I was thinking about the mass–energy equivalence, which means that for example a hot gas weighs more than a cold gas and I tried to imagine that microscopically. Lets imagine a gas in a chamber on a weighing scale. If we look at the situation microscopically, we see all the particles bouncing off the walls and every collision transfers a momentum $\Delta p_i$. The sum of all the forces that these collisions exert on the walls of the chamber is $F=\frac{\sum_i\Delta p_i}{\Delta t}$ and this has to be equal to the total weight of the gas, wich in classical mechanics is just the sum of the weights of the individual particles $F=\sum_i m_ig$.

However according to relativity, if I add energy to the gas its total mass has to increase by $\Delta m=\frac{\Delta E}{c^2}$. So i am trying to verify that via the same approach with $F=\frac{\Delta p}{\Delta t}$ as above. But before I do that I can simplify my model: The collisions on the sides cancel each other out and in the most simple case my gas consists only of one particle. So I ended up with a ball bouncing repeatedly on a scale.

So here is my experiment:
At time $t_0=0$ I release a ball at height $h_0$ above a weighing scale with an initial velocity of $v_0=0$. I want this example to be as easy as possible, so i assume a constant gravitaional field $F=mg$. At time $t_1$ the ball bounces off the scale with speed $v_1$. I assume perfectly elastic collision, so the ball will reach the same height again and the motion repeats itself with a peroid of T.
What I want to calculate is the time averaged weight $F=\frac{\Delta p}{T}$ the scale will show over one period.

Newtonian mechanics
The calculation in newtonian mechanics is easy. The speed of the ball when it reaches the scale is $v_1=\sqrt{2gh_0}$ and the falling time is $t_1=\sqrt{\frac{2h_0}{g}}$. When the ball hits the scale it's velocity changes from $v=-v_1$ to $v=+v_1$, so the change in impulse is $\Delta p=2mv_1$. Therefore the time averaged weight the scale shows is $$F=\frac{\Delta p}{T}=\frac{2mv_1}{2t_1}=mg.$$ Thats the same weight as if the ball is resting on the scale, which is what I expected in newtonian mechanics.

Relativistic mechanics
I do have experience in special relativity, but not in general relativity. So I wasn't really sure how to solve this problem. That being said, here is what i thought: It is obvious that the acceleration that I measure in my lab frame for a ball in free fall can't be constant a=g like in classical mechanics. If there would be a constant acceleration, a particle that is already falling at a speed close to c would be moving faster than c after a short time. Therefore I know that $a$ has to approach zero as $v$ approaches c.

According to this wikipedia article, I can use the usual definition of work, which means that the Integral $$\int_0^{h(t)} mgdz=mgh(t)\tag{1}$$ is equal to the difference in kinetic energy of the ball.

Conservation of energy: $$mgh_0=mgh(t)+(\gamma(v(t))-1)mc^2\tag{2}$$ $$\Rightarrow \gamma(v_1)=\frac{gh_0}{c^2}+1\tag{3}$$ and $$\Rightarrow v(t)=\frac{dh(t)}{dt}=-\sqrt{c^2-\frac{c^2}{((h_0-h(t))g/c^2+1)^2}}\tag{4}$$

The solution of that differential equation with $h(t=0)=h_0$ and $v(t=0)=0$ is $$h(t)=\frac{c^2}{g}\left(1-\sqrt{1+\frac{g^2t^2}{c^2}}\right)+h_0.\tag{5}$$

We also have $$\lim\limits_{t\rightarrow \infty}\frac{dh}{dt}=-c \qquad \text{and} \qquad \frac{d^2h}{dt^2}(t=0)=-g.$$

We get the newtonian solution back if we do a taylor approximation with $gt/c<<1$ for the square root term $$h(t)\approx -\frac{gt^2}{2}+h_0.\tag{6}$$

So everything seems fine and we can now calculate the falling time $$h(t_1)\stackrel{!}{=}0$$ $$\Rightarrow T=2t_1=2\sqrt{\frac{2h_0}{g}+\frac{h_0^2}{c^2}}.\tag{7}$$

And we can finally calculate the time averaged force $$F=\frac{\Delta p}{T}=\frac{2\gamma(v_1)mv_1}{2t_1}=\frac{\gamma(v_1)m\sqrt{c^2-\frac{c^2}{(gh_0/c^2+1)^2}}}{\sqrt{\frac{2h_0}{g}+\frac{h_0^2}{c^2}}}=\gamma(v_1)mg\frac{1}{1+\frac{gh_0}{c^2}}=mg.\tag{8}$$

However thats not the result that I expected. Since the bouncing ball has a higher total energy as a resting ball, I expected the mass to increase by $$\Delta m=\frac{\Delta E}{c^2}=\frac{mgh_0}{c^2}.\tag{9}$$

Is my calculation wrong somewhere? If so, how can I calculate the fall time and velocity of the ball properly?

  • $\begingroup$ "If there would be a constant acceleration, a particle that is already falling at a speed close to c would be moving faster than c after a short time." No, it doesn't work like that. Special relativity imposes no limits on acceleration, and you need to use the relativistic rule to combines speeds, en.wikipedia.org/wiki/Velocity-addition_formula Also see math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html $\endgroup$ – PM 2Ring Jan 31 at 16:57
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    $\begingroup$ @PM2Ring I'm talking about the observed acceleration in the lab frame. I know already how addition of velocities works in special relativity. This doesn't help me. $\endgroup$ – Azzinoth Jan 31 at 19:09
  • $\begingroup$ Not sure I follow your expectation (at the end). Yes, each bounce sees the larger mass-energy. But the bounces are further apart, because the acceleration doesn’t cause velocity to grow uniformly. $\endgroup$ – Bob Jacobsen Feb 1 at 14:22
  • $\begingroup$ @BobJacobsen Yes, and indeed $T=2\sqrt{\frac{2h_0}{g}+\frac{h_0^2}{c^2}}$ is bigger than the classical T. If this bouncing ball doesn't show a bigger weight, then multiple bouncing balls (i.e. a gas) wouldn't weigh more either (as long as I don't assume any interaction between the particles). That would mean, that a hot gas does not weigh more? $\endgroup$ – Azzinoth Feb 1 at 15:09
  • $\begingroup$ What time $t$ are you using? Is it the one at the floor? $\endgroup$ – md2perpe Feb 12 at 19:18

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