Length contraction of a pair of electrons

In summary: Still, I'm not sure how exactly it solves my quandary (I guess I'm feeling a bit obtuse today).It's possible that you're not understanding something about how the string works. You could try looking up the details of the string theory concepts you're unfamiliar with, and see if that makes your understanding clearer.
  • #36
DaveC426913 said:
Er. Is that something usually required for a Basic High School level thread?
Isn’t that what this is?
 
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  • #37
PAllen said:
Isn’t that what his is?
It is. You're asking me to do a Lorentz Transform. I picked Basic HS to limit mathematical explanations.
 
  • #38
DaveC426913 said:
It is. You're asking me to do a Lorentz Transform. I picked Basic HS to limit mathematical explanations.
I thought Lorentz transforms were ok to expect for a B level thread. If not, I don’t see how to proceed except for you to accept the relativity of simultaneity requires what I said in post #31 to be true.
 
  • #39
PAllen said:
I thought Lorentz transforms were ok to expect for a B level thread. If not, I don’t see how to proceed except for you to accept the relativity of simultaneity requires what I said in post #31 to be true.
I intuit it, just never got past HS math.
 
  • #40
DaveC426913 said:
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. Ga and Gb. Gb is one metre behind Ga and sightly offset to one side. Both are dialed down so they fire only one electron at a time.

We have two detectors, Da and Db, each 10 metres away from its gun.

To sum: , we have two identical setups, Sa, and Sb, with Sb 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.

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

The detectors should light up at exactly the same time. Right?

By "at the same time", I assume you mean "at the same time in the lab frame". You didn't specify.

You use "at the same time" a lot in your analysis, without talking about an associated frame. To me that suggests you are ignoring the relativity of simultaneity, which basically says that what you mean by "at the same time" depends on the frame you choose. You've been around enough that I thought you would have noticed this, but maybe not?

It might be helpful to draw a pair of space-time diagrams, one space-time diagram of what happens in the lab frame, another space-time diagram that happens in the frame moving with the electrons.
 
  • #41
DaveC426913 said:
I intuit it, just never got past HS math.
A Lorentz transform is high school math.
 
  • #42
OK, I guess it's time for me to do some lifting on my own.
Thanks for your help.
 
  • #43
pervect said:
By "at the same time", I assume you mean "at the same time in the lab frame".

You didn't specify.

You use "at the same time" a lot in your analysis, without talking about an associated frame.
I've always been talking about the lab frame. But I clarified here and there.

pervect said:
It might be helpful to draw a pair of space-time diagrams, one space-time diagram of what happens in the lab frame, another space-time diagram that happens in the frame moving with the electrons.
I'm interested in the lab frame. That's where the experiment is observed. The timing of the emissions, the timing of the detections and the length contraction are all observable in the lab frame.
 
  • #44
DaveC426913 said:
I've always been talking about the lab frame. But I clarified here and there.I'm interested in the lab frame. That's where the experiment is observed. The timing of the emissions, the timing of the detections and the length contraction are all observable in the lab frame.

Usually, the use of a string implies some interest in something other than the lab frame. If you never invoke any other frame, the distance between the two electrons does not change. But it's difficult to see why you'd introduce a string if you didn't care about something more than the lab frame. You might care about either the notion of the "proper length" of the string, the notion of the string as a "rigid body", and/or the closely related notion of the length of the string in its own frame. But if you don't for some reason care about any of those, it doesn't matter. But what is the purpose of the string, then, if you don't care about any of that?
 
  • #45
pervect said:
Usually, the use of a string implies some interest in something other than the lab frame. If you never invoke any other frame, the distance between the two electrons does not change. But it's difficult to see why you'd introduce a string if you didn't care about something more than the lab frame. You might care about either the notion of the "proper length" of the string, the notion of the string as a "rigid body", and/or the closely related notion of the length of the string in its own frame. But if you don't for some reason care about any of those, it doesn't matter. But what is the purpose of the string, then, if you don't care about any of that?
I've used a wire. (I know it doesn't change anything in principle.) The wire is simply a one metre long "stick" which, when we observe it moving at .87c in the lab frame, will be observed to be only .5m long. We don't need to examine it from the wire's FoR (or some point on it) to know that, in the lab, we will observe it contracted.
 
  • #46
DaveC426913 said:
I've used a wire. (I know it doesn't change anything in principle.) The wire is simply a one metre long "stick" which, when we observe it moving at .87c in the lab frame, will be observed to be only .5m long. We don't need to examine it from the wire's FoR (or some point on it) to know that, in the lab, we will observe it contracted.
If it is not attached to either electron, and it’s front end remains in touch with the front electron, do you understand that it’s rear will extend only half way to the rear electron? And this is because, in the lab frame, it’s rear accelerated more than its front.
 
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  • #47
PAllen said:
If it is not attached to either electron, and it’s front end remains in touch with the front electron, do you understand tat it’s rear will extend only half way to the rear electron?
I'm trying to figure out why the distance to the trailing electron does not likewise decrease.

I think I've figured it out (with a vast amount of greatly appreciated help).
 
  • #48
My take:

If the leading electron - right after exiting the gun - were to look back at the trailing electron, it would not immediately see the trailing electron being fired. It takes time for that event to propagate (at c) to where the leading electron experiences it. Thus the trailing electron is still accelerating even as the leading electron speeds away at its max speed.

But more: Because the leading electron is in motion relative to the trailing electron, the leading electron sees the trailing electron as time dilated, which means its acceleration is slower.

It will never catch up to the same speed as the leading electron. The leading electron will continue to pull away. And it keeps pulling away all the way to the detector.

(Not sure about that last part. They're ballistic for most of their trajectory. Shouldn't the trailing electron eventually reach the same speed? At which point, there should be no further separation.)
 
  • #49
DaveC426913 said:
My take:

If the leading electron - right after exiting the gun - were to look back at the trailing electron, it would not immediately see the trailing electron being fired. It takes time for that event to propagate (at c) to where the leading electron experiences it. ...
Relativity of simultaneity is not signal delay, but what is left after you already accounted for signal delay.
 
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  • #50
PAllen said:
Please study the posts.
DaveC426913 said:
I did.

That post is two minutes later. Your average time before responding has been something like eight minutes. I think you should stop posting for, say a day, and just read what has already been posted - you shouldn't expect to immediately understand. Firing off replies that complicate the situation isn't going to help you understand nearly as much as thinking hard about what's already been written.
 
  • #51
This thread has left me shaken and humbled.

Vanadium 50 said:
... you shouldn't expect to immediately understand.
I thought I already knew.

I sincerely thought I understood this subject better than I actually do. Bell's Spaceship Paradox, relativity of simultaneity, Lorentz contraction were all things I was comfortable with ...

It might be that I understood it a lot better 10-20 years ago, when I read up on physics heavily, but that, over time, my knowledge has rotted from the inside out. One would think, once you've learned how to ride a bike, you never "unlearn" it, but I guess use it or lose it.

Dunning-Kruger, ftw. Or onset of senility...

Thanks for being so patient, everyone.
 
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  • #52
A stick might have clocks sitting at each end - initially synchronized. As it moves along its axis, and shortens in the lab frame, the clocks are no longer found to be synchronized by lab observers, though they are still found to be synchronized by observers at each end of the stick. "Shortening" can just be restated as desynchronization.
 
  • #53
DaveC426913 said:
If we fire two electrons at the same time, one metre apart, should they length contract till they're less than a metre apart?
PAllen said:
If you force the electron acceleration profiles fo be simultaneous in the lab frame, the the string will have been stretched

I'm confused on this point. How is the electrons being "fired" in the lab frame any different than an observer being fired away from the electrons? If the 2 scenarios are equivalent, why would the string between the electrons have any tension (ignoring the electric repulsion)?
 
  • #54
metastable said:
If the 2 scenarios are equivalent
They aren't. Who/what is being acted on by a force is different.
 
  • #55
metastable said:
I'm confused on this point. How is the electrons being "fired" in the lab frame any different than an observer being fired away from the electrons? If the 2 scenarios are equivalent, why would the string between the electrons have any tension (ignoring the electric repulsion)?
They aren’t equivalent.

In one case there are inertial frames in which the lab, observer, and electron guns are moving with constant velocity throughout (the frame in which they are at rest is one of these, with the “constant velocity” being zero). There is no inertial frame in which the electrons move at constant velocity throughout.

The other case is the exact opposite. If we fire the lab and observers away from the electrons, then there are inertial frames in which the electrons are moving with a constant velocity (possibly zero) throughout there is no inertial frame in which the lab moves with constant velocity throughout.
 

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