Talk:De Broglie hypothesis

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1 Linked 'de Broglie Wavelength' to here
2 Remove added text
3 removed section on pronunciation of "de Broglie"
4 Waves of what???
5 Removed line "this equation is essentially wrong"
6 Bogus
7 Caps
8 Hypothesis?
9 There may be a mistake in the equations that were written in this page

[edit] Linked 'de Broglie Wavelength' to here

This is what his discovery is commonly called - makes more sense if it redirects here.
--Blinken 11:40, 27 Apr 2005 (UTC)

[edit] Remove added text

I removed the following text added anonymously:

According to de Broglie, an electron bound to the nucleus behaves like a standing wave, such as the vibration of a plucked guitar string. Certain points along the wave do not move at all, these points are known as nodes. There is a node at each end and sometimes between the ends. The greater the frequency of vibration, the greater the number of notes and the shorter the wavelength.

De Broglie reasoned that if an electron does behave like a standing wavfe in the hydrogen atom, the length of the wave must fit the circumference of the orbit exactly. Otherwise the wave would partially cancel itself out on each successive orbit. Eventually the amplitude of the wave would be reduced to zero, and the wave would not exist.

My reason for removal is two-fold: 1) its in the wrong place in the article, it should appear later, 2) This was not de Broglie's idea; it was proposed considerably earlier, by Bohr .. the Bohr model of the atom aka the Bohr-Sommerfeld model. It was a semi-classical theory, with action-angle variables, and a nasty perturbation theory. Google says Bohr propsed this in 1913. See wolfram article linas 03:48, 27 November 2005 (UTC)

[edit] removed section on pronunciation of "de Broglie"

It belongs in the article about him, not here. Meeve 22:30, 7 December 2005 (UTC)

[edit] Waves of what???

which kind & of what nature exactly these waves are???

i read in some books that it is waves of probability, but actually I didn't understand this well.

I wish someone inform me about it.

thanks..

In 1923 this was not known yet - I think de Broglie assumed the particles themselves exhibited wave properties. Currently, it is understood that it is the quantum mechanical wave function that oscillates. Which is perhaps just another way of saying the same thing, since a particles wave function contains all there is to know about a particle's state. You'll find more information in Introduction to quantum mechanics and articles linked from there. Shinobu 14:50, 3 December 2006 (UTC)

[edit] Removed line "this equation is essentially wrong"

If this user thinks he has falsified a Nobel Prize-winning theory he should at least post it using his user name. A particle with zero velocity shouldn't be expected to show any wave-like behaviour; if it doesn't propagate, it isn't a wave. Someone more knowledgable than I may be able to explain this better. Shayno 12:34, 12 October 2006 (UTC)

I don't know what the anon said, but I think I should at least point out that a particle with zero velocity (say, an electron in a constant potential) tends to sort of spread out over space. Compare the lowest mode in the infinite box, which is just more dense at the center, but not really localized, or the lowest mode in the continuously repeating box, which is sort of everywhere, if my memory serves me right. None of this would falsify this theory. The wavelength just goes to infinity for stationary particles, as you would expect. Shinobu 14:33, 3 December 2006 (UTC)

[edit] Bogus

This is deBogus! There seems to be something wrong with the frequency formula. 128.250.50.92 07:20, 16 October 2006 (UTC)
Shayno 18:59, 31 October 2006 (UTC): deBogus no longer! Some anonymous user deleted 'em. Accident? Halloween prank? Who knows?!

[edit] Caps

"de Broglie hypothesis": caps inconsistent on the d and the h. Someone should check what is common usage in English and fix it. Shinobu 19:53, 3 December 2006 (UTC)

It should always be "de Broglie hypothesis". It should be consistent now. — Laura Scudder ☎ 21:48, 3 December 2006 (UTC)

[edit] Hypothesis?

Why is the word "hypothesis" in the article title? Is that what it's commonly called for some reason? If not, the name (and perhaps the text) should be changed to reflect the fact that everything in the article is extremely well-verified. -- SCZenz 21:03, 22 January 2007 (UTC)

I think it's typically called the "de Broglie hypothesis" because it was just a hypothesis at the time that de Broglie proposed it. It was experimentally confirmed in 1927 by the famous Davisson-Germer experiment, which demonstrated electron diffraction.

However, so far as I know there's no general rule for when you use the word "hypothesis" in physics (as opposed to "law", "theory", etc.) In my own rough estimation, a "law" is a single idea about the relationship between quantities (e.g., Newton's law of gravitation relating gravitational force to mass and the inverse square of distance), and a hypothesis is a single idea (not necessarily an equation) like "all matter has a wave-like nature", and a theory is a collection of laws and hypothesis purporting to explain some phenomenon. But there's no strict rules, and usually the names are simply historical artifacts (e.g., someone called it the "de Broglie hypothesis" and the name stuck.) All these terms can refer to ideas both before and after experimental confirmation -- the notion of a hypothesis getting "promoted" to a law after verification is generally not correct. --Tim314 15:08, 23 January 2007 (UTC)

[edit] There may be a mistake in the equations that were written in this page

We can get Debroglie relation from the the equation of Einstein and Plank:
$E = h f = p c$
Where E is the energy possesed by an arbitrary photon relative to an intertial frame of reference, h is planck constant, f is its frequency in that frame.
$p=\frac{hf}{c}=\frac{h}{\lambda}$
Where $λ$ is the wavelength of the photon. We can then postulate that every moving body will have certain wavelength. The equation is experimentally proven to be correct for every body with non-zero rest mass. Similarly,
$f=\frac{pc}{h}=\frac{mc^2}{h}$
Where m is the relativistic mass of that photon. We can again postulate that every body with non-zero rest mass will have the frequency showing on the above equation. Since
$v p = f λ$ where $v p$ is the phase velocity. $v_p=\frac{c^2}{v}$ Apparently, frequency of a body with non-zero rest mass is not directly proportional to its kinetic energy but its total energy or total relativistic mass. Thljcl 14:23, 25 March 2007 (UTC)

The problem with your math is at

p c = m c 2

. I assume you got at it by setting

E = p c

equal to

E = m c 2

, but the first is the total energy of a massless particle, the second the total energy of a massive particle at rest.

Part of what was so unexpected about de Broglie's hypothesis was that he took equations for a massless particle and played with them until found something new that's true for massive particles. Unfortunately, it doesn't usually work that way. — Laura Scudder ☎ 21:55, 25 March 2007 (UTC)

For a particle with non-zero rest mass, when its momentum is zero relative to an inertial frame of reference, we find that

E = m 0 c 2

Where E is the total energy,

m 0

is the rest mass of the particle, c is the speed of light in vacuum. When it possesses momentum in the same frame,

$E = \sqrt{m_0^2c^4 + p^2c^2}$

= γ m 0 c 2

p = γ m 0 v

$\gamma=\; \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

When v=0,

γ

=1,

$E=m_0\;c^2$ Classically, we define momentum as a product of mass and velocity. We can still use this definition. However,mass of a particular body is no longer a constant but is frame-dependent and is a function of its velocity. Whatever frame it is, its rest mass will be the same. Therefore, many or most scientists today would like to regard rest mass is the only definition of mass because variable mass may lead to a confusion. For me, I would still use both frame-dependent mass and invariant mass. In my maths, m always means relativistic mass. I will use $m 0$ denotes rest mass. Therefore,

E = m c 2

$E_k\;=(m-m_0)c^2=(\gamma - 1)m_0\;c^2$

Where

E k

denotes kinetic energy. For a particle with zero rest mass such as photon, its kinetic energy is ill-defined. The expression

E = m c 2

is different from $E_0\;=m_0\;c^2$ where

E 0

is a particle rest energy. I use E to denotes total energy. Wherever there's energy, there's always a certain amount of mass associate with it, though there may not be the rest energy. Certainly, photon has zero rest mass. It carries both momentum and energy.

p = m c

E = h f = m c 2 = p c

Since many people dislike the term of relativistic mass, there's still an another way to derive the expression $E = p c$ . That is from $E=\sqrt{m_0^2c^4+p^2c^2}$ Let $m 0 = 0$ ,

E = p c

Therefore, the derivation of De Broglie is correct with his own postulate. He postulated that all body with non-zero rest mass also have wave-particle duality in nature just as photon does. He personally thinks that nature seems to love symmetry. Now, we discover that there is a violation of symmetery in weak interaction.

Thljcl 17:12, 26 March 2007 (UTC)

Your assertion made on 25 March 2007 is still incorrect, Thljcl. You said, "Apparently, frequency of a body with non-zero rest mass is not directly proportional to its kinetic energy but its total energy or total relativistic mass," which simply is not true, even by your own ``derivation. You state that the frequency of a body with non-zero rest mass is proportional to its total energy (including rest mass), but yet your last post assumes $m 0 = 0$ within the derivation -- so you've only shown that the frequency of a body with zero rest mass is proportional to its total energy. This is obviously true because the de Broglie relation contains the momentum term $p$ , which from the equation

$E=\sqrt{m_0^2c^4+p^2c^2}$ ,

which you state above, clearly has no dependence on rest mass $m 0$ , as you previously asserted. Dchristle 20:52, 1 April 2007 (UTC)

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Talk:De Broglie hypothesis

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Contents

[edit] Linked 'de Broglie Wavelength' to here

[edit] Remove added text

[edit] removed section on pronunciation of "de Broglie"

[edit] Waves of what???

[edit] Removed line "this equation is essentially wrong"

[edit] Bogus

[edit] Caps

[edit] Hypothesis?

[edit] There may be a mistake in the equations that were written in this page

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