# Homework Help: Special Relativity: Force applied to a bar

1. Oct 10, 2012

### Fre4k

1. The problem statement, all variables and given/known data

Two forces of equal magnitude are applied simultaneously on the ends of a bar of lenght L. In another reference frame, moving with velocity v relative to the bar (with v being parallel to the lenght L), the forces are not applied at the same time, generating momentum. I have to analyze what happens to the energy and momentum of the bar on both reference frames in two cases:
1. Considering the bar is rigid.
2. The bar is not rigid.

2. Relevant equations

Special Relativity formulas
F=kx

3. The attempt at a solution

I was able to do it for the rigid bar, but not the other case. This is what I did:
I considered F = k . dl (dl being the deformation of the bar). The bar gained an energy equal to kdl²/2, or ΔE = Fdl/2. I don't know what to do now.

Also, I'm sorry for my crappy english. I tried to explain it the best I could.

2. Oct 10, 2012

### Fre4k

Up.

The thing is, I can't see how the deformation would affect the bar in the moving reference frame.

3. Oct 12, 2012

### Staff: Mentor

The first thing to do is to solve the problem for the reference frame in which the bar is initially at rest. Why? Because if you can't solve it in that reference frame, you certainly won't be able to do it from the perspective of observers in the moving reference frame. I suggest that you do the problem for the case of the deformable bar first. That will give you some idea of what would be happening with a rigid bar. You need to solve for the kinematics of the deformation, as expressed as displacement as a function of time and position along the bar. This is not an easy problem for the uninitiated to solve.

4. Oct 15, 2012

### Fre4k

Actually, I was able to do it for the rigid bar, it's the deformed one that I'm having difficulty with. I found that the bar gained ΔE = Fdl/2, but I can't see how this energy transforms.

5. Oct 16, 2012

### Staff: Mentor

In the rigid bar example, if the two forces are equal, as stated, there should be no motion, and no energy gained. The bar remains in static equilibrium. The real issue is, what happens in the case where the system is being observed from a moving inertial reference frame, where the two forces would be reckoned to not be applied simultaneously? To help you figure this out, whoever posed the problem is trying to get you to consider the case where the bar is deformable, so that there are some kinematics and dynamics involved as reckoned from the rest reference frame. These kinematics can then be translated into the moving frame using the Lorentz Transformation. You can then look at these results, and extrapolate them to the case of very high Young's modulus to sneak up on the rigid bar solution. But the first step is solving for the dynamic deformation of the bar in its original rest reference frame, which can be accomplished using Newtonian mechanics (since the velocities are far below relativistic). You need to do a transient force balance on each differential section of the bar along its length, and include the ma term for each differential section. This should give you a partial differential equation (essentially the wave equation) involving the speed of sound in the bar (related to Young's modulus divided by the density). This should give you what you need to calculate the kinematics of the transient deformation of the bar. What happens to the bar is similar to how a golf ball deforms during contact with the club.

6. Oct 18, 2012

### Oxvillian

There is no such thing as a rigid bar in special relativity!

7. Oct 18, 2012

### Staff: Mentor

Thank you for parroting that back for us. Why do you think the OP's professor gave him this homework problem? It would have been much easier for the professor just to say that there is no such thing as a rigid bar in special relativity, and, just like you, be done with it. Would you care to analyze the problem the professor posed for the OP and use it to help explain the reasons why there is no such thing as a rigid bar is special relativity? Are you capable of articulating the reasons why by another method (of course, in your own words)?

8. Oct 19, 2012

### Oxvillian

Sorry for giving away the punchline!
I think the easiest way to see this is to notice that if you bang one end of a rigid bar, our usual understanding of the word "rigid" requires the entire bar, all the way to the other end, to respond instantaneously. Therefore the signal from the banged end telling the opposite end to start moving must be propagated instantaneously. That's inconsistent with special relativity because no signal can travel faster than light.

Therefore in light of SR we are forced to abandon the usual definition of a rigid body - or at least alter it drastically.

So part (1) of the OP's problem is rather like asking about the dating habits of a married bachelor...