Q value in nuclear and particle physics

[Hertz]

Hi, sorry if this is a novice question. I've recently been introduced to Q value in nuclear and particle physics and all the places I've looked at define it as the change in rest energy between the initial particles and final particles.

$Q=E_f-E_i=(m_f-m_i)c^2$

But this doesn't take into account kinetic energy does it? It's just rest energies of the particles involved? But isn't kinetic energy super relevant in nuclear and particle physics? I tried looking up relativistic Q value but I didn't find anything. While we're at it, it doesn't include potential energy either, does it? I understand why potential energy is most often negligible though.

 

[Orodruin]

The last step in your formula is backwards. It should be mi-mf. What you have on the left hand side are the kinetic energies. If it was not it would be zero, which would be a meaningless quantity to define.

The Q value is a measure of the mass energy released in a reaction. This is why it is important. It is the total gain of kinetic energy.

 

[Astronuc]

The Q-value is based on the different in rest masses, and if positive means that the product will have some kinetic energy. In nuclear and particle interactions, one does need to take into account the initial energy as well. A positive Q-value means the reaction is 'exothermic', and a negative value means the reaction is endothermic, i.e., it absorbs (kinetic) energy which is converted to rest mass.

 

[Hertz]

Well it wouldn't be zero because mass energy can be converted to other types of energy. The way I've written it the E's are supposed to be rest energy and in the next step I've used $E=mc^2$.

Why do they do initial minus final? I hate unconventional definitions like that it just gives you more stuff to memorize :/

 

[Orodruin]

Again, the left hand side is not the total energy. It is the difference in kinetic energy.

 

[Hertz]

Again, the left hand side is not the total energy. It is the difference in kinetic energy.

Why?

Technically I was the one who wrote it. And the way I wrote it the left side is the difference in rest energy.

I think I get it though. The difference in rest energy is where the additional kinetic energy comes from. Well, loss of mass energy and loss of potential energy is where the additional kinetic energy comes from. So $Q=E_{rest,initial}-E_{rest,final}\approx\Delta K= K_{final}-K_{initial}$ assuming the change in potential energy is negligible.

Thanks for the help fellas :)

 

[Orodruin]

echnically I was the one who wrote it. And the way I wrote it the left side is the difference in rest energy.

In that case, your definition does not comply with the definition the rest of the physics world uses and you will just cause confusion by using it. It is the difference in rest energy, but it is the initial rest masses minus the final ones, not the other way around. That this is equal to the final kinetic energy minus the initial kinetic energy is a simple consequence of energy conservation.

I think I get it though. The difference in rest energy is where the additional kinetic energy comes from. Well, loss of mass energy and loss of potential energy is where the additional kinetic energy comes from. So $Q=E_{rest,initial}-E_{rest,final}\approx\Delta K= K_{final}-K_{initial}$ assuming the change in potential energy is negligible.

Yes, this looks much more standard.