Relativistic Addition of Electron Velocities

AI Thread Summary
The discussion revolves around calculating the speed of a second electron moving in the opposite direction to the first electron, using the relativistic velocity addition formula. The initial setup involved confusion over the correct assignment of velocities and reference frames, which led to algebraic errors in the calculations. Participants suggested clarifying the variables and ensuring consistent labeling of the electrons and the lab. After some back-and-forth, the user indicated they resolved their confusion and successfully set up the equation. The conversation highlights the challenges of applying relativistic concepts and the importance of careful variable management in physics problems.
Cheezay
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Homework Statement


An electron moves to the right in a laboratory accelerator with a speed of 0.822c. A second electron in a different accelerator moves to the left with a speed of 0.424c relative to the first electron. Calculate the speed of the second electron (in c) relative to the lab. Do not enter unit.


Homework Equations


<br /> V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}<br />


The Attempt at a Solution


Relativity is difficult for me to get a handle on. The way I am setting this up, is I'm using the velocity .822c for Va/b and -.424c(since it this one goes left) for Vb/c, and Vac would be the velocity of the second electron relative to the lab, however I'm not getting the right answer. I'm guessing that I don't have the velocities substituted in the proper places in the formula, any help?
 
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Cheezay said:
… I'm using the velocity .822c for Va/b and -.424c(since it this one goes left) for Vb/c, and Vac would be the velocity of the second electron relative to the lab …

Hi Cheezay! :smile:

You're using a for the 1st electron, and b for the lab, then b for the second electron and c for the 1st electron :frown:

fiddle around with it, to get it consistent, and remember that Va/b = -Vb/a :smile:
 
Ok. So considering that a = electron 1, b = electron 2, and c = lab... I now have Va/c= .822c, and because Va/b=-Vb/a.. i use -.424c for the speed of electron 2 relative to the lab. Now i solve for Vb/c. I'm fairly confident i have my equation set up correctly now, which means I'm making an algebraic error now... any help?

Using V as a variable, in place of Vb/c

.822c=[-.424c + V]/[1-(.424c x V)/(c^2)] c's cancel...

.822c=[-.424c + V]/[1-(.424 x V)/c] i move the whole term...

.822c[1-(.424 x V)/c]=-.424c + V distribute .822c (c's cancel again)

.822c - .348528 -.822V = -.424c +V

-.348528 - .822V = -1.246c + V

-.348528 - V = -1.51582c + V

-2V = -1.16729c so V= .583c which isn't correct


Any more help would be greatly appreciated! I have been on this problem for 3 days now!
 
Hi Cheezay! :smile:
Cheezay said:
Ok. So considering that a = electron 1, b = electron 2, and c = lab... I now have Va/c= .822c, and because Va/b=-Vb/a.. i use -.424c for the speed of electron 2 relative to the lab.

(you mean relative to electron 1 :wink:)
Using V as a variable, in place of Vb/c

.822c=[-.424c + V]/[1-(.424c x V)/(c^2)] ...

However did you get an unknown on the RHS? :redface:

Choose a b and c (it might be easier if you call them 1 2 and L) so that the RHS contains your two knowns, and the unknown is on the LHS! :smile:
 
Ok.. I've figured it out. Thanks for the help!
 
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