Dropping a Ball: Special Relativity

[Quadhelix]

The question was raised here whether, according to Special Relativity, a ball dropped off the top of the Empire State Building would loose mass due to the fall (as opposed to shattering at the bottom - let us say that it is an invincible ball).

Although the issue was debated rather fiercely for a couple of pages, we never really settled the issue. Rather than necro-posting a thread to discuss something that was off-topic, I figured that it would be a better idea to raise the issue in the Science and Technology Forum. Not only is it a more appropriate location, but it would allow input from those far more knowledgeable on the subject than I.



Basically, the reasoning is this: if a ball is atop the Empire State Building, it has a certain amount of potential energy. After it has been dropped off the building, once it reaches the ground, it will have a different amount of potential energy. Of, both values are arbitrary, but the difference between is not: it is work done on the ball done as it fell.

According to Special Relativity, energy and mass are related. Therefore, the non-arbitrary change in the ball's potential energy (due to the work done on it by gravity) should result in a change of mass. Although it was argued that the potential energy is arbitrary, the work done on the ball is not - and that is were Conservation of Mass comes in.

According to this link, relativistic mass is conserved in all inertial frames of reference and rest mass is not necessarily conserved. As the ball falls, mass dilation causes its relativistic mass in increase relative to its rest mass. Since there is no source of mass to increase the ball's relativistic mass (the new mass cannot come from the Earth, as the Earth would be accelerating ever-so-subtly up to meet the - meaning that it would be experiencing mass dilation as well), we have to assume that its rest mass decreases. In short, "rest mass" is converted into "kinetic energy mass" as the ball falls.

Once the ball hits the ground, it looses the kinetic energy in the form of sound and heat. This energy is dispersed outward into the Earth, not into the ball, so the mass that that energy carries must also be transferred into the Earth, away from the ball.

As for why we cannot take the ball as our frame of reference (as was suggested in the original topic), Special Relativity applies only in inertial (that is, non-accelerating) frames of reference. However, in the ball's frame of reference, the Earth is accelerating upwards at 32 feet per second per second (9.8 m/s^2) without any apparent cause.





As a side note, for those who think that Special Relativity is wrong, my reply is this post.

Furthermore, below is an amendment (copied from here) to the original post clarifying the issue of Emitter Theory:

Spoiler:

Basically, the issue is that the hydrogen gas in the stars atmosphere is made of atoms that are moving around fairly quickly; some are moving toward us (causing the light that they emit to be blue shifted), while some are moving away from us (causing the light that they emit to be red shifted).

According to Emitter Theory, the particles that are traveling toward us should emit light that is "faster" than the light emitted by the particles moving away from us. This means that if something were to block the star, or change its luminousity in any way, the blue shifted light would reach us a fair amount of time before the red shifted light, with the rest of the spectrum filling in the gaps.


Let us take, for example, the star Algol. It is a variable star with a period of 2 days, 20 hours, and 49 minutes. During that period, it dims once for ten hours, at the end of which it returns to its normal luminousity for the remainder of the period. The explanation for this is irrelevant, but it is supposedly because the brighter star in the system is regularly eclipsed by a dimmer orbiting star.

In any case, the star that serves as the light source is supposed to have a temperature of ~12,000 K. Using the equation relating temperature to average particle kinetic energy (1/2 m * v^2 = 3/2 k * T ; found here), and a mass for the hydrogen atom 1.66*10^-27 kg, this translates to an average atomic speed of 17 kilometers per second.

That's 17 kilometers per second, which is mind-bogglingly huge compared to the Doppler shift needed to detect extrasolar planets, which is apparently on the order of "a few meters per second."

In any case, a bit of calculation reveals that that, in this case, Emitter Theory predicts that the blue shifted light should reach us roughly 94 hours before the red shifted light.

[K^2]

Your conclusion is correct, but your analysis is all wrong. Did you notice the part where it says "inertial frames of reference"? Gravity has to be taken into account, so this is inherently a General Relativity problem. Here is the full analysis.

Particle's 4-momentum infinitely far from Earth is just (m0, 0, 0, 0), where m0 is its rest mass. I'm using gravitational units, where c=G=1.

Because of time symmetry, the product (1,0,0,0)·p is a conserved quantity. (For ref. see Killing Vectors.) The metric is a Schwarzschild metric, which in spherical polar coordinates is diagonal, and its (t, r, theta, phi) components are (-(1-2M/r), 1/(1-2M/r), r², r²*sin²(theta) ), where M is mass of Earth (or any other spherical body). Since Schwarzschild metric goes to Minkowski at r->infinity [(-1, 1, r², r²*sin²(theta)) in polar], and that's where p=(m0, 0, 0, 0), we get the general expression for momentum of a free falling particle.

p = (m0/(1-2M/r), 0, 0, 0).

[Note: Spacial components of momentum is zero. But the particle is falling! That's because in GR the free fall is due to curvature, not change in momentum.]

Now all we have to do is figure out the rest mass. It is given by m²=-p·p. Again, we have to use the metric.

m²=-(m0/(1-2M/r))²*(-(1-2M/r))

m = m0*sqrt(1-2M/r)

Notice that in the limit where M/r is very small, which is a fair estimate for Earth, this becomes the quasi-classical result m0-m0*M/r, or mass of the object reduced by its change in potential energy.

So yeah, be careful when mixing Special Relativity and Gravity. If gravity is weak enough, what you'll get is probably a close estimate of what really happens, but its best to double check.