De Broglie Waves: Momentum, Wavelength & Light Speed

[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?

 

[danR]

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.

 

[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

 

[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?

 

[ibysaiyan]

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

 

[eaglelake]

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

 

[AakashPandita]

Oh i see. thanks.