Question

3. (a) Let L=L(x,y1, 41,99) where = əy/ax, i = 1, 2, be a Lagrangian satisfying the Euler-Lagrange equation which is independent of y2. Show that al constant. aya You are given that the motion in the plane of a particle of mass m has La- grangian L = (1+2 +r202) – V(r), where r and 0 are polar coordinates, V is the potential and the dots indicate derivatives with respect to time. Use the above result, identifying x as time t, =r and y2 = 0, to write down a corresponding conservation law. What, physically, is the conserved quantity in this case? (b) Let L V1 + y2/x be a Lagrangian. Show that the extremal curves are circles with centre on the y-axis. (c) Let L* = V1+ y2/y be a Lagrangian. Use Beltrami's identity and the result in (b) above, or otherwise, to find the extremal curves for this functional. and yi

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Answer 1

3. a)

For this part we need to put the given Lagrangian L in the Euler-Lagrangian equation. And use the fact that partial derivative of L with respect to y2 is zero while the independent variable here is x.

And for second part we just need to plug in the Lagrangian in Euler-Lagrangian equation and use the same fact that here L is independent of $\theta$ . Physical explanation comes from the knowledge of expression of angular momentum in 2D polar coordinate.

L2L(xy, y) ayi (21,2. Euter ta grange equation is givenby, at al du {= 1,2 ai So al dx -Boy, ) -ooo (it is only function of my Yź, yi, independent of 82) al sy a constant dr at ast, Yoar o (>1,2 14762) - In Euter -Lagrange ear here t is independent of a then, in our previous part take So if we 02Y Here 12L 2 2 - constant 2. m.p2.20 a constant. mg26 = a constant. nothing but, says that is actually Angular This is requined conservation law, which Angular momentum conserved for the Lagrangian, nho give Physically, the conserned quantity momentum of undrer (2-0) central force din polar coordinaks. It is so directed towards for away () from the origin, so Angular. momentum [anx F (r) becomes Zero of a partirlo. because central force always either case,

b)

The curve will be extremum which can be found by putting the given Lagrangian into the Euler- Lagrangian equation.

L=1&yea JE Scho dne would be For the extremal carnes extremum have Station integral I due relative to the differing infinitesimally from the path y (n). From for extremal of T equation : we tionary Euters Lagrange Caye) - Gana d

Now, IL our У aL sy ху So from Enter- Lagrange egn ( Xtre) - (ka constant) 2 k al KVHYR 1 kx12 If y12 ax du 1 and 2 ya ka yz = K²x² (1+1 912) 3 +212 (1 422) = ²x² -) Y'a & T-(Kn) 2 (la constant) [ Kr - 2 = ka VIR2x2 doa + (Sign- Vrnja >> (4- c)? (1= n² k²) k2 2 + n2 = (4-0) + (YK) 2 cirele with centre (0, e Equation of and radius (YK). are the extremal curves е и се, circles with centre on Y axis.

c) When Lagrangian is independent of independent variable i.e. here x then the Lagrangian equation reduces to the Beltrami identity. So we can find extremal curve using Beltrami identity in this case.

59 we It 412 y use * L is independent of Ne. In this ease equation reduces (becomes Beltraini identity) at e then Buter-Lagrange (es constant) Beltrami identity (If a la L - y OL ay' YG by an y JL zyl is taken from y/ HY12 y (1412) - y12 y rtyr2 "Hyr = te - yoz _ Ley 2 - 1 » Yule previous part just replacing y' in place of n') (a = constant) y - 1 dut a (cy) 2 x leyle 1 +G

N2 - 932 (gales (-c)' (tage- (y-a) 2 22 extremal +22 Required carves curves looks like Rought sketch Sketch how the above functional y graph should be Symmetrie