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

[Lotto]

TL;DR Summary

According to $E=m_0c^2$ (when $v=0$), when we add an energy to a system, its mass increases. But something like relativistic mass is just a trick. The hot coffee's particles have higher energies, but according to this logic the same masses. So, how is it?

As far as I know, something like relativistic mass is just a concept, just a trick, there is nothing like the relativistic mass. When I move faster, I have higher kinetic energy, but my mass is still the same as if I was at rest. Kinetic and potential energies do not increase object's mass.

A hotter coffee has particles with higher speeds, its energies are higher than those of a cold coffee. That is because of their higher kinetic energies. So their mass should be also the same as their "rest mass" = the only mass? But why should be then the hotter coffee's mass bigger?

Am I a bit confused about this.

 

[PeroK]

The mass of a system of particles is not the sum of the rest mass of each particle. The mass of a hydrogen atom, for example, is less that the mass of a proton plus the mass of an electron.

In your example, a system of particles in motion has a greater mass than the sum of the individual rest masses.

This is related to invariant mass, not relativistic mass.

 

[Ibix]

The four-momentum is a 4d vector. It's $t$-component is the so-called relativistic mass, and its modulus is the invariant (or "rest") mass.

The four momentum of the cup of coffee is the sum of the four momenta of the atoms. As with any other vector sum, the modulus of the sum is not the sum of the moduli, so mass is not additive. Relativistic mass is additive, but is better thought of as the energy of the system divided by $c^2$.

 

[Orodruin]

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.

 

[Lotto]

The mass of a system of particles is not the sum of the rest mass of each particle. The mass of a hydrogen atom, for example, is leas that the mass of a proton plus the mass of an electron.

In your example, a system of particles in motion has a greater mass than the sum of the individual rest masses.

This is related to invariant mass, not relativistic mass.

OK, a hydrogen atom has a smaller mass than the sum of proton's and neutron's energies because of a binding energy. I understand. So for coffee is it the same, but why? Is it because of hydrogen bonds that the water molecules create?

But still, that doesn't solve my confusion. Because when an object's real mass is independent of its speed, then how can be the hot coffee heavier? It is heavier because its particles are heavier. And they are heavier because they thave higher energies - kinetic energies. But higher kinetic energy shouldn't affect an objects mass. So where are my thoughts wrong?

 

[PeterDonis]

So for coffee is it the same

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.

when an object's real mass is independent of its speed, then how can be the hot coffee heavier?

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).

 

[PeroK]

OK, a hydrogen atom has a smaller mass than the sum of proton's and neutron's energies because of a binding energy. I understand. So for coffee is it the same, but why? Is it because of hydrogen bonds that the water molecules create?

But still, that doesn't solve my confusion. Because when an object's real mass is independent of its speed, then how can be the hot coffee heavier? It is heavier because its particles are heavier. And they are heavier because they thave higher energies - kinetic energies. But higher kinetic energy shouldn't affect an objects mass. So where are my thoughts wrong?

This had been answered by @Ibix above. The invariant mass of a system of particles is defined by:$$E^2 = P^2c^2 + M^2c^4$$Where $E$ is the total energy of the system and $P$ is the magnitude of the total momentum of the system. It is NOT defined as the sum of rest masses, nor the sum of the relativistic masses.

The example of hot and cold coffee is a poor one, since the difference is immeasurably small compared to the other factors, such as the rate of evaporation of hot coffee.

Examples, IMO, should be more relevant- like atoms.

 

[PeroK]

PS note that even a system of photons (massless particles) can have an invariant mass, using the above definition.

 

[Sagittarius A-Star]

As far as I know, something like relativistic mass is just a concept, just a trick, there is nothing like the relativistic mass.

Relativistic mass is not a trick. It is an out-dated synonym for $E/c^2$.

In it's center-of-momentum frame, the mass of an isolated system of free particles is equal to the sum of the energies of it's components, divided by $c^2$.

Source:
https://en.wikipedia.org/wiki/Mass_in_special_relativity#The_mass_of_composite_systems

 

[Lotto]

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

The total rest energy of a system is given by $E_0=m_0c^2$. What does this energy correspond to? To the sum of all kinetic energies and all potential energies of the particles? I think that it is a energy that corresponds to properties of an object.

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?

Also, when the cup has a gravitational potential energy, why does this energy have nothing to do with its total energy that we can use to calculate its mass? Why aren't objects at height 5 m heavier than objects at height 2 m?

 

[DaveC426913]

Guy are missing the forest for the trees.

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.

This can't be ignored.

(But it could be corrected by a slight clarification to the original question.)

 

[PeterDonis]

I accept that the rest mass is not additive, but why?

Because, as has already been said, things other than the rest masses of the individual constituents contribute to the rest mass of a composite system.

How to explain it simply?

It is a consequence of relativity.

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

The kinetic energy of the cup is zero in its center of mass frame, which is the frame in which the nonzero kinetic energies of the particles that contribute to the cup's overall rest mass are evaluated. This has already been stated, more than once, in this thread.

In a frame other than the cup's center of mass frame, the kinetic energy of the cup is the kinetic energy of its center of mass.

when the cup has a gravitational potential energy, why does this energy have nothing to do with its total energy that we can use to calculate its mass?

Because the gravitational potential energy you speak of is not a property of the cup, it's a property of the system Earth + cup. It has no meaning if you are considering measurements of the cup alone.

Why aren't objects at height 5 m heavier than objects at height 2 m?

First, "heavier" here has to mean mass, not weight. The weight of a cup of coffee will get smaller as its height increases (because the gravitational force gets weaker), although this difference is much too small to measure if you are going from 2 m to 5 m height.

If we mean mass, or more precisely rest mass (or invariant mass), the measurement of it involves the cup alone and has nothing to do with whatever gravitational field the cup happens to be in. You could measure the invariant mass of the cup in orbit in the International Space Station and it would still be the same. This is a consequence of the equivalence principle.

What would change if you raised the cup from 2 m to 5 m would be the overall rest mass of the Earth + cup system: to raise the cup you would have to add a very tiny amount of energy to that system, and that added energy would show up as a very tiny increase in the overall rest mass of the system. But you would not be able to see this by just measuring the cup alone. You would have to measure the whole system.

 

[PeterDonis]

This can't be ignored.

In any real case, this is true. But we can consider an idealized case where the cup is somehow thermally isolated so that it can't lose heat from evaporation or any other means. Idealized thought experiments like this are used in physics all the time to help gain an understanding of a particular effect without other effects interfering. I don't see any problem with interpreting the OP's question as referring to such an idealized thought experiment.

 

[DaveC426913]

In any real case, this is true. But we can consider an idealized case where the cup is somehow thermally isolated so that it can't lose heat from evaporation or any other means. Idealized thought experiments like this are used in physics all the time to help gain an understanding of a particular effect without other effects interfering. I don't see any problem with interpreting the OP's question as referring to such an idealized thought experiment.

That wasn't what I was suggesting, or how I interpreted Orodruin's objection.The OP is vague about what two scenarios are being compared. It does not say it is the same sample (i.e. cold coffee being heated up).

If one litre of cold coffee is compared to one litre of hot coffee , the hot coffee will mass less, because there will literally be less coffee in that litre (because it's less dense).

 

[PeterDonis]

If one litre of cold coffee is compared to one litre of hot coffee , the hot coffee will mass less, because there will literally be less coffee in that litre (because it's less dense).

Yes, agreed. I suspect that the OP was implicitly assuming the same number of "coffee molecules" in both cases; that is the case my posts were also implicitly assuming. But I agree that this should be made explicit in the scenario so we know exactly what we're discussing.

 

[DaveC426913]

Yes, agreed. I suspect that the OP was implicitly assuming the same number of "coffee molecules" in both cases; that is the case my posts were also implicitly assuming. But I agree that this should be made explicit in the scenario so we know exactly what we're discussing.

Granted. Upon review, it is apparent that the OP is specifically interested in relativistic effects and assumes (albeit implicitly) the same number of molecules in both scenarios.

 

[Lotto]

What would change if you raised the cup from 2 m to 5 m would be the overall rest mass of the Earth + cup system: to raise the cup you would have to add a very tiny amount of energy to that system, and that added energy would show up as a very tiny increase in the overall rest mass of the system. But you would not be able to see this by just measuring the cup alone. You would have to measure the whole system.

So when I do a work, I increase an object's rest mass. If I lift it on a table and let it there at rest, its rest mass is higher, although it is not moving.

So when I am pulling a box with a constant speed by using a force that is equal to a frictional force, the object's rest mass is constantly getting bigger because I am conducting a work on it? Or not because I need a total work to be non-zero?

 

[Vanadium 50]

OK, so I accept that the rest mass is not additive, but why?

Today it is 40 degrees in New York, 50 degrees in Boston and 60 degrees in Washington. In total, 150 degrees!

Hopefully this example convinces you that things do not always combine with simple addition.

 

[cianfa72]

Because, as has already been said, things other than the rest masses of the individual constituents contribute to the rest mass of a composite system.

Rest masses of each constituents are the same as their invariant masses ?

 

[PeterDonis]

So when I do a work, I increase an object's rest mass.

No, you don't. Doing work on the object will increase either its kinetic energy or its potential energy, but not its rest mass.

Heating the object up will increase its rest mass.

Doing work on an object to raise it in a gravitational field will increase the rest mass of the system consisting of the object + the source of the field--provided you put in the work from outside the system. If you do the work using something that is already present in the system--for example, you lift the object yourself--then the overall rest mass of the system does not change; all that changes is that some of the internal energy that contributes to that overall rest mass is redistributed from you to the object. But in either case the rest mass of the object itself does not change.

So when I am pulling a box with a constant speed by using a force that is equal to a frictional force, the object's rest mass is constantly getting bigger because I am conducting a work on it?

No. See above.

 

[PeterDonis]

Rest masses of each constituents are the same as their invariant masses ?

"Rest mass" and "invariant mass" are just two different names for the same thing.

 

[Sagittarius A-Star]

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

The proper time squared of an object, moving with reference to an inertial frame, is:
${d\tau}^2 = {dt}^2 - \frac{{dx}^2+ {dy}^2+ {dz}^2}{c^2}$

You can multiply this equation with the invariant quantity $\frac{m^2}{{d\tau}^2}$. Then with $\gamma = \frac{dt}{d \tau}$ you get:
$m^2 = (\gamma m)^2 - \frac{1}{c^2}(\gamma m v)^2 = (\gamma m)^2 - (\frac{p}{c})^2$

The mass squared of an isolated composite system of free particles is:
${m_{\text{System}}}^2 = (\sum_i {(\gamma_i m_i)})^2 - \frac{1}{c^2}\left \| \sum \vec p_i \right \|^2$

And in the center-of-momentum frame you get:
$${m_{\text{System}}} = \sum_{i=1}^N (\gamma_i m_i) = \gamma_1 m_1 + \gamma_2 m_2 + ... +\gamma_N m_N $$

 

[Nugatory]

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

Say that we have some stuff - maybe a chunk of lead, maybe some hot coffee, maybe some mirrors and light bouncing back and forth between them, maybe a nuclear bomb about to explode, maybe a herd of zebras being chased by a lion, maybe all of the above and a lot more as well, pretty much anything we can imagine.

But whatever it is, we can imagine that it is all inside an imaginary massless box and that we put that box on a scale to weigh it. We will find that the mass changes only when something enters or leaves the box - because the rest mass of the box itself plus its contents is related to the total energy in the box by $E=mc^2$ this is just another way of saying that energy is conserved.

Suppose that the only thing in the box is a billiard ball, and it is not moving (relative to the imaginary box, which in turn is at rest relative to the imaginary scale I’m using to weigh it). The scale will tell me that mass of the box is equal to the rest mass of the ball. Now I reach into the box and give the ball a shove. I’m bringing energy in from outside, added it to the box, so the weight of the box must increase according to $E=mc^2$, $m=E/c^2$. But…. The rest mass of the ball hasn’t changed and that’s the only rest mass we have.

Therefore we can’t just add up the rest masses to get the mass of the total system.

 

[Orodruin]

Now I reach into the box and give the ball a shove. I’m bringing energy in from outside, added it to the box, so the weight of the box must increase according to E=mc2, m=E/c2.

I was with you up until here where you also added momentum to the box, making the full system no longer be at rest. What you describe following that is essentially the relativistic mass of the system since the only contribution to the system’s 4-momentum is the ball.

 

[PeterDonis]

Now I reach into the box and give the ball a shove.

To address @Orodruin's objection, we would need to have at least two balls, so that we could shove them in opposite directions to keep the total momentum zero.

 

[Nugatory]

I was with you up until here where you also added momentum to the box, making the full system no longer be at rest. What you describe following that is essentially the relativistic mass of the system since the only contribution to the system’s 4-momentum is the ball.

You have caught me cutting corners. To do it right I need two billiard balls, and I must shove them in opposite directions to avoid changing the total momentum.

 

[cianfa72]

But whatever it is, we can imagine that it is all inside an imaginary massless box and that we put that box on a scale to weigh it.

The scale involved here should be actually a twin-pan scale in order to measure mass w.r.t. a reference mass (assumed to be unitary).

 

[Orodruin]

The scale involved here should be actually a twin-pan scale in order to measure mass w.r.t. a reference mass (assumed to be unitary).

The type of scale really doesn’t matter. It is fine to talk about an undefined ideal mass measuring device.

 

[cianfa72]

The type of scale really doesn’t matter. It is fine to talk about an undefined ideal mass measuring device.

But if we use a normal scale we actually bring in the concept of weight due to gravity, I believe.

 

[Ibix]

But if we use a normal scale we actually bring in the concept of weight due to gravity, I believe.

It's the same with a beam balance, though. It confirms that the torque exerted by one weight equals that exerted by the other. So you are comparing one weight force to another, whereas a spring balance compares a weight force to an elastic force.

The way to directly measure mass without the intervention of gravity is to apply a known force for a known amount of time, measure its velocity change, and apply Newton's second law or its relativistic equivalent.

 

[cianfa72]

It's the same with a beam balance, though. It confirms that the torque exerted by one weight equals that exerted by the other. So you are comparing one weight force to another, whereas a spring balance compares a weight force to an elastic force.

My point was take the above (using beam balance) as the operational definition of mass (w.r.t. a given unitary mass).

The way to directly measure mass without the intervention of gravity is to apply a known force for a known amount of time, measure its velocity change, and apply Newton's second law or its relativistic equivalent.

I tend to think of the Newton's second law as really a law between already defined terms (i.e. a mathematical relation between force, mass and acceleration as already defined terms/objects). In your use it is basically not a law but just the definition of mass.

 

[Orodruin]

But if we use a normal scale we actually bring in the concept of weight due to gravity, I believe.

The beam balance scale is definitely dependent on gravity. It doesn’t matter. For the system as a whole, it is a good approximation to use the energy content as gravitational mass for most cases that are not extreme.

The way to directly measure mass without the intervention of gravity is to apply a known force for a known amount of time, measure its velocity change, and apply Newton's second law or its relativistic equivalent.

Indeed. This is how astronauts measure their mass in orbit (aka “weightlessness”).

 

[Orodruin]

I tend to think of the Newton's second law as really a law between already defined terms (i.e. a mathematical relation between force, mass and acceleration as already defined terms/objects). In your use it is basically not a law but just the definition of mass.

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

 

[PeroK]

By the time we agree on how to measure its mass, the hot coffee will have got cold!

 

[Orodruin]

By the time we agree on how to measure its mass, the hot coffee will have got cold!

Time to brew another cup!

 

[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.