Sum of the forces and torques = 0

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Discussion Overview

The discussion centers on the concept of equilibrium in mechanical systems as analyzed through the lens of special relativity. Participants explore whether the conditions for equilibrium, defined as the sum of forces and torques being zero, hold true when viewed from different inertial frames, particularly in relation to Lorentz contraction and the relativity of simultaneity.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants assert that an object is at equilibrium if the sum of forces and torques equals zero, but question whether this holds in a moving frame due to Lorentz contraction.
  • Others argue that if Lorentz contraction implies no time t' exists for equilibrium in the moving frame, it contradicts the ability to transform between inertial frames without introducing acceleration terms.
  • One participant evaluates torques about the origin and suggests that in the moving system, the torque T' may not be zero, indicating potential non-equilibrium.
  • Another participant introduces the idea that the transmission of torque across a beam occurs at finite speed, which could affect equilibrium in different frames.
  • Some participants emphasize that equilibrium must be defined in the context of Lorentz covariance, suggesting that forces and torques should be evaluated at different times in moving frames.
  • There is a mention of a published paper from 1972 that claims mechanical equilibrium in one inertial frame should hold in all inertial frames, although this claim is contested by others who seek clarification and examples.

Areas of Agreement / Disagreement

Participants express disagreement regarding the application of equilibrium conditions across different inertial frames. Some maintain that equilibrium is frame-invariant, while others challenge this view, leading to an unresolved discussion on the implications of special relativity on mechanical equilibrium.

Contextual Notes

Participants note limitations in the discussion, including assumptions about the nature of forces and torques, the dependence on definitions of equilibrium, and the implications of Lorentz transformations in non-inertial frames.

Who May Find This Useful

This discussion may be of interest to those studying mechanics, special relativity, or the foundations of physics, particularly in understanding the complexities of equilibrium in different reference frames.

aranoff
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An object is at equilibrium at a certain time t if the sum of the forces and torques = 0. Look at the object from a moving frame, and apply the equations of special relativity. We find that Lorentz contraction means that in the moving frame (x',t'), there is no time t' such that the object is at equilibrium.

Is the Lorentz contraction valid, or what? Doesn't equilibrium mean that in the moving system we look at the sum of the forces and torques at a certain time t'?
 
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aranoff said:
Look at the object from a moving frame, and apply the equations of special relativity. We find that Lorentz contraction means that in the moving frame (x',t'), there is no time t' such that the object is at equilibrium.

If that were true, you could not transform between inertial frames without an acceleration term cropping up (in which case, they would not be inertial frames to begin with).

It would be helpful if you would show how you applied "the equations of special relativity" to come to this conclusion.

Regards,

Bill
 


Evaluate the torques about the origin. T=x0F2-y0F1 = 0.
In the moving system, y’=y, x’=(1/gamma)x.
Similar equations for forces.
Apply this, and at time t’ in the moving system, T’ is not 0.
 


aranoff said:
Evaluate the torques about the origin. T=x0F2-y0F1 = 0.
In the moving system, y’=y, x’=(1/gamma)x.
Similar equations for forces.
Apply this, and at time t’ in the moving system, T’ is not 0.

Maybe its not in equilibrium even in the frame in which the torque is zero at a certain instant. The torque applied at one end of the beam doesn't 'know' about the torque applied at the other end of the beam immediately, because the torque applied at one end at must 'travel' to the other end of the beam with some finite speed less than or equal to the speed of light.
 


atyy said:
Maybe its not in equilibrium even in the frame in which the torque is zero at a certain instant. The torque applied at one end of the beam doesn't 'know' about the torque applied at the other end of the beam immediately, because the torque applied at one end at must 'travel' to the other end of the beam with some finite speed less than or equal to the speed of light.

Let us apply the principles of physics as they are taught. Equilibrium means that at a time t, the sum of torques = 0 and sum of forces = 0.

You still have not clarified the difficulty. Again: Equilibrium in the rest system. At no time t' is another intertial frame do we have equilibrium.
 


Reply given in a previous thread.
 


aranoff said:
Let us apply the principles of physics as they are taught. Equilibrium means that at a time t, the sum of torques = 0 and sum of forces = 0.

You still have not clarified the difficulty. Again: Equilibrium in the rest system. At no time t' is another intertial frame do we have equilibrium.

Imagine a long lever with a fulcrum at its centre. Twosuitable weights are placed simultaneously (in the rest frame of the fulcrum) at either end, so that the lever does not rotate before or after the weights are placed on it. (The lever may bend a bit).

In a different reference frame another observer see the lever system as moving from left to right and he sees the weight applied to the left hand side of the lever before the other weight is applied to the right hand side. It seems that in his reference frame there is a period when the there is a weight on one end of the lever but not on the other and the system should be out of equilibrium. The lever should be rotating anticlockwise and the free end of the lever on the right should be moving upwards.

It turns out that due to the finite time it takes for the impulse to travel from the left end of the lever to the right end, the weight on the right hand side is applied just in time before the free end can start moving. In this way equilibrium is maintained in all inertial reference frames if there is equilibrium in the rest frame, due the relativity of simultaneity and the finite time for forces to be transmitted.
 


The observer is at the center of mass.
 
  • #10


aranoff said:
The observer is at the center of mass.

Yes, if we restrict to only making measurements and observations in the rest frame of the object under consideration, it brings about a considerable simplification of relativity.
 
  • #11


kev said:
Yes, if we restrict to only making measurements and observations in the rest frame of the object under consideration, it brings about a considerable simplification of relativity.

Now we want to do the same in an inertial frame. An observer is at the center of mass at time t'. At this time, there is no equilibrium.
 
  • #12


aranoff said:
Evaluate the torques about the origin. T=x0F2-y0F1 = 0.
In the moving system, y’=y, x’=(1/gamma)x.
Similar equations for forces.
Apply this, and at time t’ in the moving system, T’ is not 0.

Although it's not quite clear to me what your equilibrium tourque equation describes, I'll assume (for sake of argument) that F1 and F2 are orthogonal, and act at points tangential to a disc (i.e. F2 is in the the y direction acting at (x0,0), and F1 is in the x direction acting at (0,y0) - such that the corresponding torques cancel when x0 and y0 are the same sign). In the case I present, you will note that the net force on the disc is not zero - and the disc will accelerate into the first quadrant due to the net force.

As an accelerating frame, the rest frame of this disk is not an inertial frame (unless F1=F2=0). Thus, the condition of equilibrium torque is insufficient to identify an inertial frame.

Lorentz transformation into a non-inertial (accelerating) frame is invalid.

Regards,

Bill
 
  • #13


The rest frame is an inertial frame, with the forces and torques zero. I wrote all this up in a paper in 1972. It is available on my web site.
 
  • #14


aranoff said:
The rest frame is an inertial frame, with the forces and torques zero.

Unless I'm mistaken, the rest frame you refer to is only an inertial frame relative to frames accelerating at the same rate, and in the same direction.

Regards,

Bill
 
  • #15


To aranoff,

A mechanical system that is in mechanical equilibrium in one inertial reference frame must be in mechanical equilibrium in any other inertial reference frame. From your analysis, you conclude this will not be the case if one uses the Lorentz Transformation to switch reference frames.

Most of the responses you have received so far in this thread are an embarrassment to this forum. May I suggest you post an example that you believe supports you conclusion. Please provide a drwaing and numbers. I'll post a solution in a few days.
 
  • #17


I took a look at your paper. That's exactly why I asked for a single example. I have no interest in plodding through all that.
 
  • #18


My point is that if an object is at equilibrium at a time t in the inertial frame, it will also be at equilibrium in any other moving frame. The definition of equilibrium must take into account Lorentz covariance. This means that in the moving frame, we evaluate the forces at different times based upon the Lorentz transformation. The property of equilibrium is invariant; the times when we evaluate forces and torques are dependent upon Lorentz covariance.
 
  • #19


aranoff said:
I have given drawings and numbers in a published paper in 1972.

I looked at figures 1 and 2 (and some of the discussion that follows), but it is not clear to me that F3 does not contribute to the torque in the transformed case. Would you care to comment on that?

Regards,

Bill
 
  • #20


To aranoff,

I took a closer look at your paper. You are using the standard transfromation for forces and that does imply a problem. I would have used a nonstandard transformation, but since this is a forum for standard SR, your method trumps mine.

So now that I started reading your paper, its pretty interesting. I'll give you more feedback when I finish it.
 
  • #21


Antenna Guy said:
I looked at figures 1 and 2 (and some of the discussion that follows), but it is not clear to me that F3 does not contribute to the torque in the transformed case. Would you care to comment on that?

Regards,

Bill

When we evaluate torques, we must pick an arbitrary point about which to evaluate the torques. The point I chose had the torque due to F3 zero.
 

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