B Does a hot coffee have bigger mass than a cold coffee?

[PeterDonis]

In this simple case we can think of binding energy as distributed/associated to the electric field.

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]

I don't know if it actually makes sense or not: can we think of the bi-atomic molecule as a system of two particles with a spring attached to them? In this model the binding energy should be simply "stored" in spring's potential energy.

 

[PeterDonis]

can we think of the bi-atomic molecule as a system of two particles with a spring attached to them?

You can capture some phenomenological features this way, but of course there is no actual spring.

In this model the binding energy should be simply "stored" in spring's potential energy.

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.

 

[Orodruin]

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.

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]

It can be perfectly described by the energy difference of the electric field configuration of the bound system and the separated system.

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]

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?

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.

 

[cianfa72]

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)

In this case the bound system is actually "two particles plus the electric field".

 

[Orodruin]

In this case the bound system is actually "two particles plus the electric field".

"two particles plus some of the electric field"

 

[cianfa72]

"two particles plus some of the electric field"

Why some and not all of the electric field distributed over the space ?

 

[Orodruin]

Why some and not all of the electric field distributed over the space ?

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]

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.

Sorry, I didn't get your point. Which are the other field's sources other than the 2 charged particles ?

 

[Orodruin]

Sorry, I didn't get your point. Which are the other field's sources other than the 2 charged particles ?

If you have only two particles in the entire universe and they have forever been in a bound state, none.

 

[cianfa72]

If you have only two particles in the entire universe and they have forever been in a bound state, none.

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).