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

[Mister T]

You don’t really. Inertial mass is a concept of classical mechanics.

I agree. Inertia is not a well-defined quantity in SR.

In relativity ”mass” generally refers to rest energy. To me, the real beauty of $E = mc^2$ is that the rest energy is equal (with c=1) to a system’s rest frame inertia. Thereby justifying the use of the word “mass” for the rest energy.

Yes, but why are you using the term "inertia" when speaking about relativistic concepts?

 

[renormalize]

Inertial mass is a concept of classical mechanics.

I agree. Inertia is not a well-defined quantity in SR.

In relativity ”mass” generally refers to rest energy. To me, the real beauty of
is that the rest energy is equal (with c=1) to a system’s rest frame inertia. Thereby justifying the use of the word “mass” for the rest energy.

Yes, but why are you using the term "inertia" when speaking about relativistic concepts?

Well, Landau and Lifshitz in The Classical Theory of Fields (3rd ed.) clearly embrace the notion of "inertial mass" in the context of general relativity. From chap. 11 "The Gravitational Field Equations", §101 "The Energy-Momentum Pseudotensor", pg. 309:

 

[Orodruin]

I agree. Inertia is not a well-defined quantity in SR.

It is a perfectly well defined concept. It is just not invariant, nor a scalar.

Yes, but why are you using the term "inertia" when speaking about relativistic concepts?

I … wasn’t. I was answering OP with regards to the classical concept of Newton’s second law. Later I was discussing that the rest energy in relativity turns out to be the same as the inertial mass in the rest frame. In an object’s rest frame, classical mechanics works well for that object. The inertia concept I mentioned above degenerates to scalar multiplication.

 

[cianfa72]

Therefore the inertial mass from Landau and Lifshitz is the invariant/rest mass.

 

[vanhees71]

I agree. Inertia is not a well-defined quantity in SR.

Yes, but why are you using the term "inertia" when speaking about relativistic concepts?

In SR the measure of inertia is the energy-momentum-stress tensor, and that's why in the mathematical expression, this gets the "sources" of the "gravitational fields" in GR due to the equivalence principle. The universality of the corresponding coupling constant (modulo factors Newton's gravitational constant, $G$) is due to the non-Abelian nature of the local gauge symmetry underlying GR.

 

[cianfa72]

I was thinking, for example, of a gas of bi-atomic molecules. There should be a potential/potential energy for each of them.

So as @vanhees71 said in post#62, we should not include the potential energy of each bi-atomic molecule in the calculation of system's total Energy...

 

[vanhees71]

The invariant mass is the total energy of the system in the rest frame of its center of momentum (natural units, $c=1$, for simplicty). E.g., for a gas it's the total inner energy in the rest frame of the container.

 

[cianfa72]

The invariant mass is the total energy of the system in the rest frame of its center of momentum (natural units, $c=1$, for simplicty). E.g., for a gas it's the total inner energy in the rest frame of the container.

Yes, but in the of bi-atomic gas case, the total inner energy includes the rest energies of atoms plus their KE evaluated in rest frame of the system's center of momentum (i.e. the rest frame of the container) plus the potential energy of each bi-atomic molecule.

 

[vanhees71]

Of course. It's the total internal energy of the gas, $U$, in the center-momentum frame.

 

[cianfa72]

Of course. It's the total internal energy of the gas, $U$, in the center-momentum frame.

Can we therefore continue to use the concept of potential energy also in special relativity ?

 

[vanhees71]

In some circumstances yes. E.g., for a particle in a static electric field you can write down the Lagrangian (in the (3+1) non-covariant formalism with $\dot{\vec{x}}=\mathrm{d} \vec{x}/\mathrm{d} t$)
$$L=-m \sqrt{1-\dot{\vec{x}}^2} -q \Phi(\vec{x}).$$
This is of course an approximation, where you neglect radiation-reaction effects.

 

[cianfa72]

In some circumstances yes. E.g., for a particle in a static electric field you can write down the Lagrangian

Sorry, I believe this is the case for a particle in an external given/assigned external electric field.
In the above case, indeed, there is not an external field but the potential energy is due to the interaction between atoms in each bi-atomic molecule.

 

[vanhees71]

The interaction cannot be described by a potential. There's no action at a distance in relativistic physics.

 

[cianfa72]

The interaction cannot be described by a potential. There's no action at a distance in relativistic physics.

Therefore the potential energy for each bi-atomic molecule is not defined in relativity.

 

[PeterDonis]

Therefore the potential energy for each bi-atomic molecule is not defined in relativity.

"Potential energy" is the wrong term for what I think @vanhees71 was talking about for the bi-atomic molecule. The bi-atomic molecule (and an individual atom, for that matter) has binding energy: the energy it would take to separate its constituents into free particles. That binding energy makes a negative contribution to the molecule's invariant mass. This is just as true in relativity as in non-relativistic physics.

 

[cianfa72]

The bi-atomic molecule (and an individual atom, for that matter) has binding energy: the energy it would take to separate its constituents into free particles. That binding energy makes a negative contribution to the molecule's invariant mass.

AFAIK, the binding energy for a bi-atomic molecule should be negative. What is the reason ?

 

[PeterDonis]

AFAIK, the binding energy for a bi-atomic molecule should be negative. What is the reason ?

Because it is a negative contribution to the invariant mass of the molecule: the invariant mass of the molecule is less than the sum of the invariant masses of its constituent atoms.

 

[cianfa72]

That binding energy makes a negative contribution to the molecule's invariant mass. This is just as true in relativity as in non-relativistic physics.

In a simple "classical" case of 2 bonded charged particles (plus and minus) the binding energy is the work made by external forces (external w.r.t. the system of 2 particles) to bring the charged particles from infinity to the current system configuration. Such work is negative thus the binding energy as well.

 

[PeterDonis]

In a simple "classical" case of 2 bonded charged particles (plus and minus) the binding energy is the work made by external forces (external w.r.t. the system of 2 particles) to bring the charged particles from infinity to the current system configuration. Such work is negative thus the binding energy as well.

Yes. And this is perfectly consistent with what I said in post #87.

 

[cianfa72]

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

 

[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).