De Broglie Waves: Momentum, Wavelength & Light Speed

  • Context: Graduate 
  • Thread starter Thread starter AakashPandita
  • Start date Start date
  • Tags Tags
    De broglie Waves
Click For Summary
SUMMARY

The discussion centers on the relationship between momentum (p), wavelength (λ), and the speed of light (c) as described by de Broglie's equation, λ=h/p. Participants clarify that while λ and p are inversely proportional, the equation p=mv does not hold true for light, as photons possess zero mass. Instead, in the relativistic context, momentum is defined as p=E/c, where E represents energy. The conversation emphasizes that increasing energy leads to increased momentum without affecting the constant speed of light.

PREREQUISITES
  • Understanding of de Broglie's equation (λ=h/p)
  • Familiarity with the concept of momentum (p=mv)
  • Knowledge of relativistic physics and the properties of photons
  • Basic grasp of wave-particle duality in quantum mechanics
NEXT STEPS
  • Study the implications of de Broglie's hypothesis on wave-particle duality
  • Learn about relativistic momentum and its differences from classical momentum
  • Explore the relationship between energy and momentum in quantum mechanics
  • Investigate the properties of photons and their role in electromagnetic radiation
USEFUL FOR

Students of physics, particularly those interested in quantum mechanics, relativistic physics, and the behavior of light and photons.

AakashPandita
Messages
157
Reaction score
0
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?
 
Physics news on Phys.org
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 \lambda decreases, then p increases. But, this is not a cause and effect thing.

That aside, if v = c, then p \ne mv.

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 v=c, which is relativistic. Most likely you are talking about light (electromagnetic radiation). In the relativistic case p=E/c, where c is the constant speed of light. For light the photons have zero mass and when you increase E the momentum p increases but the speed of light c remains constant.

The point is that p=mv does not apply when v=c.

Best wishes
 
Oh i see. thanks.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 41 ·
2
Replies
41
Views
6K
  • · Replies 28 ·
Replies
28
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K