11
$\begingroup$

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?

$\endgroup$
21
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Measuring mass in free fall is rather difficult $\endgroup$ – user56903 Nov 3 '14 at 15:32
  • 1
    $\begingroup$ @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. $\endgroup$ – Floris Nov 3 '14 at 16:54
  • 1
    $\begingroup$ I 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. $\endgroup$ – Hot Licks Nov 3 '14 at 17:26
  • $\begingroup$ 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? :-( $\endgroup$ – Carl Witthoft Nov 3 '14 at 18:56
  • 2
    $\begingroup$ @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. $\endgroup$ – Luaan Nov 4 '14 at 9:53
18
$\begingroup$

Of course, it does, since: $$\frac{\partial E}{\partial t} = \frac{\partial }{\partial t} \left(m \cdot c^2 \right) $$ Very little, though

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user56903 Nov 3 '14 at 14:58
  • $\begingroup$ You are right. I took the liberty of calculating "very little". $\endgroup$ – Floris Nov 3 '14 at 15:20
  • 5
    $\begingroup$ Thanks for the formatting, but I think d/dt (c²) is negligible ;) $\endgroup$ – MSalters Nov 3 '14 at 20:05

protected by Qmechanic Nov 4 '14 at 10:37

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?