Hi everyone. I'm doing an experiment. In another forum I got into a heated argument regarding the definition of mass. I.e. when physicists use the term mass in relativity it means one of two things, either proper mass or relativistic mass. The objection is that if you have more than one meaning to a term mass then all hell breaks loose.

I pointed out that this is not unique to mass since the meaning of the term momentum depends on the context it's used. In Newtonian mechanics it refers to linear mechanical momentum p= mv. In relativity it can mean 4-momentum and in quantum mechanics it means conjugate momentum, aka canonical momentum. So with your kind patience and indulgence I'd like to try an experiment here.

You are all aware that the uncertainty principle relates uncertainty in momentum to the uncertaintly in position. But what you don't learn until advanced courses on quantum mechanics is that the momentum this refers to is actually what is called "cannonical momentum" (aka conjugate momentum). Its different than the momentum p = mv that you learn about in basic physics. Suppose a charged particle is moving in a magnetic field. The cannonical momentum p has the value

(B-45) p = mv + qA

where A is known as the magnetic vector potential. A graduate text in quantum mechanics by Cohen Tannoudji explains this on page 225) as follows
Since canonical momentum is a rather advanced topic and probably not known to most visitors here I'm curious as to whether such a fact is of interest to anyone, i.e. would you want to know that p is a function of the magnetic field through the quantity A?

I read an article on this subject. It appeared in the American Journal of Physics. If there is no magnetic field then p = mv and [LaTeX Error: Can't write to directory] which implies [LaTeX Error: Can't write to directory]. When there is a uniform magnetic field present then this becomes
[LaTeX Error: Can't write to directory]

This would not be apparent if one didn't know how p is actually defined. So my question is this - Do you feel that lower level quantum mechanics texts are not preparing the reader well for this fact if they don't clarify what the momentum that quantum mechanics really refers to?

Thank you all for your help.

Pete

I feel you are right else one gets a wrong picture. However when one moves from the comfortable exact world of classical mechanics to the uncertanties in quantum and concepts like operators etc, it might be advisable to introduce the concept simply at first, then pointing out that the concept of momentum is much wider and that what was previously considered is only a special case.

Do you think this is a point I should not mention to the layman?

In the above post there is something that is worth mentioning, i.e. the quantity [LaTeX Error: Can't write to directory]. This is an expecation value of a field! That can be confusing. It refers to the expectation of the magnetic field measured at the location that the particle is measured to be found.

I think you should mention it with an explanation. (I need it too.) In a text book this can definitely be introduced as a footnote.

How would I explain it to a layman who has no knowledge of math though?

well, in my book , momentum of photons is written as h/wavelength and has written that this relation gives particle nature to radiation.
whereas in the chapter atomic phy, it says (p^2/2m+V) zi= E zi where it sounds like an ordinary classical momentum to me.
but blow! in next step it removes p by a derivative!!
i am confused.
and what is canonic momentum?

I think you're referrring to canonical momentum. The uncertainty principle is the relationship between canonical momentum and position. It is not a relationship between mechanical momentum (i.e. defined by mv) and position. Canonical momentum is a term which comes from analytical mechanics where one uses the principle of stationary action as the starting point. This leads to an important quantity called the Lagrangian. The canonical momentum is defined as the partial derivative of the Lagrangian with respect to a coordinate. The resulting quantity is then called the canonical momentum conjugate to the given coordinate.

If the system is a charged particle moving in an electric field then the canonical momentum is different from mv. This is one thing to be careful about in quantum mechanics, i.e. the difference between mechanical momentum and canonical momentum. Canonical momentum is also called generalized momentum.

but since it is also momentum, there must be a relation between classical and cononical momentum?
say, i try to relate the two...
d/dx(lagarangian)=p
but from classsical mech i know d/dv(1/2 mv^2)= mv=p
but the derivatives are d/dx in former , d/dv in later.
is this the difference between canonical and classical momentum??
this is a great topic!! no doubt.
SORRY , IHAVE GOT IT . THE DERIVATIVE IS INCLUDED IN THE lagrange thing. then it is a double derivative, to be precise.
by the way what is lagrangean?
AND IT MUST BE A DERIVATIVE OF ENERGY. I HAVE GOT IT!!

NO. there is something wrong. please ignore the previous post. i will think this in a few days.
and what is (b-45) in the 1st post anyway?

d/dx(Lagrangian)= P canonical
multiplying both sides by dt, d/dv(lag)=P canonical * dt
and from familiar classical mechanics, d/dv (1/2 mv^2)=P classical
according to wiki, Lagrangian is nothing but kinetic energy of free particle.
so left hand side matches for both.
so rhs must match.
so integration P canonical dt= P classical.........(proved!!)

v = w/k (wave)=2 pi f/ 2 pi * lambda=f lambda=E/h * h/p=E/p=(1/2 mv^2)/mv=v/2
this means de broglie hypothesis is not consistent with einstein or planck quantization law.
is this where canonical momentum can save the situation?.....
please give me a few days to think this.
i am going seriously wrong somewhere.

wait!! all is not lost yet!!
if the limits of integration P canonical dt be so adjusted that right side comes Pclassical/2 then the situation can be saved.
i wonder what the limits will be?
can anyone help me here, please?

See - Lagrangian - Wikipedia, the free encyclopedia

The Lagrangian for a charged particle moving in an EM field is given by

L = mv^2/2 = (e/c)v*A - q[LaTeX Error: Can't write to directory]

The Canonical momentum conjugate to the position coordinate x is found by plugging this into the expression I mentioned above, i.e.

Usually one uses a "dot" over the variable to deterime the derivative with respect to time but since Latex is still isn't working we'll continue to use the notation q' = dq/dt.

(1) to find P_x take the partial derivative of L with respect to x' = dx/dt
(2) to find P_y take the partial derivative of L with respect to y' = dy/dt
(3) to find P_ztake the partial derivative of L with respect to z' = dz/dt

This will give the following result

P= p + qA

where p = mv. So

P= mv + qA

you can see the difference now, I hope?

I made an error in my previous posts. I have gone back and corrrected them. The error I made was to say that Canonical momentum P was defined as

P = @L/@q (wrong!!)

where q is a generalized coordinate. It is actually defined as

P = @L/@q'

Sorry for making that mistake and the confustion it costs. P is then said to be the canonical momentum conjugate to the position variable q. If q is an angle then P is an angular momentum.

For the definition of the Lagrangian see http://en.wikipedia.org/wiki/Lagrangian

well , i was happy with the previous(incorrect) one. it seemed to fit with the above discripancy. i will try to understand what you posted. just give me some time.
however, i have understood the lagrangian and its utility in classical mechanics. to tell the truth , i find it analogous to fermat principle in optics.
and what is B-45??(post 1)
many, many thanks for helping me.