Recent content by rajark

Oops , there is a typo. It should be a minus sign between the two terms, instead I had put a plus sign there by mistake ##x' = x\sqrt\frac{1+\frac vc}{1-\frac vc} - v [ \frac xc \sqrt\frac{1+\frac vc}{1-\frac vc} ] = x\sqrt{1-\frac {v^2}{c^2}}##

Enlightened, Erland :) I used these coordinates in Tom's frame to verify the current location of C2. That too gave me the location that is consistent with Length contraction ##x' = x\sqrt\frac{1+\frac vc}{1-\frac vc} + v [ \frac xc \sqrt\frac{1+\frac vc}{1-\frac vc} ] = x\sqrt{1-\frac...

i agree with you on different time corrections for Mary and Tom. Whatever time C2 that Mary concludes at in her frame is different from what Tom concludes at in his frame. The method I went by for Tom's conclusion of C2, if right, poses a question to me that should he not also correct the space...

yes the correction is different for different observers. Mary has to correct only for time, as the clock is not moving in her frame. But Tom, in addition to time correction, should he not also correct the space coordinates?

Is the problem with "zero" reading? or with the "same reading" for both of them? As I learnt, Mary and Tom when located at the "same place" in space should see same reading in C2. Do you say this is not correct?

I did some sloppy work by saying C2's reading to Mary and Tom as zero. Since light will take sometime to reach even Mary, C2 should be reading something in negative. But anyway we will end up in the same scenario of C2 at X' showing to Tom, the time what it was sometimes before when it was not...

It all started when I read that different inertial observers from the same place at the same time should see same things. Say there are two clocks C1 and C2 in a stationary frame of reference S. C1 is at X=0 and C2 at X=X (some positive X) and both are syncronized in this frame. Say there is...

ok, to calculate the length of a "stationary" object in a moving frame 1) Take the coordinates (x1, t1) of one endpoint of the object as measured in the stationary frame and using Lorentz transformation equations, calculate the equivalent coordinates (x1', t1') in the moving frame. 2) Now...

That makes it clear. Thanks, dvf and George

Ok George. Setting t'=0 gets the right answer. t' = (t-VX/(c^2))/√(1-[V/c]^2) so t = VX/(c^2) for t' to be zero Sub t in X' = (X - Vt)/√(1-[V/c]^2) gives the right answer X' = X √(1-[V/c]^2) But what does "Setting t' = 0" mean physically? I could not figure out

Suppose I am in a stationary frame of reference S and there is a lamp post at a distance X from my origin in the positive X direction. Say you move at a velocity V along that axis and the distance of the lamp post in your frame of reference S' is X'. Then by Lorentz transformation equation X'...

Well I should learn to have patience to go through these equations. It was clearly shown that <H> should be no less than ℏω\2. Thanks, cattlecattle Also the point that a state with a non-integer eigenvalue is not allowed, as commented by Bill, is noted.

Hi cattlecattle, If I have understood your comment properly, you only proved that energy eigenvalues should be positive. But my question is that why the lowest allowed energy eigenvalue of QHO should be ℏω\2. Why not some "positive" value less than that is allowed? The lowering operator...

Hi All, If there is something fundamentally wrong in my understanding of quantum mechanics, pardon me for I have just started learning it. We know that if we can come up with a solution for Schrödinger Equation of a Harmonic Oscillator, then we can generate further solutions by acting on it...

Thanks for pointing out, Doc