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 '19 at 16:57
  • 1
    $\begingroup$ @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 '19 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 '19 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 '19 at 15:09
  • $\begingroup$ What time $t$ are you using? Is it the one at the floor? $\endgroup$ – md2perpe Feb 12 '19 at 19:18

Calculations in Rindler coordinates

The Rindler coordinates describe an accelerating system in a flat space. Every point retains a constant proper acceleration, but the proper acceleration differs between heights. The space coordinates describe proper length, but the time coordinate only describes proper time at one height.

Let the coordinate system of some inertial system be denoted by $(T,X)$ and the coordinates of a Rindler system be denoted by $(t,x).$ We will also use $(\tau,\xi)$ for proper time and length.

The transformation formulas from the Rindler coordinates to the inertial coordinates are given by $$X = x \cosh \alpha t \\ T = x \sinh \alpha t$$ where $\alpha$ is a constant (a parameter generating a family of coordinate systems) and we use $c=1.$

One easily finds that $ds^2 = -dT^2 + dX^2 = -\alpha^2 x^2 \, dt^2 + dx^2.$ Thus the proper time is given by $d\tau = \alpha x \, dt$ and the proper distance by $d\xi = dx.$

An object at a constant $X = x_\max$ will be freely falling in the accelerating system, with maximum $x = x_\max$ at time $t=0$ and following $$x = \frac{x_\max}{\cosh \alpha t}.$$

Let us consider a fall down to $x = x_\min$. The falling object will arrive there at time $$t = \frac{1}{\alpha} \cosh^{-1} \frac{x_\max}{x_\min}$$ so the proper time of fall is $$\tau_{\text{fall}} = x_\min \cosh^{-1} \frac{x_\max}{x_\min}.$$

The proper speed when the object comes down is $$v = \frac{d\xi}{d\tau} = -\tanh \alpha t = - \sqrt{1 - \left(\frac{x_\min}{x_\max}\right)^2}$$ so the $\gamma$ factor is $$\gamma = \frac{1}{\sqrt{1-v^2}} = \frac{x_\max}{x_\min}.$$

The proper momentum will therefore be $$ p = \gamma m v = (-) m \sqrt{\left(\frac{x_\max}{x_\min}\right)^2-1}. $$

Thus, the "force" will be $$ F = \frac{p}{\tau_{\text{fall}}} = \frac{m \sqrt{\left(\frac{x_\max}{x_\min}\right)^2-1}}{x_\min \cosh^{-1} \frac{x_\max}{x_\min}}. $$


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