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The Mysteries of Mass
Scientific American ^ | July 2005 (that issue) | Gordon Kane

Posted on 06/30/2005 8:58:05 AM PDT by PatrickHenry

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To: Physicist
It is possible that the particles we see are all actually massless, their apparent masses corresponding to extra-dimensional momentum components we can't as yet detect.

Interesting, but I don't get it. (Not an uncommon event when I read about this stuff). So you have a bunch of massless particles in a 3d space, presumably moving in straight lines except when they collide. If you project them onto a 2d space, aren't they still going to appear to move in straight lines? Or is the projection somehow a nonlinear function that can map straight lines to orbits and other curved paths? Even if so, if the particles in the 3d space move independently of each other, how would any projection create the appearance of dependency?

I don't know if that made any sense; I find this stuff fascinating but am missing a lot of the theoretical background. Trying to get through Penrose's Road to Reality, but I start spacing out on calculus on manifolds...

41 posted on 06/30/2005 11:09:46 AM PDT by ThinkDifferent (These pretzels are making me thirsty)
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To: Physicist
Try this: take a bunch of massless, interacting particles and let them fly around in three dimensions, crashing into each other and bouncing off as they may. Now take their trajectories, and project them onto a two-dimensional plane. As viewed in the two-dimensional plane, the particles interact as if they had masses, the apparent masses being proportional to their momenta in the direction perpendicular to the plane.

It is possible that the particles we see are all actually massless, their apparent masses corresponding to extra-dimensional momentum components we can't as yet detect.

Sounds like a great simplification, but I'm having difficulty visualizing this. Can you direct me to a website that might have some diagrams of what that 2D projection might look like? All I can think of is a Mercator projection, and I'm sure that's not even in the ballpark.

42 posted on 06/30/2005 11:11:16 AM PDT by PatrickHenry (Felix, qui potuit rerum cognoscere causas. The List-O-Links is at my homepage.)
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To: macamadamia

Communication through the aesthetic medium. LOL


43 posted on 06/30/2005 11:11:53 AM PDT by RightWhale (withdraw from the 1967 UN Outer Space Treaty)
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To: willgolfforfood

1. The m is the same in both equations, yes. F is the force of interaction when a particle interacts with another. If particle 'b' perturbs the motion of particle 'a', then the amount of force 'b' exerts on 'a' is measured by F=ma. If you know the mass of 'b', the force exerted upon it by 'a', and the acceleration change from the interaction, you can determine the mass of particle 'a'.
E is the total amount of energy that could be released by converting all the mass of the particle to energy. Not a lot of ways to do this; a matter-antimatter collision of equivalent particles comes to mind. The total energy released by the destruction of the two particles would be their combined masses x c^2. If you know the total amount of energy released in such an instance, and divide by c^2, you get the total mass involved.
2. Yes. F/a is usually a small quantity over another small quantity. E/c^2 is a huge quantity over another huge quantity. It balances out.

And yes, I had to grab scratch paper to be sure :)


44 posted on 06/30/2005 11:21:11 AM PDT by Dawsonville_Doc
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To: ThinkDifferent
If you project them onto a 2d space, aren't they still going to appear to move in straight lines? Or is the projection somehow a nonlinear function that can map straight lines to orbits and other curved paths?

I'm not talking about gravity, here, I'm talking about straight Newtonian mechanics. Billiard balls. F=ma.

45 posted on 06/30/2005 11:55:58 AM PDT by Physicist
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To: PatrickHenry
Can you direct me to a website that might have some diagrams of what that 2D projection might look like?

Picture a rod with a kink in it. Picture two of them, meeting at the kinks. Now imagine their shadow on a bright, sunny day.

46 posted on 06/30/2005 12:03:57 PM PDT by Physicist
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To: Physicist
Picture a rod with a kink in it. Picture two of them, meeting at the kinks. Now imagine their shadow on a bright, sunny day.

Ah yes ... I see it clearly. It's an asterisk! (Well, three rods.)

47 posted on 06/30/2005 12:37:38 PM PDT by PatrickHenry (Felix, qui potuit rerum cognoscere causas. The List-O-Links is at my homepage.)
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To: Physicist
I'm not talking about gravity, here, I'm talking about straight Newtonian mechanics.

Ok, I might get it now. So in the hypothetical 3d space all particles have the same resistance to acceleration, but because they have differering velocities in the z direction, their projections on the xy plane will appear to accelerate at different rates given the same force. Is that the idea? If it is, then two further questions:
- Wouldn't observers in the xy plane sometimes see two particles seemingly occupying the same location (because they'd have the same x and y coordinates but different z)?
- How would this model account for gravity? It eliminates inertial mass by making it a function of velocity in the z direction, but what about gravitational mass?

48 posted on 06/30/2005 12:38:06 PM PDT by ThinkDifferent (These pretzels are making me thirsty)
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To: ThinkDifferent
So in the hypothetical 3d space all particles have the same resistance to acceleration, but because they have differering velocities in the z direction, their projections on the xy plane will appear to accelerate at different rates given the same force. Is that the idea?

Just so.

Wouldn't observers in the xy plane sometimes see two particles seemingly occupying the same location (because they'd have the same x and y coordinates but different z)?

[Geek alert: Well, for one thing, the extra dimension would likely be curled up into a very tiny circle, too small to be noticed at our scale. (Call it "periodic boundary conditions", if you prefer.) Alert readers might have wondered why, say, an electron couldn't just have any old momentum in the 5th dimension, and therefore any old mass. It's obviously not that way: electrons all have the same mass. The answer is because, in the tiny extra dimension, the quantum wavefunction of the electron would have to have an integer number of wavefunctions along the circle, to match the boundary conditions. [Super Geek Alert: This means only certain masses would be allowed. Could the higher harmonics represent the muon and the tau? The argument has been made, but we still don't know why only three.] ANYWAY, the wavefunction covers the entire space, so the Pauli Exclusion Principle still applies. [Super Geek Alert: Some theorists postulate that there are LARGE extra dimensions. In such models, it actually is theoretically possible (using polarization) to get electrons to "pass through" each other with a head-on trajectory, because they miss each other in the 5th dimension!]]

How would this model account for gravity? It eliminates inertial mass by making it a function of velocity in the z direction, but what about gravitational mass?

Rather than snow you under with handwaving about Kaluza-Klein towers of gravitons, I'll admit that that's a very technical question, which I'm not qualified to answer. I can tell you that the whole model has not been worked out.

49 posted on 06/30/2005 1:03:31 PM PDT by Physicist
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To: Physicist

Thanks, I actually understood a decent portion of that :)


50 posted on 06/30/2005 1:23:22 PM PDT by ThinkDifferent (These pretzels are making me thirsty)
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To: Physicist
an integer number of wavefunctions

Wavelengths. An integer number of wavelengths.

51 posted on 06/30/2005 1:35:54 PM PDT by Physicist
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To: Physicist
Wavelengths. An integer number of wavelengths.

Half-integer. A half-integer number of wavelengths. Sheesh!

52 posted on 06/30/2005 1:50:08 PM PDT by Physicist
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To: AMHN
E=mc^2=(mo*c^2)/sqrt(1-(v^2/c^2))=mo/(mu*epsilon*sqrt(1-(v^2*mu*epsilon))).

That would have been prettier in MathML

53 posted on 06/30/2005 1:57:33 PM PDT by zeugma (Democrats and muslims are varelse...)
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To: Physicist
Half-integer. A half-integer number of wavelengths. Sheesh!

This stuff makes my head hurt bump.

54 posted on 06/30/2005 2:30:03 PM PDT by zeugma (Democrats and muslims are varelse...)
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To: Physicist

Not to belabor what is probably obvious, but I assume the "mystery dimension" is orthogonal to the sub-space in which we observe the particles, yes?


55 posted on 06/30/2005 5:33:56 PM PDT by longshadow
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To: longshadow
That's right.
56 posted on 06/30/2005 6:10:13 PM PDT by Physicist
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To: Physicist; longshadow
If we were flatlanders, living in a 2D plane, and a 3D rod were perpendicular to our plane, we'd see it only as a 2D circular cross-section of the rod. I see no possibility for simulating mass. If the rod were in motion, passing through our plane, I still see no illusion of mass. I'm obviously missing something. Perhaps if we could, in our 2D way, handle the circle, and test its mass, it would then reveal the effect of the motion, but we wouldn't know about that motion. We'd just think it was massive, and we wouldn't know why. Is that it?
57 posted on 06/30/2005 6:25:39 PM PDT by PatrickHenry (Felix, qui potuit rerum cognoscere causas. The List-O-Links is at my homepage.)
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To: PatrickHenry
I'm obviously missing something.

Objects can have motion components in any or ALL dimensions, not just the ones we can see, or just the ones we don't see.

58 posted on 06/30/2005 6:28:39 PM PDT by longshadow
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To: longshadow
Objects can have motion components in any or ALL dimensions, not just the ones we can see, or just the ones we don't see.

These other dimensions seem very subversive. Why don't they reveal themselves? What are they afraid of? I think it's a plot to pollute our precious bodily fluids.

59 posted on 06/30/2005 6:32:07 PM PDT by PatrickHenry (Felix, qui potuit rerum cognoscere causas. The List-O-Links is at my homepage.)
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To: PatrickHenry
If we were flatlanders, living in a 2D plane, and a 3D rod were perpendicular to our plane, we'd see it only as a 2D circular cross-section of the rod.

The rod isn't perpendicular to the plane (although it could be). The extra dimension is perpendicular to the plane.

If the rod were in motion,

Full stop. The "rod" only represents the trajectory of the particle over time. The particle is pointlike and massless. The particle moves through the space as a massless object. The shadow of the particle on the plane moves on the plane as if it had mass.

60 posted on 06/30/2005 6:49:12 PM PDT by Physicist
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