What Does the Right End of an Oscillating Pen Look Like in Relativistic Motion?

[Marius311]

Imagine a pen or something of proper length L. In 2D (1 space, 1 time), if the pen is osciallating so that, to an observer at rest at x=0, the left end traces out x(t)=Sin(.7t), what will the right end look like?
This question goes along with a program that I'm writing that you can find at http://www.people.cornell.edu/pages/mm473/" in Applet form.
My first thought was to just apply the length contraction forumla (L/gamma) across the function so the right end would be x_right(t)=x(t)+L*Sqrt(1-x'(t)^2). This is the function you can see on the right. It can't be right because, for one, it goes faster than light at points.
If you right-click on the Sine function and click Traverse, the program will animate what it would look like going along the worldline of the left end of the "pen". I expected the right end of the pen to be a constant length away as viewed from this frame, and you can see as it is traversed a constant length is traced out. This however doubles back on itself (impossible). This I now realize is not true because the pen is not an inertial frame. So the length changes as viewed from an outside stationary observer, and from the non-intertial frame of the pen, but still... what does it look like?

 

[Spin_Network]

Imagine a pen or something of proper length L. In 2D (1 space, 1 time), if the pen is osciallating so that, to an observer at rest at x=0, the left end traces out x(t)=Sin(.7t), what will the right end look like?
This question goes along with a program that I'm writing that you can find at http://www.people.cornell.edu/pages/mm473/" in Applet form.
My first thought was to just apply the length contraction forumla (L/gamma) across the function so the right end would be x_right(t)=x(t)+L*Sqrt(1-x'(t)^2). This is the function you can see on the right. It can't be right because, for one, it goes faster than light at points.
If you right-click on the Sine function and click Traverse, the program will animate what it would look like going along the worldline of the left end of the "pen". I expected the right end of the pen to be a constant length away as viewed from this frame, and you can see as it is traversed a constant length is traced out. This however doubles back on itself (impossible). This I now realize is not true because the pen is not an inertial frame. So the length changes as viewed from an outside stationary observer, and from the non-intertial frame of the pen, but still... what does it look like?

Interesting, but the applet does not work for me, it may be my end? I get a... java.lang.UnsupportedClassVersionError:msketch2AppletMain (Unsuported major.minor version 49.0

but to get back to what I think your conveying, the 2-D "pen" should look like it is "reflected" about a rotation, in all directions?

 

[Marius311]

http://www.java.com/en/download/index.jsp" may help with the applet.

I'm not sure if this is what you mean, but when I say what does it "look" like I don't mean what will your eyes see. I mean, what is the world line of the other end. Although it is certainly some rotation in space-time, I hope that this problem involves only special, and not general relativity.

 

[Ich]

The problem is tricky: You can´t assume the pen to be rigid, as the information cannot go faster than c. Therefore the right end will not react until after a time lag when you accelerate the left end. Your pen will behave like a spring, being compressed or elongated even in its own momentary rest frame.
Unfortunately, I have no idea how exactly it would move. I even don´t know if the problem is stated exactly enough. Maybe others can help.

 

[Marius311]

Oh yes, I remember the gigantic scissors though-experiment. So as not to get too complicated, assume infinite-stiffness.

 

[selfAdjoint]

Oh yes, I remember the gigantic scissors though-experiment. So as not to get too complicated, assume infinite-stiffness.

Ich's point was that if you assume infinite stiffness you have already violated relativity, so you can't expect to get correct realtivistic results.

 

[Marius311]

Hm... so if it wasn't for the non-infinite stiffness, would the pen remain constant length as viewed from it's frame? Or would it still changed due to other effects?

 

[pervect]

You can model a typical bar as a distributed spring-mass system. The typical solution you get when you model a bar in this manner is a system that obeys the wave equation, with the waves propagating at the velocity of sound.

Relativity restricts this solution so that the speed of sound in the material can't exceed the speed of light. (I think that the "wave equation" model is still a reasonable model, though I'm not 100% sure about this point - it's a reasonble place to start, though).

One idealization of the ultimately rigid bar would be to have a bar where the speed of sound in the material was equal to 'c'. This isn't particularly realistic with actual solid bars out of normal matter, but you can make your "object" out of beams of light. The speed of light is 'c', and light obeys the wave equation, so if you imagine "less rigid" bars as obeying the wave equation with v_sound << c, with the "most rigid bar physically possible" as obeying the wave equation with v_sound = c, you'll have a reasonable start on answering your question.

 

[pmb_phy]

One way to solve the problem is to assume something less than maximum stiffness. This means that any compression on one end the sound wave would move at less than the speed of light.

Pete