Griffiths must have explained it. Basically, you start with an initial guess of the solution and assume that solutions look like:
$$c_a=\epsilon^0c^0_a+\epsilon^1 c^1_a+\epsilon^2c^2_a+...$$
$$c_b=\epsilon^0c^0_b+\epsilon^1 c^1_b+\epsilon^3c^2_b+...$$
Here the superscript denotes the order of...
If you go at any other place in our universe, you will always conclude the same thing as all the galaxies are receding away from you. This proves that there is no unique centre.
I have made a mistake in typing. It should be:
##t'=l/(V_a+V_b)\gamma##
This the time of the collision as observed in the referance frame of Rocket A which does equal the way I calculated time elapsed in post #1. ##x'## is 0 as expected since the event takes place at the origin in the frame of...
Ok. I think I have confused myself in taking events.
Using this, we get (Lorentz Transformation to go from A to B):
##\Delta x'= \gamma(lV_a/(V_a+V_b)-V_al/(V_a+V_b)=0##
## t'=\gamma(lV_a/(V_a+V_b)-V_a(lV_a/(V_a+V_b))/c^2 = lV_a/(\gamma (V_a+V_b)##
First, from the frame of Earth observer (whom I called A):
Event 1: t=0, Coordinates of Rocket B=0 (Only X-Coordinates as the problem is 1D), Coordinates of Rocket C= ##l##
Event 2: t=##l/(V_a+V_b)##, coordinates of B= ##V_al/(V_a+V_b)##, coordinates of C=##l(1-V_a/(V_a+V_b))##
From the frame...
The question was taken from the book A Guide to Physics Problem, Vol-1, Mechanics, Relativity, Electrodynamics by Cahn and Nadgorny. This is just 2nd question under Relativity section. The question to find the time elapsed in the reference frame of B was not asked but I thought about that...
Ok. Here is my attempt:
Let Event 1 be when the 2 rockets are separated by length l,t=0 and Even 2 be when they collided.
Now, the length observed by B is by defination happens when ##\Delta t'=0## (primes denotes measurement in B, unprimed in A). Then, we have:
$$\Delta t'=0 \Rightarrow \Delta...
Consider an observer on Earth (Neglect any effect of gravity). Call him A. Let 2 rockets be moving in opposite direction along x-axis (x-axis coincides with the x-axis of A) with uniform velocities. Call them B and C. At t=0, in A's frame, the rockets are separated by length ##l## . Let ##V_a##...
This problem can be done using geodesic equation of motion. But there is a simpler way to do using Lagrangian mechanics. The Lagrangian of the given metric is:
##L= g_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda}= -\dot t^2+ a^2(t)(\dot r^2+ r^2 \dot \theta^2+ r^2 \sin^2\theta \dot \phi^2) ##...
$$\text{E}= -K_{\mu}U^{\mu}= -g_{\mu \nu}K^{\nu}U^{\mu}=-g_{t \mu}U^{\mu}= -g_{tt}U^t - g_{t\phi}U^{\phi}= -g_{tt}\dot t - g_{t \phi} \dot \phi$$
Here only ##t## component contributes in 2nd step.
Similar mistake in computation of ##L##
EDIT:
Corrected Typos
If f and g are Scalars then:
##\nabla.(gf) ## has no meaning! One of them should be a vector. Further, in the attachment you provided in post #2, I found a lot of typos. Also, are you using any assumption for example: Coloumb Gauge(This problem can be solved without even assuming that gauge)?
Check your solution again.
It is better to write the position vectors and unit vectors using Cartesian unit vectors since they are fixed during integration whereas spherical units vary.
1) The hint was to use Geometry of problem to find ##r^2## rather then vectors. Since, you used Vectors, that is just fine.
2) The integral is easy to do in in this case.
Yes. It is better to pick up a specific frame. By Principal of equivalence, it should hold in every frame. In the comoving frame in which fluid elements is at origin, the connection coefficient simplify considerably.
Some Hints:
##\nabla_j T^{0j}=\frac {\partial T^{0j}}{\partial x_j}+...
Use the boundary conditions for continuity of potential across surface and discontinuity of electric fields across surface to solve for A and B!
The reason Gauss theorem won't work here is that problem does not possesses spherical Symmetry but Cylindrical symmetry!
Also ##P_l(1)=1## for all l!
Ok. I will do one more step for you.
##\mathcal{L}= \frac{-1}{16\pi} \eta^{\mu n}\eta^{\nu m} F_{nm}F_{\mu\nu}##
Now, taking the differential and applying product rule to above term throws a factor of 2(Due to symmetry of product) so this becomes (Dropping the metric tensors for brevity)...
Hints:
##\mathcal{L}= -\frac{1}{16\pi}F^{\mu\nu}F_{\mu\nu} ##
Now:
##F^{\mu\nu}= \eta^{\mu n}\eta^{\nu m}F_{nm}##
Now use this in above equation and:
##F_{nm}=\partial_n A_m- \partial_m A_n##
And
##F_{nm}= -F_{mn}##
We have (Taking c=1, and v and v' to be particle velocities) :
##E_{init}= M_{nucleus}##
And
##E_{final}= M_{\alpha}+ M_{daughter}+ M_{\alpha}v^2/2 + M_{daughter}v'^2/2##
By conservation of Energy:
$$E_{init}=E_{final}$$
Combine the above equations!
Gauss law is not a valid tool here simply because there is charge at the boundary (Introduction to Electrodynamics, Griffith). The field is not smooth on the surface. There is discontinuity! Hence, Gauss theorem is repealed and so does your method! By modifying the problem, Gauss law is made...
Gauss Law is applicable on surface which bound volume(This should be clear from derivation). First your solution is wrong since there is no symmetry argument that you can apply to lone cube. This is essential since you can't take out vector ##\vec E## from integration as it behaves differently...
As said, the given formula is wrong. The picture will be really helpful to you. Alternatively (which is how I actually got y-coordinate) , we have R distance from ground to centre of 1st sphere. Then there is 2Rcos##\theta## so that total distance is R+2Rcos##\theta##
Your Geometrical issue has typing mistake. Instead of ##3R-2(R-cos\theta)##, it should be ##3R-2R(1-cos\theta)## as can be seen from geometry as well as dimensions.
For Algebraic issue:
Try using Eq 2 to substitute ##\theta_1## in Eq (1)
Hmmm.. I think you are talking of involvement of non-holonomic constraint in Lagrange formulation. Not all non-holonomic constraint can be incoroporated in Lagrangian(cf. Goldstein).
Try using approximation r>>R. The magnetic field is approximately constant over the surface. Or you can be more exact if you Taylor expand the magnetic field due to wire at large distance. After that use:
## U= -m.B##
Yes. The differential equation is same as original Hydrogen atom problem with the restriction. As you have correctly determined only odd values are the solutions so as to satisfy boundary conditions. The degeneracy is now over those restricted ##l##