See-saw problem in special relativity

[chaah]

In the middle of a train carriage, there is a light source with a switch. At each end of the carriage, there is a simple apparatus which, when struck by light, releases a ball vertically onto the floor of the carriage. At the floor is a see-saw, with ends vertically beneath the falling balls. So if both balls strike the see-saw ends simultaneously, the see-saw will not tilt, but if they strike the see-saw ends at different times, the see-saw will tilt (if only momentarily).

Apparatus......Light source.....Apparatus
x<---------------------------O--------------------------->x
|...............|
|...............|
|...............|
|...............|
|...............|
|...............|
O..............O
Falling ball............Falling ball

.......See-saw.......
===============================================
........|

A man in the carriage switches on the light and finds that light strikes the ends of the carriage simultaneously, the apparatuses at the ends activate simultaneously, the balls drop to the see-saw simultaneously, and the see-saw doesn't tilt. (Assume that the reference frame of the carriage is inertial.)

But a second man (in a different reference frame) traveling past the carriage at a constant velocity (from left to right, say) will find light to strike one end of the carriage before the other. So the balls don't drop to the see-saw simultaneously, and the see-saw will tilt.

This doesn't look right? Whether the balls strike the ends of the see-saw simultaneously may depend on one's reference frame, but whether or not the see-saw tilts surely cannot depend on one's reference frame? It seems that we must say that the see-saw does not tilt for the second man even though he finds one ball to strike the see-saw first.

Is this the right conclusion to draw? But how can the see-saw fail to tilt if struck by the balls at different times? Perhaps I don't quite understand how a see-saw works!

Or does special relativity allow that whether or not a see-saw tilts can depend on one's reference frame?

 

[Ich]

When you say "the see-saw tilts", you mean that when you push one end, the other end goes up "at the same time". You assume a rigid body.
Rigid bodies violate SR. They don´t exist: every disturbance propagates at sound speed, which is necessarily lower than light speed. After all, the binding forces are of electromagnetic nature.
Therefore, the see-saw will bend rather than tilt, and even this bending will not happen instantaneously over the whole see-saw. The bending will look symmetric in the see-saw´s rest frame, and asymmetric in the other frame.
I guess that a further discussion (which will show explicitly that you can´t use this difference to trigger some event) will be rather complicated. Maybe somebody else can help you.

 

[chaah]

Ok that's cool, you can anyone else help me out a little more?

So the second man will see one end of the see-saw bend down (when the first ball hits) but will not see the other end bend up. (The see-saw will not tilt, in that sense.) The reason being that, when one end goes down, there is a time lag t before the other end can go up, for mechanical reasons, or whatever, and (for this second man) the second ball will hit that other end before this interval runs out.

This what you saying right?

But, in that case, does the first end of the see-saw bend down a little more (e.g. 1cm more) for this second man than for the first man (for whom both balls hit simultaneously?)

It seems like it's got to? If not, I'd be curious to know why. How does a see-saw work?

 

[Ich]

Ok that's cool, you can anyone else help me out a little more?

So the second man will see one end of the see-saw bend down (when the first ball hits) but will not see the other end bend up. (The see-saw will not tilt, in that sense.) The reason being that, when one end goes down, there is a time lag t before the other end can go up, for mechanical reasons, or whatever, and (for this second man) the second ball will hit that other end before this interval runs out.

This what you saying right?

Yes

But, in that case, does the first end of the see-saw bend down a little more (e.g. 1cm more) for this second man than for the first man (for whom both balls hit simultaneously?)

No, it doesn´t. You can calculate the motion of the see-saw in its rest frame and then transform the result to the other frame. 1 cm deflection will remain 1 cm. That´s how it works - Maxwell´s equations, which you would use, are compatible with SR. I´m sure this answer will not satisfy you, but I don´t know enough about it to come up with a better one.

 

[chaah]

Yes that doesn't satisfy me. I was looking for an answer directly in terms of the mechanics of a see-saw. But it helps, thanks, I guess I need to figure out how a see-saw works. There are tons of problems of this sort (which on the surface present a contradiction in SR) but resolve themselves quite nicely when you look at the details of the various contraptions involved (e.g., there's a mechanism which releases a trapdoor, etc., but, aha, how exactly does the trapdoor work?, etc.), but this particular one I haven't been able to get around. I feel like I'm in kindergarten. I don't know how a see-saw works. A standing wave must be set up along the length of the see-saw or something, which travels back and forth along the length of the see-saw ... I'll figure it out, darn.

 

[clj4]

But a second man (in a different reference frame) traveling past the carriage at a constant velocity (from left to right, say) will find light to strike one end of the carriage before the other.

Does he?
Are you sure about that? The light speed is isotropic in ALL reference frames. Are you sure you are not using a misplaced intuition?

 

[Ich]

The light speed is isotropic in ALL reference frames.

That´s why he sees light strike one end of the carriage before the other. Light speed remains constant, but the carriage is moving away from / towards the source.

 

[Ich]

A remark for chaah:
Forget about the mechanics of a see-saw. When you consider effects at this speed, the see-saw is nothing more than a cluster of atoms which try to sweep along each other.

 

[clj4]

That´s why he sees light strike one end of the carriage before the other. Light speed remains constant, but the carriage is moving away from / towards the source.

Ach, Ich

You are giving away the answer and all the fun

This problem is best solved with Mikowski diagrams.
What happens is that, from the perspective of the person watching the carriage , the light reaches the rear end of the car earlier than the front end. This is due to the fact that the rear end of the car rushes towards the light source at speed v while the front end of the car recedes at v. This is called "closing speed". So, the rear ball falls first. When exactly? This is a function of the two closing speeds (c+v) and (c-v).
Now, do the two balls have the same speed? Do they travel the same distance from the point of view of the outside observer? What would it take for them to hit symmetric positions about the center of the see-saw? (after all, they do this in the carriage frame, so they should do this in any other frame). Do we have to transform the coordinates of the see-saw center and endpoints via Lorentz transforms?

Now, suppose that all of the above raises more questions than gives answers: how about if we transformed the problem into an easier one?
Instead of the carriage with the experiment moving with speed +v wrt the experimenter let's use the perfectly equivalent situation where the experimenter moves with speed -v wrt the carriage. What will the experimenter see? Do the two balls get released simultaneously? Do they hit the see-saw simultaneously and symmetrically about the leverage point?

 

[Garth]

Remember simultaneity is relative.

Garth

 

[Ich]

Immer ich!