_B = dp_B/dt = d/dt (mB (v-(v-v_e)) = d/dt (m_B (v - u_e)) = dm_B(dv/dt - du_e/dt)  + (v - u_e)dm_B/dt = dm_Bdv/dt + (v - u_e)dm_B/dt 

Since dm_B = - dm_A 

F_B  = - dm_Adv/dt - (v - u_e)dm_A/dt = F₄ - F₃ - F₅ = -X (non-zero!)

Add F_A and F_B

F_total = F_A + F_B = dp/dt = m_Adv/dt + vdm_A/dt  - dm_Adv/dt - (v - u_e)dm_A/dt = (m_A - dm_A)dv/dt + vdm_A/dt - vdm_A/dt + u_edm_A/dt = 0

Hence 

(m_A - dm_A)dv/dt = - u_edm_A/dt 

or

F₁ + F₄ = F₅

There's a couple of differences here to your thrust equation. I've used u_e instead of v_e to show that the exhaust velocity is specific to the comoving inertial reference frame of the rocket whereas v is for any observer. I've also left the dm_A term in the left hand expression to emphasize that the thrust equation involves m_A rather than m of the overall equation. Mathematically, it's justifiable to delete this dm_A as is is overwhelmed by m_A.

Please take note how I have distinguished different forces with different subscripts. Much of your confusion appears to stem from an expectation that all Fs must be equal. Sometimes they are. Generally they are not.

[martillo]

33 minutes ago, sethoflagos said:

F_total = dp/dt = mdv/dt + vdm/dt = F₁ - F₂ = 0  For total system

Consider momentum change of body A (rocket +unused fuel)

F_A = dp_A/dt = m_Adv/dt + vdm_A/dt = F₁ - F₃ = X (non-zero!)

Consider momentum change of body B (exhaust) 

F_B = dp_B/dt = d/dt (mB (v-(v-v_e)) = d/dt (m_B (v - u_e)) = dm_B(dv/dt - du_e/dt)  + (v - u_e)dm_B/dt = dm_Bdv/dt + (v - u_e)dm_B/dt 

Since dm_B = - dm_A 

F_B  = - dm_Adv/dt - (v - u_e)dm_A/dt = F₄ - F₃ - F₅ = -X (non-zero!)

Add F_A and F_B

F_total = F_A + F_B = dp/dt = m_Adv/dt + vdm_A/dt  - dm_Adv/dt - (v - u_e)dm_A/dt = (m_A - dm_A)dv/dt + vdm_A/dt - vdm_A/dt + u_edm_A/dt = 0

Hence 

(m_A - dm_A)dv/dt = - u_edm_A/dt 

or

F₁ + F₄ = F₅

There's a couple of differences here to your thrust equation. I've used u_e instead of v_e to show that the exhaust velocity is specific to the comoving inertial reference frame of the rocket whereas v is for any observer. I've also left the dm_A term in the left hand expression to emphasize that the thrust equation involves m_A rather than m of the overall equation. Mathematically, it's justifiable to delete this dm_A as is is overwhelmed by m_A.

Please take note how I have distinguished different forces with different subscripts. Much of your confusion appears to stem from an expectation that all Fs must be equal. Sometimes they are. Generally they are not.

I'm confused with your approach and will take time to me analyze it but I will.

33 minutes ago, sethoflagos said:

Consider momentum change of body B (exhaust) 

F_B = dp_B/dt = d/dt (mB (v-(v-v_e)) = d/dt (m_B (v - u_e)) = dm_B(dv/dt - du_e/dt)  + (v - u_e)dm_B/dt = dm_Bdv/dt + (v - u_e)dm_B/dt 

Here is a mistake I think: in your ending relation does not appear m_B. It should be:

F_B = dp_B/dt = d/dt (mB (v-(v-v_e)) = d/dt (m_B (v - u_e)) = m_B(dv/dt - du_e/dt)  + (v - u_e)dm_B/dt = m_Bdv/dt + (v - u_e)dm_B/dt 

Let me say if it is right for you. If you correct this may be we can go on.

Edited February 23, 2023 by martillo

[studiot]

1 hour ago, martillo said:

May be I didn't express it correctly. Let me try again:

If we consider the total system (the rocket and exhaust together) we have not only dp/dt = 0 but also p = 0. If we consider now rocket and exhaust separately, say pr and pe, then pr = -pe and so pr + pe = 0. Is not a conserved quantity unless it means zero.

Let we see how it proceeds...

Quote

I think something is going wrong there...

 

Yes your contention that p is necessarily zero.

 

Consider the rocket (as I did) just moving along at velocity V, at time t.

If the rocket engine is not firing this sytem has non zero momentum.

If it is unimpeded by gravity or air or other resistance Newton's first law tells us this will continue indefinitely.

If the rocket engine now fires and the rocket starts to accelerate (can you see why this must then happen ?)

The rocket will gain extra new momentum equal to the negative momentum of the exhaust.

But it will still retain at least some of its original momentum.

This situation could easily occur in a series of rocket engine 'burns'.

 

This is why I did not start the conditions at t = 0 or V = 0.

These are conditions that can be added into the analysis, but many analyses start the rocket off fromstanding still. in which case its momentum will obviously be zero.

However this last case can lead the unwary into a Zeno like paradox (have you heard of them?) where you can try to argue that no rocket can ever leave the ground since it cannot move until the exhaust start to move.

 

Edited February 23, 2023 by studiot

[martillo]

8 minutes ago, studiot said:

 

Yes your contention that p is necessarily zero.

 

Consider the rocket (as I did) just moving along at velocity V, at time t.

If the rocket engine is not firing this sytem has non zero momentum.

If it is unimpeded by gravity or air or other resistance Newton's first law tells us this will continue indefinitely.

If the rocket engine now fires and the rocket starts to accelerate (can you see why this must then happen ?)

The rocket will gain extra new momentum equal to the negative momentum of the exhaust.

But it will still retain its original momentum.

This situation could easily occur in a series of rocket engine 'burns'.

 

This is why I did not start the conditions at t = 0 or V = 0.

These are conditions that can be added into the analysis, but many analyses start the rocket off fromstanding still. in which case its momentum will obviously be zero.

However this last case can lead the unwary into a Zeno like paradox (have you heard of them?) where you can try to argue that no rocket can ever leave the ground since it cannot move until the exhaust start to move.

Until the moment the rocket begins to move we can consider the exhausted fuel as just wasted or used to heat the motor to a right temperature or something like that. I think we can well consider the moment of the total system (rocket + exhaust) as p=0 at the initial state as at any state. Or you don't want to analyze this way?

[sethoflagos]

1 hour ago, martillo said:

Here is a mistake I think: in your ending relation does not appear m_B. It should be:

Over the time interval dt, m_B increases from 0 to dm_B. So in context m_B = dm_B. 

[studiot]

I said

Edit correctly attribute quotation

30 minutes ago, studiot said:

Consider the rocket (as I did) just moving along at velocity V, at time t.

end edit

So why oh why did you respond

18 minutes ago, martillo said:

Until the moment the rocket begins to move

 

I even explained in some detail why I did not start with a stationary rocket.

I also said that the analysis must nevertheless cope with the case of starting with a stationary rocket as with any other possibility.

 

A further tip.

Rocket engineers are less concerned with acceleration v momentum as they are with questions like

How to maximise final velocity ?

How to maximise acceleration ?

How to maximise burn time ?

These are not all compatible so they need a pretty flexible analysis for all that and more.

Edited February 23, 2023 by studiot

[martillo]

12 minutes ago, sethoflagos said:

Over the time interval dt, m_B increases from 0 to dm_B. So in context m_B = dm_B. 

No but m_B will be the total exhausted fuel all the time from the beginning to the end while dm_B is a differential component of it in a differential element of time dt. You cannot relace one by the other in the equations. Is something wrong.

10 minutes ago, studiot said:

I said

So why oh why did you respond

 

I even explained in some detail why I did not start with a stationary rocket.

I also said that the analysis must nevertheless cope with the case of starting with a stationary rocket as with any other possibility.

 

A further tip.

Rocket engineers are less concerned with acceleration v momentum as they are with questions like

How to maximise final velocity ?

How to maximise acceleration ?

How to maximise burn time ?

These are not all compatible so they need a pretty flexible analysis for all that and more.

Fine, I said:

2 hours ago, martillo said:

May be I didn't express it correctly. Let me try again:

If we consider the total system (the rocket and exhaust together) we have not only dp/dt = 0 but also p = 0. If we consider now rocket and exhaust separately, say pr and pe, then pr = -pe and so pr + pe = 0. Is not a conserved quantity unless it means zero. I think something is going wrong there...

Let we see how it proceeds...

Proceed now as you want with p = some constant. Doesn't mind for me. I will consider:
p = p₀ for instance and pr + pe = p₀  constant.

Fine for you?

Edited February 23, 2023 by martillo

[sethoflagos]

4 minutes ago, martillo said:

No but m_B will be the total exhausted fuel all the time from the beginning to the end while dm_B is a differential component of it in a differential element of time dt. You cannot relace one by the other in the equations. Is something wrong.

Only if you want it to be.

Once dm_B has been exhausted it plays no further part in rocket propulsion. I'm pretty sure I'm free to reset the counter back to zero again. The total amount of exhaust is of no relevance to the OP. So why bother keeping track of it.

[studiot]

11 minutes ago, martillo said:

Fine for you?

Think carefully about what you just proposed.

is p_r constant ?

If so why?

If not why not ?

[martillo]

2 minutes ago, sethoflagos said:

Only if you want it to be.

Once dm_B has been exhausted it plays no further part in rocket propulsion. I'm pretty sure I'm free to reset the counter back to zero again. The total amount of exhaust is of no relevance to the OP. So why bother keeping track of it.

You can't. The momentum of the total system of the rocket and the exhausted fuel includes all the mass exhausted from the initial state to the end. You cannot just discard all that mass already exhausted time ago in the conservation of momentum. If you don't agree with this I cannot proceed anymore.

Just now, studiot said:

Think carefully about what you just proposed.

is p_r constant ?

If so why?

If not why not ?

No, p_r is not constant, is just  p_r = p₀ - p_e . Both increasing in time with the energy of the combustion of the fuel.