Recent content by Abhishek11235

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...

So, how we will calculate the time elapsed in B's frame using B's referance frame?

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...

Can you please explain it in more detail?

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

You can't get any help from anyone here if you say this. This is just ##\textbf{put-into-the-formula}## problem

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)?

Let ##d_- \rightarrow d## (The distance between 2 wires) and ##d_+ \rightarrow a## (The radius of wire) in the formula of potential you wrote.