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Force from power at rest = infinity?

by Pharrahnox
Tags: force, infinity, power, rest
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jbriggs444
#19
Apr18-14, 02:00 PM
P: 949
Quote Quote by my2cts View Post
KE=Pt is only correct if P has been constant since t=0 and zero before that. The general expression is KE=Integral Pdt.
The relevant integral can be improper. That, in turn, allows for the possibility that P(0) is not defined.

If we are given that KE=0 at t=0 then it is the (improper) definite integral from 0 to t that matters. If we are only interested in the motion for t>0 then the values for P(t) for t<0 simply do not enter in.

If one insists on extending positive constant power into the past, we would admittedly have the problem of an imaginary KE at t<0. That would not seem to be very physical.

Edit: The point about power being possibly undefined at t=0 is irrelevant. The trajectory you get if you crank the numbers will have a well defined P at t=0 even if the force or acceleration at t=0 are undefined. It will be a one-sided limit, but that's OK as long as we are only considering a trajectory from the starting point forward and not from the starting point going into the past.
A.T.
#20
Apr18-14, 03:36 PM
P: 4,035
Quote Quote by Pharrahnox View Post
Ah, so the error was the assumption that just sticking power into something would just magically make it do something.
The problem might be the idea, that power is something "that you stick into something". Power is just a numerical value, that some observer computes based on the velocities in his reference frame. If the velocity is zero in his frame, then so is power and he cannot use F = P/v to determine F, which still can have some finite value.
DaleSpam
#21
Apr18-14, 04:40 PM
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Quote Quote by jbriggs444 View Post
The point being made (as I understand it) is that the motion of the object is determinable, even though the second derivitive of its position at t=0 may be non-existent (infinite).
Mathematically sure, but no real force can do that.

I don't know what you and dauto are going on about. Even if you can mathematically work around the infinite force, and even if other quantities do not diverge, the fact remains that an infinite force is not physical. The objections based on that fact, which point out that F is infinite or undefined at v=0 for a finite P, are entirely correct objections.
DaleSpam
#22
Apr18-14, 04:41 PM
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Quote Quote by dauto View Post
What's the force between two electrons in the limit their distance goes to zero?
I think that quantum questions are off-topic here, but the force is finite:
http://en.wikipedia.org/wiki/Electro...eracy_pressure
DaleSpam
#23
Apr18-14, 04:45 PM
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Quote Quote by my2cts View Post
F=dP/dv and will be finite at v=0.
No, P=F.v and F will be infinite if P is finite and v is 0.
jbriggs444
#24
Apr18-14, 08:59 PM
P: 949
Quote Quote by DaleSpam View Post
Mathematically sure, but no real force can do that.

I don't know what you and dauto are going on about. Even if you can mathematically work around the infinite force, and even if other quantities do not diverge, the fact remains that an infinite force is not physical. The objections based on that fact, which point out that F is infinite or undefined at v=0 for a finite P, are entirely correct objections.
An objection that F is infinite or undefined at zero is mathematically correct. But that is not a physical objection.
jbriggs444
#25
Apr18-14, 08:59 PM
P: 949
Quote Quote by DaleSpam View Post
No, P=F.v and F will be infinite if P is finite and v is 0.
That is not the only definition of P.

It is valid to define P as the derivitive of KE with respect to time. That formulation does not suffer from the problem of an undefined force multiplied by a zero velocity.
my2cts
#26
Apr19-14, 05:14 AM
P: 80
Quote Quote by DaleSpam View Post
No, P=F.v and F will be infinite if P is finite and v is 0.
I am confident that your approach is in disagreement with classical mechanics.
DaleSpam
#27
Apr20-14, 11:41 PM
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Quote Quote by jbriggs444 View Post
That is not the only definition of P.

It is valid to define P as the derivitive of KE with respect to time. That formulation does not suffer from the problem of an undefined force multiplied by a zero velocity.
That is indeed a valid definition, but it doesn't change anything. If you start with some fixed power under that definition and consider what happens for an object at rest you still find that it undergoes an infinite acceleration and therefore an infinite force.
DaleSpam
#28
Apr20-14, 11:43 PM
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Quote Quote by my2cts View Post
I am confident that your approach is in disagreement with classical mechanics.
Then provide a reference for you formula: f=dP/dv
jbriggs444
#29
Apr21-14, 05:26 AM
P: 949
Quote Quote by DaleSpam View Post
That is indeed a valid definition, but it doesn't change anything. If you start with some fixed power under that definition and consider what happens for an object at rest you still find that it undergoes an infinite acceleration and therefore an infinite force.
Yes, certainly. An infinite acceleration and a corresponding infinite force applied over a zero time interval. That makes it almost certainly non-physical and almost certainly non-measurable.


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