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

[velo city]

What is the proof of Einsteins famous equation E = mc² ?

 

[dipstik]

There is a derivation:
https://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[Matterwave]

I believe this ancillary paper is a more direct proof of what you want: https://www.fourmilab.ch/etexts/einstein/E_mc2/www/

Einstein does use results from his previous paper though, so you might want to read both.

 

[russ_watters]

In science, "proof" generally refers to evidence. Nuclear energy and particle accelerators are two good pieces of evidence.

 

[bcrowell]

There are many different possible ways of proving this. For a critique of some of the technical issues in Einstein's original derivation, see Ohanian, "Einstein's E = mc^2 mistakes," http://arxiv.org/abs/0805.1400 . In my SR book http://www.lightandmatter.com/sr/ I give a recap of the Einstein argument in section 4.2, and resolutions of the issues Ohanian complains about in sections 4.4 and 9.2. It's hard to give a really rigorous treatment without using the stress-energy tensor, which hadn't been invented when Einstein wrote his 1905 proof.

 

[stevendaryl]

There is a heuristic derivation of the relativistic expressions for energy and momentum (not a rigorous proof) that goes something like this:

1. Assume that the energy of a particle is proportional to the mass and is a function of the magnitude of the velocity: E = g(v) m (By "mass", I mean rest mass).
2. Assume that the momentum of a particle is proportional to the mass and is in the same direction as the velocity: \vec{p} = f(v) m \vec{v}. I'm also assuming that the unknown function f(v) depends only on the magnitude of the velocity, and not it's direction.
3. Assume that the expression for momentum approaches the Newtonian result, \vec{p} = m \vec{v} for low speeds. That implies that f(0) = 1
4. Assume that the expression for kinetic energy (total energy minus rest energy) approaches the Newtonian result: m g(v) - m g(0) \Rightarrow 1/2 m v^2
5. Assume that in a collision of particles, the energy and momentum are conserved.

With all those assumptions, we can prove that
f(v) = \gamma = 1/\sqrt{1-v^2/c^2}
g(v) = \gamma m c^2

To see this, consider an experiment where we collide two identical pieces of clay, each of mass m, so that they stick to form a single, larger chunk of clay of mass M. We look at the collision in three different rest frames:

1. Frame 1: Pick a frame in which one mass is traveling at velocity -v in the y-direction, and the other at velocity +v in the y-direction. Neither has any velocity in the x-direction.
2. Frame 2: Pick another frame whose spatial origin is moving at velocity u in the x-direction, relative to the first frame. Assume that u \ll v.
3. Pick a third frame whose spatial origin is moving at velocity u in the y direction. Again, choose u \ll v.

In Frame 1: By symmetry, afterward the composite mass M is stationary. Conservation of energy tells us that 2 g(v) m = g(0) M.

In Frame 2: The total momentum in the x-direction before the collision is approximately (ignoring higher-order terms in u) -2 f(v) m u. The total momentum afterward is approximately (using the Newtonian approximation, since u is small): -M u = -2 g(v) m u/g(0). Conservation of momentum in the x-direction gives: f(v) = g(v)/g(0).

In Frame 3: One mass is moving at velocity v_1 = \dfrac{v-u}{1-uv/c^2} \approx v - u (1-v^2/c^2) in the y-direction. (In the approximation, I did a Taylor expansion, and only kept the lowest order terms in u). The other mass is moving at velocity v_2 = -\dfrac{v+u}{1+uv/c^2} \approx -v - u (1-v^2/c^2). After the collision, mass M is moving with velocity -u in the y-direction.

Since we are assuming that momentum has the form p = f(v) m \vec{v}, if we change frames so that the velocity changes slightly by \delta v, the momentum will change by an amoung \delta p = m (\frac{d}{dv}(v f(v))) \delta v (ignoring higher-order terms in \delta v)

So the first particle will have momentum in the y-direction: f(v) m v - m u (\frac{d}{dv}(v f(v))) (1-v^2/c^2)

The second particle will have momentum in the y-direction: -f(v) m v - m u (\frac{d}{dv}(v f(v))) (1-v^2/c^2)

The total momentum before the collision in the y-direction is:

-2mu (\frac{d}{dv}(v f(v))) (1-v^2/c^2)

The momentum after the collision in the y-direction is:
- Mu = -2mg(v)/g(0) u = -2m f(v) u
(where I used the results from the other two frames).

So conservation of momentum in this frame gives:

-2mu (\frac{d}{dv}(v f(v))) (1-v^2/c^2) = -2m f(v) u

So we have a first-order differential equation for f(v):

(\frac{d}{dv}(v f(v))) (1-v^2/c^2) - f(v) = 0

The unique solution satisfying f(0) = 1 is

f(v) = 1/\sqrt{1-v^2/c^2}

Now going back to g(v). We showed that g(v)/g(0) = f(v) = 1/\sqrt{1-v^2/c^2}. Expanding for low-velocities gives: g(v) = g(0) (1 + 1/2 v^2/c^2 + ...). So the kinetic energy at low velocities is m (g(v) - g(0)) = m g(0) 1/2 v^2/c^2 + .... For this to agree with the Newtonian result, g(0) = c^2

So we have the two results:
\vec{p} = 1/\sqrt{1-v^2/c^2} m \vec{v} = \gamma m \vec{v}
E = g(v) m = m g(0) f(v) = m c^2 1/\sqrt{1-v^2/c^2} = \gamma m c^2

 

[bcrowell]

There is a heuristic derivation of the relativistic expressions for energy and momentum (not a rigorous proof) that goes something like this[...]

Since this is only for point particles, I think it's subject to the same criticisms that Ohanian makes of Einstein's 1905 derivation of E=mc².

 

[stevendaryl]

Since this is only for point particles, I think it's subject to the same criticisms that Ohanian makes of Einstein's 1905 derivation of E=mc².

Yeah, as I said, it's a heuristic derivation, not a proof.

 

[DrStupid]

[*]Assume that the energy of a particle is proportional to the mass and is a function of the magnitude of the velocity: E = g(v) m (By "mass", I mean rest mass).

Is there a justification for this assumption? Why not E = E_0 + g(v) m?

 

[stevendaryl]

Is there a justification for this assumption? Why not E = E_0 + g(v) m?

Right off the top of my head, I don't remember an argument for why rest energy should be proportional to mass.

 

[PeterDonis]

Right off the top of my head, I don't remember an argument for why rest energy should be proportional to mass.

If energy and momentum form a 4-vector, then rest energy has to be proportional (equal, if you choose units properly) to the invariant length of the 4-vector, i.e., to invariant (rest) mass.

 

[PeterDonis]

Is there a justification for this assumption? Why not E = E_0 + g(v) m?

$E_0$ is just $m$ (in units where $c = 1$, which you can always choose), so this equation reduces to stevendaryl's equation (you just redefine $g(v)$ to include the constant $1$).

 

[DrStupid]

I don't remember an argument for why rest energy should be proportional to mass.

Me too.

Here is a derivation of the relativistic momentum without any assumptions apart from the basics of classical mechanics and Lorentz transformation (and without energy at all):

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

Starting with the result of this derivation I get

dE = F \cdot ds = v \cdot dp = \frac{{m \cdot v \cdot dv}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} ^3 }}

The integration results in

E - E_0 = \frac{{m \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} - m \cdot c^2

With your assumption #1 you simply set the constant of integration to zero resulting in Eo=m·c². That turns your derivation into a circular argumentation.

 

[Fredrik]

...without any assumptions apart from the basics of classical mechanics and Lorentz transformation (and without energy at all):

This is a bit of an overstatement. You're using that $F=\frac{dp}{dt}$ and that $p=\gamma m v$. These are non-trivial assumptions. They can be considered definitions of the left-hand sides, but you can also take $E^2=p^2c^2+m^2c^4$ to be the definition of E, and then you don't have to do any calculations at all. This definition needs no justification other than the fact that the theory makes very accurate predictions (as already mentioned by Russ).

 

[DrStupid]

$E_0$ is just $m$

Remember the topic! Eo=m·c² has to be proofed. Thus you can't take it as a given.

 

[DrStupid]

You're using that $F=\frac{dp}{dt}$ and that $p=\gamma m v$. These are non-trivial assumptions.

$F=\frac{dp}{dt}$ is Newton's second law and $p=\gamma m v$ the result of the derivation.

 

[Fredrik]

$F=\frac{dp}{dt}$ is Newton's second law

Right, but so is $F=m\frac{d^2x}{dt^2}$, and this formula doesn't lead to the desired result. If you want the calculation to be considered a proof, or just a fairly solid argument, you have to explain why you're using that specific statement of Newton's 2nd. If it's just because it leads to the result we want, you might as well just write down the result we want and call it a definition.

and $p=\gamma m v$ the result of the derivation.

Can you justify the last equality in the first line without this?

 

[DrStupid]

Right, but so is $F=m\frac{d^2x}{dt^2}$

No. That results from the second law in classical mechanics but it is not identical with the second law. In SR you get another result.

you have to explain why you're using that specific statement of Newton's 2nd

The use of the original statement of Newton's second law does not need to be explained. It is the special case you mentioned above that would need a justification.

If it's just because it leads to the result we want, you might as well just write down the result we want and call it a definition.

Stick to facts!

 

[Matterwave]

If the argument is regarding whether Newton's second law states $F=ma$ or $F=\frac{dp}{dt}$, the answer is actually that his language is explicitly given as that of the second formula, although he explicitly assumes that the momentum in the second formula is given by $p=mv$.

Reading the Principia you find that Newton stated his second law thus (taken from the translation by Andrew Motte):

"The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed".

Here "motion" is roughly Newton's word for "momentum". He also only ever used the word "proportional" and not equal to, so there could be some as yet defined constant. So his language technically says $\frac{dp}{dt}\propto F$. However, earlier in the text, Newton defined "motion" (momentum) thus:

"The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly".

He states that if you double the quantity of matter (his word for mass) you double the "motion", and if you double the velocity you also double the motion. THIS explicitly assumes $p \propto mv$ and NOT $p=\gamma mv$ (obviously he would have no idea about the second case as that arose 300 years after him).

So, although his language technically might agree with DrStupid, I think the spirit of his language agrees with Frederik.

 

[Fredrik]

No. That results from the second law in classical mechanics but it is not identical with the second law. In SR you get another result.

The use of the original statement of Newton's second law does not need to be explained. It is the special case you mentioned above that would need a justification.

Is the stuff you feed into the calculation (of the work required to accelerate a massive point particle in SR) motivated by analogy with non-relativistic classical mechanics, or defined in the framework of SR? If it's the former, you certainly do have to explain why you favor dp/dt over ma. If it's the latter, then you have to explain why you consider your definitions (of momentum, work, etc) more fundamental than the end result, which can also be taken as a definition.

 

[DrStupid]

He states that if you double the quantity of matter (his word for mass) you double the "motion", and if you double the velocity you also double the motion. THIS explicitly assumes $p \propto mv$ and NOT $p=\gamma mv$ (obviously he would have no idea about the second case as that arose 300 years after him).

I can't follow your argumentation. Could you please explain why his statement excludes $p=\gamma mv$?

 

[DrStupid]

Is the stuff you feed into the calculation (of the work required to accelerate a massive point particle in SR) motivated by analogy with non-relativistic classical mechanics, or defined in the framework of SR?

Did you read the link above?

 

[Matterwave]

I can't follow your argumentation. Could you please explain why his statement excludes $p=\gamma mv$?

He says explicitly that if you double the mass you double the motion. If you double the velocity you double the motion. If you double both mass and velocity you quadruple the motion. This is not compatible with $p=\gamma mv$ because $\gamma$ depends on $v$ and his statements will no longer be true. If you double v, you certainly more than double $\gamma mv$ (assuming v<.5c).

 

[DrStupid]

If you double the velocity you double the motion.

I read that as "If you double the velocity [and keep quantity of matter constant] you double the motion."

Without this assumption it won't even be correct in classical mechanics. In SR you will need an open system (or two different systems) to meet this condition.

Edit: I added the possibility of two different systems because the explanation could also mean: If you have a system A with quantity of matter q and velocity v and a system B with the same quantity of matter but the velocity 2·v, than system b will have twice the motion of system A.

 

[Matterwave]

I read that as "If you double the velocity [and keep quantity of matter constant] you double the motion."

Without this assumption it won't even be correct in classical mechanics. In SR you will need an open system to meet this condition.

? Yes, keep m constant, $\gamma m v$ still more than doubles when you double v...I don't see how you can conclude $\gamma m v$ is a valid possible expression of the "motion" as defined by Newton.

 

[stevendaryl]

He says explicitly that if you double the mass you double the motion. If you double the velocity you double the motion. If you double both mass and velocity you quadruple the motion. This is not compatible with $p=\gamma mv$ because $\gamma$ depends on $v$ and his statements will no longer be true. If you double v, you certainly more than double $\gamma mv$ (assuming v<.5c).

I don't quite understand the point of this argument. Obviously, Newton's laws of physics don't imply SR. What I was looking for with my derivation, and Dr. Stupid presumably was looking for in his derivation, was an alternate theory of energy, momentum and force that reduces to the Newtonian case in the low-velocity limit and which is invariant under Lorentz transformations. The distinction between \dfrac{dp}{dt} = F and m \dfrac{dv}{dt} = F doesn't matter in the non-relativistic limit.

In my derivation, I didn't explicitly mention forces at all, but instead just assumed that in collisions there were two quantities that were conserved: (1)A scalar-valued quantity, energy, and (2)a vector-valued quantity, momentum.

 

[DrStupid]

? Yes, keep m constant, $\gamma m v$ still more than doubles when you double v

Remember: Motion is defined the product of quantity of matter and velocity. That means in SR quantity of matter is not m but $\gamma m$.

 

[Matterwave]

Remember: Motion is defined the product of quantity of matter and velocity. That means in SR quantity of matter is not m but $\gamma m$.

This is a stretch...and is certainly NOT in the spirit of Newton. Saying things such as "quantity of matter" is $\gamma m$ is certainly NOT something Newton would agree with in his time. If you want to say it NOW, you are ASSUMING special relativity is right. Even in special relativity, there is very limited utility of defining the relativistic mass, as this approach only works for forces which are co-linear with the object's motion.

But this is beyond the point.

My point is that, as formulated by Newton, the second law should read, given modern notation, $F=\frac{dp}{dt}$ but it MUST be appended with the explicit assumption that $p=mv$.

If you want to explore what Newton meant by "m", then you need to read his definition I in his Principia. A problem arises here; however, because mass was mysterious to Newton himself. Newton basically defined mass as density times volume, which just begs the question of defining the density.

But given Newton's definition, there is no way to define quantity of matter as $\gamma m$, both the density and volume should be measured in the rest frame.

 

[stevendaryl]

Remember: Motion is defined the product of quantity of matter and velocity. That means in SR quantity of matter is not m but $\gamma m$.

That's a possible interpretation. The first derivation of E = mc² that I saw used relativistic mass (that is, it assumed that \vec{p} = m(v)\vec{v}, and derived the form of m(v)). I didn't like that derivation, since I tended to think of "relativistic mass" as an old-fashioned concept that was no longer used. However, I have seen some respectable physicist (on the sci.physics.research newsgroup---I can't remember his name) argue that there are good mathematical reasons to have a relativistic mass. I don't remember the details, but there was some group-theoretic reason showing that both SR and galilean relativity were special cases of a more general kind of relativity.

 

[DrStupid]

What I was looking for with my derivation, and Dr. Stupid presumably was looking for in his derivation, was an alternate theory of energy, momentum and force that reduces to the Newtonian case in the low-velocity limit and which is invariant under Lorentz transformations.

My derivation also works for relativistic velocities. E=m·c² doesn't make much sense in the non-relativistic case.

 

[stevendaryl]

My derivation also works for relativistic velocities.

Of course. I wasn't saying that the derivation depended on the assumption of low velocities, but that the correspondence with Newtonian physics applies at low velocities.

 

[DrStupid]

Saying things such as "quantity of matter" is $\gamma m$ is certainly NOT something Newton would agree with in his time.

Of course not. In his time Galilei transformation was out of discussion and as I demonstrated in my link above it leads to an invariant quantity of matter. But this changes with replacement of Galilei transformation by Lorentz transformation. As Newton wasn't aware of that it is just a coincidence that the laws of motion in their original wording remain valid in SR. However, that doesn't make any difference.

My point is that, as formulated by Newton, the second law should read, given modern notation, $F=\frac{dp}{dt}$ but it MUST be appended with the explicit assumption that $p=mv$.

Why should we change the original definitions just to make them invalid? That doesn't make any sense.

But given Newton's definition, there is no way to define quantity of matter as $\gamma m$, both the density and volume should be measured in the rest frame.

I already explained why I disagree. It seems we will not come to a consensus.

 

[Matterwave]

Why should we change the original definitions just to make them invalid? That doesn't make any sense.

Why do you think I've changed the original definition, rather than just restated it in modern notation? Do you think that Newton did NOT mean $p=mv$ when he defined the "motion"? I am only saying that Newton defined the "motion" to be $p=mv$ and obviously NOT $p=\gamma mv$.

 

[AlephZero]

My point is that, as formulated by Newton, the second law should read, given modern notation, $F=\frac{dp}{dt}$ but it MUST be appended with the explicit assumption that $p=mv$.

I can't think of anywhere in Newton's writing where he analyzes the motion of an object whose mass changes with time (for example a rocket, assuming classical mechanics). So I think you are putting ideas into Newton's head with the benefit of hindsight, with no firm evidence that he ever thought about the momentum of a system with variable mass, let alone a system whose mass might be function of its velocity.

 

[Matterwave]

I can't think of anywhere in Newton's writing where he analyzes the motion of an object whose mass changes with time (for example a rocket, assuming classical mechanics). So I think you are putting ideas into Newton's head with the benefit of hindsight, with no firm evidence that he ever thought about the momentum of a system with variable mass, let alone a system whose mass might be function of its velocity.

Doesn't this just cement my statement more? If Newton never even considered a velocity dependent mass, then are we not free to say that Newton said $p=mv$? In fact, I argue that $p \propto mv$ is exactly his statement when he defines the "motion". How can you make the argument that Newton's second law is only $F=\frac{dp}{dt}$ but not look at what is meant by $p$?

 

[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&#039; \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&#039; \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.