Recent content by Furbishkov

So from my lorentz equation, t' = γ(t-(vx/c2)) I set t' = 0 and solve for v from there...

So I put frame S at a point where the origin is where the sun explodes so that the separation between the two events is just described by the Earth event. Then, for frame S, x = distance from sun to earth. t = 4 minutes. In S' I want t' = 0 and x'=γ(x-vt)?

So I want to take my frame S (t) to be the sun-earth frame where the events are separated by 4 minutes, and then my S' to be the sun-earth frame(t') in which the two events are simultaneous.

So from a Lorentz Transformation, t' = γ(t - vx/c2) where t'=t and solve for v from this equation? If I set this up, solving for v seems quite difficult. My other thought it to use the velocity transformation, Ux' = Ux - v / (1-(vUx/c2)) But what is my Ux and v in this equation?

1. The problem: If the Sun blows up at some instant and four minutes later we on Earth sit down to eat lunch, these two events are separated by a spacelike interval. The explosion of the sun cannot have influenced us at the time we sat down because it takes 8 minutes for light to reach us from...

That change of variable makes sense when I look on it on a graph, thanks. Now for my bounds I get, 1≤v≤2 and -v≤u≤v . I worked out the integral to just become sin(v). Thanks for the help!

I still am completely stuck. Any way I look at it I get stuck with something I can't integrate with the sin in the integral. I am working off the assumption that v = x+y is correct. This would make my integral something like ∫ sin(v)/v . My last thought is to make u = 1/x+y. But this still seems...

Looking at my graph is seems to be convenient if I define u to be a constant, possibly u = 2? ... When I learned change of variables I was told I would never need to come up with the change of variable myself so that is why I'm having some troubles. I appreciate the help.

I sketched D and put in the constant v=x+y but still can not see any good change of variable for u. The sin is really causing me problems. I constantly think I need to make my "u' have the sin term in it but can not come up with anything that works

Homework Statement Evaluate the integral ∫∫sin(x+y)/(x+y) dydx over the region D whereD⊆R2 is bounded by x+y=1, x+y=2, x-axis, and y-axis. Homework EquationsThe Attempt at a Solution I think that I need to use a change of variables but can not find any change of variables that work. One thing...

General form is : y(x,t) = Aei(kx-ωt) = A[cos(kx-ωt) + isin(kx-ωt)] So if I notice the exponents are the same under the condition α=-1 and β=1 how does that exactly translate to it being a solution of the wave equation?

Homework Statement Is the function y(x,t) = Ae(−β2x2−2βxt−t2) + Be(−x2+2αxt−α2t2) a solution to the wave equation ∂2y / ∂t2 = v2 (∂2y / ∂x2) Homework Equations ∂2y / ∂t2 = v2 (∂2y / ∂x2) The Attempt at a Solution I have found the solution through finding the partial derivatives (∂2y / ∂t2...

I have looked online and understand the general idea of solving a second-order differential equation, but I don't know what I am trying to solve for here and how it relates to the Steady-state equation.

Thanks for the replies. My equation for KVL I wrote, q = cVocosωt -cLq(double dot) - cRq(single dot) is the same as yours with some rearranging. I understand what the steady state represents as a non-transient state of oscillations. But, I am just stuck as to what am I suppose to do to get a...

Homework Statement An ideal AC voltage source generating an emf V (t) = V0 cosωt is connected in series with a resistance R, an inductance L, and a capacitance C. a) Find the steady-state solution for the charge, q(ω,t), which is of the form q0(ω)cos(ωt− δ(ω)). b) Find the steady-state...