I have a tricky problem involving an electron and photon.

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Homework Statement


The question I have is simple enough: at what energy do an electron and a photon have the same wavelength?

Homework Equations


I know that for a photon, λ=[itex]\frac{hc}{E}[/itex] and for an electron, λ=[itex]\frac{hc}{\sqrt{E^{2}-mc^{2}}}[/itex]

The Attempt at a Solution


I can't just equate the two, as that results in the absurd result mc^2=0. I can't think of a method that works. Are there any relativistic corrections that I've missed?
 

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  • #2
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At the same energy, photon and electron cannot have the same momentum, their corresponding wavelength cannot be equal.
 
  • #3
rude man
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Homework Statement


The question I have is simple enough: at what energy do an electron and a photon have the same wavelength?

Homework Equations


I know that for a photon, λ=[itex]\frac{hc}{E}[/itex] and for an electron, λ=[itex]\frac{hc}{\sqrt{E^{2}-mc^{2}}}[/itex]

The Attempt at a Solution


I can't just equate the two, as that results in the absurd result mc^2=0. I can't think of a method that works. Are there any relativistic corrections that I've missed?
The problem is with your energy of the electron, I believe.

What is the kinetic energy K of the electron as a function of its deBroglie wavelength?
(Hint: K = K(p,E0), p = momentum, E0 = rest energy, then substitute deBroglie λ for p and equate this K(λ) to the photon energy hc/λ.)

Remember that when we speak of the energy of a particle with finite rest mass we mean its Kinetic energy, not its total Energy.

EDIT: mfb may be right, probably is, but try to solve for lambda anyway. You'll find out soon enough!

EDIT EDIT: The answer is lambda = infinity.
 
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  • #4
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λ=[itex]\frac{hc}{\sqrt{E^{2}-mc^{2}}}[/itex]
Check the units under the square root. Do you think they are considtent?
 
  • #5
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The problem is with your energy of the electron, I believe.

What is the kinetic energy K of the electron as a function of its deBroglie wavelength?
(Hint: K = K(p,E0), p = momentum, E0 = rest energy, then substitute deBroglie λ for p and equate this K(λ) to the photon energy hc/λ.)

Remember that when we speak of the energy of a particle with finite rest mass we mean its Kinetic energy, not its total Energy.

EDIT: mfb may be right, probably is, but try to solve for lambda anyway. You'll find out soon enough!

EDIT EDIT: The answer is lambda = infinity.
Thanks for the help. Will the momentum need a γ correction?

Check the units under the square root. Do you think they are considtent?
It's from E2=(pc)2+(mc2)2

EDIT:
Is the equation [itex]\lambda=\frac{hc}{\sqrt{2E_{k}mc^{2}}}[/itex] valid?
 
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  • #6
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It's from E2=(pc)2+(mc2)2
It does not follow from that. Which is not surprising, because it is inconsistent and cannot follow from anything meaningful.
 
  • #7
rude man
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Thanks for the help. Will the momentum need a γ correction?

No, but it turns out the only solution is infinite lambda and zero energy.

It's from E2=(pc)2+(mc2)2
You should write the rest mass as m0.
 
  • #8
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Check the units under the square root. Do you think they are considtent?
Good catch, I missed that, even if it does not change the result.

@rude man: m is used for mass ("rest mass") everywhere in physics, the usage of m0 (as notation) or a relativistic mass are deprecated.
 
  • #9
rude man
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Good catch, I missed that, even if it does not change the result.

@rude man: m is used for mass ("rest mass") everywhere in physics, the usage of m0 (as notation) or a relativistic mass are deprecated.
No.

When dealing in relativity, m is relativistic mass. As in E = mc2.
Rest mass is then distinguished by giving it the symbol m0.. Check any textbook, e.g. Weidner and Sells.

Obviously, in dealing with Newtonian mechanics, m is rest mass.
 
  • #10
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No.

When dealing in relativity, m is relativistic mass. As in E = mc2.
Rest mass is then distinguished by giving it the symbol m0.. Check any textbook, e.g. Weidner and Sells.

Obviously, in dealing with Newtonian mechanics, m is rest mass.
mfb is right, the relativistic correction is on the momentum, not the mass.
 
  • #11
rude man
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mfb is right, the relativistic correction is on the momentum, not the mass.
That's what I said, and why I insisted (and insist) that the mass here be denoted m0 instead of m, to clarify the fact that it is not the relativistic but the rest mass.

The correct expression thus becomes, as I said, E2 = (pc)2 + (m0c2)2.
 
  • #12
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There is no such thing as relativistic mass.
 
  • #14
Dick
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Gee, I hope you're kidding ... :surprised
Of course there is a notion of relativistic mass. It just isn't used much anymore. So more modern books just use m as rest mass and don't talk about 'relativistic mass'. It causes more confusion than it is useful.
 
  • #15
rude man
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Of course there is a notion of relativistic mass. It just isn't used much anymore. So more modern books just use m as rest mass and don't talk about 'relativistic mass'. It causes more confusion than it is useful.
Thanks for the clue-in, Dick. Amazing!

So you can't say m = γm0 any more? So how do they e.g. write kinetic energy? Used to be mc2 - m0c2. What is it now?

Oh my.

So when the OP wrote E2 = (pc)2 + (mc2)2, that was right because m is now rest mass? And what symbol do you use for (may God forgive me for saying it) relativistic mass?

And E = mc2 is now "energy = rest mass times c2"?

I am impressed. I hope Einstein knows about this ...
 
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  • #16
Dick
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Thanks for the clue-in, Dick. Amazing!

So you can't say m = γm0 any more? So how do they e.g. write kinetic energy? Used to be mc2 - m0c2. What is it now?

Oh my.

So when the OP wrote E2 = (pc)2 + (mc2)2, that was right because m is now rest mass? And what symbol do you use for (may God forgive me for saying it) relativistic mass?

And E = mc2 is now "energy = rest mass times c2"?

I am impressed. I hope Einstein knows about this ...
Kinetic energy is just ##mc^2 (\gamma-1)##. I, personally, have always taken E=mc^2 to mean equivalence between rest mass and energy. Energy in a boosted frame is ##\gamma mc^2##. I didn't make the choice to avoid the use of 'relativistic mass'. But that seems to be the way the market is choosing. I don't think Einstein would be upset. It's just a change in notation.
 
  • #17
rude man
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Kinetic energy is just ##mc^2 (\gamma-1)##. I, personally, have always taken E=mc^2 to mean equivalence between rest mass and energy. Energy in a boosted frame is ##\gamma mc^2##. I didn't make the choice to avoid the use of 'relativistic mass'. But that seems to be the way the market is choosing. I don't think Einstein would be upset. It's just a change in notation.
I am rendered speechless. And I do think Einstein would seriously not be pleased. But OK, we bend with the wind ...
 
  • #18
Dick
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I am rendered speechless. And I do think Einstein would seriously not be pleased. But OK, we bend with the wind ...
I don't see why. You sprinkle a few more gamma's around. Drop some m0's. It's still just classical special relativity. There's no conceptual change, just notation. If Einstein were rendered speechless by this, I'd say he had a pretty thin skin. :)
 
  • #19
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If you think about it, an objects mass cannot increase as it's an extensive property based on its internal structure, which doesn't change with velocity.
 
  • #20
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@rude man: There is no need to get sarcastic. Check any recent (after 1980-1990) publication, you will not find the concept of a relativistic mass there.
 
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I'm not being sarcastic, it's just a reason why relativistic mass doesn't make sense.
 
  • #22
rude man
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No more F = d/dt (mv) (which was correct even relativistically)..
No more centripetal force = mv^2/R.
And, best of all ... no more p = mv!

All those traditional formulas rendered erroneous ... reminds me of the 'improvements' in Windows!

OK, I promise, this is my last comment on the subject. Apparently some folks are not happy with them.
 
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  • #23
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##F=\frac{dp}{dt}##
Centripetal force ##\frac{\gamma m v^2}{R}##
##p=\gamma m v##
Where is the issue?

@gsurfer04: I did not mean you in my post.
 
  • #24
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There are a few papers by Okun on the whole "relativistic mass" issue. Quite an interesting read.
 
  • #25
rude man
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##F=\frac{dp}{dt}##
Centripetal force ##\frac{\gamma m v^2}{R}##
##p=\gamma m v##
Where is the issue?

@gsurfer04: I did not mean you in my post.
You just stated 'where is the issue'. All the traditional formulas now have to be changed by multiplying by gamma.

And I wasn't sarcastic. Just flabbergasted at the way the physics community has dumbed-down.
 

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