Physics Asked by rim on December 29, 2020
Does energy and momentum also dilate and contract as time and length do respectively, since energy and time and momentum and length are complementary quantities both in relativity & QM?
In a way, yes, one can roughly say so. This can be seen from the way their formula "change". The non-relativistic momentum (dropping vector signs) $$P=mv$$ turns into $$P=gamma mv$$ and the non-relativistic (kinetic) energy $$E=1/2mv^2$$ turns into $$E=(gamma-1)mc^2.$$ These changes in the formulas for momentum and energy are similar to the changes for time dilation $$Delta t=gamma Delta t_0$$ and length contraction $$ell =ell _0/ gamma$$ $gamma$ is defined as $$frac{1}{sqrt{1-frac{v^2}{c^2}}}$$ This is also very clearly visible from their plots.
Source for the images.
Answered by physicopath on December 29, 2020
Energy and momentum together form a "four vector" just like time and space do. So we find that $(E/c,p_x,p_y,p_z)$ transforms the same way that $(c t, x, y, z)$ transforms. However, time dilation and length contraction are more specific.
Time dilation is the time in a "moving" frame for a clock which is not moving in the "rest" frame. So it is the transform $(c Delta t,0,0,0) rightarrow (c gamma Delta t, gamma v Delta t,0,0)$. This would correspond, for energy and momentum to a case where an object has (rest) energy but not momentum. In other words $(E/c,0,0,0) rightarrow (gamma E/c,gamma v E,0,0)$. So indeed there is energy dilation.
Length contraction is the length in a "moving" frame for a rod which is not moving in the "rest" frame. So you would start with $(0,Delta x,0,0)$. For energy and momentum this would correspond to $(0,p_x,0,0)$. However, this does not make sense for momentum because this would represent a particle with momentum but no energy. Such a particle would have an imaginary mass, which is not possible. So there is no direct equivalent of length contraction for energy and momentum, i.e. no "momentum contraction".
That said, together energy and momentum do transform as time and space so in general $(E/c,p,0,0) rightarrow (gamma E/c + gamma v p, gamma p + gamma v E/c^2,0,0)$. It is only because there is no such thing as a particle with momentum but not energy that there is no direct equivalent for "momentum contraction".
Answered by Dale on December 29, 2020
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