The Relativistic Force-Norm and the Proof of E=mc2

[rubi]

By the way, it's not even true in classical mechanics that $F =\frac{\mathrm d p}{\mathrm d t}$ if $m = m(t)$. It's just a coincidence (an unfortunate one in my opinion) that it happens to work in the case of a rocket. If the fuel would be exhausted in any other direction, then $F=\frac{\mathrm d p}{\mathrm d t}$ would yield the wrong answer. The correct treatment of the situation requires the usage of $F = ma$ together with a force that describes the thrust, which happens to be given by $F_\mathrm{thrust} = v_\mathrm{rel}\frac{\mathrm d m}{\mathrm d t}$, where $v_\mathrm{rel}$ is the relative velocity of the exhausted fuel to the rocket, and it coincides with the formula that can be derived from $F =\frac{\mathrm d p}{\mathrm d t}$ only in the case that $v_\mathrm{rel} = - v$.

 

[PAllen]

By the way, it's not even true in classical mechanics that $F =\frac{\mathrm d p}{\mathrm d t}$ if $m = m(t)$. It's just a coincidence (an unfortunate one in my opinion) that it happens to work in the case of a rocket. If the fuel would be exhausted in any other direction, then $F=\frac{\mathrm d p}{\mathrm d t}$ would yield the wrong answer. The correct treatment of the situation requires the usage of $F = ma$ together with a force that describes the thrust, which happens to be given by $F_\mathrm{thrust} = v_\mathrm{rel}\frac{\mathrm d m}{\mathrm d t}$, where $v_\mathrm{rel}$ is the relative velocity of the exhausted fuel to the rocket, and it coincides with the formula that can be derived from $F =\frac{\mathrm d p}{\mathrm d t}$ only in the case that $v_\mathrm{rel} = - v$.

What?! F=dp/dt is exactly correct in all cases in classical mechanics. It is certainly true for e.g. a water balloon with two holes at random angles, undergoing chaotic motions from water escaping in two directions and mass of the balloon constantly changing. Whatever the case, the momentary net force on the balloon is always dp/dt, with p expressed in terms of momentary mass of the balloon and the velocity of the balloon COM.

 

[rubi]

What?! F=dp/dt is exactly correct in all cases in classical mechanics. It is certainly true for e.g. a water balloon with two holes at random angles, undergoing chaotic motions from water escaping in two directions and mass of the balloon constantly changing. Whatever the case, the momentary net force on the balloon is always dp/dt, with p expressed in terms of momentary mass of the balloon and the velocity of the balloon COM.

Consider a water balloon with two holes at opposite angles, moving at a constant velocity $v(t=0)\neq0$. The velocity stays constant since an equal amount of water is exhausted in both directions. Constant velocity implies that the net force on the balloon is $F=0$ by Newtons first law. However $\dot p = \dot m v + m \dot v = \dot m v(t=0) \neq 0$, so $F\neq\dot p$.

Edit: Wikipedia agrees with me.

 

[PAllen]

Consider a water balloon with two holes at opposite angles, moving at a constant velocity $v(t=0)\neq0$. The velocity stays constant since an equal amount of water is exhausted in both directions. Constant velocity implies that the net force on the balloon is $F=0$ by Newtons first law. However $\dot p = \dot m v + m \dot v = \dot m v(t=0) \neq 0$, so $F\neq\dot p$.

Edit: Wikipedia agrees with me.

Ok, I oversimplified, but it is still true if you include the ejecta as well, and sum all the momenta; then F = dp/dt holds.

 

[rubi]

Ok, I oversimplified, but it is still true if you include the ejecta as well, and sum all the momenta; then F = dp/dt holds.

Well, obviously if you treat the system as a many-particle system (consisting of the empty balloon and the individual water molecules) with each of these "particles" evolving according to $F_i=m_i a_i$, then the sum of all momenta (the total momentum) is conserved ($\dot p_\mathrm{total} = 0$) or more generally, it is the sum of all the forces that act on the individual "particles". This just confirms the idea that in classical mechanics, only $F=ma$ is valid and every corresponding formula for variable-mass systems is just an effective version of $F=ma$.

 

[Delta2]

Amazing and i thought F=dp/dt holds everywhere since it is used in many books to prove conservation of momentum. What is the formula for F in this case where mass varies with time?

 

[DrStupid]

Why do you think I've changed the original definition, rather than just restated it in modern notation?

In my derivation of the relativistic momentum (see my link above) I started from the original definition and got another result. How did you get your "modern notation"?

Do you think that Newton did NOT mean $p=mv$ when he defined the "motion"?

The question is, what m means.

 

[DrStupid]

If the fuel would be exhausted in any other direction, then $F=\frac{\mathrm d p}{\mathrm d t}$ would yield the wrong answer.

Not if you do it correctly.

Constant velocity implies that the net force on the balloon is $F=0$ by Newtons first law.

The first law says F=0 -> a=0. That's not equivalent to a=0 -> F=0.

 

[stevendaryl]

By the way, it's not even true in classical mechanics that $F =\frac{\mathrm d p}{\mathrm d t}$ if $m = m(t)$. It's just a coincidence (an unfortunate one in my opinion) that it happens to work in the case of a rocket.

Now that I think about it, I don't see how it is true for a rocket, either. In a rocket, let's assume for computational purposes that exhaust is expelled in discrete amounts $\delta m$ at a characteristic relative velocity $v_{rel}$ at intervals of $\delta t$. If $m$ is the mass of the rocket, and $v$ is its velocity, then by conservation of momentum:

$m\ \delta v - \delta m\ v_{rel}= 0$

So the equation of motion for the rocket is:

$m \dfrac{dv}{dt} + \dfrac{dm}{dt} v_{rel} = 0$

(Note, since the rocket is losing mass, $\dfrac{dm}{dt} = - \dfrac{\delta m}{\delta t}$)

On the other hand, the rate of change of the momentum of the rocket is given by:

$\dfrac{dp}{dt} = m \dfrac{dv}{dt} + \dfrac{dm}{dt} v$

The first expression uses $v_{rel}$ while the second uses $v$, so I don't see how $\dfrac{dp}{dt}$ is relevant to solving the problem.

 

[stevendaryl]

So the equation of motion for the rocket is:

$m \dfrac{dv}{dt} + \dfrac{dm}{dt} v_{rel} = 0$

On the other hand, the rate of change of the momentum of the rocket is given by:

$\dfrac{dp}{dt} = m \dfrac{dv}{dt} + \dfrac{dm}{dt} v$

The first expression uses $v_{rel}$ while the second uses $v$, so I don't see how $\dfrac{dp}{dt}$ is relevant to solving the problem.

Actually, it occurs to me that there is one case in Newtonian physics involving variable mass for which it is true that $\dfrac{dp}{dt} = 0$.

Imagine a "reverse rocket" that starts off empty, traveling at a high velocity through the atmosphere, and as it goes, it scoops up air (initially at rest) and stores it. Assume that there is no friction on the outside of the rocket.

In that case, we have:

Initially:

$p = m v$

After scooping up an amount of air $\delta m$ through an interval $\delta t$

$p = (m + \delta m) (v + \delta v)$

By conservation of momentum:

$m\ \delta v + \delta m\ v = 0$

So we would have: $\dfrac{dp}{dt} = m \dfrac{dv}{dt} + \dfrac{dm}{dt} = 0$

This would be true of the regular rocket in the (unrealistic) case that the exhaust is always expelled at the right relative velocity that its final velocity is zero, relative to the initial rest frame.

 

[rubi]

Not if you do it correctly.

You can't solve such problems without taking the relative exhaust velocity of the ejected matter into account. It should be obvious that it matters, because the resulting motion of the balloon/rocket/.. must depend heavily on both the direction and the magnitude of the relative exhaust velocity. The equation $F=\dot p$ can't take the relative exhaust velocity into account, since the term that shows up if you apply the product rule ($\dot m v$) doesn't depend on the relative exhaust velocity at all! Thus $F=\dot p$ must necessarily be wrong in almost all cases where $\dot m \neq 0$, except those in which $F_\mathrm{thrust}= \dot m v_\mathrm{rocket}$ happens to be true by coincidence.

Here's yet another argument: If $\dot m\neq 0$, then $\dot p$ isn't invariant under the Galilean transform $x\rightarrow x+v_0 t$ anymore since only a second derivative can get rid of the $v_0 t$ term. $\dot p$ contains also a first derivative, so $\dot p\rightarrow \dot p + \dot m v_0$ under a Galilean transform.

There's really no way to save $F=\dot p$. It is just inconsistent.

The first law says F=0 -> a=0. That's not equivalent to a=0 -> F=0.

The first law doesn't even matter. It follows already from a many-particle description of the situation (empty balloon + water molecules) that there is no force acting on the balloon and the water molecules that are still contained in the balloon, since they are moving at constant velocity and the fundamental description $F=ma$ of the situation really implies $F=0$. If $F=\dot p$ were consistent for a variable-mass system, then it would have to reproduce this fact.

Actually, it occurs to me that there is one case in Newtonian physics involving variable mass for which it is true that $\dfrac{dp}{dt} = 0$.

Unfortunately there are some coincidental situations in which it works. That's the reason for why it is such a wide spread misconception that even some textbook authors get it wrong.

It should be clear, that if we have a microscopic theory consisting only of particles of constant mass, then any effective description of a composite system like a variable-mass system requires a derivation from first principles, i.e it requires to be derived from a theory with only particles of constant mass. So even if it were true, we couldn't simply postulate $F=\dot p$ for $\dot m\neq 0$.

 

[DrStupid]

The equation $F=\dot p$ can't take the relative exhaust velocity into account

Of course it can:

The momentum of the rocket is
$$p_R = m_R \cdot v_R$$
As the exhausted particles are moving with individual but constant velocities the total momentum of the exhausted fuel is
$$p_F = \int {\dot m_F \cdot v_F \cdot dt}$$
According to the second law the corresponding forces are
$$F_R = \dot m_R \cdot v_R + m_R \cdot \dot v_R$$
and
$$F_F = \dot m_F \cdot v_F$$
With the third law
$$F_R + F_F = 0$$
the conservation of mass
$$\dot m_R + \dot m_F = 0$$
and the velocity of the currently exhausted fuel
$$v_F = v_R + v_{rel}$$
this results in the rocket equation
$$\dot v_R = \frac{{\dot m_R }}{{m_R }} \cdot v_{rel}$$

Here's yet another argument: If $\dot m\neq 0$, then $\dot p$ isn't invariant under the Galilean transform

As force doesn't need to be frame independent this is not a problem so far. There are other problems with $\dot m\neq 0$ and Galilei transformation (see my derivation of the relativistic momentum).

There's really no way to save $F=\dot p$. It is just inconsistent.

I provided a counterexample for SR. What is wrong with it?

So even if it were true, we couldn't simply postulate $F=\dot p$ for $\dot m\neq 0$.

Why not?

 

[rubi]

Of course it can: [...]

I provided a counterexample for SR. What is wrong with it?

I really don't want to search for errors in lots of individual example calculations now. I have already said that $F=\dot p$ happens to work for the rocket equation by coincidence, so if you arrive at the correct result, it doesn't mean your reasoning was right. If you want to discuss an example, let's stick to the balloon example above, because it describes a situation which can't be derived from $F=\dot p$ by coincidence.

As force doesn't need to be frame independent this is not a problem so far. There are other problems with $\dot m\neq 0$ and Galilei transformation (see my derivation of the relativistic momentum).

Unless there are external forces, the equation of motion should transform according to the Galilean transformations. Otherwise there would be a preferred inertal frame of reference, which isn't even true for classical mechanics.

Why not?

I had already explained it in my posting. If you have a microscopic theory and a macrosopic theory that describe the same physical situation, then the macroscopic theory must emerge from the microscopic theory by the correspondence principle.

Here's another argument that just came to my mind: Consider a box of mass $m$, moving at constant velocity $v$. Now divide the box mentally into two pieces of mass $\frac{m}{2}$. Acoording to $F = \dot p$, the very act of dividing the box mentally into two pieces would have exerted a force on the box, which is ridiculous.

If you want a reference, look at this: http://articles.adsabs.harvard.edu/full/seri/CeMDA/0053//0000227.000.html

I didn't want to hijack this thread and discuss variable-mass systems. I only wanted to give another argument for why Matterwaves criticism of your derivation is perfectly valid.

 

[DrStupid]

If you want to discuss an example, let's stick to the balloon example above, because it describes a situation which can't be derived from $F=\dot p$ by coincidence.

The calculation is similar as demonstrated above for the rocket. You just need to consider the momentum of the second jet. If you still think this is impossible please provide a proper justification.

Unless there are external forces, the equation of motion should transform according to the Galilean transformations.

In special relativity it should transform according to Lorentz transformation.

I had already explained it in my posting. If you have a microscopic theory and a macrosopic theory that describe the same physical situation, then the macroscopic theory must emerge from the microscopic theory by the correspondence principle.

In order to turn that into an explanation you need to proof that this is not possible with F=dp/dt.

Here's another argument that just came to my mind: Consider a box of mass $m$, moving at constant velocity $v$. Now divide the box mentally into two pieces of mass $\frac{m}{2}$. Acoording to $F = \dot p$, the very act of dividing the box mentally into two pieces would have exerted a force on the box, which is ridiculous.

I considered this problem in my derivation of relativistic momentum and solved it by m(v)=mo·f(v).

If you want a reference, look at this: http://articles.adsabs.harvard.edu/full/seri/CeMDA/0053//0000227.000.html

This paper claims

"If we consider the simple case of a variable mass, and write Newton's second law as:

$\vec F = m \cdot \frac{{d\vec v}}{{dt}} + \vec v \cdot \frac{{dm}}{{dt}}\quad \quad \quad \quad \quad \left( 2 \right)$

we can easily see that it violates the relativity principle under Galilean transformations. When F is zero, in particular, Equation (2) implies that the particle will remain at rest in a system where it is originally at rest, but it will be accelerated by the "force" -vdm/dt in a system where the particle moves with velocity v!"

This is wrong. If m is a function of velocity than equation (2) turns into

$\vec F = m \cdot \vec a + \vec v \cdot \left( {m' \cdot \vec a} \right)$

(I actually used that equation in my derivation.) As acceleration is frame-invariant under Galilean transformation there is no force in all frames of reference. If m is not a function of velocity (only possible in open systems) than there might be a force but it does not accelerate the particle. After this massive fault I stopped reading. Is there something else in this paper that might be relevant for this thread?

 

[rubi]

The calculation is similar as demonstrated above for the rocket. You just need to consider the momentum of the second jet. If you still think this is impossible please provide a proper justification.

You are the one who needs to provide justification. I have already shown that it doesn't work, so it is your turn now.

In special relativity it should transform according to Lorentz transformation.

But I'm talking about classical mechanics and there it needs to transform according to Galilean transformation and I've shown that it doesn't, so this is a 100% proof that $F=\dot p$ is inconsistent in classical mechanics, which can only be wrong if you find an error in that particular argument. Please note that giving lots of individual examples that work by coincidence does not consitute a refutation of this argument.

In order to turn that into an explanation you need to proof that this is not possible with F=dp/dt.

No, that's not how physics works. If you claim that a macroscopic theory gives the same results as a more fundamental microscopic one, you need to prove that. You can't simply postulate it and wait for others to disprove it. If you are unable to provide such a proof, then the macroscopic theory is just a heuristic theory that might work in some situations and might fail in others.

I considered this problem in my derivation of relativistic momentum and solved it by m(v)=mo·f(v).

Again, I'm only talking about classical mechanics here. And I'm doing so in order to prove that $F=\dot p$ requires justification and isn't valid in general, i.e. it is invalid in classical mechanics.

This paper claims

"If we consider the simple case of a variable mass, and write Newton's second law as:

$\vec F = m \cdot \frac{{d\vec v}}{{dt}} + \vec v \cdot \frac{{dm}}{{dt}}\quad \quad \quad \quad \quad \left( 2 \right)$

we can easily see that it violates the relativity principle unter Galilean transformations. When F is zero, in particula, Equation (2) implies that the particle will remain at rest in a system where it is originally at rest, but it will be accelerated by the "force" -vdm/dt in a system where the particle moves with velocity v!"

This is wrong. If m is a function of velocity than equation (2) turns into

$\vec F = m \cdot \vec a + \vec v \cdot \left( {m' \cdot \vec a} \right)$

(I actually used that equation in my derivation.) As acceleration is frame-invariant under Galilean transformation there is no force in all frames of reference. If m is not a function of velocity (only possible in open systems) than there might be a force but it does not accelerate the particle. After this massive fault I stopped reading. Is there something else in this paper that might be relevant for this thread?

This just shows that you haven't properly read it. The paper considers a function $m(t)$ that depends only on time. Thus the claim of the paper is completely valid. It's just an application of the product rule.

 

[Matterwave]

In my derivation of the relativistic momentum (see my link above) I started from the original definition and got another result. How did you get your "modern notation"?



The question is, what m means.

I don't think we will come to a consensus. So I'll just drop the issue. You can take this as "you win" if you like.

 

[stevendaryl]

this results in the rocket equation
$$\dot v_R = \frac{{\dot m_R }}{{m_R }} \cdot v_{rel}$$

That's the correct equation of motion. But it doesn't seem consistent with $F_R = \dot{p_R} = \dot{m_R} v_R + m_R \dot{v_R}$

 

[DrStupid]

You are the one who needs to provide justification. I have already shown that it doesn't work, so it is your turn now.

Sorry, not this way! It was your claim that it doesn't work. Therefore you need to provide the corresponding justification. In regard to the rocket I disproved your claim instead of waiting for your proof. But that does not mean that I will do that again and again.

But I'm talking about classical mechanics

The thread is about E=mc². Therefore it makes no sense to limit the discussion to classical mechanics. This link shows what I'm talking about:

https://www.physicsforums.com/showpost.php?p=4737048&postcount=4

and there it needs to transform according to Galilean transformation and I've shown that it doesn't, so this is a 100% proof that $F=\dot p$ is inconsistent in classical mechanics

I have also shown that for closed systems (see my link above). Your proof for open systems is still missing. It's not sufficient just to claim it and wait for a disproof.

If you claim that a macroscopic theory gives the same results as a more fundamental microscopic one, you need to prove that. You can't simply postulate it and wait for others to disprove it.

As I didn't wrote something about macroscopic or microscopic theories I don't need to provide a corresponding proof. You raised that topic. You need to provide the proof.

Again, I'm only talking about classical mechanics here.

Again, that's off-topic.

And I'm doing so in order to prove that $F=\dot p$ requires justification and isn't valid in general, i.e. it is invalid in classical mechanics.

I'm still waiting for this proof.

The paper considers a function $m(t)$ that depends only on time. Thus the claim of the paper is completely valid.

I already explained why it is not valid. I will not repeat it. You can read it above.

PS: This discussion starts to get out of control and I'm not interested in meta discussions. Thus I will limit my replies to statements with justification and ignore everything else.

 

[stevendaryl]

Again, I'm only talking about classical mechanics here. And I'm doing so in order to prove that $F=\dot p$ requires justification and isn't valid in general, i.e. it is invalid in classical mechanics.

The one place where $F = \dot{p}$ is preferable to $F = m \dot{v}$ is in Lagrangian mechanics. We define the canonical momentum corresponding to a coordinate $q$ to be $p_q = \dfrac{\partial L}{\partial \dot{q}}$, and we define the generalized force corresponding to coordinate [iteq]q[/itex] to be: $F_q = \dfrac{\partial L}{\partial q}$. In that case, the equations of motion are:

$\dot{p_q} = F_q$

not

$m \ddot{q} = F_q$

The latter is only true if the coordinate is cartesian and the force is due to a scalar potential.

 

[DrStupid]

That's the correct equation of motion. But it doesn't seem consistent with $F_R = \dot{p_R} = \dot{m_R} v_R + m_R \dot{v_R}$

This equation is part of the derivation. If it is not consistent with the result than there would be an error in the calculation. Do you see such an error?

 

[rubi]

Ok, I'm not going to waste any more time on this. Your claim in Post #18 was that $F=\dot p$ is to be favoured over $F=ma$ and I have provided several arguments that prove that the contrary is true, and you haven't addressed a single one of them. This makes the discussion entirely pointless.

 

[rubi]

The one place where $F = \dot{p}$ is preferable to $F = m \dot{v}$ is in Lagrangian mechanics. We define the canonical momentum corresponding to a coordinate $q$ to be $p_q = \dfrac{\partial L}{\partial \dot{q}}$, and we define the generalized force corresponding to coordinate [iteq]q[/itex] to be: $F_q = \dfrac{\partial L}{\partial q}$. In that case, the equations of motion are:

$\dot{p_q} = F_q$

not

$m \ddot{q} = F_q$

The latter is only true if the coordinate is cartesian and the force is due to a scalar potential.

This is true, but a variable-mass system can't be described with a Lagrangian like this. The paper I posted above addresses this.

Or in other words: One can't simply put $m=m(t)$ and use the same equations that are known to work for the constant mass case. Neither in Newtonian mechanics nor in the Lagrangian formulation nor in the Hamilton formalism. The resulting equations of motion don't describe experimentally testable facts in general.

 

[PAllen]

This is true, but a variable-mass system can't be described with a Lagrangian like this. The paper I posted above addresses this.

Or in other words: One can't simply put $m=m(t)$ and use the same equations that are known to work for the constant mass case. Neither in Newtonian mechanics nor in the Lagrangian formulation nor in the Hamilton formalism. The resulting equations of motion don't describe experimentally testable facts in general.

You could go all the way to using a Lagrangian density and the techniques of continuous systems. A bit of overkill, though.

 

[stevendaryl]

This equation is part of the derivation. If it is not consistent with the result than there would be an error in the calculation. Do you see such an error?

Well, I would say that the force on the rocket is

(1) $F_R = -\dot{m}_F\ v_{rel}$

rather than

(2) $\tilde{F}_R = -\dot{m}_F\ v_F$

And I would also say that

(3) $F = m\ \dot{v}$
rather than

(4) $F = m\ \dot{v} + \dot{m}\ v$

Interestingly, in this case, if you assume (1) and (3), (which I do), you get the same equations of motion as if you assume (2) and (4) (which you do):

$F_R = m_R\ \dot{v}_R$
$\Rightarrow -\dot{m}_F\ v_{rel} = m_R\ \dot{v}_R$

versus

$\tilde{F}_R = m_R\ \dot{v}_R + \dot{m}_R\ v_R$
$\Rightarrow -\dot{m}_F\ v_F = m_R\ \dot{v}_R + \dot{m}_R\ v_R$
$\Rightarrow -\dot{m}_F\ v_F - \dot{m}_R\ v_R = m_R\ \dot{v}_R$

Since $\dot{m}_R = -\dot{m}_F$ and $v_F = v_R + v_{rel}$, the expression $-\dot{m}_F\ v_F - \dot{m}_R\ v_R$ is equal to $-\dot{m}_F\ v_R$

So the question is: Is this two different ways of thinking about things, equally legitimate, or is it a matter of cancelling errors?

 

[DrStupid]

Well, I would say that the force on the rocket is

(1) $F_R = -\dot{m}_F\ v_{rel}$

Why would you say that? Can you derive this equation from fundamental laws or definitions?

And I would also say that

(3) $F = m\ \dot{v}$
rather than

(4) $F = m\ \dot{v} + \dot{m}\ v$

Force is defined as F=dp/dt and momentum as p=m·v. Thus F=m·a is valid for constant mass only. How do you justify the use of that equation for a system with variable mass?

So the question is: Is this two different ways of thinking about things, equally legitimate, or is it a matter of cancelling errors?

That depends on your justification for the equations you started with. Your calculation is acceptable if you know why you can do that. (I know it, but I will wait for your explanation )

 

[stevendaryl]

Why would you say that? Can you derive this equation from fundamental laws or definitions?

I would say so. Let me do a discrete approximation (which is appropriate, since matter is actually discrete).

So assume that you have a rocket of mass $m_R$ initially traveling at speed $v_R$ Some of the mass is made of fuel. Imagine a discrete amount of fuel $\delta m$ being burnt and thrown out the back at relative velocity $v_{rel}$ (this will actually be a negative value). Let $\delta t$ be the time interval for this process.

To avoid the controversial step of considering variable mass, I will consider this system to be composed of two objects with constant mass (at least during time interval $\delta t$):

1. A rocket of mass $m_R - \delta m$.
   This has an initial velocity of $v_R$ and a final velocity of $v_R + \delta v_R$. The change in momentum is $\delta p_R = (m_R - \delta m)\ \delta v_R$.
2. A quantity of fuel of mass $\delta m$.
   This has an initial velocity of $v_R$ and a final velocity of $v_R + v_{rel}$. The change in momentum is $\delta p_F = \delta m\ v_{rel}$.

By conservation of momentum,

$\delta p_R = -\delta p_F$

So

$\delta p_R = -\delta m\ v_{rel}$

The average force on the rocket is $\delta p_R/\delta t = -v_{rel}\ \delta m/\delta t$.

In the continuum limit, $F_R = - v_{rel} \dfrac{dm}{dt}$

 

[OmCheeto]

It appears velo city has run away. If I were in high school, I'd probably run away too. I would answer the question, with evidence:

Binding Energy
...
In 2005, Rainville et al. published a direct test of the energy-equivalence of mass lost in the binding-energy of a neutron to atoms of particular isotopes of silicon and sulfur, by comparing the new mass-change to the energy of the emitted gamma ray associated with the neutron capture. The binding mass-loss agreed with the gamma ray energy to a precision of ±0.00004 %, the most accurate test of E=mc2 to date.

Original article

I know this is a bit simplistic, but it was a simple question. Kind of like; "Give me a proof that the sun is shining". I don't think I'd even bother with such a question, much beyond, pointing at it, and claiming it does. Which is what Rainville et al, appear to have done, with E=mc².

 

[DrStupid]

To avoid the controversial step of considering variable mass, I will consider this system to be composed of two objects with constant mass (at least during time interval $\delta t$):
[...]
The average force on the rocket is $\delta p_R/\delta t = -v_{rel}\ \delta m/\delta t$.

This result is limited to the conditions above. You derived a force that would act on a "rocket" (it actually is a cannon) with constant mass during the time interval $\delta t$. As a rocket cannot change its momentum without changing its mass this condition cannot be fulfilled. If you consider the different mass at the begin and at the end of the time intervals you will get different equations.

However, you finally get the correct result because the error goes to zero for infinite small time intervals $\delta t$. If you forget the forces and return to the momentum you get

$\frac{{\delta p_R }}{{\delta t}} = \left( {m_R - \delta m} \right) \cdot \frac{{\delta v_R }}{{\delta t}} = - \frac{{\delta p_F }}{{\delta t}} = - \frac{{\delta m}}{{\delta t}} \cdot v_{rel}$

For infinite small steps this results in

$\frac{{\delta v_R }}{{\delta t}} = - \frac{{\delta m}}{{\delta t}} \cdot \frac{{v_{rel} }}{{m_R }}$

(due to $\delta m \to 0$)

Your use of the term "force" is at questionable but your balance of momentum is correct.


But there is a possibility to justify your calculation even for variable mass. With the original definition of Force (Lex 2) and momentum (Definition 2 in Newton's Principia) I initially get

$F_R = m_R \cdot a_R + \dot m_R \cdot v_R$
and
$F_F = \dot m_F \cdot \left( {v_R + v_{rel} } \right)$

In order to get your equations for the forces I just need to transform these equations into an inertial system with the same velocity of the rocket. In classical mechanics forces transform according to

$F' = m' \cdot a' + \dot m' \cdot v' = m \cdot a + \dot m \cdot \left( {v - u} \right) = F - u \cdot \dot m$

With $u = v_R$ this results in

$F'_R = m_R \cdot a_R$
and
$F'_F = \dot m_F \cdot v_{rel}$

and with Lex 3 finally in

$- \dot m_F \cdot v_{rel} = m_R \cdot a_R$

This corresponds to your first calculation. As the resulting equation is frame-independent under Galilean transformation you may use it for the derivation of the rocket equation in all inertial systems. But you must not use the equations of force in other frames of reference or at different times.

 

[stevendaryl]

It appears velo city has run away. If I were in high school, I'd probably run away too. I would answer the question, with evidence:

This is a terminological issue. To me, there is a distinction between asking: "What is the empirical evidence that X is true?" and "What is the proof that X is true?"

For example, if someone asked for a proof that $sin(45^o) = 1$, it would not be sufficient to just construct a $45^o$ triangle and measure the lengths of the sides. Proof requires a logical or mathematical argument, or demonstration. Proof is more certain than empirical evidence, but that certainty comes at cost, which is that proof is always relative to a set of assumptions, and those assumptions could be wrong.

(Actually, empirical demonstrations also have hidden assumptions, but they are different kinds of assumptions.)

 

[DrStupid]

I know this is a bit simplistic

I wouldn't call it simplistic but basic because in physics experiments always have the final say. But it's always nice to have a theoretical background that predicts such experimental observations. That's where its starts to be complicate.

 

[stevendaryl]

This result is limited to the conditions above. You derived a force that would act on a "rocket" (it actually is a cannon) with constant mass during the time interval $\delta t$. As a rocket cannot change its momentum without changing its mass this condition cannot be fulfilled. If you consider the different mass at the begin and at the end of the time intervals you will get different equations.

What you're saying doesn't make a lot of sense to me. There is no difference between saying:

Description A
The rocket initially has mass $M_R$. The exhaust initially has mass $0$. The rocket's mass decreases by $\delta m$ and the exhaust's mass increases by $\delta m$

Description B
There are two objects, one of mass $M_R - \delta m$ and another of mass $\delta m$. Initially, the two objects are traveling together, but the second object is burned and sent out the rear.

In Description A, the objects change mass. In Description B, they don't. The only difference is how you divide the boundary around what counts as the objects. There is no conceptual difference between a rocket and a cannon--in both cases, you have a large mass splitting up into two unequal masses, and the smaller of the two is ejected.

You seem to be saying that the force depends on how you draw the boundary of two objects.

There is no fundamental reason in Newtonian physics to ever consider variable-mass objects. Since mass is conserved, you can always break the problem up into a many-particle problem involving tiny chunks of matter. So there is no need, ever, for a law of physics to describe what happens with variable mass. "Variable mass" is not a different situation, it is a different way of describing the same situation. So it's weird that the force would depend on how you describe things.

 

[DrStupid]

The only difference is how you divide the boundary around what counts as the objects. There is no conceptual difference between a rocket and a cannon--in both cases, you have a large mass splitting up into two unequal masses, and the smaller of the two is ejected.

There would be no conceptual difference if you would consider the transfer of momentum due to the change of the system boundaries after each step in Description B. But this is not included in your equations.

For the case that this explanation was too short:

In description A we start with a rocket of mass $m_R$ and velocity $v_R$ and no exhausted fuel (previus exhausted fuel can be neglected because it does not interact with the rocket) and we end up with a rocket of mass $m_R-\delta m$ and velocity $v_R + \delta v$ as well as with exhausted fuel with the mass $\delta m$ and the velocity $v_R + v_{rel}$. The corresponding changes of momentum are

$\delta p_R = \left( {m_R - \delta m} \right) \cdot \left( {v_R + \delta v_R } \right) - m_R \cdot v_R = m_R \cdot \delta v_R - \delta m \cdot \left( {v_R + \delta v_R } \right)$

$\delta p_F = \delta m \cdot \left( {v_R + v_{rel} } \right)$Description B consits of 2 parts. In the first part the rocket with initial mass $m_R$ and velocity $v_R$ is split into the remaining part of the rocket with the mass $m_R-\delta m$ and fuel with the mass $\delta m$, both traveling with the initial velocity of the rocket. This results in the following changes of momentum:

$\delta p_R = \left( {m_R - \delta m} \right) \cdot v_R - m_R \cdot v_R = - \delta m \cdot v_R$

$\delta p_F = \delta m \cdot v_R$

In the second step the velocity of the rocket is changed to $v_R + \delta v$ and the velocity of the fuel to $v_R + v_{rel}$, leaving the mass of both parts constant. The corresponding changes of momentum are

$\delta p_R = \left( {m_R - \delta m} \right) \cdot \left( {v_R + \delta v_R } \right) - \left( {m_R - \delta m} \right) \cdot v_R = \left( {m_R - \delta m} \right) \cdot \delta v_R$

$\delta p_F = \delta m \cdot \left( {v_R + v_{rel} } \right) - \delta m \cdot v_R = \delta m \cdot v_R$

Both steps of description B result in the same equations as description A, but you described the second step of B only and neglected the first one. This is sufficient if you are interested in the acceleration only because the first step doesn't change the velocity. But as it changes the momentum this is not sufficient if you try to calculate the force.

 

[stevendaryl]

There would be no conceptual difference if you would consider the transfer of momentum due to the change of the system boundaries after each step in Description B. But this is not included in your equations.

Ah! Okay, I think I understand what you're saying.

It's the old continuity equation business. If you have a surface representing the boundary of an "object", there are two ways that momentum and energy can change:

(1) The particles within the boundary might change their energy or momentum.
(2) Particles may enter or leave through the boundary (taking energy and momentum with them).

So I think that this is just a terminological difference. I would not consider effect (2) to be a "force", at all. Effect (2) is present even for noninteracting particles. For example, suppose that I have a collection of noninteracting particles, initially confined within a region shaped like a cube (whose faces point in the x, y, and z directions). Half the particles have velocity zero, and the other have a nonzero velocity $v$ in the x-direction. Initially, the stationary particles and the moving particles are both distributed uniformly within the cube.

If we consider the cube to be the object, then the object will be losing mass at a rate of $\dfrac{dm}{dt} = -\rho v A$ where $\rho$ is the density of the moving particles near the face pointing in the x-direction, $v$ is the speed normal to the face, and $A$ is the area of the face.

The particles that the cube is losing all have velocity $v$ in the x-direction, so the momentum of the cube is changing with time according to

$\dfrac{dp}{dt} = \dfrac{dm}{dt}\ v$

I would not call this rate of change of momentum a "force" at all. To me, noninteracting particles by definition cannot exert forces.

However, I don't think it makes any difference. If you want to use the word "force" to describe the changes of momentum due to particles flowing across subjective boundaries, I don't care. It's a matter of accounting. I don't like it, because I think "force" should reflect something real, rather than an artifact of our modeling, but that's just aesthetics.

The nice thing about this boundary-dependent force is that it obeys Newton's laws of motion, just like "real" forces. So I'm convinced that it is harmless to consider it a force, although I wouldn't.

 

[DrStupid]

The particles that the cube is losing all have velocity $v$ in the x-direction, so the momentum of the cube is changing with time according to

$\dfrac{dp}{dt} = \dfrac{dm}{dt}\ v$

I would not call this rate of change of momentum a "force" at all.

That's not a matter of taste. According to Newton's second law this is a force. The only choice you have is avoiding forces completely for open systems.

To me, noninteracting particles by definition cannot exert forces.

The particles cross a system boundary and carry momentum from one system to another. That actually is an interaction.

 

[stevendaryl]

Ok, I'm not going to waste any more time on this. Your claim in Post #18 was that $F=\dot p$ is to be favoured over $F=ma$ and I have provided several arguments that prove that the contrary is true, and you haven't addressed a single one of them. This makes the discussion entirely pointless.

I think I've gotten to the bottom of this dispute, and I'm fairly sure that it's just a terminological difference. Dr. Stupid's position simply amounts to calling the expression $\dfrac{dm}{dt}\ \vec{v}$ a "force". There are no consequences to this choice. It's not a different theory, it's just a different way of accounting for things.

The way I prefer doing things is to reserve the word "force" to mean a physical interaction, a push or a pull exerted on particles. With this notion of force, Newton's equations of motion are:

(1) $\vec{F}_{interaction} = m \dfrac{d\vec{v}}{dt}$

Dr. Stupid's approach is to say that both the left-hand side and the right-hand side should be augmented:

$\vec{F}_{total} = \vec{F}_{interaction} + \vec{F}_{boundary}$

where $\vec{F}_{boundary}$ reflects the rate at which momentum is passing across the object boundary.

In terms of $\vec{F}_{total}$, we have:

(2) $\vec{F}_{total} = m \dfrac{d\vec{v}}{dt} + \dfrac{dm}{dt} \vec{v}$

Assuming that (by definition?) $\vec{F}_{boundary} = \dfrac{dm}{dt} \vec{v}$, it's clear that equation (2) has exactly the same content as equation (1). You're just adding equal terms to both sides of an equation.

So in the case of the water balloon with equal leaks on each end, the Dr. Stupid approach would say that the force is nonzero, but that the force only contributes to $\dfrac{dm}{dt} \vec{v}$, not to $m \dfrac{\vec{v}}{dt}$