Acceleration through PD, relativistic momentum

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Discussion Overview

The discussion revolves around the calculation of momentum for an electron accelerated through a potential difference (PD) and the implications of relativistic effects on this calculation. Participants explore both non-relativistic and relativistic approaches to determine the momentum resulting from the acceleration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that accelerating an electron through a PD of 10^3 volts results in a momentum of 1 MeV/c, questioning the role of relativity.
  • Another participant corrects this to a PD of 10^6 volts, stating that the non-relativistic formula would yield a momentum of approximately 1 MeV, while the relativistic result would be around 1.4 MeV.
  • A participant acknowledges the correction regarding the PD and seeks clarification on how to arrive at the relativistic momentum result.
  • One participant describes their method of calculating energy using K=Vq and K+m(c^2)=E, noting discrepancies when converting energy to momentum units (MeV to MeV/c).
  • The same participant expresses uncertainty about their calculations, suggesting they may have arrived at the correct answer but were confused about the units involved.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations, as there are differing interpretations of the potential difference and its implications for momentum. The discussion includes corrections and clarifications, but no final agreement is established.

Contextual Notes

There are limitations regarding the assumptions made about the potential difference and the application of relativistic formulas. The discussion also reflects uncertainty in unit conversions and the relationship between energy and momentum in relativistic contexts.

Stickybees
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When accelerating an electron through a PD of 10^3, that will give a momentum of 1 MeV/c right? Or is there something I'm not taking into account with relativity?

Thanks :D
 
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You probably meant PD=10^6 volts. Then, the NR formula would give p-1 MeV.
But the relativistic result is 1.4 MeV.
 
Meir Achuz said:
You probably meant PD=10^6 volts. Then, the NR formula would give p-1 MeV.
But the relativistic result is 1.4 MeV.

Ah yes sorry I did, but how would I would out the relativistic result like you did?
 
So I've used K=Vq, then K+m(c^2)=E to get the energy and I've got approximately the right answer, but when I divide by c to get units from MeV to MeV/c for momentum I'm of course not even in the same order of magnitude.
 
Stickybees said:
So I've used K=Vq, then K+m(c^2)=E to get the energy and I've got approximately the right answer, but when I divide by c to get units from MeV to MeV/c for momentum I'm of course not even in the same order of magnitude.

Nevermind, I think I must have the right answer I was just thinking that /c had already happened.
 

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