A certain quasar recedes from Earth at 0.55 c. A jet of material ejected from the quasar toward the Earth moves at 0.13 c relative to the quasar. How do you find the speed of the ejected material relative to Earth?

1 Answer
Cosmic Defect
Mar 4, 2016

\vec{u^'} = (u_x^', u_y^', u_z^'); \qquad \vec{u}=(u_x, u_y, u_z) = ?
v=0.5c; \qquad u_x^'=-0.13c; \qquad u_y^'=0; \qquad u_z^'=0
u_x=(-0.13c-0.5c)/(1-(-0.13c*0.5c)/c^2)= - 0.59c; \qquad u_y=u_z=0;

Explanation:

Suppose if v is the relative velocity between two frames X (earth) and X' (quasar). The relative velocity of an object in the two frames are related to each other as given by the velocity-addition rule of special relativity :

u_x^' = (u_x+v)/(1+(u_x^'v)/c^2); \qquad u_y^' = u_y/(\gamma(1+(u_x^'v)/c^2)); \qquad u_z^' = u_z/(\gamma(1+(u_x^'v)/c^2));

\gamma \equiv 1/\sqrt{1-v^2/c^2}

The inverse transforms are :
u_x = (u_x^'-v)/(1-(u_x^'v)/c^2); \qquad u_y = u_y^'/(\gamma(1-(u_x^'v)/c^2)); \qquad u_z = u_z^'/(\gamma(1-(u_x^'v)/c^2));

We can choose the orientation of the coordinate systems such that the X axes are parallel to the velocity of the quasar jet. This would simplify the problem by rendering the Y and Z components of the velocity vector zero.

\vec{u^'} = (u_x^', u_y^', u_z^'); \qquad \vec{u}=(u_x, u_y, u_z) = ?
v=0.5c; \qquad u_x^'=-0.13c; \qquad u_y^'=0; \qquad u_z^'=0

Using the inverse transform,
u_x=(-0.13c-0.5c)/(1-(-0.13c*0.5c)/c^2)= - 0.59c; \qquad u_y=u_z=0;

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