Assume we have a red hot cannonball in space. It starts off with mass M at 1000K. Later it has cooled by radiation to 100K. Has the mass decreased?
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http://physics.stackexchange.com/questions/69704/increase-in-cloths-mass – dmckee --- ex-moderator kitten Nov 03 '14 at 15:11
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Also, you should link to that other thread. – dmckee --- ex-moderator kitten Nov 03 '14 at 15:15
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http://physics.stackexchange.com/q/144514/ – dmckee --- ex-moderator kitten Nov 04 '14 at 00:37
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MSalters already said "yes". I would like to expand on that by computing the change.
Let's take a 10 kg cannon ball, made of lead. Heat capacity of 0.16 J/g/K means that in dropping from 1000 K to 100 K it has lost $10000\cdot 900 \cdot 0.16 \approx 1.4 MJ$. This corresponds (by $E=mc^2$) to a mass of $1.6 \cdot 10^{-11} kg$ or one part in $6\cdot 10^{11}$.
I cannot think of an experiment that will allow you to measure that mass change on an object in outer space.
UPDATE if you think of temperature as "a bunch of atoms moving", I was wondering whether the relativistic mass increment would be sufficient to explain this mass change.
The velocity of atoms in a solid is hard to compute - so I'm going to make the cannonball out of helium atoms (just because I can) in a thin shell. The mean kinetic energy is $\frac32 kT$, so mean velocity $v = \sqrt{\frac{3kT}{m}}\approx 2500 m/s$. When things cool down to 100 K, velocity will drop by $\sqrt{10}$ to about 800 m/s.
Now at 2500 m/s the relativistic factor $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \approx 1 + \frac{v^2}{2c^2}$. It is encouraging that this scales with $v^2$, just as $T$ scales with $v^2$. Writing this all for one atom:
$$\Delta m = m (\gamma - 1) = m \frac{v^2}{2c^2}$$
Now putting $\frac12 m v^2 = \frac32 kT$, we get
$$\Delta m = \frac{3kT}{2c^2}\\ \Delta m c^2 = \frac32kT$$
The change in mass really does scale with temperature! So even though I was using the average velocity of the atoms, it seems that this is sufficient to explain a real (if hard to measure) change in mass... relativity works. I love it when that happens.
Floris
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1@DirkBruere - in general it is difficult. Doing it to better than 1 part in $10^{12}$ when the object changes from 1000 K to 100 K makes "rather difficult" an understatement... I find 0.01% experiments in lab conditions hard enough. – Floris Nov 03 '14 at 16:54
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1I was just hearing on NPR this AM that scientists in Colorado have a clock that is accurate to something like 10 to the -15th power, and hence can detect relatively minor changes in gravity. They probably don't want to set fire to it, though. – Hot Licks Nov 03 '14 at 17:26
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This is interesting: I'm guessing the hot cannonball has a lot of atomic kinetic energy (that's what temperature is measuring, after all), and the atoms' speeds are non-relativistic. Does the mass of an atom really change just because the kinetic energy is given off by transferring energy to electron orbital jumps and thence released as photons as the electrons fall back (and the cannonball cools) ? <-- or did I just forget something fundamental again? :-( – Carl Witthoft Nov 03 '14 at 18:56
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@HotLicks - I bet it's a retelling of the story I read and referred to a couple of days ago. Even then I'm not sure that a 1 part in 10^12 change in mass would result in measurable change in gravity - not with a 10 kg cannonball. The thermal expansion would make it hard to keep the distance accurately constant... – Floris Nov 03 '14 at 19:15
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@HotLicks Moderately famous quote: "never measure anything but frequency". – dmckee --- ex-moderator kitten Nov 03 '14 at 19:20
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@dmckee - I am more familiar with "the only thing you can measure accurately is zero". That's more of a principle than a quote - but it's what many accurate measurements ultimately try to do. – Floris Nov 03 '14 at 19:24
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Actually in many cases measuring zero accurately is hard. That is that setting really tight bounds on the non-existence of things is very demanding. Proton decay, various symmetry violations and so on. The advantage of measuring frequency is that you do it by counting cycles, so the error bound drops by the inverse of the number of cycles you've counted. Very nice alternative to counting statistics. – dmckee --- ex-moderator kitten Nov 03 '14 at 19:25
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@dmckee - OK. I mean that in setting up an experiment, if you can return it to "no difference" between two quantities you usually get rid of nonlinear terms. Wheatstone bridge, conventional balance, force-restoring balance, two signals beating against each other, ... each time you look for zero, or the "intercept with zero" if you measure several points close to zero. I think we're on the same page: accurate measurements are hard; and this is getting off topic. Moving on. – Floris Nov 03 '14 at 19:29
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Sure, that's a powerful way to design an experiment: measure the feedback needed to achieve zero residual. – dmckee --- ex-moderator kitten Nov 03 '14 at 19:31
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2@CarlWitthoft Yup, you're missing the part where nothing really is non-relativistic. Even walking means you're getting extremely tiny relativistic effects relative to a standing observer, even though the effects are way too small to measure. That's what the discussion here is all about - relativity does predict a mass loss, but is the loss big enough to actually be measurable? Noone is arguing whether kinetic energy (and thus "heat") has an effective mass - it does, that's quite an important point of special relativity. The photons are carrying this mass away with them. – Luaan Nov 04 '14 at 09:53
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@HotLicks: That's a strontium clock, and it was first publicised back in January. – Lightness Races in Orbit Nov 04 '14 at 10:24
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@Luaan I'm very puzzled by the claim that a photon has no mass, and by the idea that mass means "invariant mass" (and not rest mass) in modern physics. So I was glad to see "The photons are carrying this mass away with them" because I thought that means that I was right in thinking that photons do have mass. So is it invariant mass that is lost by the object as it cools? Do photons have invariant mass? – Matthew Christopher Bartsh Nov 22 '21 at 21:48
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@MatthewChristopherBartsh Ultimately, there is no distinction. Spacetime is curved by energy, and it doesn't matter in what form that energy is. Photons have momentum, and they have energy. They do cause spacetime to curve. But photons are still massless particles - that means they always travel at the speed of light (in a vacuum), and can't be slowed down or accelerated. In QFT, it's tied to other properties of the underlying field (if you're interested, there's different "classes" of wave equations - for photons, all waves have the same speed, but can have different frequencies). – Luaan Nov 23 '21 at 07:42
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@MatthewChristopherBartsh A simpler way to think about this might be that mass is a property of a system. A lone photon doesn't have mass (and inertia and all that). But put that photon in a magical perfectly reflecting box, and suddenly you have both mass and inertia (e.g. when you try to move the box, there is a force resisting that). Much the same way, a proton is formed out of a huge amount of energy in massless fields - but the proton itself is very massive indeed (though be careful treading there, because there's many ways to interpret how the inside of a proton looks like). – Luaan Nov 23 '21 at 07:45
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That's helpful, but I still am not entirely clear about this. Could we talk in a chat room about this, since discussions in the comments are discouraged? – Matthew Christopher Bartsh Nov 25 '21 at 15:34
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Of course, it does, since: $$\frac{\partial E}{\partial t} = \frac{\partial }{\partial t} \left(m \cdot c^2 \right) $$ Very little, though
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Yes. There seems to be some dispute in another thread about whether a fire loses mass by (non nuclear) radiation. My answer was yes, it does. – Nov 03 '14 at 14:58
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