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# Is this derivation RIGHT(physics)? Answered

Hi to whoever is reading this.Yesterday me and my friend derived an expression relating mass of a photon and its frequency .Can anyone tell me if its right?Ps I'm only 15 years old so i may not be right .
Here is the link for it: the folder name isPMF.docx and anyone can view it

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## 11 Replies

From the way you phrased your question, I'm going out on a limb and saying "No." I'm still waiting for the document, but photons don't have mass. Ever. Period. They have energy, which is given by E = hf (where h is Planck's constant); and they have momentum, which is given by p = hλ (where λ is wavelength). The energy and momentum are related by E = pc (notice that there is no "m" or "1/2" or any other Newtonian stuff in there).

...And there's your document. Your problem is at the top of page 2. You assumed non-relativistic momentum (i.e., you assumed p = mv), which is just wrong. Light is purely relativistic, so the Newtonian approximation is as wrong as it can ever be :-) Everything after that is just wrong, because your starting assumption is wrong.

Strontium (author)2013-12-05

Why should photons have a relativistic mass?

kelseymh (author)2013-12-05

They shouldn't, and don't. Their rest mass is zero. If you try to apply the usual Lorentz factor, what you get is zero divided by zero (from γ = 1/sqrt(1-1)), which is undefined.

Jack A Lopez (author)2013-12-05

I think you're missing a division operator there.  The momentum carried by a single photon is Plank's constant divided by wavelength, meaning red (longer wavelength) photons carry less momentum than blue (shorter wavelength) photons.

p  = h/λ  =  h/(c/f)  =  h*f/c  =  E/c

Of course there are  also typos in the reply I wrote.  All the 2s seem to be missing from my c-squareds.
So where it says (c^), it should say (c^2)
;-P

kelseymh (author)2013-12-05

Hi, Jack. Yes, there is a "/" missing in my post. I wonder if I ate it when I pasted in the lambda :-(

Josehf Murchison (author)2013-12-05

Dude you just gave me a reason not to tinker with a six inch portable TV I have.

The newer TVs are lousy at explaining Big Bang theory red shift and TV static.

Digital TVs don’t give you the static.

Yea I've been waiting 10 minuts and I still cant get the doc.

Isn't it nice to be 15 and full of wonder.

Joe

kelseymh (author)2013-12-05

Heh. I'd much rather have a 15-year-old exploring this stuff and trying to come up with answers!

...And my response was pretty harsh, considering. I'm going to post a followup apologizing and clarifying.

Josehf Murchison (author)2013-12-05

Dude I have had them more than one.

I say build it you learn.

First it is cool to watch it blow up.

Second it teaches you to finish a job.

Third if it doesn’t work it doesn’t kill you unless it blows up.

Keep a straight face that is the hard part.

As a parent don’t get mad and don’t laugh no matter how funny it is.

Most important do not let your wife see you laugh, you won’t be laughing long.

kelseymh (author)2013-12-05

Hey, I just re-read my previous response to you, and I should apologize. My tone was harsher than it needed to be. Unintentionally or not, I should be more polite, and will try to be next time.

Having read your note, I very much like the way you laid everything out -- you defined your notation and explained each step. That made it very easy to follow, and to identify the point where you went off.

Please feel free to follow up with questions, or post questions to my Orangeboard, if you wish. I will try not to be as snarky as I was this time. Good luck!

steveastrouk (author)2013-12-06

Yes, let he who is without a calculator or slide-rule cast the first stone.

Jack A Lopez (author)2013-12-05

It looks to me like the math is right.  Although some of the assumptions are a little funny.

By using the classical formulas for momentum, p=m*v, and kinetic energy, E=0.5*m*v^2, you are implicitly assuming your photon is a classical particle with some mass.

Coincidentally, you get the same answer as that predicted by the notion of mass-energy equivalence, that is m=E/(c^).

The trouble is photons are not supposed to have mass, well rest mass that is.   If they did this would contradict special relativity,
http://en.wikipedia.org/wiki/Special_relativity
which essentially says nothing with mass will be observed moving faster than the speed of light.  Since photons move at the speed of light, they cannot have any rest mass.

However if you've got a box with some photons bouncing around in it, supposedly that box
will be observed to have slightly more mass than it would without the photons in it.  This is just going back to this notion of mass-energy equivalence.
http://en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence

That is to say, energy U stored in a box, no mater what kind of energy, makes the box heavier, that is more massive, by an amount U/(c^).

That's the story.  If true, it means compressed springs weigh more than uncompressed springs, charged capacitors weigh more than discharged ones, and hot coffee weighs more than cold coffee, etc.

I don't know if I've done a clear job of explaining this or not.  Basically what's going on here is, as a consequence of Special Relativity, the mass some thing is observed to have is a function of how fast that thing is moving relative to the observer.

If the thing is not moving, then just its "rest mass", m0, is observed.

But if that thing is moving at some significant fraction of c, relative to the observer, then it's mass will be measured to be something much larger.

m(as a function of v)  = m0*(  1/ (1-v^2/c^)^(1/2)  )

where that  expression (  1/ (1-v^2/c^)^(1/2)  ) is given a special name, the Lorentz factor.  The Greek letter lowercase-gamma, γ, is often used for the Lorentz factor, for some reason or another.
http://en.wikipedia.org/wiki/Lorentz_factor