Does the De Broglie equation relate to all particles?

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Offline jeffreyH

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Can the relationship [tex]\rho\,=\,\frac{h}{\lambda }[/tex] apply to all particles? If so then does [tex]\rho[/tex] still represent mv?

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Offline chiralSPO

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Re: Does the De Broglie equation relate to all particles?
« Reply #1 on: 03/10/2015 16:07:27 »
I believe the relationship holds for all particles. ρ = mv for all particles except for a photon, for which mv doesn't make any sense... so it's just ρ = h/λ...

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #2 on: 03/10/2015 16:24:21 »
That is what I thought. Just confirming it. So then particles with mass have the relationship [tex]\frac{h}{\rho \lambda}\,=\,1[/tex] which cannot be true for massless particles traveling with a set velocity.

So then we should be able to say since [tex]v\,=\,\frac{h}{m \lambda}[/tex] then kinetic energy can be expressed as [tex]Ke\,=\,\frac{1}{2}m\left( \frac{h}{m \lambda} \right)^2[/tex]. This will then reduce to [tex]Ke\,=\,\frac{1}{2m}\left( \frac{h}{\lambda} \right)^2[/tex]. If I am wrong I would like to know.

It is interesting to consider that when c=G=1 then 2m describes the radius of the event horizon of a black hole.

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Offline chiralSPO

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Re: Does the De Broglie equation relate to all particles?
« Reply #3 on: 03/10/2015 17:57:57 »
That is what I thought. Just confirming it. So then particles with mass have the relationship [tex]\frac{h}{\rho \lambda}\,=\,1[/tex] which cannot be true for massless particles traveling with a set velocity.

So then we should be able to say since [tex]v\,=\,\frac{h}{m \lambda}[/tex] then kinetic energy can be expressed as [tex]Ke\,=\,\frac{1}{2}m\left( \frac{h}{m \lambda} \right)^2[/tex]. This will then reduce to [tex]Ke\,=\,\frac{1}{2m}\left( \frac{h}{\lambda} \right)^2[/tex]. If I am wrong I would like to know.

This algebra looks correct to me.  [tex]\frac{h}{\rho \lambda}\,=\,1[/tex] is true for light, but it just reduces to 1 = 1, so it's just not a useful expression on its own. You can't get from there to the v of a massless particle, but the rest of the math works just fine for particles with mass.


It is interesting to consider that when c=G=1 then 2m describes the radius of the event horizon of a black hole.

I don't know what c and G mean here, they are not mentioned previously in this post.
« Last Edit: 03/10/2015 18:00:40 by chiralSPO »

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #4 on: 03/10/2015 18:01:24 »
I mean when c and G are eliminated for convenience when working out gravitational equations. Sorry for the confusion. Please ignore that reference. It was an afterthought.

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #5 on: 03/10/2015 18:19:36 »
That is what I thought. Just confirming it. So then particles with mass have the relationship [tex]\frac{h}{\rho \lambda}\,=\,1[/tex] which cannot be true for massless particles traveling with a set velocity.

So then we should be able to say since [tex]v\,=\,\frac{h}{m \lambda}[/tex] then kinetic energy can be expressed as [tex]Ke\,=\,\frac{1}{2}m\left( \frac{h}{m \lambda} \right)^2[/tex]. This will then reduce to [tex]Ke\,=\,\frac{1}{2m}\left( \frac{h}{\lambda} \right)^2[/tex]. If I am wrong I would like to know.

This algebra looks correct to me.  [tex]\frac{h}{\rho \lambda}\,=\,1[/tex] is true for light, but it just reduces to 1 = 1, so it's just not a useful expression on its own. You can't get from there to the v of a massless particle, but the rest of the math works just fine for particles with mass.


It is interesting to consider that when c=G=1 then 2m describes the radius of the event horizon of a black hole.

I don't know what c and G mean here, they are not mentioned previously in this post.

I think you may have missed my point. I was referring th the form [tex]\frac{h}{mv \lambda}\,=\,1[/tex]. It depends upon whether this expression can be true for massive particles. I may have gotten it the wrong way round. If it is true then other things follow. It requires the wavelength to be constant for massive particles in the local F o R.

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #6 on: 03/10/2015 18:37:08 »
To illustrate the point the equation can be reformulated as [tex]\lambda\,=\,\sqrt{\frac{1}{2m}} \frac{h}{\sqrt{Ke}}[/tex] showing the spread in the wavelength due to kinetic energy and therefore momentum. What, however, happens relativistically?
« Last Edit: 03/10/2015 18:38:57 by jeffreyH »

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Offline chiralSPO

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Re: Does the De Broglie equation relate to all particles?
« Reply #7 on: 03/10/2015 18:53:43 »
p = mv is a simplified equation that breaks down at relativistic speeds, gets supplanted by another more complex equation:

https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation
http://www.microscopy.ethz.ch/properties.htm

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #8 on: 03/10/2015 20:14:24 »
Well that makes it more interesting and a bit of a challenge.

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #9 on: 04/10/2015 00:00:49 »
OK so using the three-velocity where U = (ux, 0, 0) the particle can be constrained to move along the x-axis. So that [tex]P\,=\,\gamma(u_x)m_0U[/tex]. I will have to think a bit more how to incorporate this.

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #10 on: 04/10/2015 01:28:34 »
I'll have a stab at a first step. So [tex]\lambda\,=\,\frac{h}{\gamma(u_x)m_0U}[/tex]. Then [tex]U\,=\,\frac{h}{\gamma(u_x)m_0 \lambda}[/tex] so [tex]Ke\,=\,\frac{1}{2}\gamma(u_x)m_0 \left(\frac{h}{\gamma(u_x)m_0\lambda}\right)^2[/tex]. The resulting equation should now be [tex]Ke\,=\,\frac{1}{2\gamma(u_x)m_0} \left(\frac{h}{\lambda}\right)^2[/tex]. I haven't rigorously checked this though.
« Last Edit: 04/10/2015 01:31:55 by jeffreyH »

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Offline PmbPhy

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Re: Does the De Broglie equation relate to all particles?
« Reply #11 on: 08/10/2015 16:19:59 »
Quote from: jeffreyH
Can the relationship [tex]\rho\,=\,\frac{h}{\lambda }[/tex] apply to all particles? If so then does [tex]\rho[/tex] still represent mv?
Yes.

Quote from: chiralSPO
I believe the relationship holds for all particles. ρ = mv for all particles except for a photon, for which mv doesn't make any sense...
If you're speaking about mass defined as proper mass then you're. However, if you were speaking about mass as the more useful concept relativistic mass (RM) then you'd be quite wrong.

Three texts which use RM are and calculate the mass of a photon are listed at:
http://home.comcast.net/~peter.m.brown/ref/relativistic_mass.htm

However, in my opinion, you should have made it clear to Jeff which mass you had in mind.

Most SR texts which use the concept of relativistic mass readily define mass as m = p/c or as m = hf/c2. I created list of such textbooks and placed it in a webpage on my old website. The texts I have listed there are

Relativity: Special, General and Cosmological by Wolfgang Rindler, Oxford Univ. Press, (2001).
From Introducing Einstein's Relativity by Ray D'Inverno, Oxford Univ. Press, (1992).
Special Relativity by A. P. French, MIT Press, (1968).

Those are just three. There are many else of course. These are just examples to illustrate the point.

Jeff: If you really want to read an article which makes all of this quite clear then you can read the article I wrote. It was published in an Indian Journal and is now in a book too. :)  It's online at:
http://arxiv.org/abs/0709.0687

Essentially the definition of inertial mass is quite simple and works in all possible situations of closed systems. I.e. mass is defined as the quantity m such that p = mv is a conserved quantity.

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Offline chiralSPO

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Re: Does the De Broglie equation relate to all particles?
« Reply #12 on: 08/10/2015 17:22:10 »

If you're speaking about mass defined as proper mass then you're [ RIGHT??? ]. However, if you were speaking about mass as the more useful concept relativistic mass (RM) then you'd be quite wrong.
...
However, in my opinion, you should have made it clear to Jeff which mass you had in mind.
...

I mean proper mass. I rarely have to address relativistic mass, so my default has always been proper mass. I will always specify "relativistic" if that is what I mean. I suppose that for someone who constantly has to consider relativity, the opposite would be true, and proper mass would need to be specified, while relativistic mass would make a reasonable default...

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Offline lightarrow

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Re: Does the De Broglie equation relate to all particles?
« Reply #13 on: 08/10/2015 17:36:15 »

If you're speaking about mass defined as proper mass then you're [ RIGHT??? ]. However, if you were speaking about mass as the more useful concept relativistic mass (RM) then you'd be quite wrong.
...
However, in my opinion, you should have made it clear to Jeff which mass you had in mind.
...

I mean proper mass. I rarely have to address relativistic mass, so my default has always been proper mass. I will always specify "relativistic" if that is what I mean. I suppose that for someone who constantly has to consider relativity, the opposite would be true, and proper mass would need to be specified, while relativistic mass would make a reasonable default...
It's the other way round.
Nuclear physicists, elementary particle physicists, high energies physicists and theoretical physicists always use "mass" as "invariant mass".
("Proper" mass is not correct for particles like photons).

--
lightarrow
« Last Edit: 08/10/2015 17:40:51 by lightarrow »

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #14 on: 09/10/2015 06:03:22 »
Jeff: If you really want to read an article which makes all of this quite clear then you can read the article I wrote. It was published in an Indian Journal and is now in a book too. :)  It's online at:
http://arxiv.org/abs/0709.0687

I have read the paper before but I will go back and review it again.

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #15 on: 11/10/2015 17:46:59 »
I have been reading through Griffith's Introduction to Quantum Mechanics. It struck me that the kinetic equation from above [tex]Ke\,=\,\frac{1}{2\gamma(u_x)m_0} \left(\frac{h}{\lambda}\right)^2[/tex] was in a form similar to [tex]H\,=\,\frac{1}{2m}\left[\rho^2\,+\,\left(m \omega x)^2 \right) \right][/tex]. Since [tex]\rho[/tex] and x are operators I am wondering how to modify the kinetic energy equation so that it is complex. Is it nonense to try?

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Offline jeffreyH

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Re: Does the De Broglie equation relate to all particles?
« Reply #16 on: 11/10/2015 22:27:04 »
AS a side note but related in simple harmonic motion "the period of oscillation is independent of both the amplitude and gravitational acceleration". See

https://en.wikipedia.org/wiki/Simple_harmonic_motion#Mass_on_a_spring

If in this case the wavelength does not vary due to gravity then maybe the wave function is also unaffected by gravity and cannot be the cause of time dilation. An increase in relativistic mass however may affect the wave function. It may only apply to binding energy and not quark mass. So the quark masses should be neglected when determining relativistic mass. This would indicate a gluon/gluon type interaction if the force carrier for gravitation is at all gluon-like. THis increase in binding energy itself may slow down interactive processes.