- #51

Nugatory

Mentor

- 12,768

- 5,368

Either you've misunderstood the article or it's wrong.I have found an article where the author says the following about einstein's SR: If we have two reference frames, S and S', with S stationary and the S' moving along the 'x' axis with speed 'v' of S, then as per SR there is length contraction in the 'x' direction but no change in the 'y' or the 'z' directions of objects in S'. That is y=y' and z=z'. If we take just the y=y', then the length y2-y1 = y'2-y'1. However, SR says that the time is slower in S' than in S. This means t2-t1 is < t'2-t'1. So,if we have a beam of light traveling along the y axis in Sthen the speed of the light is C= y2-y1/t2-t1. For the observer in S' who is also looking at the same beam of light it will be C'= y'2-y'1/t'2-t'1. Now, the numerators are equal but the denominator in S' is larger than in S due to time dilation. This means C'<C, which contradicts the constancy of the speed of light postulate of SR!!!!Also, for light traveling in x-axis directionwe have from length contraction x2-x1 > x'2-x'1 and from time dilation we have t2-t1 < t'2-t'1, then the speed of light in S is C=x2-x1/t2-t1 and in S' it is C'= x'2-x'1/t'2-t'1. So, for C' we have the numerator that is smaller than in S and the denominator that is greater than that in S. This again leads to C' < C !!!!. So, how do we resolve this? thanks.

A beam of light traveling along the y-axis in one frame is not travelling along the y-axis in the other frame - it has an x-axis component as well, and therefore travels a longer distance. The time dilation and length contraction is exactly enough to balance that effect, keep the speed of light the same.

Although a beam of light traveling along the x-axis in one frame is also traveling along the x-axis in the other frame, the distance covered from the origin of one frame as viewed from the other is different in the two frames. Again, length contraction and time dilation together exactly balance that effect.

You can see this for yourself if you choose either reference frame, imagine two pulses of light leaving the origin of that frame at time zero. After one second, the (x,y,t) coordinates of the pulse directed along the y-axis will be (0,1,1) and the coordinates of the pulse directed along the x-axis will be (1,0,1). Use the Lorentz transforms (not the contraction and dilation formulas!) to convert these into coordinates in the other frame, then calculate the distance covered by the light in one second in that frame. It'll come out to be c.