Suppose there's a box with one face cold, and the opposite face hot. So when the air molecules hit the cooler face, it will transfer its momentum and energy to the wall, bouncing back with less momentum. And when molecules hit the hotter face, it will bounce back faster. And the hotter side will be hit more frequently by air molecules. So will the box feel any force due to the temperature difference?


2 Answers 2


You do not mention what is happening on the outside of the box. If the outsides are also cold one side hot the other, and it is in air, you will get a force. See Crooke's radiometer

If this is all internal to the box, no net force:

Remember the Dr Who episode, where The Doctor is adrift between his TARDIS and some spaceship? The Doctor throws a cricket ball at the spaceship. This gives him a small momentum towards the TARDIS. The cricket ball bounces off the spaceship (we do not see the small change in velocity of the spaceship) and (this being The Doctor) returns directly towards him. He catches the cricket ball, his velocity towards the TARDIS doubles, and he escapes, looking very smug.

OK. Now, say, the Doctor is in a big box, still stranded. He throws his ball at the hot wall, and as a result gains a velocity towards the cold wall. No net motion of the box, until the ball strikes the hot wall. The whole box now has 1 cricket ball's worth of momentum towards the TARDIS. The ball bounces off the hot wall, with a bit of momentum added (perhaps a Vogon caught it and threw it back with force). The box now has two cricket ball's worth, plus the heat energy. The Doctor catches the returned ball, and drifts with double (and a bit of heat energy) towards the cold wall. Until he strikes the cold wall, the whole box will be moving towards the TARDIS. But when he strikes the cold wall, all his momentum, exactly opposite to the box momentum, is given back to the box, and it stops.

No net change in the box momentum.

event      box     ball     Dr
start      0        0
Throw      0        1       -1
StickHot   1(&ball) 0       -1
BounceHot  2+h      -(1+h)  -1
CatchHot   2+h      0       -(2+h){Dr&Ball}
StickCold  0  = 2+h  -  (2+h)

Too many other events to list. Does the Dr stick? Does he stick & toss? Does he toss back and fourth? All, of course, will not result in an average net momentum change of the box. However, if the Dr changes position in the box, that changes the center of gravity, so the box will move a bit. In this way, he escapes by moving the box to where Ramona can grab it. She, being an outside force, can pull the whole contraption in to safety.

Say it is just the ball, bouncing between hot and cold. Moving from hot to cold, it will have Mbounce + h. Moving from cold to hot, it will have Mbounce + c. Ah-ha! They are different! But it travels faster with Mbounce+h, T = D*(Mbounce+h)/Mass, so averaging based on time in each state gives Mbounce+h/(Mbounce+h) = Mbounce+c/(Mbounce+c).

To translate this to air in a box, set Dr Who to the air mass, the cricket ball to an air molecule, the Vogon to the heat on the hot wall. Average over a billion different directions and positions.

  • 1
    $\begingroup$ So conflicted... on the one hand I don't think this is correct - you're assuming that the cricket ball's momentum is the same when it hits the cold wall as it was when it hit the same wall, whereas in the case of the box of gas this isn't true, so I think there is a net transfer of momentum from the hot to the cold side. On the other hand, you based your answer on an episode of Doctor Who that was first shown in 1982, which makes me really want to give you a +1. $\endgroup$
    – N. Virgo
    Jun 16, 2013 at 15:15

The cold or hot surfaces are in fact balancing the forces if there is local equilibrium. For example, the hot surface has momentum of particles coming from both sides. One sides is the apparatus that creates heat, and the other side is the gas in the box. Same thing for the cold side.

If there is no local equilibrium, then yes, there is a possibility to have non-zero momentum. I have no example on how that could happen though ...

  • $\begingroup$ It could happen if the gas is very diffuse. If you imagine the extreme case where there's only one particle, it's fairly obvious that momentum has to be transferred from the hot side to the cold one, because the particle is (on average) moving faster when it's moving in one direction than when it's moving the opposite direction. $\endgroup$
    – N. Virgo
    Jun 16, 2013 at 15:30

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