de Broglie's relation derivation

Question

While going through de Broglie's relation in my textbook I was stuck by a derivation:

Here $$\ E=mc^2 $$ has been applied to photons. That confused me as I thought that mass of photon=0 but a photon does have energy. So I came across this post on SE. But this has confused me more. If the equation does not apply for photons then why have we used it here to derive the Broglie's relation?

Having seen a couple of your questions now: you really, really need to buy a new book. Please don't keep trying to decipher this one. — knzhou, Jul 10 '16 at 18:06
@knzhou this is a chem book and I had to ask this ques on the chem SE but no one answers there so I asked it here. What could I refer to for chem...? — oshhh, Jul 10 '16 at 18:14
Try Chemical Principles, by Atkins. Or, try the free lecture notes and videos for the course 5.111 on MIT OCW. — knzhou, Jul 10 '16 at 18:17
@knzhou Thank you...will refer to the lectures and notes for sure :) — oshhh, Jul 10 '16 at 18:20

score 4 · Answer 1 · edited Apr 18 '20 at 17:09

4

Okay, so first, here's another good explanation for a photon's "mass":

...the word "mass" has been used in two different ways in physics. One was the way Einstein used it in $E=mc^2,$ where mass is really just the same thing as energy (E) but measured in different units. This is the same "m" that you multiply velocity by to find momentum (p), and thus is sometimes called the inertial mass. It's also the mass that provides the source of gravitational effects. Light has this "$m$" because it has energy.$^1$

This can be briefly summarized as follows:

There are two types of mass, the type you and I think about normally, like how much mass an apple has. The second type is what light has - inertial mass. This is the sort of mass Einstein used in $E=mc^2$, that really is just the same thing as energy, but in different units.

The equation does apply for photons. A way you can see this is true is the fact that light is affected by gravity - bent by the mass of stars/planets, and trapped by black holes. It isn't "against the rules" to use the definition of mass described above.

However, as Daniel Kerr says in the comments, de Broglie's derivation is for matter, so $E = mc^2$ definitely applies. The photon part isn't really related to what the book is talking about (at the very least, it makes it unclear).

I hope this helps!

$^1$This quote is from this website.

edited Apr 18 '20 at 17:09

Glorfindel

1,424

answered Jul 10 '16 at 12:28

auden

7,027

Great answer... so this mass is simply E/c^2 and has no relation with the other mass? – oshhh Jul 10 '16 at 12:40
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I believe so, yes. – auden Jul 10 '16 at 12:40
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Hi Heather. Photons don't react to gravity because of some hypothetical mass. In Newtonian gravity the gravitational acceleration is independent of the mass, i.e. all objects fall at the same speed, and this applies even in the limit of zero mass. In GR things are a bit more complicated but it still remains the case that a particle does not need a mass to move in a curved trajectory. So you cannot use the fact that light is bent by gravity to infer it has a mass. It does not. There is no useful sense in which a photon has a mass. – John Rennie Jul 10 '16 at 19:25
@JohnRennie, I'm not saying it has a mass the way we think about a mass...as I was researching this question, I found that there were two types of mass as detailed at the website I cite, the mass like how much I weigh, and the mass that is involved in Einstein's equation $E=mc^2$. It's the second type of mass that light has, which isn't really like normal mass at all (from what I've read). – auden Jul 10 '16 at 19:28
If you start talking about photons as having a mass in any sense physicists around you will faint away in sheer horror. It is not a useful concept and there is no useful sense in which a photon has a mass. – John Rennie Jul 10 '16 at 19:37
@JohnRennie, okay, then how would you explain the fact that photons have energy but no mass...especially considering $E=mc^2$. – auden Jul 10 '16 at 19:40
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That fact requires no explanation. We are used to associating energy and momentum with massive particles from our everyday experience. But this experience lets us down in the weird world of quantum mechanics. Massless particles can have both energy and momentum because, well, that's the way QM works. If it's any consolation generations of physics students have made the same complaint about the implausibility of this, and no doubt future generations will continue to do so :-) – John Rennie Jul 10 '16 at 19:44
@JohnRennie, should I delete my answer, or severely edit it, or what? – auden Jul 10 '16 at 19:46
You can't delete an accepted answer. I know this to my cost as I have an answer that is wrong but that I can't delete because it's accepted. My strategy is just to hope that no-one notices :-) I'm not sure I would worry too much about it. In any case you shouldn't take my views as gospel. Searching this site will find lots of stuff on photon mass from people far more experienced than me, and my guess is you'll find it interesting reading. – John Rennie Jul 10 '16 at 19:50
@JohnRennie, thanks for your advice! Learned something new today. =) – auden Jul 10 '16 at 19:51
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I think the answer is mostly fine I would just edit it with the clarification that de Broglie's derivation is for matter, so $E=mc^2$ definitely applies. The way the text describes it is misleading as they seem to be "perturbing" the photon into matter. It's not a mathematical procedure, just a semantic one. – Daniel Kerr Jul 10 '16 at 20:13
@DanielKerr, thank you, I edited it as you suggested. – auden Jul 10 '16 at 20:17

score 4 · Accepted Answer · answered Jul 10 '16 at 19:36

The argument in your book is heuristic, that is it presents a reasonable justification for the de Broglie wavelength but it is not a proof.

The correct expression for the de Broglie wavelength is:

$$ \lambda = \frac{h}{p} \tag{1} $$

For a massive particle in the non-relativistic limit the momentum is given by:

$$ p = mv $$

so we can write equation (1) as:

$$ \lambda = \frac{h}{mv} $$

For a photon we can use the relativistic equation linking energy and momentum:

$$ E^2 = p^2c^2 + m^2c^4 $$

And since for a photon $m=0$ this gives us $E=pc$ and therefore equation (1) becomes:

$$ \lambda = \frac{hc}{E} $$

Rearranging this gives the usual expression:

$$ E = \frac{hc}{\lambda} = h\nu $$

By comparison with a massive particle it is tempting to write the momentum of a photon as:

$$ p = mc $$

where $c$ is the photon velocity and $m$ is some hypothetical mass given by $E/c^2$. Indeed if we substitute $m=E/c^2$ in the above equation we get back the correct relativistic result:

$$ p = \frac{E}{c^2}c = \frac{E}{c} $$

But this is just numerology. The photon does not have a mass in any useful sense of the word.

score 1 · Answer 3 · answered Jul 10 '16 at 15:05

The text "any material particle … by analogy with photons" shows that the author is talking about massless photons in contrast to mass particles.

That means that his following formulations are for the best imprecise. $ E = mc^2$ applies to mass which is rest energy, and it does not apply to other types of energy. Apparently, the author wanted to say:

"if the energy of the photon was rest energy, the following formula would apply"

and his objective was the derivation of the De-Broglie equation.

de Broglie's relation derivation

3 Answers3