I The Simultaneity Paradox: Investigating Net Rotation of a Balanced Beam

[e2m2a]

TL;DR Summary

Paradox concerning the effect of two lightning bolts striking a balanced beam

Suppose there is a beam balanced on a fulcrum at its center of mass. One observer at rest with respect to the beam sees two lightning bolts strike the ends of the beam simultaneously, such that there is no net rotation of the beam. However, a moving inertial observer sees one bolt strike one end first and another bolt strike the other end. Why doesn't the moving observer see the beam disturbed from its equilibrium position?

 

[Ibix]

Both observers will note that the beam flexes as a result of the strikes. Both will observe that the shock waves reach the fulcrum simultaneously. Thus both will conclude that (beyond the flexing of the beam) the beam does not fall.

 

[e2m2a]

Another words, you are saying both observers will not see a net torque on the beam?

 

[Ibix]

Another words, you are saying both observers will not see a net torque on the beam?

Different frames are just different descriptions of the same physics. Symmetry in the rod frame requires that the torque around the fulcrum be zero in this frame, and this is a direct observable so there will be zero torque at the fulcrum in all frames, yes. The flexing of the rod will be asymmetric in the moving frame, though.

 

[Ibix]

The point is that both frames will see the beam disturbed from equilibrium. In both frames its ends will be pushed down by the strikes, and this will start waves of distortion propagating along the rod at the speed of sound. But the speed of sound is direction dependent in the moving frame, and this will conspire with the time between the strikes in this frame and the motion of the fulcrum so that the waves arrive at the fulcrum simultaneously. This means that there is no net rotation at the fulcrum.

The rod is not, and cannot be, a rigid object. It will be disturbed from equilibrium by being made to flex. But, given your setup, this cannot lead to tipping at the fulcrum.

 

[PeroK]

Summary:: Paradox concerning the effect of two lightning bolts striking a balanced beam

Suppose there is a beam balanced on a fulcrum at its center of mass. One observer at rest with respect to the beam sees two lightning bolts strike the ends of the beam simultaneously, such that there is no net rotation of the beam. However, a moving inertial observer sees one bolt strike one end first and another bolt strike the other end. Why doesn't the moving observer see the beam disturbed from its equilibrium position?

This is a good question. It's also worth pointing out that for a normal sized beam the lack of simultaneity will be small, even in a frame where the beam is moving at near light speed. At most a time lapse of $L/c$, where $L$ is the length of the rod. For a $30m$ beam this would be about $10^{-7}s$.

In other words, the loss of simultaneity is limited. You can't have a situation where in one reference frame there is a $1s$ delay, say, between the ends being loaded. Not unless the beam itself is $1$ light second long, which is, of course, $3 \times 10^8 m$.

In fact, if you wanted to do this experiment, you might ask how easy it would be to load both ends of the beam to within $10^{-7}s$ of each other in the first place?. I.e. in practical terms, for a $30m$ beam this lack of simultaneity would be within the bounds of experimental error!

That said, it still needs a theoretical explanation, which @Ibix has provided.

 

[jartsa]

So, let's say that in our frame a beam moves in space and gets hit at its both ends by dust particles. If the sound waves generated by the hits arrive at the middle of the beam at the same time, and if the middle point receives a hit from a dust particle at that time, and the momentum of the last hit cancels out the momentum of the two other hits, then the beam does not tilt in our frame. Right?The beam is rotating during the short time between the two hits. Which causes a small tilt. Which is then somehow corrected by the third hit.(The dust particles hit the beam the same way as the lightning hits the beam in post #1, which is in that way that a vertical lightning hits a horizontal beam, I suppose. And the third particle does the same job as the fulcrum, which has an infinite mass, or is bolted to ground, I guess)

 

[PeroK]

So, let's say that in our frame a beam moves in space and gets hit at its both ends by dust particles. If the sound waves generated by the hits arrive at the middle of the beam at the same time, and if the middle point receives a hit from a dust particle at that time, and the momentum of the last hit cancels out the momentum of the two other hits, then the beam does not tilt in our frame. Right?The beam is rotating during the short time between the two hits. Which causes a small tilt. Which is then somehow corrected by the third hit.

You could model the beam as a set of simple harmonic oscillators. In the frame where the beam is moving, the oscillators will be out of sync, compared to the rest frame of the beam. The oscillations, however, will be less out of sync close to the middle, and the middle point will be influenced equally from both sides.

Perhaps a better illustration is to imagine the beam as an elastic spring that gets pulled from both ends simultaneously in its rest frame.

The waves on either side will be at asynchronous in the moving frame, but the centre of the spring will not accelerate in either frame.

 

[Janus]

Another words, you are saying both observers will not see a net torque on the beam?

As @Ibix has noted, the eand of each beam will slightly flex under the impact. The flex at from each end will travel along the beam at the speed of sound. In the bar beam frame, this means they leave at the same time, travel at the same speed, meet at the midpoint at the same time, and cancel each other out.
In the moving inertial observer's frame, you have to apply relativistic addition of velocities to the moving flexes.
If v is the relative velocity between the observer and the beam and s the velocity of the flex relative to the beam as measured from the beam's frame then, If the relative motion of the beam as measured by the observer is left to right, then:
The velocity of the flex moving starting at the left end will be # \frac{ (v+s)}{1+\frac{vs}{c^2}}# relative to the observer, and this minus v with respect to the beam
and
The velocity of the flex moving starting at the right end will be # \frac{ (v-s)}{1-\frac{vs}{c^2}}# relative to the observer, and this minus v with respect to the beam

So, for example if v= 0.8c and s where 0.01c, then the observer would measure the left to right flex as moving at 0.80357c relative to himself and 0.00357 c with respect to the beam. He would measure the right to left flex as moving at 0.7964c relative to himself and -0.00363c with respect to the beam ( the minus indicates right to left motion vs. left to right) Note that with respect to the beam, the left to right flex travels slower with respect to the beam. According to the inertial observer, due to relativity of simultaneity, the Left flex started earlier.
It is the combination of the left flex having a lower speed relative to the beam and starting sooner that results in it arriving at the midpoint at the same time as the right flex. Ergo, both the beam and moving observer agree that the effects of the impacts meet at the center of the beam and cancel each other out in terms of net effect on the balance of the beam.

 

[Ibix]

the beam does not tilt in our frame. Right?

The problem is that "tilt" is too simplistic. You can get away with it in Newtonian mechanics because rigid rods are fine. And in relativistic mechanics you can often gloss over the issues (as @PeroK points out).

But the reality is that the rod is flexing. Is a C-shaped rod tilting? Is a J-shaped one tilting? Depending on your answer to that, you may conclude that the rod does tilt for some time during its reaction to the collision. But if the shockwaves from identical impacts meet in the middle of the uniform rod then (once the vibration has settled down) it will not be rotating, no.

 

[jartsa]

The problem is that "tilt" is too simplistic. You can get away with it in Newtonian mechanics because rigid rods are fine. And in relativistic mechanics you can often gloss over the issues (as @PeroK points out).

But the reality is that the rod is flexing. Is a C-shaped rod tilting? Is a J-shaped one tilting? Depending on your answer to that, you may conclude that the rod does tilt for some time during its reaction to the collision. But if the shockwaves from identical impacts meet in the middle of the uniform rod then (once the vibration has settled down) it will not be rotating, no.

In this video we can see a triangle-shaped spaceship, seen from a frame where it moves to the right, turning all its engines on, simultaneously in its own frame, there are a large number of rocket motors at the bottom of the ship.

We can see that the rockets give the ship angular momentum.

A single rocket at the middle would do the same thing. Right? Well, I think the video is correct, and my common sense says that a spaceship with rocket at the middle flies like a speceship with rocket at each corner.

This is related to the current problem in many ways, like for example a rocket at the center pushes the ship like the fulcrum pushes the beam.

 

[PeroK]

In this video we can see a triangle-shaped spaceship, seen from a frame where it moves to the right, turning all its engines on, simultaneously in its own frame, there are a large number of rocket motors at the bottom of the ship.

We can see that the rockets give the ship angular momentum.

A single rocket at the middle would do the same thing. Right? Well, I think the video is correct, and my common sense says that a spaceship with rocket at the middle flies like a speceship with rocket at each corner.

This is related to the current problem in many ways, like for example a rocket at the center pushes the ship like the fulcrum pushes the beam.

That's called a Wigner rotation. The difference between that and the current scenario is that the centre of the beam, in its own rest frame, remains at rest.

 

[Ibix]

We can see that the rockets give the ship angular momentum

I don't think so. The ship is under thrust the whole time. If it turns off the engine it will remain (kind of) rotated, but will not continue to rotate. If it was gaining angular momentum it would continue to rotate.

Edit: note that your original scenario stopped the rod by cancelling the momentum from the end strikes with a middle strike. The same applies to the balanced beam, but not to the rocket.

 

[Ibix]

A single rocket at the middle would do the same thing. Right?

Depends how precise you want to be. If the rocket is rigid enough that it doesn't noticeably distort in its rest frame, then yes. Otherwise, no, because the back corners of the rocket won't be in the same place as those of the rocket in the video.

 

[pervect]

I haven't read this thread in detail, but the Lewis-Tolman "lever paradox" might be relevant here.

See for instance https://aapt.scitation.org/doi/abs/10.1119/1.1976326?journalCode=ajp

(This isn't the original LT paper, it's a quote from the abstract of another paper on the topic).

When a rigid right-angled lever experiences equal and opposite torques as measured by an observer O′ at rest with respect to the lever in frame S′, the usual transformation of these torques to another Lorentz frame S makes it appear that the torques are not balanced for an observer 0 at rest in S. This paradox was first pointed out by Lewis and Tolman. The consensus among textbook authors is that a net torque does exist on the lever according to the general Lorentz observer.

So, my take-away from this is that the concept of balanced torques is generally considered to be frame-dependent.

 

[Ibix]

So, my take-away from this is that the concept of balanced torques is generally considered to be frame-dependent

But that abstract goes on to say

The basic approach herein is to construct for observer O a four-tensor $I_{\mu\nu}$ to represent the relative four-torque about a point P′ fixed in frame S′ which has a constant speed ν = βc along the +x axis of S. For some unknown reason, this tensor seems never to have been defined before. The space part of $I_{\mu\nu}$ reduces to the corresponding classical definition of torque in three-space if ν = 0. Based on $I_{\mu\nu}$ a comparison of the torque components for two different Lorentz observers shows that no torque exists on the square or on the lever for any Lorentz observer if no net torque exists for the observer at rest with respect to the square or lever.

That suggests that Butler, at least, thinks the previous approaches were incorrect - or, at least, that there exists a covariant approach that previous authors did not use. Unfortunately I can't download the paper...

 

[pervect]

But that abstract goes on to say

That suggests that Butler, at least, thinks the previous approaches were incorrect - or, at least, that there exists a covariant approach that previous authors did not use. Unfortunately I can't download the paper...

I haven't really studied Butler's paper, to be honest. I'd need more than the abstract to really understand his approach, but it appears to me as if he's suggesting solving the problem of torque being frame-dependent by redefining it. He only claims that his tensor matches the usual definition of torque when v=0. When v is not zero, presumably it doesn't match.

I'm just pointing out that it's known that with standard definitions, torque is frame dependent. And I can suggest some useful keywords to start researching the topic, the "Lewis-Tollman paradox". This isn't something I've taken a great deal of time to study, but I know it exists.

Whether or not Butler's solution is the best solution, I can't really say. Certainly for communication purposes, it's probably better to stick with the standard textbook approach, and Butler pretty much states that his approach isn't the standard textbook approach. But it does look interesting. The meta point is that the issue the OP is raising is a real one, and that it has been discussed in the literature - while the scenarios discussed in the literature aren't exactly the same as what the OP is interested in, the motivation is similar.

 

[PeroK]

In this video we can see a triangle-shaped spaceship, seen from a frame where it moves to the right, turning all its engines on, simultaneously in its own frame, there are a large number of rocket motors at the bottom of the ship.

We can see that the rockets give the ship angular momentum.

A single rocket at the middle would do the same thing. Right? Well, I think the video is correct, and my common sense says that a spaceship with rocket at the middle flies like a speceship with rocket at each corner.

This is related to the current problem in many ways, like for example a rocket at the center pushes the ship like the fulcrum pushes the beam.

Another answer.

In the rocket scenario there are three frames involved. The horizontally moving frame; the original rest frame of the rocket; and, the rocket frame after acceleration in the vertical direction.

In the beam scenario, there are only two frames: the original rest frame of the beam and the horizontally moving frame. These remain related throughout by the same Lorentz transformation, regardless of the motion of the beam. If the beam ends up at an angle in the moving frame, then it must end up at an angle in the orignal rest frame and vice versa. Albeit at a different angle in each.

This is the heart of the original poster's question. The beam cannot rotate in one frame and not in the other. A y-coordinate in one frame transforms to the same y-coordinate in the other frame, using the Lorentz transformation. Only the synchronicity of movements in the y-direction are different in the two frames

 

[jartsa]

Another answer.

In the rocket scenario there are three frames involved. The horizontally moving frame; the original rest frame of the rocket; and, the rocket frame after acceleration in the vertical direction.

In the beam scenario, there are only two frames: the original rest frame of the beam and the horizontally moving frame. These remain related throughout by the same Lorentz transformation, regardless of the motion of the beam. If the beam ends up at an angle in the moving frame, then it must end up at an angle in the orignal rest frame and vice versa. Albeit at a different angle in each.

This is the heart of the original poster's question. The beam cannot rotate in one frame and not in the other. A y-coordinate in one frame transforms to the same y-coordinate in the other frame, using the Lorentz transformation. Only the synchronicity of movements in the y-direction are different in the two frames

Funny, I was planning to tell how the video and post #1 are similar, if we use our imagination.

The circle in the video is a fulcrum floating in space. The stuff coming off the spaceship consists of small particles flying off as the bottom of the ship is hit by lightnings.

The lightnings accelerate and tilt the ship. Later the force exerted by the fulcrum accelerates and tilts the ship back to the original orientation.

In some earlier scenario the circle was the ship's destination planet. Now it's a fulcrum. Don't ask how there can be lightnings in space, I have no idea.
By the way, I understand that the beam cannot become tilted in one frame and not in the other. Except temporarily.

This scenario seems quite clear to me, even the question whether the beam has angular momentum seems clear. it has, temporarily, in some frames. There is just a small problem: Where does that angular momentum go at the end?

 

[jartsa]

Maybe I could try to write a complete story based on post #1So, there's a horizontal beam moving to the right.
A vertical lightning from above strikes on the beam's left end.
That gives the beam downwards momentum.
As the downwards momentum is on the left side, the beam has angular momentum.
As the left side is moving downwards, while the right side isn't, the beam is turning.
As the downwards momentum starts to spread into the beam, beam's angular momentum starts decreasing.
A vertical lightning from above strikes on the beam's right end.
That gives the beam new downwards momentum.
As the fresh downwards momentum is on the right side, the beam has almost no angular momentum now.
After all downwards momentum has left the beam, it's tilted and has no angular momentum.
The beam is now bent, as it just collided with the fulcrrum. So the fulcrum exerts an upwards force on the beam.
The beam's middle point receives upwards momentum, which spreads into the rest of the beam.
As the upwards momentum travels faster to the left than to the right, the beam is gaining angular momentum and starting to turn.
The beam tilts back to horizontal direction, overshoots, starts going back ... and so on.
In this story "beam's angular momentum" does not include the angular momentum the beam has because of the motion of the beam's center of mass.

 

[Karl Coryat]

Suppose, instead of a beam flexing due to imparted momentum, the lightning strikes enter the ends of a fiber optic cable, at the center of which are two detectors very close together, or equidistant from the ends and staggered a smidge. Do both detectors click at the same time in all frames, which seems to contradict light-speed invariance? Or do moving observers see the detectors clicking at different times?

 

[Ibix]

Do both detectors click at the same time in all frames,

Yes (assuming we can treat the two detectors as being in the same place). If they are in the same place and click at the same time in one frame then they must do so in all.

which seems to contradict light-speed invariance

Light speed in vacuum is invariant. And in any case, in a moving frame, the light does not travel the same distance along each optical fiber, since the "middle" is moving forwards. And the flashes did not occur simultaneously. The difference in their timing will be compensated for by the difference in travel times due to the different distances.

 

[PeterDonis]

Do both detectors click at the same time in all frames, which seems to contradict light-speed invariance?

If two co-located detectors click at the same time in one frame, they will click at the same time in all frames. But the times at which the lightning strikes enter the fiber optic cables at the two ends will be different in different frames. So there is no contradiction with the invariance of the speed of light.

 

[Karl Coryat]

in a moving frame, the light does not travel the same distance along each optical fiber, since the "middle" is moving forwards.

I don't get this. Are the relative distances from the detectors to the ends of the cable frame-dependent?

 

[Ibix]

I don't get this. Are the relative distances from the detectors to the ends of the cable frame-dependent?

No. But the cable is moving, so the midpoint is running away from one flash and running to meet the other. So where the flashes meet (in the moving frame) ends up being further from where one flash entered the cable than the other, despite both halves of the cable being the same length.

 

[Ibix]

You can see this in Newtonian physics. Imagine you have a 2m long channel. You simultaneously insert two marbles from each end, rolling at 1m/s. They collide in the middle 1s later.

Now imagine that you walk past the apparatus at 1m/s. One of the marbles is stationary in your frame, and the other is doing 2m/s. They still collide in the middle of the channel 1s later, but one marble didn't move and the other came all the way to it.

The effect is qualitatively similar in relativity. It's just made more complicated by the less trivial velocity addition rule and the relativity of simultaneity.

 

[Karl Coryat]

I'm terribly sorry, but I'm still stuck.

One of the marbles is stationary in your frame

But we're talking about light. I assume we won't observe the light from the two flashes as having different speeds, as we do with the marbles.

So where the flashes meet (in the moving frame) ends up being further from where one flash entered the cable than the other

How can we then meaningfully say one cable isn't longer than the other in our moving frame? Assuming, that is, that the flashes still meet at the detectors — which I assume is a necessary condition for the detectors to click together.

What assumption am I getting wrong?

 

[Janus]

Suppose, instead of a beam flexing due to imparted momentum, the lightning strikes enter the ends of a fiber optic cable, at the center of which are two detectors very close together, or equidistant from the ends and staggered a smidge. Do both detectors click at the same time in all frames, which seems to contradict light-speed invariance? Or do moving observers see the detectors clicking at different times?

Light travels at ~ 0.66c through a fiber optic cable. The speed of light is only invariant in a vacuum.
Light traveling through the fiber optic cable would be subject to the relativistic addition of velocities.
So for example, if the fiber optic cable is moving left to right at 0.5c relative to some observer, he would measure the light leaving the left end of the cable as moving at (0.5c+0.66c)?(1+ 0.5c(0.66c)?c^2) = ~0.87c to the right relative to himself and ~0.37 c to the right relative to the cable.
Light leaving the right end would be measured as moving at ~0.24c to the left relative to himself and ~0.74c to the left relative to the cable.
If the strikes were simultaneous in the fiber frame, then, in this external observer frame, the lightning strikes the left end first and then the right end. The light traveling along the cable takes longer according to him to reach the center of the cable if it starts at the left end then it does if it starts at the right end. This difference in travel time will exactly cancel out the head start the light starting to the left had, and the pulses reach the center of the cable according to our external observer just like it does in the cable frame.

 

[Karl Coryat]

Light travels at ~ 0.66c through a fiber optic cable.

My mistake. Let's take that out. Mirrors rather than the ends of cables. Nothing else changes. I'm still missing something.

 

[PeterDonis]

where the flashes meet (in the moving frame) ends up being further from where one flash entered the cable than the other

This might be confusing to the OP as it is stated. The point is that "where the flashes entered the cable" is frame-dependent. In the cable's rest frame, the points "where the flashes entered the cable" are at rest relative to the cable. In the moving frame, those points are moving relative to the cable.

Making this concrete is one reason why it helps to include the embankment in the classic Einstein train thought experiment. If you think of the cable as like the train in that experiment, and the points "where the flashes entered the cable" in the frame in which the cable is moving as being points on the embankment, corresponding to the points where the lightning strikes hit the embankment in Einstein's version, it's easier to see how the point where the flashes meet, in the frame in which the cable is moving, is not equidistant from the points where the flashes entered the cable.

How can we then meaningfully say one cable isn't longer than the other in our moving frame?

See my response to @Ibix just above.

 

[Karl Coryat]

the point where the flashes meet, in the frame in which the cable is moving, is not equidistant from the points where the flashes entered the cable.

So I must be wrong in my assumption that such equidistance is a condition of the detectors clicking together.

 

[Janus]

My mistake. Let's take that out. Mirrors rather than the ends of cables. Nothing else changes. I'm still missing something.

Then you just have a variation of Einstein's train experiment. Light hits the mirrors. In the frame in which the mirrors are at rest, the light leaves the Mirrors at c, relative to the mirrors in both directions and meet at the center. In the frame where the Mirrors of moving to the right, the light travels at c relative to this frame, and hits/leaves the left mirror first. If the mirrors are moving at 0.5c, then the left mirror's light has a speed of 0.5c with respect to the mirrors and the right mirror's light has a speed of 1.5c with respect to the Mirror.
Same result. Light leave the left mirror first and takes longer to "catch up" to the center than the later leaving right mirror light, and they meet at the midpoint between mirrors.

 

[Karl Coryat]

I guess my faulty assumption was that 0.5c and 1.5c (with respect to moving objects) were not things that could exist in this universe. But, if I were moving along with these mirrors, my understanding is that I would not measure the speed of light at those values, is that correct?

 

[PeterDonis]

So I must be wrong in my assumption that such equidistance is a condition of the detectors clicking together.

More precisely, you were wrong in your assumption that "equidistance" is a frame-invariant condition. It isn't. You would need to specify the condition as equidistance in the cable's rest frame.

 

[PeterDonis]

If the mirrors are moving at 0.5c, then the left mirror's light has a speed of 0.5c with respect to the mirrors and the right mirror's light has a speed of 1.5c with respect to the Mirror.

This is a very misleading way of stating it; the "speeds" you describe here are not speeds in the usual sense of the term and are not subject to the rule that speeds cannot exceed the speed of light. Nor do they obey the relativistic velocity addition rule. Nor are they what an observer, in any frame, would actually measure as the speeds of the light beams--those would all be measured to move at c.

I don't know if there is a standard term in relativity for the things you are calling "speeds", but I would call them "rate differences"--they are what you get if you just subtract the mirror speed from the light speed, paying appropriate attention to signs, and now using "speed" in its correct relativistic sense, as the speeds that would actually be measured. So the left mirror and the light are moving in the same direction, thus their speeds have the same sign and the rate difference is $1 - 0.5 = 0.5$, while the right mirror and the light are moving in opposite directions, thus their speeds have opposite signs and the rate difference is $1 - (- 0.5) = 1.5$.

I guess my faulty assumption was that 0.5c and 1.5c (with respect to moving objects) were not things that could exist in this universe.

The things @Janus called "speeds" are not properly called that. See above.

 

[Karl Coryat]

Thanks, guys. That was very enlightening.

 

[Ibix]

Not to scale, but here's a quick sketch of @Karl Coryat's two fibers connected to a black box detector, made at three different times in a frame where the apparatus is moving. The fibers are of equal length.

In the top diagram a pulse of light (coloured red) enters one end of the apparatus. In the second diagram the red pulse has traveled some way along the fiber, and the other pulse (coloured blue) enters the other end of the apparatus. In the final diagram both pulses arrive at the detector simultaneously. I've added vertical grey lines showing the points in space where the light pulses entered the fibers and where they meet at the detector. You can see that the distances are very different, although the distances from detector to fiber end are equal at all times. The point is that, in a frame where the apparatus is moving, "where the light entered the fiber" and "where the fiber end is now" are two different things in general.

Since this isn't drawn to scale, the same diagram works just as well for the "light moving in free space" variant.