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Author Topic:   Funny Comic and Newtons Laws (in a non-inertial reference frame)
JustinC
Member (Idle past 4844 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 1 of 5 (388655)
03-06-2007 10:27 PM


First, the comic xkcd: Centrifugal Force
Now, the question: how do you construct Newton's Law in a non-inertial reference frame?
I did the coordinate transformation with two frames at an angle two each other, but how do you I represent that fact that the coordinate system is constantly changing. Do I make it a function of time?
Help would be greatly appreciated.
Edited by Adminnemooseus, : Added the "(in a non-inertial reference frame)" to the topic title.

Replies to this message:
 Message 3 by cavediver, posted 03-07-2007 4:29 AM JustinC has not replied
 Message 4 by Son Goku, posted 03-07-2007 6:30 AM JustinC has not replied
 Message 5 by AnswersInGenitals, posted 03-09-2007 4:42 AM JustinC has not replied

  
Adminnemooseus
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Joined: 09-26-2002


Message 2 of 5 (388662)
03-06-2007 11:02 PM


Thread moved here from the Proposed New Topics forum.

  
cavediver
Member (Idle past 3643 days)
Posts: 4129
From: UK
Joined: 06-16-2005


Message 3 of 5 (388669)
03-07-2007 4:29 AM
Reply to: Message 1 by JustinC
03-06-2007 10:27 PM


how do you I represent that fact that the coordinate system is constantly changing. Do I make it a function of time?
Yes, absolutely. I'm too busy to go through it now but check out Wiki on rotating reference frame and all the maths is there.
One interesting fact about the casual dismissal of centrifugal force is that everything said about it applies equally to gravity... that is why we talk about gravitation. Gravity is a coordinate dependent commodity.

This message is a reply to:
 Message 1 by JustinC, posted 03-06-2007 10:27 PM JustinC has not replied

  
Son Goku
Inactive Member


Message 4 of 5 (388674)
03-07-2007 6:30 AM
Reply to: Message 1 by JustinC
03-06-2007 10:27 PM


Rotating frames
Basically yes. Change the angle from being theta, to something like omega*time.
As an interesting side problem, see what a free particle in the inertial frame looks like in the rotating frame.
You could also make the time dependence more complicated, in which case you should notice a new force.
As cavediver said, it is interesting to note that General Relativity basically says that gravity is similar to the centrifugal force in being related to coordinate transformations.

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AnswersInGenitals
Member (Idle past 151 days)
Posts: 673
Joined: 07-20-2006


Message 5 of 5 (388938)
03-09-2007 4:42 AM
Reply to: Message 1 by JustinC
03-06-2007 10:27 PM


Simplify the problem.
The biggest secret to doing science, the first step in the scientific process, is that when you confront a problem that is in any way challenging, don't start out by trying to solve the problem. Start out by trying to simplify the problem. See if you can eliminate any elements of the problem or replace any of those elements with simpler models, or replace the problem with a simpler one with fewer, conceptually easier, or more intuitive elements. In this case, you are trying to get a handle on interactions in an accelerating frame of reference. First, change the problem to one with just one dimension. The easiest problem for your little man in his accelerating frame is to put him in a little box on the x axis that is accelerating to the right at a constant rate of acceleration. In fact, let's simplify it further and have the box just sit there, lets say at position X0 on the x axis. Applying Newton's law: F = Ma where a is the second derivative of position with respect to time, we see that d2(X0)/dtt = 0 since X0 is a constant. (Sorry to write the derivative so funny, but I want to avoid special characters that always get screwed up in transmission. I'm assuming you know a little calculus and can figure out what I'm saying.) Since the acceleration is 0 the force is 0 and the little man just floats around in his little box getting nauseous. Pretty simple and pretty intuitive.
For the next step in complexity, lets put the box in a frame thats moving to the right a a constant velocity, V. The box's frame of reference (as seen by us observing the box) is given by x = X0 + Vt. Note that, yes, time does explicitly enter into the frame position, or frame of reference transformation equation, even for this simple example. The second derivative is still 0 and the little man feels no force. He has no way of knowing that his situation is any different than in the first constant position case. Now we put him in an accelerating frame: x = X0 + Vt + Att. We now do get a non-zero second derivative and A is the box's constant acceleration. The force the man feels = mA against the left wall of the box. If he stands on this wall and holds two balls in his hands, one a ten pound lead ball and the other a ten once wood ball, at the same height from the left wall (which seems to him to be his 'floor') and drops them, they will 'fall' at the same rate and hit the floor at the same time. I have 'fall' in quotes because to us on the outside looking at the box (did I mention it has glass walls), the balls don't seem to fall, they just don't continue to accelerate with the little man. The man feels like there is a force pushing him down (to the left) against the box's left wall. But it is not pushing down on his shoulders or on the outside of his body. It is pushing down on every internal part, on every atom of his body (and anything else in the box. From our prospective, it is really the box pushing up on his feet since it is accelerating to the right and his body's inertia is resisting being accelerated.
Ok, now for the wheel spinning in two dimensions. I won't go into the mathematical details, which I'm sure are covered in cavedivers reference, but at any instance, the man is at distance R from the center of the wheel and at some angle b = Bt from his starting position and this angle is increasing at a constant rate in time (i. e., the wheel is rotating at a constant rate = B degrees or radians per second). His change in position (his coordinate transformation) is given by all those sin(b) and cos(b) terms, but we can simplify since we are interested in his instantaneous change. Since sin(b) is approximately equal to b for small angles, his instantaneous change in position tangent to the wheel is Rb =RBt, i. e., he seems at each instant to be moving at a constant velocity = RxB tangent to the wheel and his acceleration tangent to the wheel is zero so that he feels no force in that direction. Since cos(b) is approximately 1 - bxb/t = 1 - BBtt/2, and the second derivative of this is BB, he feels an acceleration (and thus a force) perpendicular to the tangent to the wheel with magnitude of RBB (feet per second squared).
In our frame of reference watching him from the outside, we see him deviating from his straight, constant velocity path tangential to the edge of the wheel by this amount and interpret that as an acceleration and force on him towards the center of the of the wheel, which we call a centripetal force. In his frame of reference, he feels a force pushing him away from the center of the wheel and against the rim, which he calls a centrifugal force. Since this is the only force he feels, he can't tell if he is accelerating along a straight line as in our first example, or being rotated on a spinning wheel, or just standing on the floor in a gravitational field.
There is a way for him to distinguish which situation he is in if his little box has an upper and a lower floor connected by a stairway so that he can move up and down (where 'up' and 'down' are defined by the direction of the force he feels). With the uniform acceleration along a line, as in our first example, he will feel the same force on both floors. With the rotating wheel, the force will increase if he goes 'upstairs', i. e., away from the center of the wheel. With a gravitational force, the force will decrease if he goes upstairs, further from the center of mass of the gravitating body. You should keep your scale on the second floor of your house if you want to minimize your apparent weight.
Because of the similarity of the gravitational force to these forces derived from time dependent geometric transformations, theoreticians tried to derive gravity from such transforms long before Einstein's success. In particular, Bernhard Riemann, who invented the system of differential geometry that is essential for the mathematical treatment of general relativity, attempted to derive gravity as a result of transformations about thirty years before Einstein publish his solution, but failed because he limited himself to transformations in three spatial dimensions and was unaware of the need to consider a four dimensional space-time continuum. Since he was not successful, he did not publish this work and it is unclear if Einstein was aware of it.
Sorry to be so long winded and hope this helps.

This message is a reply to:
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