Post by: jeffreyH on 03/10/2015 12:35:51
apply to all particles? If so then does 
still represent mv?
Post by: chiralSPO on 03/10/2015 16:07:27
Post by: jeffreyH on 03/10/2015 16:24:21
which cannot be true for massless particles traveling with a set velocity.So then we should be able to say since
then kinetic energy can be expressed as 
. This will then reduce to 
. 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.
Post by: chiralSPO on 03/10/2015 17:57:57
That is what I thought. Just confirming it. So then particles with mass have the relationship
So then we should be able to say since
This algebra looks correct to me.
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.
Post by: jeffreyH on 03/10/2015 18:01:24
Post by: jeffreyH on 03/10/2015 18:19:36
That is what I thought. Just confirming it. So then particles with mass have the relationship
So then we should be able to say since
This algebra looks correct to me.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
. 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.
Post by: jeffreyH on 03/10/2015 18:37:08
showing the spread in the wavelength due to kinetic energy and therefore momentum. What, however, happens relativistically?
Post by: chiralSPO on 03/10/2015 18:53:43
https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation
http://www.microscopy.ethz.ch/properties.htm
Post by: jeffreyH on 03/10/2015 20:14:24
Post by: jeffreyH on 04/10/2015 00:00:49
. I will have to think a bit more how to incorporate this.
Post by: jeffreyH on 04/10/2015 01:28:34
. Then 
so 
. The resulting equation should now be 
. I haven't rigorously checked this though.
Post by: PmbPhy on 08/10/2015 16:19:59
Quote from: jeffreyH
Can the relationshipYes.
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.
Post by: chiralSPO 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...
Post by: lightarrow on 08/10/2015 17:36:15
It's the other way round.
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...
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
Post by: jeffreyH 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.
Post by: jeffreyH on 11/10/2015 17:46:59
was in a form similar to 
. Since and x are operators I am wondering how to modify the kinetic energy equation so that it is complex. Is it nonense to try?
Post by: jeffreyH on 11/10/2015 22:27:04
https://en.wikipedia.org/wiki/Simple_harmonic_motion#Mass_on_a_spring (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.