- #1

- 5

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter sam_p_r
- Start date

- #1

- 5

- 0

- #2

- 136

- 1

Well the relations between energy/momentum and frequency/wavelength hold generally:

[tex]E=h\nu[/tex], [tex]p=h/\lambda[/tex]. It's the relation between E and p which is relevant in going from non-relativistic to relativistic to ultra-relativistic limits (as determined by the ratio [tex]pc/mc^{2}[/tex]). You could always use [tex]E^{2}=(pc)^{2}+(mc^{2})^{2}[/tex]. We ignore the second term in the ultra-relativistic limit, while in the non-relativistic limit, it simplifies to [tex]E=p^{2}/2m[/tex]. Otherwise, use the full expression.

[tex]E=h\nu[/tex], [tex]p=h/\lambda[/tex]. It's the relation between E and p which is relevant in going from non-relativistic to relativistic to ultra-relativistic limits (as determined by the ratio [tex]pc/mc^{2}[/tex]). You could always use [tex]E^{2}=(pc)^{2}+(mc^{2})^{2}[/tex]. We ignore the second term in the ultra-relativistic limit, while in the non-relativistic limit, it simplifies to [tex]E=p^{2}/2m[/tex]. Otherwise, use the full expression.

Last edited:

- #3

- 371

- 1

Well the relations between energy/momentum and frequency/wavelength hold generally:

[tex]E=h\nu[/tex], [tex]p=h/\lambda[/tex]. It's the relation between E and p which is relevant in going from non-relativistic to relativistic to ultra-relativistic limits (as determined by the ratio [tex]pc/mc^{2}[/tex]). You could always use [tex]E^{2}=(pc)^{2}+(mc^{2})^{2}[/tex]. We ignore the second term in the ultra-relativistic limit, while in the non-relativistic limit, it simplifies to [tex]E=p^{2}/2m[/tex]. Otherwise, use the full expression.

Could you elaborate a little more about the ultra-relativistic case?

- #4

- 371

- 1

What specific problem do you see?.

- #5

- 5

- 0

What specific problem do you see?.

That E does not equal cp, it equals the square root of (cp)^2 + (mc^2)^2. So for de broglie equation, you have to neglect mc^2, which I can see as reasonable if it is very small in comparison to cp, but there are many occasions where it wont be.

de broglie combines E=hc/lambda and E=cp,

So if you dont have E=cp you cant get the de broglie relation as it is.

- #6

- 5

- 0

[tex]E=h\nu[/tex], [tex]p=h/\lambda[/tex]. It's the relation between E and p which is relevant in going from non-relativistic to relativistic to ultra-relativistic limits (as determined by the ratio [tex]pc/mc^{2}[/tex]). You could always use [tex]E^{2}=(pc)^{2}+(mc^{2})^{2}[/tex]. We ignore the second term in the ultra-relativistic limit, while in the non-relativistic limit, it simplifies to [tex]E=p^{2}/2m[/tex]. Otherwise, use the full expression.

So what you're saying is that it only applies (in the exact lambda=h/p) for relativistic particles? (as its the only time where you can neglect mc^2).

Because in many examples in textbooks it applies the formula to non-relativistic conditions.

- #7

- 371

- 1

So what you're saying is that it only applies (in the exact lambda=h/p) for relativistic particles? (as its the only time where you can neglect mc^2).

Because in many examples in textbooks it applies the formula to non-relativistic conditions.

I think it applies to both relativistic and non-relativistic conditions. The m that you list is actually m0, the rest mass. P should be the relativistic momentum and E is the relativistic energy. These both have a correction due to speed of the particle (or refference frame).

In the limit where v=c this correction disappears.

I don't think the rest mass energy can be neglected in either the relativisti or non-relativistic case. But I may be wrong.

I think the de Broglie waves have been superceeded by the QM wave funtion, but I haven't looked at the difference. I know that the Schrodinger equation is not Lorentz-invariant and that the Dirac equation is. I would assume that there is no problem with the wave function itself. It would be nice to compare the de Broglie wave with the wave function. This must be done in many books but I can't find it in the ones I have. Maybe I can find something on the Web.

- #8

- 371

- 1

Now, reading again about de Broglie's waves, these are faster than light. I don't remember any mention in QM about the wave funtion of a free particle being faster than light. Maybe I just didn't pay attention to that.

Maybe that's where the energy corresponding to the rest mass comes into play!

If anybody can make this clear for us I'll appreciate it.

Share:

- Replies
- 2

- Views
- 463