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

[cianfa72]

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

So if use Newton's second law to define inertial mass (i.e. by defining the ratio ##F/a## as the inertial mass ##m## of a body for a given force ##F## involved in the definition) then Newton's great discovery might be that if we apply another force ##F_1## on the same object then the ratio ##F_1/a_1## does not change.

 

[BeedS]

On a related note: The effect of the mass increase due to higher energy is absolutely dwarfed by the thermal expansion of the coffee. Warm coffe will therefore have lower density than cold coffee.

And the coffee will have a greater buoyancy in the atmosphere, so if you try to weigh the coffee will the buoyancy cause the coffee to weigh less?

 

[Dale]

But what is the difference between kinetic energy of a moving cup of coffee and kinetic energy of the particles that it consists of?

The difference is the momentum. Take two cups of coffee, a hot one and a cold one with the same energy. The difference will be that the cold one has a greater momentum and therefore a lower invariant mass.

 

[cianfa72]

The difference is the momentum. Take two cups of coffee, a hot one and a cold one with the same energy.

Do you mean the cold one with the same energy as the hot one (at rest) when the cold cup's macroscopic kinetic energy of motion is included ?

 

[Dale]

Do you mean the cold one with the same energy as the hot one (at rest) when the cold cup's macroscopic kinetic energy of motion is included ?

Yes. The cup’s macroscopic KE is part of ##E## in the formula ##m^2 c^2=E^2/c^2-p^2##.

You have to include the KE.

 

[cianfa72]

Yes. The cup’s macroscopic KE is part of ##E## in the formula ##m^2 c^2=E^2/c^2-p^2##.
You have to include the KE.

So the macroscopic KE of the cup (due to its macroscopic motion) contributes to its total Energy. Furthermore its total squared momentum ##p^2## includes the macroscopic momentum of the cup (due to its macroscopic motion) as well.

Hence, as you said, the invariant/rest mass ##m## of the cold cup of coffee will be lower than the invariant/rest mass of the hot cup with the same Energy.

 

[cianfa72]

Relativistic mass is additive, but is better thought of as the energy of the system divided by ##c^2##.

Ah ok, so relativistic mass is just the total energy of the system divided by ##c^2##. It is basically just another name for Energy.

 

[vanhees71]

No, it isn't. A hot cup of coffee has a larger invariant mass than the sum of the rest masses of its individual molecules, because the kinetic energy associated with its temperature also contributes to the overall invariant mass. Invariant mass is not additive.An object's invariant mass is independent of its speed. But the invariant mass of a system composed of multiple objects is not the sum of the invariant masses of the individual objects. As has already been pointed out, invariant mass is not additive.

For a system composed of multiple objects, the kinetic energies of the individual objects, in the overall rest frame of the system, contribute to the system's total invariant mass. The molecules in the hot cup of coffee have more kinetic energy in the cup's rest frame than the molecules in a cold cup of coffee. That extra kinetic energy increases the invariant mass of the overall system (the cup of coffee).

One should just add that the usual definition of the invariant mass of a composite system is that it's the total energy of the system as given in its center-momentum frame, i.e., in the frame, where the total momentum is 0.

In this sense the invariant mass of the hot coffee is larger by ##\Delta Q/c^2## than set of the cold coffee, where ##\Delta Q## is the heat added to heat up the cold coffee.

It's just a very convenient "natural" convention to measure internal properties of composite systems in appropriate rest frames. This also resolves all the quibbles about the Lorentz-transformation properties thermodynamic quantities in the theory of relativity. E.g., temperature is always defined in the rest frame of the "heat bath", i.e., it's measured by a thermometer co-moving with the medium, etc.

 

[cianfa72]

Just to be clear: for the cup of coffee at rest in an inertial frame the ensemble average ##\left \| \sum \vec p_i) \right \|^2## of its constituents actually vanishes. Yet its invariant mass is the sum of its constituents' energy included their individual kinetic energies -- all divided by ##c^2##.

 

[PeterDonis]

for the cup of coffee at rest in an inertial frame the ensemble average ##\left \| \sum \vec p_i) \right \|^2## of its constituents actually vanishes.

Just to be clear, you mean the vector sum of the momenta of all the constituents in the cup's rest frame, correct? If so, there is no need to put the absolute value sign around the sum and square it. The vector sum itself is zero.

 

[cianfa72]

you mean the vector sum of the momenta of all the constituents in the cup's rest frame, correct? If so, there is no need to put the absolute value sign around the sum and square it. The vector sum itself is zero.

Yes, it is zero since there is not a net momenta in the cup's rest (inertial) frame.

 

[cianfa72]

In the energy-momentum formula for the system, the momentum ##p_i## of each system's constituent is given by $$\gamma (v_i)m_i v_i$$ where ##m_i## is the invariant mass of each constituent and ##v_i## its velocity in the picked inertial frame ?

 

[Sagittarius A-Star]

In the energy-momentum formula for the system, the momentum ##p_i## of each system's constituent is given by $$\gamma (v_i)m_i v_i$$ where ##m_i## is the invariant mass of each constituent and ##v_i## its velocity in the picked inertial frame ?

That's the norm of the 3-momentum of each system's constituent.

The 3-momentum of each system's constituent is a vector, like the velocity:
$$\vec p_i = \gamma (v_i)m_i \vec v_i$$
The 4-momentum ##\mathbf P = (\frac{E}{c}, \vec p) = (\frac{E}{c},p_x, p_y, p_z) ## is additive, related to an isolated system of free particles.

 

[vanhees71]

Just to be clear: for the cup of coffee at rest in an inertial frame the ensemble average ##\left \| \sum \vec p_i) \right \|^2## of its constituents actually vanishes. Yet its invariant mass is the sum of its constituents' energy included their individual kinetic energies -- all divided by ##c^2##.

That's the definition of "at rest". In general the invariant mass by definition is the internal energy of the system divided by ##c^2##.

 

[cianfa72]

The 3-momentum of each system's constituent is a vector, like the velocity:
$$\vec p_i = \gamma (v_i)m_i \vec v_i$$

Yes, in the given inertial frame the ##p^2## that enters in the energy-momentum equation $$m^2 c^2=E^2/c^2-p^2$$ is actually the square of the norm of the vector sum ##\vec p## of 3-momentum system's constituents.

 

[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.

 

[Mister T]

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.