Recent content by newTonn

Yes I agree there is something confusing in my calculation.the equation should be L= L' x gamma.But the result is as per that. Basicaly the actual length(or distance travelled) in the moving frame will be more than that of observed from stationary frame.The actual time elapsed in the moving...

Why can't We use the same principle for measuring and predicting the velocity of the object in other frame? For example as in the previous example, Observed velocity = V' = 0.9c observed distance traveled per second L'= 2.7E+8 meters. Actual distance traveled per second L = L'/\gamma =...

Length contraction, according to Hendrik Lorentz, is the physical phenomenon of a decrease in length detected by an observer in objects that travel at any non-zero velocity relative to that observer. This contraction (more formally called Lorentz contraction or Lorentz-Fitzgerald contraction)...

Ok let us stick to one stationary frame from where the observation are made. Consider another object moving with a velocity of 0.9c (all movements are in x direction)w.r.t the frame. For this statement to be true, the object,after 1 seconds should be at a distance of 0.9 x (3E+8) =(...

In fact the space itself is contracted isn't it?. If only the rod was contracted,the ends would not have touch B and D. The contraction of space leads to an ambiguity.Because the formula for velocity(derived from time dilation) doesn't take into consideration of this reduced distance.

Actualy i mean the centre to centre distance of molecules(or atoms of the rod) I will try to explain my question with another example.A,B,C and D are four persons.Initialy they are all at rest with respect to each other. B,Cand D are in a line in x direction. C is in the middle of B and D C...

Ok.i will come to the point.Consider A,B and C are at rest to each other.B & C are each carrying a rod of length 'L'.This rods are made of 'n' number of molecules in x direction.Now let us say the spacing from centre to centre of molecules are 's'.Let us mark it on C's Rod. Now B,with the rod...

This shortening will be from one end or it would be symmetrical from centre of Rod?

Ok what about if B and C are long rods with length 'c' in their frames?

Is there any length contraction to be considered or not?

Sorry i din't mean it.Initally let all of them have same x co-ordinates =0; According to A,what will be the positions of B & C in after 1 second in A's frame?

Ok i looked at the book and got an answer of 0.9944c(approx). How can i justify this with the relative displacement after one second?

With reference to the sketch attached, if A and B is having a relative velocity of 0.9c,and B and C having a relative velocity 0.9c What will be the relative velocity between A and C? (all in x direction)

You have ignored part of my post where i stated"if the observer,measure the distance from him to the beam at each minute fraction of a second and plot the position of the beam,he will get a diognal line." to show that i am unfamiliar with the co-ordinate systems. Does this directly implies...

If i agree with your last statement,If the distance is not d,why should i say time is dilated,instead of saying the basic principle of measurement of time in the moving clock is wrong?