Why is interstellar cosmic radiation more energetic than radiation from our sun?

Question

I just watched an interesting minidoc about cosmic radiation. The video, and other resources I've read afterward, explains that the most energetic cosmic radiation comes from supernovae and black holes. Our sun also bombards us with particles, but these are lower energy. These sources imply that interstellar particles are higher energy because they come from high energy events (stars exploding or imploding).

I wonder if the high energy of these interstellar particles might also be a consequence of our solar system's movement relative to other stars. If the origin of a cosmic ray is a star that is traveling very fast relative to our own system, perhaps this would significantly increase the energy of the particle as we experience it. Alternatively, the relative movement of distant stars to our own may be a drop in the bucket compared to the energy imparted to a ray by the implosion/explosion of a star.

My question is this: does the relative movement of the point of origin for a cosmic ray have a significant impact on that ray's energy as we experience it here on earth, or is this effect dwarfed by the power imparted on particles by the effects of the black holes and super novae that generate them?

Answer 1

My question is this: does the relative movement of the point of origin for a cosmic ray have a significant impact on that ray's energy as we experience it here on earth, or is this effect dwarfed by the power imparted on particles by the effects of the black holes and super novae that generate them?

The second idea is the correct one. The Earth moves around the Sun at a velocity of about 30 km/s, which revolves around the center of the Milky Way at about 200 km/s. Andromeda is getting closer to us at a speed of about 300 km/s. We know from the microwave background that the total speed of the Earth with respect to the CMB frame is about 370 km/s.

How fast are cosmic rays coming from a supernova? They move at (or slightly below) the speed of light, which is 300000 km/s. You see that the small velocity of the Earth with respect to other celestial objects is 1000 times smaller. Therefore the velocity of the Earth is definitely negligible. The energy of the cosmic rays will be almost the same, whether they smash with the Earth head-on, or they hit it from behind.

If you want to calculate exactly how tiny is this effect, you can use the following approximate formula:

$${E_{em} \over E_{obs}} = \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$

where $E_{em}$ is the energy of a cosmic ray in the frame where it is emitted by the supernova, $E_{obs}$ is the energy of the cosmic ray as seen here in Earth's frame, $v$ is the speed of the supernova w.r.t. the Earth and $c$ is the speed of light.

For a relative velocity $v=-300$ km/s, you get $E_{obs} = 1.001 E_{em}$.

This formula will work with highly relativistic particles (where $p \gg mc$) and when the supernova is moving mostly radially with respect to the Earth. (For transverse velocity the formula is different, see Redshift)

Edit:

If the cosmic rays come from an very distant source like a quasar, the cosmological redshift comes into play. In that case

$${E_{em} \over E_{obs}} = 1+z$$

Where $z$ is the redshift of the source. Since $z>0$, this effect can only reduce the energy of the cosmic ray, not increase it.

It can be interpreted as the result of the expansion of the universe, that moves very far objects away from us at superluminal speed. For the most distant quasars we know today, $z \approx 7$.