Recent content by crimsonidol

I don't think that's right. you have torque T=Fa since it applied perpendicular. since it gives an impulse to balance the torque. if torque is zero no impulse. then a=0 is the point you are looking for i assume. Your langrangian should have the constraint term too coming from the applied force.

i got the point.. I see it was a silly question .Thanks for the answer.

why do we use light signals to measure time? In flat spacetime Let's say we have two observers in one's rest frame the other one has velocity v. We suppose each of them sent light signals to other one with time separation of Δt(their proper time) and measure time accordingly.we see time...

Yeah Tsny you are right. I think i got confused.Sorry for the confusion. in Jacobian always two indices same if we choose u's as basis vectors in some direction.I was thinking something else probably , but when i read again I got it. But there is one thing confusing me. Should not we use a...

the thing is that , in question it asks you to use the geodesic equation to find the components.then you do not need to use the identity of symmetric part of jacobian curvature. Using basis vectors in the geodesic equation doing just fine. he just made a mistake in indices.the second equation...

no. You gone wrong in the part J^{\mu}_{\alpha\alpha\gamma} part.If you do it like that you will have same index 4 times in an expression. (nμ;αuα);βuβ+Jμαβγuαuβnγ=0 this is correct. just think every u component =1 but keep u indices different. so you will can find every component of J...

if you read the section 3.3 Length contraction part of Rindler's book Relativity: Special, General, and Cosmological you will understand it well. question based on that information of hitting wall propagates at speed of light and so reaching information of hitting wall to the other end of...

I have the same problem. No answers yet. off diagonal components of the antisymmetric 4 tensors in special relativity involves 3 vectors and we can form 4 vectors from them. If any component of that 3 vector is zero under LT the 4-vector is zero then all the off-diagonal terms are zero. This is...

ok i thing I figured it out. Thanks e.bar.goum I look for quasar and found this https://docs.google.com/viewer?url=http%3A%2F%2Fassets.cambridge.org%2F97805218%2F49609%2Fexcerpt%2F9780521849609_excerpt.pdf too many assumptions in it by the way. you should take the speed of the stellar...

Sorry I'm not comfortable with the codes so I wrote a paper and uploaded it. Hope some1 can tell me where I have gone wrong. http://imgur.com/ZN50J

it is not a homework but i need to solve this. it is killing me slowly and I'm in a closed loop coming to the same point everytime. please help me about it.

I need a spark :S If anything is unclear I may rewrite it.

https://www.physicsforums.com/showthread.php?t=139465

you need to look at contour integration and use xt instead of x there. By using residue and appropriate contour you'll be able to find t^-1/2

you can do the proof and also find part b when you understand the first part I assume.look hildebrand page 561 to be exact