The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Is this true?

ImageUploadedByStudent Room1399929688.309335.jpg


Posted from TSR Mobile
Original post by Ben6597


No it is not true. The error is when you say velocity = wavelength*frequency.

A particle or "wave packet" in quantum mechanics is composed of a superposition of many simple sine/cosine waves, each with a particular wavelength, frequency and velocity. For each individual wave, the above equation holds. But when you superimpose all of these individual waves to form a wave packet, you end up with 2 possible measures of velocity, phase velocity and group velocity.

http://en.wikipedia.org/wiki/File:Wave_group.gif

The above link shows an image that is useful for visualizing the difference. Basically the wave packet moves at the group velocity, but the individual oscillations within the packet move at the phase velocity. Ultimately the physics is in the group velocity, because this is the velocity of the wave packet and hence the velocity of the apparent particle.

1. Phase velocity

Phase velocity is calculated using the usual formula, vp=λfv_p = \lambda f

Now we can re-write this as vp=ωkv_p = \frac{\omega}{k}

Where w represents angular frequency (w = 2pi*f), and k is the wavenumber, which is related to wavelength in the same way period is related to angular frequency, that is k = 2*pi/lambda.

Now to calculate kinetic energy, we use the group velocity below, not the phase velocity.

2. Group velocity

Unlike phase velocity, group velocity is calculated as ∂ω∂k\frac{\partial \omega}{\partial k}
It can be shown using the debroglie relation, and E=hf, that this is equivalent to:
∂E∂p\frac{\partial E}{\partial p}

Where E is energy and p is momentum.

Now kinetic energy, E=0.5mv2=p22mE = 0.5mv^2 = \frac{p^2}{2m}

Therefore, group velocity is, ∂E∂p=pm=v\frac{\partial E}{\partial p} = \frac{p}{m} = v

If you run through the analysis using group velocity instead of phase velocity, you will get E = KE, which makes sense if there is no external potential. If there is an external potential, E = KE + U, where U is the external potential.
(edited 9 years ago)
Original post by Ben6597
No. Because you have mixed both classical (Newtonian) mechanics (KE = 1/2 mv2) with relativitistic mass (E = mc2) in the same derivation.

v is limited to c (speed of light = 3x108 ms-1) in all cases.

Einstein's equation says that mass and energy are the same thing and one may be converted to the other.

1/2mv2 only holds when v is a small fraction of lightspeed and is dependent on the reference frame of the observer.
(edited 9 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you