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Question on: Classical disintegration of particles, Landau-Lifshitz series on Physics

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MManuel Abad
#1
Jan21-11, 11:19 AM
P: 41
Hi there.

I'm a second-year physics student. I'm from Mexico (so please, forgive my english), and I'm on winter holidays. I've been studying Mechanics with the first volume of the "Course of Theoretical Physics", by Landau and Lifgarbagez, that awesome though hard book series. My study plan is to do all exercises and to explain every mathematical detail Landau skips in his exposition. Two days ago, everything was alright, but then I found, in chapter IV "Collisions between particles", something I cannot fully comprehend.

In the first section (from the last paragraph in page 43 in the 3rd english edition), Landau considers the disintegration of many identical particles, isotropically oriented in space and wants to know the distribution of the resulting particles in direction etc.

Please, I can't get the picture of the problem. Can someone explain it to me?
I think it's this:

You have this bunch of particles, all identical, randomly oriented. Each of them disintegrates in two resulting particles of masses m1 and m2, with equal momentums in the C (center of mass) frame of reference, and the direction of m1 making the same angle phi (say) with that of m2 for every original particle. As a result of this, there's a uniform distribution of the directions of the resulting particles. But then there's the problem:

So far in the text, the angle [tex]\theta[/tex]0 was the angle the velocity (seen from the C frame) of one of the particles made with the direction of the velocity V of the original particle as seen from the LAB frame of reference (see fig. 14). Now, what does [tex]\theta[/tex]0 mean?? When you consider that system of particles disintegrating in the LAB system, are they moving with velocity V? In what direction? Is V the velocity of the center of mass of the system of particles as seen from the LAB reference frame?? Then [tex]\theta[/tex]0 is the angle, say, particle 1 makes with the velocity of the center of mass? But then [tex]\theta[/tex]0 varies with the resulting particle of mass m1 you choose, cuz they're resulting particles from different original particles differently oriented. If this is so, then how do they measure the solid angle d[tex]\omega[/tex]??? And another question: how do they obtain eq. 16.8?? I mean, I know they differentiate v2 and all that (paragraph between eq 16.7 and 16.8), but d(cos([tex]\theta[/tex]0)) is -sin([tex]\theta[/tex]0)d[tex]\theta[/tex]0!!! WITH A MINUS SIGN!!! When you substitute in eq. 16.7 and obtain eq 16.8, what happened to the sign???

Please, I need your help!!!
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torquil
#2
Jan21-11, 02:26 PM
P: 641
Caveat: I may also have misunderstood key points, but I gave it a shot:

Quote Quote by MManuel Abad View Post
You have this bunch of particles, all identical, randomly oriented. Each of them disintegrates in two resulting particles of masses m1 and m2, with equal momentums in the C (center of mass) frame of reference, and the direction of m1 making the same angle phi (say) with that of m2 for every original particle. As a result of this, there's a uniform distribution of the directions of the resulting particles.
I agree with your understanding of the sitation described in the book.

So far in the text, the angle [tex]\theta[/tex]0 was the angle the velocity (seen from the C frame) of one of the particles made with the direction of the velocity V of the original particle as seen from the LAB frame of reference (see fig. 14). Now, what does [tex]\theta[/tex]0 mean??
[tex]\theta[/tex]0 is now the angle of a chosen disintegration particle with respect to [tex]\mathbf{V}[/tex], the centre-of-mass velocity of the system of primary particles.

When you consider that system of particles disintegrating in the LAB system, are they moving with velocity V? In what direction? Is V the velocity of the center of mass of the system of particles as seen from the LAB reference frame??
Yes, boldface [tex]\mathbf{V}[/tex] is the velocity of the c-o-m frame of the system of primary particles and V is its abolute value.

Then [tex]\theta[/tex]0 is the angle, say, particle 1 makes with the velocity of the center of mass?
Yes

But then [tex]\theta[/tex]0 varies with the resulting particle of mass m1 you choose, cuz they're resulting particles from different original particles differently oriented.
Yes

If this is so, then how do they measure the solid angle d[tex]\omega[/tex]???
I assume you mean what is called [tex]do[/tex]0 in the book, the solid angle differential element for distribution of disintegration particles in the centre-of-mass frame? By symmetry, you already know they will be isotropically oriented in the c-o-m frame, so in that frame it is simply as stated in the text. In the c-o-m frame, [tex]\theta[/tex]0 is simply an angular coordinate in [tex][0,\pi)[/tex], so you can find the distribution with respect to that coordinate as in the text.


And another question: how do they obtain eq. 16.8?? I mean, I know they differentiate v2 and all that (paragraph between eq 16.7 and 16.8), but d(cos([tex]\theta[/tex]0)) is -sin([tex]\theta[/tex]0)d[tex]\theta[/tex]0!!! WITH A MINUS SIGN!!! When you substitute in eq. 16.7 and obtain eq 16.8, what happened to the sign???
They are looking for the probability distribution, which is supposed to give positive probabilities when integrated. The difference in sign just comes from the face that as [tex]\theta_0[/tex] increases, T decreases. Therefore an extra sign has been introduced, so that you don't have to do all integrations in T-space in the negative direction to get positive probabilities.

Please comment on this, e.g. I'd like to know if I am wrong on any point

I hope my latex expressions are not jumbled/incorrect. Seems to be quite problematic to construct a post containing several latex-expressions, since they are mixed togethereach time i change the post and click on preview. Seems to not even help using the browser reload
lalbatros
#3
Jan21-11, 02:45 PM
P: 1,235
MManuel,

Please observe that this paragraph contains text together with formulas.
To translate that in pure mathematical notations, you need to translate properly the words "the fraction is" and "the required distribution" that appear in front of equations 16.7 and 16.8 .

To do that you will write something like

Prob(T < T' < T+dT) = Prob(T < mv/2 < T+dT) = ...

When you finally arrive at the last expression involving the speeds, you need to extract what is called "a distribution" in the proper way. To do that remember that a distribution f(x) is defined by:

Prob(x < x' < x+dx) = f(x) dx for dx>0

The sign that you are looking for is hidden in the fact that the above definition is valid for dx>0.

If you have other questions, please be a little bit more specific. (I have a headache because of another thread here)
At first sight, this is just a question changing coordinates from L to C and assuming isotropic disintegration in the C system for each particle. It is indeed interesting that the energy distribution in L can be derived and that the energy of the disintegration can be determined by statistical observations in L.

I would be interrested to know a practical application of this.
I don't see an example of disintegration leading to two equal mass fragments.
Spontaneous fission is already much more complicated.
The "gamma rays disintegrate" by electron-positron creation, but do that by transfering momentum to an existing charged particle. Could the Landau example be extended to this pair creation process??
Anyway, as usual, Landau is an excellent reading.
Solving all the exercices is an excellent idea, but you may need to postopone some to use your time more efficiently, unless you are very gifted.

MManuel Abad
#4
Jan21-11, 09:33 PM
P: 41
Question on: Classical disintegration of particles, Landau-Lifshitz series on Physics

Thank you so much!!! You both were very helpful, indeed!!!

Torquil:

Yes, I did the integration of the distribution in terms of T, and, as you say, I saw that when [tex]\theta[/tex]0=0 then T=Tmx, and when [tex]\theta[/tex]0=[tex]\pi[/tex] then T=Tmn; then when you do the integration, the minus sign changes the limits of integration and voil: you get +1, that is the whole distribution :D THANK YOU!!! And well, then!! I wasn't that far from the right explanation of the situation!!! :D Yeah, what you explained to me makes so much sense!!!

And lalbatros:

Thank you so much too!! You're explanation was something didn't occur to me, and a very clever one!!! And yeah, solving all the problems is indeed very good, but I think I'll take your advice and skip/postpone some of them: I've been "losing" so much time in them!!! D:

As you say, there might be practical applications of this, but I think it's too simplified, so it's more like "introductory". Not to mention that this is Classical Mechanics, and disintegration is a quantic process. As you say, particle disintegration is actually much more complicated. I don't know if Landau explains those phenomena deeper, but I believe he does, in his 3th volume: Quantum Mechanics. And about that electron-positron creation, I'm pretty sure Landau and Lifgarbagez have something to say in their 4th volume of these series: Quantum Electrodynamics.

Thank you so much again :)


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