Binding energy is a property of the bound system. It cannot be assigned to any particular part of it. The "electric field" isn't even a part of the bound system, properly speaking, since it extends to infinity.cianfa72 said:In this simple case we can think of binding energy as distributed/associated to the electric field.
You can capture some phenomenological features this way, but of course there is no actual spring.cianfa72 said:can we think of the bi-atomic molecule as a system of two particles with a spring attached to them?
Yes, but as noted above, there is no actual spring so this model does not mean that the binding energy can actually be localized this way.cianfa72 said:In this model the binding energy should be simply "stored" in spring's potential energy.
It can be perfectly described by the energy difference of the electric field configuration of the bound system and the separated system. The energy density of the electric field being proportional to the field strength squared. The electric field is certainly an important part of the bound system, without it the system would not be bound.PeterDonis said:Binding energy is a property of the bound system. It cannot be assigned to any particular part of it. The "electric field" isn't even a part of the bound system, properly speaking, since it extends to infinity.
Wouldn't this have to cover an unbounded region, since it would have to include radiation emitted to infinity during the process of forming the bound system?Orodruin said:It can be perfectly described by the energy difference of the electric field configuration of the bound system and the separated system.
The radiation is not part of the bound system. It is of course a matter of definition how you draw the boundary of your system (particularly as there is only one EM field - so separating different contributions is a bit arbitrary in the first place), but the bound state itself is stationary and electrically neutral so the bound state field is a dipole at worst - falling off as 1/r^3.PeterDonis said:Wouldn't this have to cover an unbounded region, since it would have to include radiation emitted to infinity during the process of forming the bound system?
In this case the bound system is actually "two particles plus the electric field".Orodruin said:The radiation is not part of the bound system. It is of course a matter of definition how you draw the boundary of your system (particularly as there is only one EM field - so separating different contributions is a bit arbitrary in the first place)
"two particles plus some of the electric field"cianfa72 said:In this case the bound system is actually "two particles plus the electric field".
Why some and not all of the electric field distributed over the space ?Orodruin said:"two particles plus some of the electric field"
Because the electric (and magnetic) field can have other sources than the particles in the bound state in addition to containing the radiation emitted when the bound state formed.cianfa72 said:Why some and not all of the electric field distributed over the space ?
Sorry, I didn't get your point. Which are the other field's sources other than the 2 charged particles ?Orodruin said:Because the electric (and magnetic) field can have other sources than the particles in the bound state in addition to containing the radiation emitted when the bound state formed.
If you have only two particles in the entire universe and they have forever been in a bound state, none.cianfa72 said:Sorry, I didn't get your point. Which are the other field's sources other than the 2 charged particles ?
You mean that in real universe where there are multiple particles, the current field distribution depends on the two particles in the bound system acting as sources as well on all other charged particles (not included in the two particle's bound system).Orodruin said:If you have only two particles in the entire universe and they have forever been in a bound state, none.