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

[cianfa72]

What if the system's constituents are not free ? In other words if there is a field due to their interactions do we need to add in the total 4-momentum's vector sum also the 4-momentum associated to the field?

Note that in this case the system includes the system's constituents and the field as well.

 

[Sagittarius A-Star]

What if the system's constituents are not free ? In other words if there is a field due to their interactions do we need to add in the total 4-momentum's vector sum also the 4-momentum associated to the field?

Note that in this case the system includes the system's constituents and the field as well.

The contribution of each constituent is only uniquely determined for a system of free particles, like an ideal gas. In coffee, there is also binding energy between the constituents.

 

[cianfa72]

In coffee, there is also binding energy between the constituents.

So the only non vanishing component of the field 4-momentum vector is the "time" component -- i.e. the field Energy.

 

[vanhees71]

What if the system's constituents are not free ? In other words if there is a field due to their interactions do we need to add in the total 4-momentum's vector sum also the 4-momentum associated to the field?

Note that in this case the system includes the system's constituents and the field as well.

The mass is given by the entire internal energy of the composite system, not only the sum of the kinetic energies + rest energies of their constituents.

An extreme example are the baryons, e.g., the proton: it's mass is almost completely due to gluon-field energy.

 

[Nugatory]

So the only non vanishing component of the field 4-momentum vector is the "time" component -- i.e. the field Energy.

Not necessarily. Depending on the exact moment that we look at the system we may find a charged particle moving in one direction while the field carries momentum in the other direction, although a moment later we might find the field momentum has been transferred to another particle.

Conservation of energy is a powerful concept for many reasons; one is that it tells us that we can safely ignore all of these complexities as long as we keep track of the energy entering and leaving the system.

 

[Sagittarius A-Star]

So the only non vanishing component of the field 4-momentum vector is the "time" component -- i.e. the field Energy.

If a system contains field energy, then also the stress caused by the field must be considered.
Example:
https://www.researchgate.net/public...lativistic_paradox_about_electrostatic_energy

See also:
https://www.physicsforums.com/threads/four-momentum-of-a-field.180779/

 

[vela]

OK, so I accept that the rest mass is not additive, but why? What is the reason? How to explain it simply?

One answer is that "because it doesn't work that way." The universe works a certain way, and just because you think it should work or want it to work another way doesn't really matter.

The burden here is really on you to justify why rest mass should be additive. Your reason or intuition is likely based on everyday experience, and relativity is teaching you that your intuition isn't generally true. It only works as an approximation.

But what is the difference between kinetic energy of a moving cup of coffee and kinetic energy of the particles that it consists of? Because the kinetic energy of the moving cup doesn't affect its mass, but the kinetic energies of its particles do. So what is the difference?

The kinetic energy of the particles is stored within internal degrees of freedom in the cup. It's literally part of the cup, so it contributes to the mass of the cup. A moving cup has kinetic energy attributed to external degrees of freedom, so it doesn't affect the cup's mass.

 

[Herman Trivilino]

It is essentially the definition of inertial mass. How else would you define inertial mass if not as the resistance to acceleration?

I don't know how to define inertial mass in special relativity, but resistance to acceleration gives a weird result. The resistance to acceleration gives different results depending on the direction of the force.

SI provides a definition but I no longer understand it since they did away with the standard kilogram. I suppose I could figure out with a moderate amount of effort.

 

[Nugatory]

The resistance to acceleration gives different results depending on the direction of the force.

That's an artifact of using frame-dependent coordinate acceleration instead of proper acceleration, and it goes away if we work with four-forces and four-velocities instead.

 

[vanhees71]

The mass is given by the entire internal energy of the composite system, not only the sum of the kinetic energies of their constituents.

I don't know how to define inertial mass in special relativity, but resistance to acceleration gives a weird result. The resistance to acceleration gives different results depending on the direction of the force.

It doesn't. As soon as you describe everything in a manifestly covariant way, it gets as simple (or complicated ;-)) as in Newtonian physics. Particularly mass is the same in both theories, i.e., the invariant mass,
$$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=K^{\mu}(x,\mathrm{d}_{\tau} x,\tau).$$
By construction, the "Minkowski force" has to fulfill
$$\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} K_{\mu}=0,$$
because by definition of the proper time, $\tau$, you have
$$(\mathrm{d}_{\tau} x^{\mu})(\mathrm{d}_{\tau} x_{\mu})=c^2.$$

SI provides a definition but I no longer understand it since they did away with the standard kilogram. I suppose I could figure out with a moderate amount of effort.

The SI defines the units. Since 2019 everything is defined by defining the fundamental constants of Nature, with the only exception of $G$, the Gravitational constant, because it's not accurate enough to determine. That's why the second, as one of the base units (in some sense it's indeed The base unit of the SI) is still defined via the Cesium standard and not implictly by the definitions of the natural constants.

Having the second, you define the metre as the unit of length by giving the speed of light in vacuum a definite value.

For the kg you need in addition $h=2 \pi \hbar$.

The Coulomb is simply defined by setting the value of the elementary charge (the charge of a proton or the negative of the charge of an electron).

Then the mol is defined by defining the Avogadro constant.

And finally the Kelvin is defined by setting the value of the Boltzmann constant. That's it.

 

[cianfa72]

The mass is given by the entire internal energy of the composite system, not only the sum of the kinetic energies + rest energies of their constituents

Therefore, if there is a potential describing the interaction between system's constituents, we can just include in the system's total internal energy also the total potential energy without the need to introduce force field with its 4-momentum.

 

[vanhees71]

There is no potential describing the interaction between the system's constituents. There's a famous no-go theorem that any attempt to create such a theory (at least if you assume that it can be formulated in terms of the action principle) of interacting point particles you fail. The really successful relativistic theories are all field theories. On the classical level matter should be described by continuum rather than point-particle mechanics to get a fully consistent theory.

There was a lot of progress in the recent years concerning relativistic hydrodynamics, derived via relativistic kinetic theory from quantum field theory.

 

[Orodruin]

I don't know how to define inertial mass in special relativity, but resistance to acceleration gives a weird result. The resistance to acceleration gives different results depending on the direction of the force.

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

 

[vanhees71]

My point was to emphasize that if you try to make sense of point-particle mechanics (which is not so easy, if not impossible, by the way) the quantity mass in (special) relativity is the same as in Newtonian physics, if you use the right description in terms of a manifestly covariant way. The key is to introduce the concept of proper time and Minkoski force.

Of course, point-particle mechanics in relativity is a bit limited, because what you can really describe to some extent is the motion of a free particle and a particle in an external field, neglecting the inevitable interaction with its own field, i.e., in electrodynamics, the electromagnetic Coulomb field around the point particle + the radiation field when it's accelerated and the back reaction of the latter to the particle, aka radiation damping.

If you include the latter inevitable consequences of relativity in point-particle mechanics, you come to the conclusion that you get not much further than to the Landau-Lifshitz approximation of the Lorentz-Abraham-Dirac equation.

A consistent description of interacting point particles is even less convincing. There you don't get further than some corrections to the Newtonian limit.

 

[cianfa72]

There is no potential describing the interaction between the system's constituents

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

 

[Sagittarius A-Star]

It doesn't. As soon as you describe everything in a manifestly covariant way, it gets as simp
$$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=K^{\mu}(x,\mathrm{d}_{\tau} x,\tau).$$By construction, the "Minkowski force" has to fulfill$$\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} K_{\mu}=0,$$

This is only valid for a pure force, that means a mass-preserving force. A counter-example exists during an elastic collision of objects. The more general definition of force is:
$$\mathbf F=\frac{d}{d \tau} \mathbf P=\frac{d}{d \tau}(m \mathbf U)=m \mathbf A+\frac{dm}{d \tau}\mathbf U$$
Source, see equation (70):
http://www.scholarpedia.org/article/Special_relativity:_mechanics#Four-force_and_three-force

 

[Sagittarius A-Star]

Time to brew another cup!

We can also put a heating pad under the cup to make the cold coffee warm again. This required a special cup.

The heating pad impresses on the coffee a heatlike 4-force $\mathbf F =\frac{d}{d \tau}(m \mathbf U)$. In the common rest-frame of coffee and heating pad, it has only a temporal component.

Source:​

Rindler's book "Introduction to Special Relativity", 2nd edition, chapter 35 "Three-force and four-force"​

Short version online:​

http://www.scholarpedia.org/article/Special_relativity:_mechanics#Four-force_and_three-force​

The above used definition of 4-force extends Newton's 2nd law to 4 dimensions.

DEFINITION II.​

The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjunctly.
...

https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)/Definitions

LAW II.​

The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.
...

https://en.wikisource.org/wiki/The_...l_Philosophy_(1846)/Axioms,_or_Laws_of_Motion

 

[Orodruin]

We can also put a heating pad under the cup to make the cold coffee warm again.

Eeeeew.
(While safe, it makes coffee taste more bitter.)

 

[vanhees71]

This is only valid for a pure force, that means a mass-preserving force. A counter-example exists during an elastic collision of objects. The more general definition of force is:
$$\mathbf F=\frac{d}{d \tau} \mathbf P=\frac{d}{d \tau}(m \mathbf U)=m \mathbf A+\frac{dm}{d \tau}\mathbf U$$
Source, see equation (70):
http://www.scholarpedia.org/article/Special_relativity:_mechanics#Four-force_and_three-force

I was talking about the formulation of a point-particle theory in special relativity. I've never heard the expression "pure force". What do you mean by that? As I tried to explain the most clear formulation of relativistic point-particle mechanics (which is of way less applicability than in Newtonian physics by the way) is in terms of the manifestly covariant equation with the Minkowski four-force.

 

[Sagittarius A-Star]

I've never heard the expression "pure force". What do you mean by that?

This expression is used in Rindler's book "Introduction to Special Relativity", 2nd edition, chapter 35 "Three-force and four-force". It means a mass-preserving force and therefore $\mathbf U \cdot \mathbf F = 0$.

In Wikipedia, they write a disclaimer before defining a pure force:

For a particle of constant invariant mass $m>0$ ...
${\mathbf {F} =m\mathbf {A} =\left(\gamma {\mathbf {f} \cdot \mathbf {u} \over c},\gamma {\mathbf {f} }\right).}$

Source:
https://en.wikipedia.org/wiki/Four-force#In_special_relativity

The temporal component gets zero in the rest-frame of the particle.

 

[Herman Trivilino]

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.