De Broglie Waves: Momentum, Wavelength & Light Speed

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    De broglie Waves
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Discussion Overview

The discussion revolves around the relationship between momentum, wavelength, and the speed of light as described by de Broglie's hypothesis. Participants explore the implications of increasing wavelength on momentum and velocity, particularly in the context of light and relativistic physics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants state that according to de Broglie's equation, an increase in wavelength (λ) implies an increase in momentum (p), leading to a potential increase in velocity (v).
  • Others clarify that λ and p are inversely proportional, suggesting that an increase in λ should actually correspond to a decrease in p.
  • One participant notes that if v equals the speed of light (c), then the classical relation p=mv does not hold, as it applies to non-relativistic particles.
  • Another participant points out that for light, which consists of photons with zero mass, the relationship between energy (E) and momentum (p) is given by p=E/c, indicating that while energy can increase, the speed of light remains constant.
  • There is a mention of the relationship c = fλ, linking wavelength and frequency in the context of light.

Areas of Agreement / Disagreement

Participants express differing views on the implications of increasing wavelength on momentum and velocity, with no consensus reached on the correct interpretation of these relationships, particularly in the context of light and relativistic effects.

Contextual Notes

Some statements rely on classical mechanics, while others introduce relativistic concepts, highlighting the limitations of applying classical equations to light. The discussion also reflects varying levels of understanding regarding quantum mechanics and the nature of photons.

AakashPandita
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According to Broigle,

λ=h/p
where
p=momentum,
h=planck constant, and
λ=wavelength

But that means,
if λ increases, then p will increase
p=mv
and so v will increase along with the wavelength

But what if the v is that of light,i.e, c?
 
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AakashPandita said:
According to Broigle,

λ=h/p
where
p=momentum,
h=planck constant, and
λ=wavelength

But that means,
if λ increases, then p will increase
p=mv
and so v will increase along with the wavelength

But what if the v is that of light,i.e, c?

For starters, why increase? λ and p are inversely proportional. Or it's New Year's morning and I'm not thinking straight.
 
danR is correct. If [tex]\lambda[/tex] decreases, then [tex]p[/tex] increases. But, this is not a cause and effect thing.

That aside, if [tex]v = c[/tex], then [tex]p \ne mv[/tex].

Best wishes
 
when p increases ,i.e, mv (that of light)
then either m or v (of light) should increase...
then what will be that would increase?
 
AakashPandita said:
According to Broigle,

λ=h/p
where
p=momentum,
h=planck constant, and
λ=wavelength

But that means,
if λ increases, then p will increase
p=mv
and so v will increase along with the wavelength

But what if the v is that of light,i.e, c?
There clearly is an inverse relationship between the two parameters of your interest. Also Lambda can't increase as your post suggests , in Q.M as you may know light comes in packets ' photons' which have specific energy levels.

For a given wave , it's wavelength and speed can be related by : c = fλ
 
AakashPandita said:
when p increases ,i.e, mv (that of light)
then either m or v (of light) should increase...
then what will be that would increase?

This is true only for non-relativistic particles! You want [tex]v=c[/tex], which is relativistic. Most likely you are talking about light (electromagnetic radiation). In the relativistic case [tex]p=E/c[/tex], where [tex]c[/tex] is the constant speed of light. For light the photons have zero mass and when you increase [tex]E[/tex] the momentum [tex]p[/tex] increases but the speed of light [tex]c[/tex] remains constant.

The point is that [tex]p=mv[/tex] does not apply when [tex]v=c[/tex].

Best wishes
 
Oh i see. thanks.
 

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