- #1

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λ = h/p

v/f = h/mv as f*λ=v

then

f = mv^2/h

but it should f = E/h and E=1/2*mv^2

- Thread starter ChAshutosh
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- #1

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λ = h/p

v/f = h/mv as f*λ=v

then

f = mv^2/h

but it should f = E/h and E=1/2*mv^2

- #2

Suraj M

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- #3

bhobba

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http://hyperphysics.phy-astr.gsu.edu/hbase/debrog.html

Thanks

Bill

- #4

Suraj M

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- #5

bhobba

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I don't know what you mean by that.SHM of each particlealong the wave, to the KE?

Like most I have read about the De-Broglie theory and seen the equations. They are an inconsistent mish-mash eg what's the wavelength of a stationary particle - and by frame jumping you can always go to a frame where its stationary. Because of that I tend to not get worried about manipulations involving them - it's wrong anyway.

In QM the KE of a free particle is 1/2 mV^2 where V is the velocity operator.

Interestingly one can actually prove that from symmetry in QM - but that is another story.

Thanks

Bill

- #6

Suraj M

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I mean the particles would have an extra amount of energy of ##½m \omega^2 A^2## can we use that?I don't know what you mean by that.

- #7

bhobba

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You lost me.I mean the particles would have an extra amount of energy of ##½m \omega^2 A^2## can we use that?

Can you explain what your terms mean?

Thanks

Bill

- #8

Suraj M

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The regular SHM terms! ##\omega## = 2##\pi /T## ; A = amplitude of the wave

- #9

bhobba

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The energy in QM is actually based on the Galilaen transformations while QFT is based on the Lorentz transformations.

Thanks

Bill

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