## Lorentz transformed pressure for a rod with two masses at its ends

hello,
Please see attached Ex and its Answer. I'm surprised that the pressure Txx has the same value as that in the instantaneous frame of the rotating rod Tx'x'. I believe this could be a typo isn't it ?
Isn't the Lorentz transform the following matrix for the rotating rod around the z axis ?

1 0 0 0
0 sqrt(1-v*v) v 0
0 -v sqrt(1-v*v) 0
0 0 0 1

where v = aw

thanks,
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 Recognitions: Gold Member Science Advisor Staff Emeritus Apart from your question, there are some things about the question that don't quite make sense to me. One is that they say that the rod is infinitesimally thin, but then I think p would have to be infinite. Also, the rod is massless, so I would think $\rho$ would be zero. I don't know, maybe this is just my mistake or there's something I'm not understanding properly. Anyway, in the matrix you wrote out for the Lorentz transformation, what is the order of the coordinates? The transformation should mix t and y, not x and y. -Ben
 Indeed I also had the same questions as you...it is indeed weird... the order of coordinates is t,x,y,z . I'm wondering, why you mentioned that the transformation should mix t and y ? what would differentiate x from y since the rotation is around z? Also , why in the matrix for the rotation, we do not find cos and sin ? Thanks,

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## Lorentz transformed pressure for a rod with two masses at its ends

 Quote by zn52 I'm wondering, why you mentioned that the transformation should mix t and y ? what would differentiate x from y since the rotation is around z? Also , why in the matrix for the rotation, we do not find cos and sin ? Thanks,
The transformation is not a rotation, it's a Lorentz transformation. Let's call the inertial frame co-moving along with that piece of the rod K, and the lab frame K'. They're transforming from K to K'. K is not a rotating frame, it's a nonrotating frame that is moving along a straight line parallel to the y axis. In K, the piece of the rod appears *momentarily* at rest, but it doesn't remain at rest. The transformation from K to K' is a Lorentz transformation for a boost parallel to the y axis, so it mixes y and t.

 I think , please correct if I'm wrong that there is a misunderstanding perhaps . The rod is not in a translational motion along the y axis, but rather it is rotating around its center, the z axis. In this case the frame K of the rod has its axis also rotating around the lab frame K'. In this case one can say that there is no boost since the axis are not parallel at all times. But perhaps I'm wrong or I'm I confused ?
 Recognitions: Gold Member Science Advisor Staff Emeritus They refer to the instantaneous rest frame of an element of the rod. That's a description of what I'm calling K. If they were describing the frame rotating with the rod, they wouldn't use the words "instantaneous" or "element" to describe the frame, because the frame would be permanently at rest relative to the entire rod, not just one element at one time. Looking at the matrix you gave in #1, and comparing with your later posts, it seems like you may be getting confused because you've seen a statement that Lorentz transformations are like rotations, and you're taking that analogy too literally. The matrix you gave in #1 isn't a valid rotation (because it has gammas and stuff in it instead of sines and cosines), and it isn't a valid Lorentz transformation (because it mixes x and y without changing t). -Ben
 I thank you so much Ben. Now everything falls in place including the pressure term above and got my confusion cleared up since I had not payed much attention to that instantaneous frame indeed. I'm now so happy and I can sleep at night :) My warm regards.