6
$\begingroup$

Inspired by reading some recent questions about going to space and torchships and such, I've come to the sad conclusion that space is hard and rocket equations suck.

So what if we made advances in a different direction? Instead of getting more efficient fuels and engines, what about reducing mass instead?

This is definitely not a new idea; I remember that the starship Andromeda Ascendant worked this way - by using anti-gravity fields it reduced the mass of itself and all of its contents to less than 1kg which then was easy to maneuver.

So how realistic is this? Ok, apart from the method itself. Let's imagine that we have invented a handwavium-powered device that produces an anti-Higgs-field-or-something which reduces the mass of all particles within 2m radius to about 0.1% of what it was originally. All other particle properties stay the same - charge, spin, polarization, whatever. Just the mass magically changes.

What would happen when the switch was thrown? Would everything around ir suddenly become featherweight? Or would everything around it disintegrate, because the kinetic energies of the particles would sudddenly be large enough to escape molecular bounds? Or something else?

If a human would be standing next to it, would they live to tell the tale?

I know that gravity at quantum scale is negligible, but mass also affects acceleration and maybe some other things, so I'm uncertain of the final (macroscopic) effect.

Improve this question
$\endgroup$
20
  • 1
    $\begingroup$ With much handwavium, yes. With hard science, no. Particle mass is quite a fundamental value, our bodies will cease to exist if we change it too much. E = mc^2 holds true for all reactions, including chemical ones. $\endgroup$
    – Alexander
    May 3, 2022 at 19:09
  • $\begingroup$ @Alexander - I don't mean to break the laws of physics. So E=mc^2 stays, just that m suddenly becomes a lot smaller for... everything. $\endgroup$
    – Vilx-
    May 3, 2022 at 19:12
  • $\begingroup$ Then, if c stays constant, E becomes much smaller too. Or conversely, a bigger E (from light, room temperature etc.) would wreak havoc on your m. $\endgroup$
    – Alexander
    May 3, 2022 at 19:17
  • $\begingroup$ @Alexander - That's kinda what I was afraid of. $\endgroup$
    – Vilx-
    May 3, 2022 at 19:18
  • $\begingroup$ We all know that $E = mc^2$, but we rarely stop to think that this means that $m = E / c^2$. By far the largest part (that is, about 99%) of the mass of the objects around us is actually the binding energy of the quarks which make up the protons and the neutrons in the nuclei of the atoms of which those objects are made. The contribution of the Higgs field is minor. I would not want to be in the same country as a device which lowers the binding energy of the quarks inside neutrons by a factor of 1000. $\endgroup$
    – AlexP
    May 3, 2022 at 20:39

2 Answers 2

Reset to default
7
$\begingroup$

It would reduce the mass of the W- and Z-bosons in the atoms that make up your body by a factor of 1000. Which would - due to the uncertainty principle - automatically increase the range of the weak nuclear interaction by a comparable factor. The range would be long enough for the weak force to compete with the electromagnetic forces that hold your atoms together. I have no idea how you would die, if you would explode or desintagrate, but whatever would happen in your body it would have nothing to do anymore with any physical processes we rely on to exist.

Improve this answer
$\endgroup$
2
$\begingroup$

A regular human has about 5 kg of blood, pumped to a pressure of about 120/80 mmHg.

Suddenly you have 1/1000th of the blood mass being pumped to the same pressure. That outta hurt. The air around you also has a very rapid drop in mass, and hence a linearly proportional drop in pressure. Last I read about something similar was an incident in 1983, in which some divers went from 9 atm to 1 atm instantly due to someone disregarding safety measures. This would still be more gentle than your method:

Medical investigations were carried out on the remains of the four divers and of one of the tenders. The most notable finding was the presence of large amounts of fat in large arteries and veins and in the cardiac chambers, as well as intravascular fat in organs, especially the liver. This fat was unlikely to be embolic, but must have precipitated from the blood in situ. The autopsy suggested that rapid bubble formation in the blood denatured the lipoprotein complexes, rendering the lipids insoluble. The blood of the three divers left intact inside the chambers likely boiled instantly, stopping their circulation. The fourth diver was dismembered and mutilated by the blast forcing him out through the partially blocked doorway and would have died instantly.

Coward, Lucas, and Bergersen were exposed to the effects of explosive decompression and died in the positions indicated by the diagram. Investigation by forensic pathologists determined that Hellevik, being exposed to the highest pressure gradient and in the process of moving to secure the inner door, was forced through the crescent-shaped opening measuring 60 centimetres (24 in) long created by the jammed interior trunk door. With the escaping air and pressure, it included bisection of his thoracoabdominal cavity, which resulted in fragmentation of his body, followed by expulsion of all of the internal organs of his chest and abdomen, except the trachea and a section of small intestine, and of the thoracic spine. These were projected some distance, one section being found 10 metres (30 ft) vertically above the exterior pressure door.

If the drop in pressure does that to humans, it also does that to objects. There are plenty of videos in Youtube of people playing with vacuum chambers. Just imagine those things happening much faster.

Improve this answer
$\endgroup$
11
  • $\begingroup$ Hmm, I'm not sure I follow. Pressure depends on volume, not mass. Let's say we're looking at just air. Why would a sudden drop in its mass mean that there's less of it? Remember - I'm not removing stuff, I'm making stuff lighter. You still have the same number of atoms in there with the same electromagnetic forces between them. $\endgroup$
    – Vilx-
    May 4, 2022 at 13:21
  • $\begingroup$ @Vilx- the ~1 atm pressure on your body right now exists due to a column of ~10km of air on top of you. That column weights about 1 kg over an area of a square centimeter. What do you think the pressure will be if you replace all that mass of gas with something similar, but 1000x less massive? $\endgroup$
    – The Square-Cube Law
    May 4, 2022 at 13:24
  • $\begingroup$ The same, because I'm not affecting that column. The machine has only 2m of range. $\endgroup$
    – Vilx-
    May 4, 2022 at 13:25
  • $\begingroup$ @Vilx- the divers in the article I mentioned were in a diving bell, which is really small and tight. You'd still get a bang if you only affected 2m around yourself, but in your case it would be an implosion if you are exposed to outer atmosphere. If you are inside a diving bell, you are suddenly in a vacuum thinner than Martian atmosphere, so you would pop. $\endgroup$
    – The Square-Cube Law
    May 4, 2022 at 13:28
  • $\begingroup$ Yes, because the bell was highly pressurized and then a channel was opened to an area with vastly smaller pressure. You seem to have come to the conclusion that once my 2m-radius sphere loses 99.9% of its mass, its pressure must also become 1000 times smaller. And after that you get an implosion. However this link is not at all clear to me. Why would reduction of mass in this sphere lead to a reduction in pressure? Hell, let's simplify it even further: enclose the whole 2m sphere in a steel shell. Now there's no external pressure, it all comes from the air within. [Continued] $\endgroup$
    – Vilx-
    May 4, 2022 at 13:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .