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Lagrangian question |
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| Aug13-12, 05:45 PM | #1 |
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Lagrangian question
1. The problem statement, all variables and given/known data
Right I've got a relativistic particle in D dimensional space interacting with a central potential field. Writing out the entire lagrangian is a bit complicated on this but I'm sure you all know the L for a free relativistic particle. The potential term is Ae-br where r is the position vector. (i) Find the momentum p as a function of the velocity v. (ii) Find Lagrange's equations of motion for the particle. (iii) Find the velocity v as a function of P and r. 2. Relevant equations 3. The attempt at a solution (i) That would be mvi/root(1 - v2/c2) (ii) ∂L/∂xi = d/dt(∂L/∂vi) , Pi = ∂L/∂vi So dPi/dt = Abe-brxi/r (iii) This is the part I'm having a problem with. I have that v2 = P2c2/(m2c2 + P2) But I have no idea how to get r into the equation. Advice please? |
| Aug14-12, 01:37 PM | #2 |
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Hi Maybe_Memorie!
![]() Did you know that f(x,y)=g(x) for any y, is not only a function of x and y, but also a function of x? Yeah... I know, it sounds a bit lame, but it's true nonetheless. |
| Aug14-12, 03:47 PM | #3 |
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Hey I Like Serena!
So are you saying that since I need to find V(r,p), my answer as a function of p is also a function of r for all r? |
| Aug14-12, 03:51 PM | #4 |
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Lagrangian question
Yes. I believe that you have v(r,p)=v(p).
It's fairly typical that p and v are independent of position in potential fields. |
| Aug14-12, 04:18 PM | #5 |
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So my answer is correct then?
Also I have another question. How do the coordinates xi and momenta Pi transform under an infinitesimal rotation in the x2x4-plane? Well, under this transformation, I know that xi, i not equal to 2 or 4 will just go to xi. In other words, invariant. If i = 2,4, then xi -> xi + εijxj where εij is an infinitesimal parametrisation of the rotation. Same for Pi. Is this correct? |
| Aug14-12, 04:23 PM | #6 |
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| Aug14-12, 04:38 PM | #7 |
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But if I need to be more rigorous, if I take my other statement of xi -> xi + εijxj which I know to be true since it was proved in class, we will have in this case xi -> xi + εjkxk and the second term just wouldn't make sense, so it would be zero. So I guess in general xi -> xi + δijεjkxk where δij is the kronecker delta. |
| Aug15-12, 11:09 AM | #8 |
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Seems like a very badly worded question
![]() eg. presumably r is the length of the position vector. Bearing in mind the spherical symmetry of the problem, it would make more sense if they asked for say the speed of the particle as a function of r. [tex]\epsilon_{1i3j567...}[/tex] to generate this rotation, where ε is the completely antisymmetric symbol. Note that the i and the j go in places 2 and 4. |
| Aug17-12, 05:58 PM | #9 |
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Okay, well... did you figure it out?
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| Aug17-12, 06:40 PM | #10 |
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I'm not sure if my argument 3 posts up is correct or not. It makes sense to me, but I'm not sure if it's right.
Any kind of point in the right direction? |
| Aug17-12, 10:08 PM | #11 |
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But I can heartily recommend the matrix I suggested 3 posts back
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| Aug18-12, 03:33 AM | #12 |
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Vectors themselves are independent of coordinate system. It's only their representation into coordinates that depends on a coordinate system. If you have a formula to represent the vector x in coordinates xi, or rather a formula to transform its coordinates to another coordinate system (as you do), then that exact same formula applies to represent the vector P in coordinates of the same coordinate system. So yes, you are correct. |
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