# "Motion is impossible" claims modern Zeno

• Nathanael
In summary, the conversation discusses the claim that it is impossible to move a piano on a frictionless surface if you are standing on a frictional surface. This is based on two statements: (1) the kinetic energy of the piano is equal to the work done on it, and (2) work cannot be done on the piano unless it is moving. However, the flaw in this logic is that it assumes a prerequisite relationship between these two quantities which change simultaneously. This is not the case, as demonstrated by the fact that you can apply a force on the piano without doing work, and once it is moving, work can be done on it. The conversation also touches on the concept of changing variables in a quantitative relationship and how it can be
Nathanael
Homework Helper
Let me set up a situation:

A piano rests on a frictionless surface. I am standing next to the piano (on a frictional surface) and I claim that the following two statements prove it is impossible for me to move the piano:

(1) ... The kinetic energy of the piano is equal to the work I've done on it.
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

"Therefore the piano is immovable," I claim.

(1) is equivalent to the work-kinetic-energy theorem $W=\Delta E$
(2) is a special case (where $\frac{d\vec s}{dt}=0$) of the definition of work $dW=\vec F\cdot d\vec s=\vec F\cdot \frac{d\vec s}{dt}dt$

Please explain where and why my logic is flawed (assuming I don't know Newton's laws).

olgerm
Nathanael said:
Please explain where and why my logic is flawed
The assumption of a prerequisite relationship between two quantities which change simultaneously.

DocZaius
Nathanael said:
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

Push it. You'll both move.

AgentSmith
A.T. said:
The assumption of a prerequisite relationship between two quantities which change simultaneously.
Sounds reasonable, but I have to say I'm not satisfied (probably because I don't entirely understand you.) Can you say more about this?

Drakkith said:
Push it. You'll both move.
Well, I probably won't move (I said my ground was frictional) but yes, it will move... but the only reason you know this is from F=ma.
But if you consider only work and energy you might come to the paradox I proposed.
I was curious if there is any way around this paradox without invoking F=ma.

You can apply a force on the piano without doing work (the piano is stationary).

If there is no friction to counteract the force you apply the piano MUST accelerate.

Once it's moving you can doe work on it.

Nathanael said:
Sounds reasonable, but I have to say I'm not satisfied (probably because I don't entirely understand you.) Can you say more about this?
When you have a quantitative relationship between two variables like y = x2, then changing one of them implies changing the other. If you misinterpret this correlation as a two way causation, you get: A change of y requires a change of x which requires a change of y... so no change of either is possible.

Dale and CWatters
Let me say, I do know that it will move. I do not believe my claim in the OP. (Perhaps I should have emphasized this.)

But my question is this:
CWatters said:
Once it's moving you can doe work on it.
If I didn't do work on it then how did it get to be moving?

I understand from the F=ma perspective: I push on it therefore it accelerates.

But from an energetic perspective: it must have energy (motion) before I can give it any energy.

A.T. said:
When you have a quantitative relationship between two variables like y = x2, then changing one of them implies changing the other. If you misinterpret this correlation as a two way causation, you get: A change of y requires a change of x which requires a change of y... so no change of either is possible.
I feel like you're getting at the heart of it. Yet... it still doesn't resonate with me in this particular situation.
Somehow I can't get past this statement: "it must have energy (of motion) before I can give it any energy."

I will sleep on your explanation. Any different ways of explaining it are welcome. Goodnight.

Nathanael said:
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

Does it have to be moving, or does it just have to move?

A.T. said:
When you have a quantitative relationship between two variables like y = x2, then changing one of them implies changing the other. If you misinterpret this correlation as a two way causation, you get: A change of y requires a change of x which requires a change of y... so no change of either is possible.
Sorry, I can't sleep. What you say sounds nice, but I can't understand how it is relevant:
We don't have something like y=x2, we have $\frac{dE}{dt}=k\sqrt{E}$ with E(0)=0. If E=0 then dE/dt=0 therefore the energy is in a steady state w.r.t. to time (i.e. it can't change).
($k=F\sqrt{\frac{2}{m}}$ not that it matters.)
Drakkith said:
Does it have to be moving, or does it just have to move?
I see what you're trying to say, but tell me if this is true or false: The only way to move is to, at some point in time, be moving.

[edited because I said "relative" instead of "relevant" ... I was tired haha]

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Nathanael said:
I see what you're trying to say, but tell me if this is true or false: The only way to move is to, at some point in time, be moving.

At some point in time, but not every point in time.

Nathanael said:
I see what you're trying to say, but tell me if this is true or false: The only way to move is to, at some point in time, be moving.
For every real number x greater than zero there is a real number y that is between x and zero. For every time when the piano is moving there is a prior time when the piano was moving.

One key to your conundrum is that you seem to want to imagine a first real number greater than zero -- a first instant when the piano is moving. But there is no such thing.

olgerm, darpan, sophiecentaur and 1 other person
Nathanael said:
If I didn't do work on it then how did it get to be moving?

I understand from the F=ma perspective: I push on it therefore it accelerates.

But from an energetic perspective: it must have energy (motion) before I can give it any energy.
So what? It sounds like you are falsely equating force and energy. f=ma does not necessarily require energy.

If you apply a 1N force to a 1kg object, starting at t=0, at t=0 the object is stationary, but accelerating at 1m/s and at that time the rate of expenditure of energy is 0. There is no conflict.

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Nathanael said:
We don't have something like y=x2, we have $\frac{dE}{dt}=k\sqrt{E}$ with E(0)=0. If E=0 then dE/dt=0
This is exactly like y=x2 where at x = 0 you have dy/dx = 0.

Nathanael said:
I see what you're trying to say, but tell me if this is true or false: The only way to move is to, at some point in time, be moving.
Natural language is ambiguous and another reason for apparent paradoxes.

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darpan
Nathanael said:
Let me set up a situation:

A piano rests on a frictionless surface. I am standing next to the piano (on a frictional surface) and I claim that the following two statements prove it is impossible for me to move the piano:

(1) ... The kinetic energy of the piano is equal to the work I've done on it.
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

"Therefore the piano is immovable," I claim.

(1) is equivalent to the work-kinetic-energy theorem $W=\Delta E$
(2) is a special case (where $\frac{d\vec s}{dt}=0$) of the definition of work $dW=\vec F\cdot d\vec s=\vec F\cdot \frac{d\vec s}{dt}dt$

Please explain where and why my logic is flawed (assuming I don't know Newton's laws).

Why can't the piano have zero kinetic energy without you doing work on it?

What you seem to have proven is that you and the piano can't have a fixed position and zero momentum. Perhaps that's true instead?

Nathanael said:
Let me set up a situation:

A piano rests on a frictionless surface. I am standing next to the piano (on a frictional surface) and I claim that the following two statements prove it is impossible for me to move the piano:

(1) ... The kinetic energy of the piano is equal to the work I've done on it.
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

"Therefore the piano is immovable," I claim.

(1) is equivalent to the work-kinetic-energy theorem $W=\Delta E$
(2) is a special case (where $\frac{d\vec s}{dt}=0$) of the definition of work $dW=\vec F\cdot d\vec s=\vec F\cdot \frac{d\vec s}{dt}dt$

Please explain where and why my logic is flawed (assuming I don't know Newton's laws).

Classically, this fails for the same reason Zeno's paradox fails. It confuses differentials with discrete steps. Think about it this way: at 0 seconds, there is still a force, f (which you are applying to the piano). Now, imagine that a differential of time, dt, has passed. Then, the piano will undergo a change in momentum, f*dt. This, after all, is the definition of force. Now that the momentum is nonzero, the instantaneous velocity is also non-zero (though infinitesimal). By definition, velocity is the rate of change of position, so now, the position will be f*dt*dt at time 2dt.

Now, if you are familiar with calculus, then to get real numbers instead of infinitesimals, you just need to integrate, and voila! The piano is now moving, and work has been done.

TheDemx27
thefurlong said:
Classically, this fails for the same reason Zeno's paradox fails. It confuses differentials with discrete steps. Think about it this way: at 0 seconds, there is still a force, f (which you are applying to the piano). Now, imagine that a differential of time, dt, has passed. Then, the piano will undergo a change in momentum, f*dt. This, after all, is the definition of force.
This imagining just pushes the problem back into infinitesimals. One is still imagining a situation where after infinitesimal time dt the piano has infinitesimal but non-zero energy. Infinitesimal work must have been done. All of which is true enough.

As you say, this confuses differentials with discrete steps. The unstated intuition is that there must be steps and that there must be a first step.

Nathanael said:
If I didn't do work on it then how did it get to be moving?
It got to be moving by accelerating, and acceleration requires non zero impulse not non zero work. The work thing is a "red herring".

Cybella and sophiecentaur
To be honest, I don’t really understand the “problem” but:

Maybe this can be resolved by considering that the displacement of the piano does not happen instantaneously when you apply a force to it. I am thinking along the lines of a mass suspended from a spring and the mass is at the top end of travel. It is instantaneously at rest, but it is under acceleration from the force of gravity. That is, there is a phase difference between the acceleration and velocity but that certainly does not prevent the gravitational force from doing work on the mass.

I think the piano on a frictionless surface will behave in a similar fashion, with inertia causing a delay between acceleration and velocity.

Nathanael said:
Sorry, I can't sleep. What you say sounds nice, but I can't understand how it is relative:
We don't have something like y=x2, we have $\frac{dE}{dt}=k\sqrt{E}$ with E(0)=0. If E=0 then dE/dt=0 therefore the energy is in a steady state w.r.t. to time (i.e. it can't change).
($k=F\sqrt{\frac{2}{m}}$ not that it matters.)

I agree with you Nathanael. An object with no kinetic energy cannot move -- ever. At least not under the physical model you have chosen. There are other solutions to this equation, but the trivial solution does not allow change.

In a more complete model, energy is better defined by Noether's Theorem. Plus the incompleteness theorem indicates that no particle can have zero energy. Time is not as simple as your model implies either I think. I thought these sorts of paradoxes led to the development of Quantum Mechanics?

A.T. said:
This is exactly like y=x2 where at x = 0 you have dy/dx = 0.
No it's not exactly like that! arhgghgharhg I'm sorry but I stand by what I said, your analogy is different and irrelevant: You have dy/dx = x =0 whereas I have dy/dx = y =0... That is importantly different! See, you're saying that I'm thinking y can't change because x is zero (which is false of course, because x is an independent variable) but I'm saying y can't change because y is zero! Isn't this importantly different?? Or have I just gone dumb...

A.T. said:
Natural language is ambiguous and another reason for apparent paradoxes.
The paradox was is in the mathematics... dE/dt=ksqrt(E)

Jeff Rosenbury said:
I agree with you Nathanael. An object with no kinetic energy cannot move -- ever. At least not under the physical model you have chosen. There are other solutions to this equation, but the trivial solution does not allow change.
Yes there are other solutions, but this is the only solution for which it starts at rest. However, I think I found the issue with this DE (see below)

jbriggs444 said:
One key to your conundrum is that you seem to want to imagine a first real number greater than zero -- a first instant when the piano is moving. But there is no such thing.
This is a good point, but at this point it's not even at about imagination... It's about a differential equation which is in a steady state if the energy is zero.

DaleSpam said:
It got to be moving by accelerating, and acceleration requires non zero impulse not non zero work. The work thing is a "red herring".
Okay, it's a red herring... but it's a red herring which leads to a differential equation to back it up! At this point, I want to know why the math is wrong. (Anyway I think I've got it.)

russ_watters said:
If you apply a 1N force to a 1kg object, starting at t=0, at t=0 the object is stationary, but accelerating at 1m/s and at that time the rate of expenditure of energy is 0. There is no conflict.
To me there seems to be a conflict in the mathematics, let me try to emphasize what my problem is (and what I think the problem with the problem is)

From F=ma you can derive the following equation: $\frac{dE}{dt}=F\sqrt{\frac{2E}{m}}$ ... where E=0.5mv2

This differential equation is equivalent to Newton's F=ma (just plug in E in your mind and you will see it reduces to F=ma) but they seem to be saying two very different things... This equation predicts that if E is zero it stays zero (it's in a steady state) whereas Newton's law obviously predicts that it will move. This is what the whole thread was about... (I think my verbal statement of the problem was a red herring to everyone else!)

So how can two equivalent equations make two different predictions?

My suspicion is that it has to do with dividing by zero... If you want to get from my equation to F=ma then plug in E=0.5mv2 and you will get Fv=mav but then to get to Newton's F=ma you have to divide by v which is zero. Therefore, since you can't divide by zero, these equations are equivalent except at v=0.Sorry for being so stubborn, but I just felt like most replies were missing the point. A few were insightful, though. Thanks for all replies regardless.

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When you solve a diff
Nathanael said:
From F=ma you can derive the following equation: $\frac{dE}{dt}=F\sqrt{\frac{2E}{m}}$ ... where E=0.5mv2

This differential equation is equivalent to Newton's F=ma (just plug in E in your mind and you will see it reduces to F=ma)
Except when v = 0. The differential equation has a singularity there. It fails to be predictive. So it is NOT equivalent to F=ma.

This equation predicts that if E is zero it stays zero (it's in a steady state).
No, it does not. It predicts that dE/dt = 0 when E=0. But that's not the same thing as predicting that E is identically zero.

darpan
jbriggs444 said:
Except when v = 0. The differential equation has a singularity there. It fails to be predictive. So it is NOT equivalent to F=ma.
I know, I pointed this out in that post. It's equivalent to f=ma except at v=0.

jbriggs444 said:
No, it does not. It predicts that dE/dt = 0 when E=0. But that's not the same thing as predicting that E is identically zero.
I don't understand... If E=0 (the initial condition) and dE/dt = 0 then doesn't this mean that (according to the DE) E will be forever zero?
(The DE is of course wrong at E=0 but I still don't understand your statement.)

Nathanael said:
I'm saying y can't change because y is zero!
That makes even less sense to me.

TheDemx27
Nathanael said:
I don't understand... If E=0 (the initial condition) and dE/dt = 0 then doesn't this mean that (according to the DE) E will be forever zero?
(The DE is of course wrong at E=0 but I still don't understand your statement.)

No. E(0)=0 and E'(0)=0 does not entail that E(x) = 0 for all x.

olgerm
I don’t believe I have ever seen the expression dE/dt = Fv before anywhere.

I see you had dW = F∙ds/dt and since W = E then dE/dt = Fv But since W = F∙s how can you treat F as a constant when differentiating W?

Since F = ma, wouldn’t you need to differentiate the ma term with respect to t also?

AgentSmith, darpan and Jeff Rosenbury
Nathanael said:
If E=0 (the initial condition) and dE/dt = 0 then doesn't this mean that (according to the DE) E will be forever zero?
No, because d2E/dt2 ≠ 0

Same with y=x2:
y(0) = 0
y'(0) = 0
y''(0) = 2

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russ_watters and jbriggs444
Nathanael said:
II don't understand... If E=0 (the initial condition) and dE/dt = 0 then doesn't this mean that (according to the DE) E will be forever zero?
Not at all. If E and all of its derivatives are zero at some point and E is a smooth function, then you can say that E is identically equal to zero. Otherwise, all bets are off.

Here $E=\frac 1 2 mv^2$, so $\frac{dE}{dt} = m \vec v \cdot \vec a$ and $\frac{d^2E}{dt^2} = m (\vec v \cdot \dot{\vec a} + a^2)$. At a point where $\vec v = 0$, both $E=0$ and $\frac{dE}{dt} = 0$, but the second derivative is zero only if the acceleration is zero.

darpan, Tom_K and Nathanael
Nathanael said:
Okay, it's a red herring... but it's a red herring which leads to a differential equation to back it up!
The differential equation does NOT back it up. The energy is quadratic in v and therefore the derivative is 0 for small v. Your Zeno paradox (like most) is resolved by the differential equations, not confirmed.

D H said:
Not at all. If E and all of its derivatives are zero at some point and E is a smooth function, then you can say that E is identically equal to zero. Otherwise, all bets are off.

Here $E=\frac 1 2 mv^2$, so $\frac{dE}{dt} = m \vec v \cdot \vec a$ and $\frac{d^2E}{dt^2} = m (\vec v \cdot \dot{\vec a} + a^2)$. At a point where $\vec v = 0$, both $E=0$ and $\frac{dE}{dt} = 0$, but the second derivative is zero only if the acceleration is zero.

Thank you. For some reason (because of my conservation with A.T.) I was thinking that y'=y implies that y''=y' which implies that all derivatives are zero. But of course it's not analogous to y'=y because of the acceleration term hidden inside of F.

Nathanael said:
This equation predicts that if E is zero it stays zero (it's in a steady state)
No. This equation does not predict that at all! ##E(t)=0## is a solution, but certainly not the only solution.

EDIT: If I did the math correctly then for a constant force:
$$E(t)=E_0 - f \sqrt{\frac{2E_0}{m}}t+\frac{f^2}{2m} t^2$$
So ##E(t)=0## corresponds to the solution where ##E_0=f=0##, but there are an infinite number of other solutions also. If ##E_0=0## then ##E(t)=\frac{f^2}{2m} t^2## which is not equal to 0 except if ##f=0##.

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Nathanael said:
For some reason (because of my conservation with A.T.) I was thinking that y'=y implies that y''=y'
Not sure how I gave you that idea, since my example y=x2 actually disproves the above.

A.T. said:
Yeah, I had one of those brain freezes going on.
dE/dt = mva is what I was trying to think of.

darpan
A.T. said:
Not sure how I gave you that idea, since my example y=x2 actually disproves the above.
Because I was trying to tell you y=x2 is irrelevant to the example because this situation is more like y'=2y (not y'=2x) which is why I was thinking about y'=y

It's not you that made me think that, just my conversation with you.

Anyway sorry for dragging this thread out so long.

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