1
$\begingroup$

Warm air has more energy than cold air. This means that according to the Einstein equation $E = mc^2$ the warmer air has a greater mass than the cold one. Why is the warm air rising, if it has a greater mass, which means that the attraction of gravity between the Earth and the warm air is greater?

$\endgroup$
  • $\begingroup$ What context did you hear about the equation $E=mc^2$ in? You seem to have applied it somewhat... strangely. Consider thinking about what each term in that represents and adding that information to your question. $\endgroup$ – user191954 Nov 19 '18 at 14:28
3
$\begingroup$

As other answers point out, the reason is buoyancy. This post is to show just how small the opposing (relativistic) effect of increased gravitational force is.

The increase in the gravitational attraction associated with kinetic energy is proportional to the Lorentz factor, $\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$.

We could look up molecular masses, and get into the Maxwell-Boltzmann distribution which describes the range of speeds of gas molecules, but for guesstimation purposes, let's use the rule of thumb that typical molecular speeds in a gas are on the same order as the speed of sound through the gas--for air near the surface of the Earth, roughly 300 m/s.

Let's say we increase the temperature of our parcel of air by 10%, from 300 K (a warm day) to 330 K (roughly the hottest day in the hottest desert). That means we increase the average molecular velocity by 5%, from our fudged 300 to 315 m/s. This takes $\gamma$ from $1+5 \times 10^{-13}$ to $1+6 \times 10^{-13}$, an increase of 1 part in $10^{13}$.

That is how much the gravitational force increases. You could cancel that increase by moving half a part in $10^{13}$ further from the center of Earth. That's a third of a micron.

$\endgroup$
1
$\begingroup$

Even though $E=mc^2$ is only for objects that are not moving and we really should be using $E^2=p^2c^2+m^2c^4$, you are conceptually correct in that objects with more energy have more mass. The issue is that c is a huge number so it takes a ton of energy to actually give something mass in a significant way. c is 299,792,458m/s and $c^2$ is 89,875,517,873,681,764$m^2/s^2$. The reason hot air rises is due to a combination of the ideal gas law and the buoyant force. According to the ideal gas law, an increase in temperature will also increase in volume (assuming pressure and # of air molecules are held constant of course). You can think of the air molecules all bouncing against each other. The hotter they are the faster they are moving and the more they push up against the air around it and expand. Now it occupies more space but is still the same mass(extremely close). The buoyant force is to the air around it also having weight. That air is "trying" to get underneath anything that takes up space. The more space it takes up the more surrounding air there is exerting a force on it. This force exceeds gravity when the air is less dense than the surrounding air causing it to rise. For more information consult the following sources:

http://physics.bu.edu/~duffy/sc527_notes01/buoyant.html http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/idegas.html

$\endgroup$
  • 1
    $\begingroup$ "a ton of energy".... $\endgroup$ – JEB Nov 19 '18 at 15:27
  • $\begingroup$ I was speaking idiomatically. $\endgroup$ – Nathan Rotta Nov 20 '18 at 19:31
  • $\begingroup$ I know, otherwise you would have included a factor of $c^2$. $\endgroup$ – JEB Nov 23 '18 at 5:39
-1
$\begingroup$

$E=mc^2$ is only valid for particles that are not moving. The full expression should be

$$E^2=p^2c^2+m^2c^4$$ where $p$ is the momentum of the particle (which is $0$ when at rest, and we recover the famous $E=mc^2$).

The reason warm air rises is to do with the fact that "warm" air has faster moving particles. This means the air becomes less dense, and so it will rise above the slower, colder, more dense air.

$\endgroup$
  • $\begingroup$ Then, particles that don't move don't have energy, which means they don't have mass? $\endgroup$ – Alyosa Dimitriyev Nov 19 '18 at 14:15
  • $\begingroup$ If a particle is not moving, then $p=0$ and so the energy becomes $E=mc^2$ $\endgroup$ – Garf Nov 19 '18 at 14:17
  • $\begingroup$ Note in my expression for $E$ I am using $m$ to denote the particle's constant rest mass $\endgroup$ – Garf Nov 19 '18 at 14:18
  • $\begingroup$ @AlyosaDimitriyev It should be noted that you really don't need to consider relativistic effects in this scenario. $\endgroup$ – Aaron Stevens Nov 19 '18 at 14:21
  • $\begingroup$ This answer kind of misses the point, which is that the OP needed to have it pointed out that relativistic effects were negligible. $\endgroup$ – Ben Crowell Nov 19 '18 at 16:39
-1
$\begingroup$

Buoyancy and the ideal gas law.

PV = nRT

P is pressure V is volume
n is number
R is a constant
T is temperature

In a closed container if you increase T then P goes up.

In the open (atmosphere) V goes up

With same mass and more V buoyancy then takes over.

$\endgroup$
  • $\begingroup$ This answer kind of misses the point, which is that the OP needed to have it pointed out that relativistic effects were negligible. $\endgroup$ – Ben Crowell Nov 19 '18 at 16:40
  • $\begingroup$ @BenCrowell Sorry you missed that volume and buoyancy part. $\endgroup$ – paparazzo Nov 19 '18 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.