# Sum of the forces and torques = 0

1. Aug 18, 2008

### aranoff

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'?

2. Aug 19, 2008

### Antenna Guy

Re: 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

3. Aug 19, 2008

### aranoff

Re: Equilibrium

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.

4. Aug 19, 2008

### atyy

Re: Equilibrium

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.

5. Aug 19, 2008

### Ich

Last edited by a moderator: Apr 23, 2017
6. Aug 19, 2008

### aranoff

Re: Equilibrium

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.

7. Aug 19, 2008

### aranoff

Re: Equilibrium

8. Aug 19, 2008

### yuiop

Re: 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.

9. Aug 19, 2008

### aranoff

Re: Equilibrium

The observer is at the center of mass.

10. Aug 19, 2008

### yuiop

Re: Equilibrium

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. Aug 19, 2008

### aranoff

Re: Equilibrium

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. Aug 20, 2008

### Antenna Guy

Re: Equilibrium

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. Aug 20, 2008

### aranoff

Re: Equilibrium

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. Aug 20, 2008

### Antenna Guy

Re: Equilibrium

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. Aug 21, 2008

### MikeLizzi

Re: Equilibrium

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.

16. Aug 21, 2008

### aranoff

17. Aug 21, 2008

### MikeLizzi

Re: Equilibrium

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. Aug 21, 2008

### aranoff

Re: Equilibrium

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. Aug 21, 2008

### Antenna Guy

Re: Equilibrium

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. Aug 21, 2008

### MikeLizzi

Re: Equilibrium

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. Aug 22, 2008

### aranoff

Re: Equilibrium

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.