de Broglie's equation

[jenn91]

de Broglie's equation which has a larger wavelength, a slow-moving proton or a fast moving golf ball

[big314mp]

Your best bet would be to plug in some theoretical numbers, and see what sort of wavelengths you will get.

[hermanntrude]

If you don't know the de broglie relationship you can derive it like this:

 

E=mc^2

E=hv

 

first, equate the two right hand sides.

second, divide by c to leave mc on one side.

third, replace mc with p, which is momentum.

[Klaynos]

If you don't know the de broglie relationship you can derive it like this:

 

E=mc^2

E=hv

 

first, equate the two right hand sides.

second, divide by c to leave mc on one side.

third, replace mc with p, which is momentum.

 

This derivation is flawed.

 

E=mc² is not valid for a moving particle.

 

mc can't infact = momentum as it'd be the momentum of a massive particle moving at the speed of light which in invalid.

 

For a photon

 

[math] p = \frac {h} {\lambda}[/math]

[hermanntrude]

it's flawed, but yet it's the one which is shown in most textbooks.

 

I was attempting to keep the answer simple.

[timo]

So how do these textbooks justify equating the p with mc and what do these letters stand for? It looks completely ridiculous to me.

[ydoaPs]

A better yet similar explanation from Stanford

about half way through.

[swansont]

[math]\lambda = \frac{h}{p}[/math]

 

Since the particles have mass, you use p = mv

 

Compare the masses and the speeds of the proton and golf ball.

 

(if you're familiar with some experiments, you would know that neutron diffraction has been observed. Golf ball diffraction probably has not*)

 

 

 

*the famous event at the 12th hole at Sherman Oaks Putt-Putt Emporium on Aug 12 1992 has yet to be adequately explained without invoking a wave nature of the ball.

[hermanntrude]

So how do these textbooks justify equating the p with mc and what do these letters stand for? It looks completely ridiculous to me.

 

m is the mass of the particle and c is the speed of light. A speed multiplied by a mass is a momentum. It is true that there is a flaw in the reasoning but it gives the correct result, and in a very simple manner. It's how my students are expected to learn the derivation.

[NeonBlack]

I don't think there is a derivation which is not flawed.

[Norman Albers]

The construction is of a wavefield where momentum is represented by wavelength and energy is represented by frequency, according to massive possibilities in special relativity.

[NeonBlack]

Swanson: Could you provide some more details on the mini-golf incident?

 

Another side note:

If you use p=mc, this is called the Compton wavelength.

[swansont]

Swanson: Could you provide some more details on the mini-golf incident?

 

The windmill vane was clearly blocking the hole. There's no way the ball could have gone through — diffraction is the only explanation.

 

 

(I can probably make up more details, too.)

 

m is the mass of the particle and c is the speed of light. A speed multiplied by a mass is a momentum. It is true that there is a flaw in the reasoning but it gives the correct result, and in a very simple manner. It's how my students are expected to learn the derivation.

 

In physics we generally don't give points to correct answers arrived at by flawed reasoning. Otherwise one tends to reinforce invalid concepts.

[hermanntrude]

The windmill vane was clearly blocking the hole. There's no way the ball could have gone through — diffraction is the only explanation.

 

 

(I can probably make up more details, too.)

 

 

 

In physics we generally don't give points to correct answers arrived at by flawed reasoning. Otherwise one tends to reinforce invalid concepts.

 

as neonblack said, there isn't really a derivation which is perfect anyway, so perhaps the simplest is best

[NeonBlack]

Maybe there is no one "right" derivation, but some are definitely more wrong than others. I think that saying p=mc is more confusing than instructive. Why use p=mc in derivation but p=mv in application? Making this simple is not hard.

For light:

Planck sez: [math]E=h \nu[/math]

Einstein sez: [math]E=pc[/math] This comes from [math]E^2=(mc^2)^2+(pc)^2[/math] in relativity with m=0. (Maxwell's electromagnetic theory gives the same result for light, but not as tersely.)

Now all you have to do is imagine that this also works for small massive objects as well.

[Klaynos]

as neonblack said, there isn't really a derivation which is perfect anyway, so perhaps the simplest is best

 

There is a difference between not perfect, and physically wrong. Your derivation has the assumption of a massive photon which is not physical.

[hermanntrude]

your derivation has the assumption of a photon moving at a speed other than that of the speed of light.

[Klaynos]

your derivation has the assumption of a photon moving at a speed other than that of the speed of light.

 

Empirical evidence wins in this case.

[Mr Skeptic]

Wasn't the original derivation also flawed (or rather a guess)? Something along the lines of [math]{\lambda} = h/p[/math] works for a photon, what if it worked for other particles too? Then using that, test emission spectra, etc., to show that it is true.

[Royston]

Wasn't the original derivation also flawed (or rather a guess)? Something along the lines of [math]{\lambda} = h/p[/math] works for a photon, what if it worked for other particles too?

 

Yes, that was the reasoning behind the de Broglie formula, it starts with considering a photon, so zero rest mass, and (as Neon Black has posted) uses the relationship from SR i.e [math]E^2 = p^2c^2 + m^2c^4[/math] which clearly simplifies to [math]E = pc[/math] the energy of a photon. Then using the Planck relationship, [math]E = hf = \frac{hc}{\lambda}[/math] and where [math]p = \frac{h}{\lambda}[/math]...giving [math]\lambda = \frac{h}{p}[/math].

 

I realize this has already been shown, but there's clearly a leap, it's derived from a particle with no mass, and then plugs mass back into the final formula, as [math]p = mv[/math] which is a bit silly (i.e the Newtonian approximation of momentum is used), but as Klaynos has pointed out, experimentally it works just fine. Maybe I'm missing something, but it does seem like a guess (as you said) that the relationship will hold for particles with mass. Obviously with something like a golf ball, the wavelength will be tiny, and so observing diffraction effects with large objects is clearly impossible.

Edited November 22, 2008 by Snail

[Norman Albers]

I read the concept as saying that we can tie energy to frequency even in massive objects. Then further, we connect momentum to wavelength. Thus we describe a phase velocity which is E/p and thus has speed of light as a lower limit. Group velocity gives us behavior of mass.