Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

What should be the de Broglie's wave-length?

+ 0 like - 0 dislike
93 views

I have learnt that the de Broglie's as λ=h/p (where h is Planck's constant,p is the momentum of the particle), we can derive it from equating Einstein's mass-energy equivalence and the energy of a photon, E=hν, then replacing 'c' with the velocity of any particle.

Waves are associated with moving particle, so as the wave length. The equation $E=mc^2$ is applicable for particles at rest; I recently found that the general eqn is $E^2=(mc^2)^2+(pc)^2$  Why do we use $E=mc^2$, when $E^2=(mc^2)^2+(pc)^2$is the eqn for moving particles? A particle at rest cannot have a wave length - isn't that so?

Closed as per community consensus as the post is undergraduate-level
asked Oct 14, 2014 in Closed Questions by RogUE (0 points) [ revision history ]
recategorized Oct 14, 2014 by Dilaton

@RogUE Welcome to PO. Try to use LaTeX in your formulae. I have done so in part of your question correcting a missing power in the formula. Yes, $E^2=(mc^2)^2+(pc)^2$ is indeed the correct relativistic energy-momentum (aka dispersion) relation. I like to joke that this should be formula that appears in the T-shirt that has three lines which read: E=ma^2E=mb^2; E=mc^2. Of course, it is obvious that $E=mc^2$ holds for a particle of mass $m$ at rest and is a special case of $E^2=(mc^2)^2+(pc)^2$.  For massless particles like the photon, one obtains $E=pc$ on setting $m=0$. I will let someone else answer the de Broglie wavelength bit.

This is an undergraduate level question, more suitable for physics stackexchange. No downvote, but vote to close.





user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...