Q theory, Relativistic Mass and Einstein

[pmb_phy]

In another thread dextercioby posted the http://https://www.physicsforums.com/showpost.php?p=406225&postcount=13

And to answer your question,YES,people working in Q theory never use the concept of "relativistic mass".They use "energy" (zero-th component of the energy-momentum 4-vector)

I want to address this interesting comment here so as not to divert the other thread.

dextercioby - What's Q theory? I assume that it pertains to quantum mechanics??

I thought you made a good point, one I'm quite familiar with. But there is something I wanted to comment on regarding this point. When someone speaks of the mass of a particle (i.e. what you referred to as relativistic mass) they are speaking about classical particle dynamics. Not all of classical particle dynamics carries over to quantum mechanics (which I assume Q theory pertains to). Many often used concepts in relativity can't be used in the quantum domain. For instance, there is no concept of a worldline in quantum theory (or at least its quite different than in classical relativity). I don't believe that there is even a 4-velocity in quantum theory. These are imporant points to keep in mind when applying relativity to quantum mechanics.

Einstein is held responsable for that.Though he had 0 contribution to QFT.

I'm sorry but I don't understand your point here. Responsible for what? If you mean that Einstein didn't "like" relativistic mass after he completed relativity (in 1915) then that is a common misconception. What Einstein did not believe in a was a velocity dependant mass for a "material particle" (his term, not mine). That is often misconstrued to refer to relativistic mass. It might be said that Einstein did use the concept of (relativistic) mass in his own work such as his 1916 GR paper (in section 16) in his text The Meaning of Relativity (TMR) (pages 100-102) and in his book Evolution of Physics.

Einstein never used the term relativistic mass so one can't say for sure. He simply referred to "mass", or "inertial mass". In his text TMR he spoke of the inertial mass of a particle as being altered when there is a gravitational field present. What he was referring to was the quantity $m = m_0 dt/d\tau$ when v << c. This is not the same quantity as $m_0$ since $dt/d\tau$ does not equal one when v << c and there is a gravitational field present.

Pete

 

[Stingray]

When someone speaks of the mass of a particle (i.e. what you referred to as relativistic mass) they are speaking about classical particle dynamics. Not all of classical particle dynamics carries over to quantum mechanics (which I assume Q theory pertains to).

There are concepts of mass in quantum mechanics, but of course they are somewhat different than the classical ones. (4-)Momenta are used in quantum field theory, so mass is often defined as the magnitude of that vector. This is of course an analogy to proper mass. I haven't often heard people talk about "relativistic mass" in QFT or particle physics (although it does occasionally happen). They usually use the word "energy" instead, which is basically the same thing. It's annoying to always distinguish which type of mass you're talking about, so I think this is a good practice.

As to Einstein, he seems to have had many different views over the years on how to interpret his theories. You can usually find any (reasonable) viewpoint from Einstein's writings if you look hard enough.

 

[dextercioby]

In another thread dextercioby posted the http://https://www.physicsforums.com/showpost.php?p=406225&postcount=13
I want to address this interesting comment here so as not to divert the other thread.

dextercioby - What's Q theory? I assume that it pertains to quantum mechanics??

I thought you made a good point, one I'm quite familiar with. But there is something I wanted to comment on regarding this point. When someone speaks of the mass of a particle (i.e. what you referred to as relativistic mass) they are speaking about classical particle dynamics. Not all of classical particle dynamics carries over to quantum mechanics (which I assume Q theory pertains to). Many often used concepts in relativity can't be used in the quantum domain. For instance, there is no concept of a worldline in quantum theory (or at least its quite different than in classical relativity). I don't believe that there is even a 4-velocity in quantum theory. These are imporant points to keep in mind when applying relativity to quantum mechanics.

I'm sorry but I don't understand your point here. Responsible for what? If you mean that Einstein didn't "like" relativistic mass after he completed relativity (in 1915) then that is a common misconception. What Einstein did not believe in a was a velocity dependant mass for a "material particle" (his term, not mine). That is often misconstrued to refer to relativistic mass. It might be said that Einstein did use the concept of (relativistic) mass in his own work such as his 1916 GR paper (in section 16) in his text The Meaning of Relativity (TMR) (pages 100-102) and in his book Evolution of Physics.

Einstein never used the term relativistic mass so one can't say for sure. He simply referred to "mass", or "inertial mass". In his text TMR he spoke of the inertial mass of a particle as being altered when there is a gravitational field present. What he was referring to was the quantity $m = m_0 dt/d\tau$ when v << c. This is not the same quantity as $m_0$ since $dt/d\tau$ does not equal one when v << c and there is a gravitational field present.

Pete

Yep,Pete,i was referring actually to Quantum Theory,either Quantum Mechanics or Quantum Field Theory.I'm sorry for using that letter instead of the more familiar acronyms:QM&QFT.Quantum Mechanics uses the concept of 'mass' with the same meaning as in classical Hamilton (nonrelativistic) dynamics,so here,no problem whatsoever.As for QFT,constructed on (special)relativistic grounds,the word "mass" means that $$m_{0}$$ from any theory of relativity and especially the celebrated formula $$m=\gamma m_{0}$$.While for the concept "relativistic mass",they use the word "energy" and "Einstein is responsible for that".:tongue2: After all,he put an equality sign between 'relativistic mass' and 'energy'.$$E=mc^{2}$$.
I guess (maybe Einstein's formula,again) the other formula is more useful to QFT:$$E^{2}=m_{0}^{2} c^{4}+ \vec{p}^{2}c^{2}$$ and with the last 2 formulas we're given the explanation why QFT teachers never use the concept of 'relativistic mass'.Neither in their research,nor in their teaching.

As for 'poor' Mr.Einstein,he discovered some formulas as part of his theories that people found useful in other theories (like QFT) with which A.Einstein had nothing to do (e.g.QFT).Whether he liked those concepts he introduced,that's relevant only for history of physics,but not for physics and its evolution.

Daniel.

 

[pmb_phy]

There are concepts of mass in quantum mechanics, but of course they are somewhat different than the classical ones. (4-)Momenta are used in quantum field theory, so mass is often defined as the magnitude of that vector. This is of course an analogy to proper mass. I haven't often heard people talk about "relativistic mass" in QFT or particle physics (although it does occasionally happen).

For similar reasons you won't find people talk about velocity in QFT either.

One can speak of 4-momentum in quantum theory because energy and momentum are compatible observables so that proper mass is a well defined quantity. However since speed/velocity is not a well defined quantity one can't defined m = p/v. If one always means "proper mass" then its best so make that clear and stick to it. Otherwise it can get tedious.

They usually use the word "energy" instead, which is basically the same thing.

If it were basically the same then they'd have the same value in all possible cases. However that is not the case. For example, consider two inertial frames of reference in standard configuration. S' moves in the +x direction relative to S. Let there be a rod at rest in S'. In S' there are two forces on the rod. One force acts on the +x end acting in the -x direction. The other force acts on the -x end and is acting in the +x direction. The forces have the same magnitude as measured in S'. The rod therefore remains at rest in S' and remains moving at constant velocity in S. The mass of the rod is defined as the magnitude of the rod divided by the rods speed, i.e. m = p/v. But, in this particular case, the mass is not related to the total inertial energy E by E = mc². Therefore one can't say that (relativistic) mass and energy are the same thing. This is a fact that is all to often overlooked. Not by Einstein though. He figure that out back in 1907 in a round about way. I.e. in 1907 he didn't speak of the mass of a stressed body. He spoke of the kinetic energy of a stressed body. To be precise he showed that in such circumstances the kinetic energy is different when a body is under stress. I.e. if the energy of a stressed body as measured in the rest frame is E₀ then the kinetic energy is not $K = (\gamma - 1)E_0$.

It's annoying to always distinguish which type of mass you're talking about, so I think this is a good practice.

I would guess that's why particle physicists use the term "lifetime" rather than "proper lifetime". But they certainly don't mean the same thing.

As to Einstein, he seems to have had many different views over the years on how to interpret his theories. You can usually find any (reasonable) viewpoint from Einstein's writings if you look hard enough.

His TMR is his most popular and noted work on SR/GR.

Pete

 

[pmb_phy]

Yep,Pete,i was referring actually to Quantum Theory,either Quantum Mechanics or Quantum Field Theory.I'm sorry for using that letter instead of the more familiar acronyms:QM&QFT.Quantum Mechanics uses the concept of 'mass' with the same meaning as in classical Hamilton (nonrelativistic) dynamics,so here,no problem whatsoever.

If you're speaking about non-relativistic quantum mechanics then the mass is an intrinsic property (what SR calls proper mass) which is quite well defined. I was speaking of the same definition of mass. I don't know relativistic quantum mechanics so I'll have to leave it rest here. But I would suspect that one doesn't use the same definition of mass. In fact I'd have to say that momentum is even defined differently. In classical mechanics momentum is defined as $p = mv = \gamma m_0 v$. But one can't use this definition of mass since v is undefined.

While for the concept "relativistic mass",they use the word "energy" ..

See my comment to Stingray on this point.

After all,he put an equality sign between 'relativistic mass' and 'energy'..

That is only valid in certain special cases. Einstein showed that the correct and complete description of mass is given by the stress-energy-momentum tensor. The E = mc^2 relation holds all the time only when the object is a free-object or a point object with no internal structure. In other cases the relation does not hold. I gave Stingray an example. I didn't calculate it out but will if requested. Its a good exercise for the curious.

Pete

 

[jcsd]

I'm not kind of authority of relativstic QM, but there is a four-velocity operator (as indeed there is a velocity operator in bog-standard QM) as it is one of the postulates of QM that operators have the same functional relationship to the momentum operator as the observables they represent have to momentum in classical physics, so it is fairly obvious how to define a velcoity operator. Mass in relativstic quantum mechanics is not a dynamical quantity (I think tho' that's not true in quantum flavordynamics) so really it's exactly the same as it is in classical mechanics i.e. it's simply the inertial mass of the particle in it's rest frame.

 

[pmb_phy]

I'm not kind of authority of relativstic QM, but there is a four-velocity operator ...

Sorry. Never heard of it. However it sounds like you're speaking about a totally different animal than 4-velocity to me.

Do you know of an online source so I can see the definition of such a quantity? Thanks.

(as indeed there is a velocity operator ...

Nope. Never heard of that either.

..in bog-standard QM)

..nor that. Perhaps in this case you're referring to non-relativistic qm?

...as it is one of the postulates of QM that operators have the same functional relationship to the momentum operator as the observables they represent have to momentum in classical physics, so it is fairly obvious how to define a velcoity operator.

Which postulate is that? Do you mean an implication of one ore more postulates? I've never seen that in any list of postulates of QM that I've ever seen (e.g. as given in, say, Quantum Mechanics Cohen-Tannoudji, Diu and Laloe, Modern Quantum Mechanics, Sakuri etc.)

Here's a list of of the postulates of quantum mechanics
http://www.upei.ca/~physics/polson/course/P322/notes/ch9.pdf

Which one are you referring to?

In any case that doesn't sound right since you're assuming that all quantities have corresponding operators and I don't see that as being true.

Pete

 

[Stingray]

To be precise he showed that in such circumstances the kinetic energy is different when a body is under stress. I.e. if the energy of a stressed body as measured in the rest frame is E₀ then the kinetic energy is not $K = (\gamma - 1)E_0$.

Particle physicists do not allow for internal stresses, so it is the same thing to them.

 

[dextercioby]

I'm not kind of authority of relativstic QM, but there is a four-velocity operator (as indeed there is a velocity operator in bog-standard QM) as it is one of the postulates of QM that operators have the same functional relationship to the momentum operator as the observables they represent have to momentum in classical physics, so it is fairly obvious how to define a velcoity operator. Mass in relativstic quantum mechanics is not a dynamical quantity (I think tho' that's not true in quantum flavordynamics) so really it's exactly the same as it is in classical mechanics i.e. it's simply the inertial mass of the particle in it's rest frame.

I have only one question for you:What is the fundamental observable for a quantum mechanical system...?If u know the answer,you might as well reconsider some of your remarks.If not,please let me know.I'm willing to tell u.

Daniel.

 

[pmb_phy]

Particle physicists do not allow for internal stresses, so it is the same thing to them.

It is an equality, not an identity.

Recall what I was saying - If it were basically the same then they'd have the same value in all possible cases. dextercioby was talking about an equality, i.e. E = mc². I was pointing out that this is an equality (in some cases) between m and E and not a definition of m given E.

When the object is a point particle (i.e. zero dimensions, no internal structure) then the equality holds.

Question: The energy E of a photon is related to its frequency f by E = hf. Since this holds in all cases do you consided E and f to be the "same thing"?

Pmb

 

[dextercioby]

Here's a list of of the postulates of quantum mechanics
http://www.upei.ca/~physics/polson/course/P322/notes/ch9.pdf
Which one are you referring to?
Pete

Those postulates were written by an amature.They were definitely intedended for the nontheorist part of phyiscs community.To draw a parallel,it's like postulating invariance of phyisical laws without knowing a bit about tensors.You can say that,but if somebody asks u about the significance and the consequences of this fundamental principle of nature,u'd raise your shoulders and tell the courious:"If you're really interested,go and search for answers yourself.I don't have them."

Daniel.

PS.It really makes me sick when somebody mocks at the Postulates of QM. :yuck:

 

[jcsd]

Sorry. Never heard of it. However it sounds like you're speaking about a totally different animal than 4-velocity to me.

Yes it does appear to be different in some ways (and I don't think just the ways that the dynmaical variables of QM are different from their classical counterparts)

Do you know of an online source so I can see the definition of such a quantity? Thanks.

It's defined here: http://www.iop.org/EJ/article/0305-4470/17/12/015/jav17i12p2433.pdf

Nope. Never heard of that either.

Briefly discussed here:
http://www.lns.cornell.edu/spr/2000-08/msg0027589.html

..nor that. Perhaps in this case you're referring to non-relativistic qm?
Which postulate is that? Do you mean an implication of one ore more postulates? I've never seen that in any list of postulates of QM that I've ever seen (e.g. as given in, say, Quantum Mechanics Cohen-Tannoudji, Diu and Laloe, Modern Quantum Mechanics, Sakuri etc.)

Here's a list of of the postulates of quantum mechanics
http://www.upei.ca/~physics/polson/course/P322/notes/ch9.pdf

Which one are you referring to?

In any case that doesn't sound right since you're assuming that all quantities have corresponding operators and I don't see that as being true.

Pete

Here is one which lists it:

http://www-ec.njit.edu/~venanzi/chem658/lecture/notes_Ch3.pdf

 

[pmb_phy]

Here is one which lists it:

http://www-ec.njit.edu/~venanzi/chem658/lecture/notes_Ch3.pdf

Where? I don't see it. Do you have a page number or something?

The first link doesn't work for me by the way.

Pete

 

[Stingray]

It is an equality, not an identity.

I agree. I was just being a little loose with my wording .

 

[dextercioby]

Yes it does appear to be different in some ways (and I don't think just the ways that the dynmaical variables of QM are different from their classical counterparts)
It's defined here: http://www.iop.org/EJ/article/0305-4470/17/12/015/jav17i12p2433.pdf

I can't comment,as i didn't have access to the article.

Briefly discussed here:
http://www.lns.cornell.edu/spr/2000-08/msg0027589.html

You can define $\frac{d\hat{x}}{dt}=:\hat{v}$ in the Heisenberg picture,but inserting this definition in the time evolution equation for a quantum observable will lead u to an inconsistency with the formalism which requires the Hamiltonian be written in terms of the canonical momntum operators,a description much more useful for describing quantum systems in interaction.Think at the case when an H atom is placed in EM field (described classically).U'd find that the Hamiltonian description involving the canonical momentum operator (and hence the Schroedinger picture) is more useful.

Here is one which lists it:
http://www-ec.njit.edu/~venanzi/chem658/lecture/notes_Ch3.pdf

Bad idea to give a link to text written for chemists.We all know what chemists make of QM.

Daniel.

 

[pmb_phy]

I agree. I was just being a little loose with my wording .

I understand. I mention it because there are those who believe that E = mc² holds in all concievable cases. I know since I was one of them until 4-5 years ago. So I stress this point when it arises. Its not that I'll just being a nitpicking weenie.

I'm going to make a web page on this point soon to illustrate this point. Meanwhile if you have Rindler's Intro to SR text he explains this in the chapter in his chapter on relativistic mechanics of continua. Schutz also mentions something to this effect in his new text.

Meanwhile this page illustrates the point quite nicely
http://www.geocities.com/physics_world/sr/rd_paradox.htm

Griffiths wrote an article sometime ago which illustrated a nice point - that the momentum of a (non-isolated) body is not always parallel to the velocity of the body. Tolman mentions this in his text but does not illustrate it. Of course the total momentum of a closed isolated system is always parallel to the velocity of the system.

(more later; system problem and must reboot)

You can define $\frac{d\hat{x}}{dt}=:\hat{v}$ in the Heisenberg picture,..

Okey dokey! I see where this is comming from now. There is an equation in QM which in the Heisenberg picuture can be expressed as

$\frac{d}{dt}\hat{X}_H(t) = \frac{1}{m}\hat{P}_H(t)$

where $\hat{X}_H(t) = U^{+}(t,t_0)\hat{X}_S U(t,t_0)$

...but inserting this definition in the time evolution equation for a quantum observable will lead u to an inconsistency with the formalism ..

What is this "inconsistencey"?

..which requires the Hamiltonian be written in terms of the canonical momntum operators,..

The Hamiltonian is always written in terms of the canonical momentum. If its not then its not called the Hamiltonian.

I guess if you want you can measure a momentum eigenvalue and then divide by m and call that "velocity". But if you do that with a photon then the "m" has to be relativistic mass.

Pete