B Length contraction of a pair of electrons

DaveC426913

Gold Member
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

Gold Member
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.

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

Nugatory

Mentor
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

Gold Member
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

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

Gold Member
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

Mentor
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

Gold Member
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.

PAllen

Er. Is that something usually required for a Basic High School level thread?
Isn’t that what this is?

DaveC426913

Gold Member
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.

PAllen

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.

DaveC426913

Gold Member
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.

pervect

Staff Emeritus
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.

PAllen

I intuit it, just never got past HS math.
A Lorentz transform is high school math.

DaveC426913

Gold Member
OK, I guess it's time for me to do some lifting on my own.

DaveC426913

Gold Member
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.

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.

pervect

Staff Emeritus
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?

DaveC426913

Gold Member
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.

PAllen

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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DaveC426913

Gold Member
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).

DaveC426913

Gold Member
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.)

A.T.

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.

Staff Emeritus
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.

"Length contraction of a pair of electrons"

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