De Broglie Waves

1. Jan 1, 2012

AakashPandita

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?

2. Jan 1, 2012

danR

For starters, why increase? λ and p are inversely proportional. Or it's New Year's morning and I'm not thinking straight.

3. Jan 1, 2012

eaglelake

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

4. Jan 1, 2012

AakashPandita

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?

5. Jan 2, 2012

ibysaiyan

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

6. Jan 2, 2012

eaglelake

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

7. Jan 3, 2012

AakashPandita

Oh i see. thanks.