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

[Lorentz Jr]

1 hour ago, martillo said:

What I claim is precisely that they use F = ma and not F = dp/dt for the rocket.

It's the same thing, martillo. F = dp/dt is derived from F = ma. It's just a simpler form for when the mass of the object changes.

Edited February 21, 2023 by Lorentz Jr

[martillo]

3 hours ago, Lorentz Jr said:

It's the same thing, martillo. F = dp/dt is derived from F = ma. It's just a simpler form for when the mass of the object changes.

A rocket is may be the best example for variable mass  and the equation F = ma is what is applied and not dp/dt. This show that the real equation for force is F = ma and not dp/dt.

There are ones with high variation in their mass, missiles for instance. They all work very fine. So they are the proof that the equation F = ma applies and not dp/dt even for variable mass. This changes it all in Physics.

[Lorentz Jr]

1 minute ago, martillo said:

This show that the real equation for force is F = ma and not dp/dt.

F = ma doesn't work in complicated situations involving relativity and quantum mechanics, and nobody cares whether it's the "real" equation for rockets. F = dp/dt is just as correct and easier to use.

[martillo]

42 minutes ago, Lorentz Jr said:

F = ma doesn't work in complicated situations involving relativity and quantum mechanics, and nobody cares whether it's the "real" equation for rockets. F = dp/dt is just as correct and easier to use.

What I'm saying is that the formula that really applies is F = ma and not dp/dt. This implies that even the classical formulation is wrong. Newton second law is formulated as F = dp/dt. I'm saying that is wrong and that the right formula is F = ma even when mass varies. Newton didn't have the opportunity to take missiles in consideration to verify the formula. They didn't exist in that epoch, right? 

42 minutes ago, Lorentz Jr said:

nobody cares whether it's the "real" equation for rockets.

That's the problem. I do care.

Edited February 21, 2023 by martillo

[Genady]

10 minutes ago, martillo said:

What I'm saying is that the formula that really applies is F = ma and not dp/dt. This implies that even the classical formulation is wrong. Newton second law is formulated as F = dp/dt. I'm saying that is wrong and that the right formula is F = ma even when mass varies.

There is no time dilation (split from The twin Paradox revisited) - Speculations - Science Forums

The same applies here.

[martillo]

24 minutes ago, Genady said:

There is no time dilation (split from The twin Paradox revisited) - Speculations - Science Forums

The same applies here.

No, you cannot say that. The evidence I'm presenting is the rockets' motion working properly and that what is found to be applied for them is F = ma and not dp/dt. This can be well verified on the entire web. In Wikipedia for instance it is found the same approach using F = ma and not dp/dt.

May be you want the thread to be discarded and sent to trash for it to not be discussed anymore, that would not be fair...

Edited February 21, 2023 by martillo

[studiot]

15 minutes ago, martillo said:

That's the problem. I do care.

Then consider the possibility that you have a fundamental misunderstanding of Mechanics, that goes back nearly to your first studies of the subject.

 

15 minutes ago, martillo said:

What I'm saying is that the formula that really applies is F = ma and not dp/dt. This implies that even the classical formulation is wrong. Newton second law is formulated as F = dp/dt. I'm saying that is wrong and that the right formula is F = ma even when mass varies. Newton didn't have the opportunity to take missiles in consideration to verify the formula. They didn't exist in that epoch, right? 

 

7 hours ago, martillo said:

I'm saying that because of rockets work very well the equation of force is F = ma and not F = dp/dt.

Your basic misunderstanding is this:

A rocket is a mechanical system.

There is no 'the equation' for this system ie one single solitary equation.

A very common situation in Mechanics is that you need to consider more than one equation to solve a system.

You need a set of simultaneous equations.

 

This is the case with the rocket system (rocket for short).

All that is happening is that you have an equation for the whole system and another one for part of the system.

 

When studying Mechanics you learn this very early on when calculating loads on a static system and reactions via forces and moments.

[martillo]

6 minutes ago, studiot said:

All that is happening is that you have an equation for the whole system and another one for part of the system.

That's a problem here.

7 minutes ago, studiot said:

When studying Mechanics you learn this very early on when calculating loads on a static system and reactions via forces and moments.

This is not a static system. Seems you are suggesting I didn't studied Mechanics, I did. This thread is all about Mechanics. You can develop what you think it must apply.

[studiot]

3 minutes ago, martillo said:

That's a problem here.

This is not a static system. Seems you are suggesting I didn't studied Mechanics, I did. This thread is all about Mechanics. You can develop what you think it must apply.

I am trying to help you, not criticise so try not to react in haste.

 

I didn't say you didn't study Mechanics.

Quite the opposite in fact.

I said you missed something important, when you did study Mechanics.

 

No, I agree a rocket is not static, but it can be quasi static otherwise known as steady state.

But that was not the point.

Static systems are in general simpler than dynamic ones, indeed one way to solve synamic ones is to reduce them to static or quasi staic ones.

 

Please read carefully my point about simultaneous equations before you reply.

Even the simpler static systems may need simultaneous equations for solution.

[Markus Hanke]

7 minutes ago, martillo said:

This thread is all about Mechanics. You can develop what you think it must apply.

Newton’s second law in the form F=dp/dt (which is equivalent to F=ma iff m=const) isn’t just made up, but follows directly from the principle of least action, and is thus mathematically derivable in a self-consistent manner. I suggest for starters you might take a look here. Your claim that this is “wrong” somehow is thus pretty meaningless.

[martillo]

31 minutes ago, Markus Hanke said:

Newton’s second law in the form F=dp/dt (which is equivalent to F=ma iff m=const) isn’t just made up, but follows directly from the principle of least action, and is thus mathematically derivable in a self-consistent manner. I suggest for starters you might take a look here. Your claim that this is “wrong” somehow is thus pretty meaningless.

Well, I studied Mechanics but not Variational Calculus. Do you think it is really needed to treat this problem? If so actually I would not be able to discuss anymore, of course.

34 minutes ago, studiot said:

Even the simpler static systems may need simultaneous equations for solution.

But here there is just one simple linear equation of motion. The problem is which formula is applied on it. F = ma or F = dp/dt? That gives the right equation to apply for objects with variable mass. F = ma is applied.

Edited February 21, 2023 by martillo

[studiot]

23 minutes ago, martillo said:

Well, I studied Mechanics but not Variational Calculus. Do you think it is really needed to treat this problem? If so actually I would not be able to discuss anymore, of course.

Er...?

Variational calculus ?

That's second year university stuff.

I said simultaneous equations.

In the UK you start learning about these around 12 or 13, actually before you learn any mechanics.

23 minutes ago, martillo said:

But here there is just one simple linear equation of motion. The problem is which formula is applied on it. F = ma or F = dp/dt? That gives the right equation to apply for objects with variable mass. F = ma is applied.

But it isn't one simple equation.

You can't calculate anything with one simple equation as you have more an one variable involved.

How many equations do you require to solve for 3 variables ?

Edited February 21, 2023 by studiot

[martillo]

13 minutes ago, studiot said:

Er...?

Variational calculus ?

That's second year university stuff.

I said simultaneous equations.

In the UK you start learning about these around 12 or 13, actually before you learn any mechanics.

Well, I did Electrical Engineering and I didn't see Variational Calculus (UNICAMP - Brasil). Not at college at all. I have a book given me as present because I was good in mathematics but never read it (Variational Calculus Krasnov, Makarenko, Kiseliov). I just took a look. Is something not for me.

Edited February 21, 2023 by martillo

[studiot]

Just now, martillo said:

Well, I did Electrical Engineering and I didn't see Variational Calculus. Not at college at all. I have a book given me as present because I was good at mathematics but never read it (Variational Calculus Krasnov, Makarenko, Kiseliov). Never read it. Is something not for me.

 

So what's your response to what I actually said please ?

The calculus of variations is totally unnecessary for this.

It barely reaches into ordinary differential equations and can even be done without them.

[martillo]

20 minutes ago, studiot said:

 

So what's your response to what I actually said please ?

The calculus of variations is totally unnecessary for this.

It barely reaches into ordinary differential equations and can even be done without them.

Well, my response is that or you don't get my point or I don't get your point. 

34 minutes ago, studiot said:

But it isn't one simple equation.

You can't calculate anything with one simple equation as you have more an one variable involved.

How many equations do you require to solve for 3 variables ?

Well, right. We have two equations to apply. One is the Thrust Equation (mdv/dt = v_rel(dm/dt)), the second is the problem, which equation to apply: F = ma or F = dp/dt. F = ma is applied and gives right results for rockets (they work isn't it?). Then follows that the right equation of motion is F = ma and not F = dp/dt, even when mass varies. Don't you get it?

Edited February 21, 2023 by martillo

[joigus]

46 minutes ago, martillo said:

Well, I studied Mechanics but not Variational Calculus. Do you think it is really needed to treat this problem? If so actually I would not be able to discuss anymore, of course.

1 hour ago, studiot said:

It's actually needed if what you mean is to argue that 'physics is in trouble.' You see, for some physical systems the concepts of force, or acceleration for that matter, or even position, are ambiguous or cannot be defined at all. It's also possible for them to have a surprising and totally non-intuitive definition that nevertheless makes mathematical sense. Take the case of a particle in a constant magnetic field. These systems are better treated in terms of a Lagrangian. And when you do that, lo and behold, the momentum of the system is not what you would expect, and includes the vector potential, turning out to be mv-qA, instead of just mv.

In the case that you guys are discussing, it's because the definition of 'the system' is what's ambiguous. The rocket is losing fuel by the second, so part of 'the system' is going away. What do you do? You drop the definition that happens to be useless for this particular case --of F=ma--, and stick to the more secure, more general one of F=dp/dt, as knowledgeable members are telling you. Then you have two terms mdv/dt, and vdm/dt. The same equation that's been applied for decades to investigate how rockets move. 

 

[studiot]

15 minutes ago, martillo said:

Well, my response is that or you don't get my point or I don't get your point. 

Well, right. We have two equations to apply. One is the Thrust Equation (mdv/dt = vrel(dm/dt)), the second is the problem, which equation to apply: F = ma or F = dp/dt. F = ma is applied and gives right results for rockets (they work isn't it?). Then follows that the right equation of motion is F = ma and not F = dp/dt. Don't you get it?

Start at the beginning and try to see where all these websites you refer to have hidden the analysis.

You have a complete rocket standing there on the ground ready for launch.

What do you know and what do you want to calculate ?

In other words what are you variables ?

[martillo]

23 minutes ago, joigus said:

In the case that you guys are discussing, it's because the definition of 'the system' is what's ambiguous. The rocket is losing fuel by the second, so part of 'the system' is going away. What do you do? You drop the definition that happens to be useless for this particular case --of F=ma--, and stick to the more secure, more general one of F=dp/dt, as knowledgeable members are telling you. Then you have two terms mdv/dt, and vdm/dt. The same equation that's been applied for decades to investigate how rockets move. 

It's not me applying the equation F = ma. It is what is actually applied in Science and in rockets development and fabrication and what works when they move, right? Everybody applies F = ma and not F = dp/dt. That's the point, F = ma is the valid one.

22 minutes ago, studiot said:

Start at the beginning and try to see where all these websites you refer to have hidden the analysis.

You have a complete rocket standing there on the ground ready for launch.

What do you know and what do you want to calculate ?

In other words what are you variables ?

No, the system is not the rocket ready to launch. Is the rocket already in the air at velocity v and expelling fuel at constant velocity -v_e. Please take a look at the link in the first post to have a better view with good simple graphics (http://www.braeunig.us/space/propuls.htm).

I don't want to calculate anything, I just observe which equations are used by everybody for the rockets. F = ma is used.

Edited February 21, 2023 by martillo

[joigus]

4 minutes ago, martillo said:

It's not me applying the equation F = ma. It is what is actually applied in Science and in rockets development and fabrication. Everybody applies F = ma and not F = dp/dt. That's the point, F = ma is the valid one.

Can I interest you in,

\[ f=ma \]

\[ f=F-v\frac{dm}{dt} \]

? with \( f \) being the rocket-generalised force?

That's similar to what you're doing when you go to a non-inertial frame with acceleration \( A \) and re-write,

\[ F=ma+mA \]

as,

\[ F-mA=ma \]

\( f=F-mA \) being the force and \( -mA \) the Euler force.

To be honest, I'm not nearly as much concerned on how we call it as I am in preventing the possibility of rockets falling on my head because people don't apply the equation correctly.

[studiot]

22 minutes ago, martillo said:

No, the system is not the rocket ready to launch. Is the rocket already in the air at velocity v and expelling fuel at constant velocity -v_e.

I asked you what you know.

You haven't quoted a single number.

So tell me just how you know what v is ?

And you are telling me that i should use f = ma instead of some other equation.

What exactly will this give me and where will I get the numbers to put into this equation ?

And where exactly does your f = ma fit into your claim that you know v and -v_e ?

 

Why do you think I specified a rocket ready to launch ?

 

22 minutes ago, martillo said:

I don't want to calculate anything, I just observe which equations are used by everybody for the rockets. F = ma is used.

If you don't want to calculate anything why are we going through all this rigmarole ?

Edited February 21, 2023 by studiot

[martillo]

28 minutes ago, joigus said:

Can I interest you in,

 

f=ma

 

 

f=F−vdmdt

 

? with f being the rocket-generalised force?

That's similar to what you're doing when you go to a non-inertial frame with acceleration A and re-write,

 

F=ma+mA

 

as,

 

F−mA=ma

 

f=F−mA being the force and −mA the Euler force.

To be honest, I'm not nearly as much concerned on how we call it as I am in preventing the possibility of rockets falling on my head because people don't apply the equation correctly.

I don't know if I get your point. The Euler Force is a transversal force in non uniform motion when transversal acceleration applies. This is not the case. May be it applies if gravity is considered on the rocket for instance, but this is not what is considered to obtain the Rocket Thrust equation.

20 minutes ago, studiot said:

Why do you think I specified a rocket ready to launch ?

Because you said:

"

Start at the beginning and try to see where all these websites you refer to have hidden the analysis.

You have a complete rocket standing there on the ground ready for launch.

"

20 minutes ago, studiot said:

If you don't want to calculate anything why are we going through all this rigmarole ?

I'm analyzing what other ones, everybody, calculate to find the force on the rocket. I just observe, I'm not calculating anything on my own.

20 minutes ago, studiot said:

I asked you what you know.

You haven't quoted a single number.

So tell me just how you know what v is ?

And you are telling me that i should use f = ma instead of some other equation.

What exactly will this give me and where will I get the numbers to put into this equation ?

And where exactly does your f = ma fit into your claim that you know v and -v_e ?

Come on, now seems you don't know what are we talking about in all this discussion, how can it be after all that was already posted?

I suggest you to read with attention the first post of the thread describing with much care the problem discussed here.

Edited February 21, 2023 by martillo

[joigus]

4 minutes ago, martillo said:

I don't know if I get your point. The Euler Force is a transversal force in non uniform motion when transversal acceleration applies. This is not the case. May be it applies if gravity is considered on the rocket for instance, but this is not what is considered to obtain the Rocket Thrust equation.

You don't, for all I can see. Euler forces* are transversal (centrifugal) or tangential (thrust experienced from inside the rocket, when measured in a non-inertial system), or a combination of the two. They're kinematical in nature. I agree it's not the same case. My comment was about maths and definitions, and what you decide to call 'the force on the system.' This requires to define what you mean by 'the system.' If 'the system' is both the rocket, the fuel and the exhaust, then all forces implied are internal and there is no net force. If it's just the rocket with the fuel that's left, well... You see? 

Thrusts are not kinematical, but sure enough passengers on the rocket will experience these forces as very much real. That's why astronauts must buckle themselves pretty tightly.

* Sorry I may have used the term 'Euler forces' in a sense that's a bit more general. I meant ficticious forces in general, centrifugal and Coriolis included.

[swansont]

9 hours ago, martillo said:

Where is the consideration of dp/dt = d(mv)/dt with the mass of the rocket? Nowhere.

Right. Because that’s not how they derived the equation.

Since dp/dt is zero (there is no net force) it means momentum is conserved. They solve the problem by applying that principle.

 

7 hours ago, Lorentz Jr said:

F = dp/dt is derived from F = ma

Other way around

 

[Markus Hanke]

1 hour ago, martillo said:

Do you think it is really needed to treat this problem?

It depends what you mean by “problem”. You were asserting in your OP that F=dp/dt isn’t the correct formulation of Newton’s 2nd law, so I tried to provide you with the bigger picture of where this relation comes from. My point was to show you that it isn’t just made up (and thus potentially wrong), but that it follows from more general principles. It’s essentially a special case of the principle of least action. F=ma then is in turn a special case of F=dp/dt - it works well for many simple systems (which is why we all learn it in school), but its domain of validity is limited. F=dp/dt is more general, and also works in some cases where F=ma doesn’t. And then of course there are systems where neither of these work (as joigus has pointed out), and then you have to use the full Lagrangian formulation of mechanics. 

This isn’t to say that F=dp/dt can’t be derived in other ways too, which is what studiot is trying to show you.

Either way, just claiming that F=dp/dt is wrong and must always be replaced by F=ma - which is what you seem to be saying - is silly, because that’s manifestly false. 

[martillo]

2 minutes ago, joigus said:

My comment was about maths and definitions, and what you decide to call 'the force on the system.' This requires to define what you mean by 'the system.' If 'the system' is both the rocket, the fuel and the exhaust, then all forces implied are internal and there is no net force. If it's just the rocket with the fuel that's left, well... You see? 

Well, let me summarize the problem here for you although I would recommend to see the problem as presented in the already posted link: http://www.braeunig.us/space/propuls.htm. The total system is the rocket with its expelled fuel presenting a total constant mass. It is applied dP/dt = 0 for this system to derive the force on the rocket alone, I mean to derive what force the expelled fuel exerts on the rocket. Then at a second place it is considered the rocket alone accelerated by a force which is the Thrust Force on the rocket. The problem is that what is applied in this derivation is F = a and not F = dp/dt.