Length contraction of a pair of electrons

[DaveC426913]

TL;DR Summary

If we fire two electrons at the same time, one metre apart, should they length contract till they're less than a metre apart?

I've managed to stump myself in attempting to answer a member on another forum.

He is attempting to demonstrate length contraction using two electrons a known distance apart and moving at a known velocity.My thought experiment is based on his proposal; that's why I've chosen the components I have.

We have two parallel electron guns. G_a and G_b. G_b is one metre behind G_a and sightly offset to one side. Both are dialed down so they fire only one electron at a time.

We have two detectors, D_a and D_b, each 10 metres away from its gun.

To sum: , we have two identical setups, S_a, and S_b, with S_b just happening to be displaced by one metre along the direction of travel.We fire both guns simultaneously. The electrons are fired with enough energy to achieve relativistic velocities, say, 0.87c.

Electron_a and Electron_a leave their gun at 0.87c, and travel to their detector 10m away.

The detectors should light up at exactly the same time. Right?
OK, now I add a bit of magic. I use a piece of nigh-infinitely thin, nigh-infinitely light wire "tied" to both electrons. We now have a stick that is one metre long. We repeat the experiment: and both ends of the stick have been accelerated to .87c.

The stick will be length contracted by a factor of 2. It will be observed as only .5m in length.

Meaning when it arrives 10m away and the detectors detect their electrons, they will not light up simultaneously, since the detectors are 1m apart.

Why the difference?

I am certain it has to do with the initial acceleration and relativity of simultaneity, but I'm stumped.

 

[PAllen]

Please look up, starting with our FAQ, the Bell Spaceship paradox. You just have an over complicated version of it.

https://www.physicsforums.com/insights/what-is-the-bell-spaceship-paradox-and-how-is-it-resolved/

 

[DaveC426913]

Please look up, starting with our FAQ, the Bell Spaceship paradox. You just have an over complicated version of it.

I guess its been a while since I've visited the Spaceship Paradox. so I didn't think of it.

Still, I'm not sure how exactly it solves my quandary (I guess I'm feeling a bit obtuse today).

Am I correct that, in the first case, both detectors will flash simultaneously, whereas in the second they will not?

 

[Nugatory]

We fire both guns simultaneously...The detectors should light up at exactly the same time. Right?

"At the same time" using which frame? Most likely you mean the frame in which the emitters and detectors are at rest. If, however, you choose to use the frame in which the electrons are at rest after they've been fired (so the emitters and detectors are moving backwards at .87c, and were even before the emitters fired, and will be even after the detection events) the two emission events did not happen at the same time.

OK, now I add a bit of magic. I use a piece of nigh-infinitely thin, nigh-infinitely light wire "tied" to both electrons. We now have a stick that is one metre long. We repeat the experiment: and both ends of the stick have been accelerated to .87c.

This is a variant of Bell's spaceship paradox. Using the frame in which the electrons are at rest after they've been fired, the two emission events do not happen at the same time; the lead electron is fired first so the string stretches and breaks as the lead electron starts moving while the trailing electron stays put. Using the frame in which the emitters and detectors are at rest, the distance between the two electrons doesn't change but the string is length contracted so is stretched and breaks. If the string breaks before it's exerted enough force to change the trajectory of either electron we'll get the same simultaneous (using the frame in which the t and emitters are at rest) triggering of the detectors that we had without the string.

Or you can assume that the string is infinitely strong so will neither stretch or break (a completely unphysical assumption, but let's go with it). In this case, the distance between the two electrons is constrained to be one meter in the frame in which the string and electrons are at rest and the detectors and emitters are moving backwards at .87c. Using this frame, the infinitely strong unbreakable string will tug on the trailing electron and start it moving as soon as the lead electron is fired and before the trailing electron gun has kicked the trailing electron. Conversely, using the frame in which the detectors and emitters are at rest, we will say that the string length-contracts and pulls the two electrons closer so that their separation is no longer one meter. Either way, we have the string exerting a force on the two electrons so that they do not remain one meter apart using the frame in which the detectors and emitters are at rest, and the detectors do not trigger simultaneously.

Google for "Bell's spaceship paradox" and look at the many threads here about that problem.

 

[Nugatory]

Am I correct that, in the first case, both detectors will flash simultaneously...

Using which frame? You've seen enough threads here to know better than to say "simultaneously" without specifying a frame

 

[PAllen]

I guess its been a while since I've visited the Spaceship Paradox. so I didn't think of it.

Still, I'm not sure how exactly it solves my quandary (I guess I'm feeling a bit obtuse today).

Am I correct that, in the first case, both detectors will flash simultaneously, whereas in the second they will not?

Yes, and it depends. If the string is impervious to tension, it will force the electron acceleration profiles to not be simultaneous in the lab frame. If you force the electron acceleration profiles fo be simultaneous in the lab frame, the the string will have been stretched longer than its original length in the post acceleration electron rest frame. All of this is exactly the same as the Bell spaceship paradox.

Please study the FAQ I linked earlier carefully.

 

[DaveC426913]

"At the same time" using which frame? Most likely you mean the frame in which the emitters and detectors are at rest.

Yes. That is the frame in which we should observe length contraction.

Using the frame in which the emitters and detectors are at rest, the distance between the two electrons doesn't change but the string is length contracted so is stretched and breaks.

Not to be pedantic, but it's not exactly the same as BSP. (I know it's the same principle, but I'm having trouble converting.)

It's not a string it's a wire. A stick, one metre long. And the stick should get shorter due to length contraction. The two electrons it is "tied" to should be getting closer, because
the whole thing is a stick with an electron at each end.

 

[PAllen]

Yes. That is the frame in which we should observe length contraction.Not to be pedantic, but it's not exactly the same as BSP. (I know it's the same principle, but I'm having trouble converting.)

It's not a string it's a wire. A stick, one metre long. And the stick should get shorter due to length contraction. The two electrons it is "tied" to should be getting closer, because the the whole thing is a stick with an electron at each end.

It is exactly the same. String or wire doesn’t change anything.

 

[DaveC426913]

It is exactly the same. String or wire doesn’t change anything.

I'm sorry for being dense. I'm usually pretty good with relativity.

Why is the one metre long wire not observed to be length contracted to .5m?

 

[PAllen]

Think about this - if an accelerating rod is contracting, can its front and back have the same acceleration profile in that frame?

 

[PAllen]

I'm sorry for being dense. I'm usually pretty good with relativity.

Why is the one metre long wire not observed to be length contracted to .5m?

I already explained all the possibilities. Please study the posts.

 

[DaveC426913]

Think about this - if an accelerating rod is contracting, can its front and back have the same acceleration profile in that frame?

They'll be different.

Yet the setups of the guns are identical.

I already explained all the possibilities. Please study the posts.

I did. If that had cleared it up I wouldn't still be stumped. But thanks anyway.

 

[DaveC426913]

Using which frame? You've seen enough threads here to know better than to say "simultaneously" without specifying a frame

Sorry. I thought it was obvious that the experiment is being observed in the rest frame of the guns/detectors. That's where the length contraction will be apparent.

 

[Nugatory]

The two electrons it is "tied" to should be getting closer, because the the whole thing is a stick with an electron at each end.

Using the frame in which the emitters and detectors are at rest: Either the two electrons get closer together as the stick length-contracts, or the stick breaks while they remain the same distance apart. Which happens depends on the mechanical strength of the stick as compared with the forces required to pull the two electrons together.

Using the frame in which the detectors are moving backwards at .87c and the electrons are at rest after they are fired: The lead electron starts moving first, so either the stick pulls the trailing electron into motion before the trailing gun fires, thus maintaining the one meter separation, or the stick breaks because one end is moving with the lead electron and the other end is still attached to the still-stationary trailing electron. Which happens depends on the mechanical strength of the stick as compared with the forces required to pull the two electrons together.

If the stick breaks the two electrons maintain a constant separation in the frame in which the detectors are at rest; if the stick does not break the electrons maintain a constant separation in the frame in which the detectors are moving. Either way, the physics is the same: either the forces are sufficient to break the stick or they aren't, and that determines in which frame the separation remains fixed at one meter.

 

[DaveC426913]

So I'm trying to bring this back around to the other members' hypothesis.

The two independent electrons hit their detectors simultaneously. This I know.
This experiment will not demonstrate length contraction. This too I know.
The reason is because the electrons - each in their own rest frames - see the other receding, not staying at a fixed distance.

 

[PAllen]

They'll be different.

Yet the setups of the guns are identical.I did. If that had cleared it up I wouldn't still be stumped. But thanks anyway.

Please especially see my post #6, and think about it. Also, look at that picture in the FAQ with the stressed and unstressed string next to each other. I doubt it is true, but it sure seems like you are not trying.

 

[DaveC426913]

Yes, and it depends. If the string is impervious to tension, it will force the electron acceleration profiles to not be simultaneous in the lab frame. If you force the electron acceleration profiles fo be simultaneous in the lab frame, the the string will have been stretched longer than its original length in the post acceleration electron rest frame. All of this is exactly the same as the Bell spaceship paradox.

Please study the FAQ I linked earlier carefully.

OK, so if we don't force them to maintain a fixed distance in the lab frame (say in the initial experiment, where there is no string/wire), then when they reach the detectors, they will not be one metre apart (in the lab frame), which means the detectors will not flash at the same time.

 

[PAllen]

So I'm trying to bring this back around to the other members' hypothesis.

The two independent electrons hit their detectors simultaneously. This I know.
This experiment will not demonstrate length contraction. This too I know.
The reason is because the electrons - each in their own rest frames - see the other receding, not staying at a fixed distance.

This is true. And if the wire stretches easily, it will expand in the electron rest frame, as has been already stated several times. On the other hand, if the wire resists stretching, it will necessarily change the acceleration profile of the electrons, and you will see contraction in the lab frame. All of this has been said a few times already.

Or, if it is only attached to e.g. the leading electron, you will see an enlarging gap between the back of the wire and the rear electron, as pictured in the faq.

 

[PAllen]

OK, so if we don't force them to maintain a fixed distance in the lab frame (say in the initial experiment, where there is no string/wire), then when they reach the detectors, they will not be one metre apart (in the lab frame), which means the detectors will not flash at the same time.

Sorry, this is almost the opposite of what I clearly stated. Without the wire, they will maintain constant distance in the lab frame and arrive simultaneously in the lab frame.

 

[DaveC426913]

This is true. And if the wire stretches easily, it will expand in the electron rest frame, as has been already stated several times.

Again, at the risk being appearing to be difficult, this is not addressing what I'm asking in a way I can convert. I'm not looking at the electron frame; I'm looking at the lab frame. And I'm not concerned about the tensile strength of the wire. It's just an object of a given length. Any object, such as a wire, should shorten, as seen in the lab frame.

Argh. All right. I will have to ponder on this. You guys have pointed me in the right direction several times.

Thanks.

 

[PAllen]

Again, at the risk being appearing to be difficult, this is not addressing what I'm asking in a way I can convert. I'm not looking at the electron frame; I'm looking at the lab frame. And I'm not concerned about the tensile strength of the wire. It's just an object of a given length. Any object, such as a wire, should shorten, as seen in the lab frame.

Argh. All right. I will have to ponder on this. You guys have pointed me in the right direction several times.

Thanks.

An object of a given length in which frame? For an accelerating object there is no such thing as being a given length in different frames. If it maintains a given length in a starting frame, it is lengthening in the frame of either end. If it is constant length from the frame of one of the ends, it’s length is shrinking in the starting frame.

 

[DaveC426913]

An object of a given length in which frame?

Again, I am referring to the lab frame. That's where the length contraction is observed, and where the detectors are measured.

 

[PAllen]

I've stated several times that I am referring to the lab frame.

If the wire maintains a given length in the lab frame, then it is stretching, in an invariant sense - its components are moving apart in their rest frames. If it is not stretching, then it will contract in the lab frame. The independent electrons can only maintain contact with both ends of a stretching accelerating wire - which will be constant length in the lab frame.

 

[DaveC426913]

If the wire maintains a given length in the lab frame, then it is stretching, in an invariant sense - it components are moving apart in their rest frames. If it is not stretching, then it will contract in the lab frame. The independent electrons can only maintain contact with both ends of a stretching accelerating wire - which will be constant length in the lab frame.

Yup. That goes without saying (or should, if my brain were firing on a bangers).

 

[Nugatory]

I'm not looking at the electron frame; I'm looking at the lab frame.

Working in the lab frame, the two electrons are fired at the same speed at the same time from points one meter apart, so they will maintain their one meter separation if we don't apply some other force to them. If we tie them together with a string, then the possibilities are:
1) The string is so stretchy and elastic that it exerts no force on the electrons. As far as the electrons are concerned, it's as if the string wasn't there. The string stretches happily to counteract the effects of length contraction and the electrons maintain their one-meter separation in the lab frame.
2) The string is not stretchy and elastic, but is also so weak that when it length-contracts it breaks before it exerts any significant force on the electrons. This is the standard form of Bell's paradox, in which the string breaks while the electrons maintain their one-meter sepration in the lab frame.
3) The string is not stretchy and elastic, but is strong enough that it doesn't break. As it length-contracts it pulls the two electrons together so that they no longer maintain their one-meter separation in this frame.

The same three possibilties are described in the electron frame as:
1) The string is so stretchy and elastic that it exerts no force on the electrons. The lead electron starts moving before the trailing electron so the separation increases to be more than one meter, and the string stretches accordingly to keep them connected. The distance between the two electrons increases while the striung stretches.
2) The string is not stretchy and elastic but also is so weak that it breaks when it is stretched, without exerting any significant force on the electrons. The leading electron starts moving before the trailing electron and the string snaps at once as the distance between the two electrons increases.
3) The string is not stretchy and elastic, but is strong enough that it doesn't break. The leading electron starts moving before the trailing electron is fired, but because the two are tied together the trailing electron is dragged into motion anyways. The one meter separation is maintained in this frame, but not in the lab frame.

Cases #1 and #2 result in simultaneous detections in the lab frame. In case #3 the detections are not simultaneous in the lab frame.

 

[DaveC426913]

The lead electron starts moving before the trailing electron...
...
The leading electron starts moving before the trailing electron ...
...
The leading electron starts moving before the trailing electron is fired

OK. Why does this happen?

They were fired simultaneously in the lab frame. They just happen to be one metre apart.

 

[PAllen]

OK. Why does this happen?

They were fired simultaneously in the lab frame. They just happen to be one metre apart.

And you think simultaneity is frame independent? If it is true in the lab frame, it is not true in the frame of either electron.

 

[DaveC426913]

And you think simultaneity is frame independent? If it is true in the lab frame, it is not true in the frame of either electron.

You didn't answer the question.

They were fired simultaneously in the lab frame. Why would the leading electron get a head start?

 

[Nugatory]

OK. Why does this happen?

They were fired simultaneously in the lab frame. They just happen to be one metre apart.

That's relativity of simultaneity at work (and surely you've noticed Orodruin's sig - if not, find one of his posts and look at the sig). The two events "gun 1 fires" and "gun 2 fires" are simultaneous in the lab frame, therefore not simultaneous in the electron frame.

 

[DaveC426913]

That's relativity of simultaneity at work (and surely you've noticed Orodruin's sig - if not, find one of his posts and look at the sig). The two events "gun 1 fires" and "gun 2 fires" are simultaneous in the lab frame, therefore not simultaneous in the electron frame.

So, all of those statements you made were referring to which frame?

BTW, post 1:

I am certain it has to do with the initial acceleration and relativity of simultaneity, but I'm stumped.

 

[PAllen]

You didn't answer the question.

They were fired simultaneously in the lab frame. Why would the leading electron get a head start?

I did answer. Simultaneous in the lab frame cannot be simultaneous for either electron.

Consider the simplest case of each electron instantly changing from speed 0 to speed v at time 0 in the lab frame. Then, in a frame moving at v with respect to the lab frame, in which the electrons are soon both at rest, the event of the the lead electron coming to rest occurs before the event of the trailing electron coming to rest. As a result, the distance between them has increased. To force you to seriously consider this, I think no one should respond until you exhibit a Lorentz transform of this simplified case.

 

[PAllen]

So, all of those statements you made were referring to which frame?

BTW, post 1:

Negatory gave two sets of descriptions, one for each frame. I don’t see how he could have been more explicit.

 

[DaveC426913]

To force you to seriously consider this, I think no one should respond until you exhibit a Lorentz transform of this simplified case.

Er. Is that something usually required for a Basic High School level thread?

 

[Nugatory]

So, all of those statements you made were referring to which frame?

The first set of #1-#3 are using the lab frame; the second #1-#3 are using the electron frame. However, both #1 statements are describing the same physical situation (stretchy string), and likewise both #2 statements are describing the same physical situation (breaking string) and both #3 statements are describing the unbreakable unstretching string case.

 

[DaveC426913]

Negatory gave two sets of descriptions, one for each frame. I don’t see how he could have been more explicit.

Oops. Sorry. I zeroed in on those statements too eagerly, and forgot the context.