# 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.

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