# Hypervolume Quantum Shape

1. Apr 21, 2015

### Helios

Certainly particles are not shaped like anything. However shapes themselves are informative.
If particles were spheres, then they would trace out tubes in 4-d. After this though, we have lost the particle, so we have to chop the tube back up to pieces. However this piece has now become a "hypervolume" instead of a volume. This piece I'll call a bullet as a quantum of hypervolume out of the world tube. It is 4-d object, spheroidic in space but rectangular or parallelogramical by other cross-sections. It is a tube bound by two oblate spheroids as end pieces.
It must be easy enough to fashion a bullet to be
1) an invariant quantum of hypervolume
2) a shape which recovers kinematic variables
It seems I can formulate this shape as necessary. As I use hyperbolic functions, I omit the angle notation.
major axis = D
minor axis = D sech
bullet 4-height = i D cosh
bullet hypervolume = [ i D cosh ] x ( bullet end volume ) = = i ( pi / 6 ) D^4
So this shape does not reside in Euclidean 4-space. Hypervolume here is reckoned to require an imaginary component. Furthermore, velocity = c tanh, momentum = mc sinh, and energy = mc^2 cosh are familiar and all spaned by the inner geometry. The ellipse aspect provides an interfocal distance proportinal to the velocity. The "i" provides an inner transverse diagonal proportinal to the momentum.
So beyond this, I then have the question whether "hypervolume" requires the imaginary component "i", or if there is such a thing as actual hypervolume, or if this is a special kind of hypervolume described by a mathematical classification.

2. Apr 21, 2015

### Staff: Mentor

What do you mean by this?

3. Apr 21, 2015

### Helios

It means that since a world line or in this case a world tube is infinite, there is no event we can distinguish. Particles are used as test particles because they reside at events. If a four dimensional paradigm intends to supplant the picture of a particle as a volume parcle, the new thing can only be a discrete 4-d object in the locality of an event, one piece of the tube, as if it were chopped up into uniform pieces, that purport to be a kind of quanta.
Basically, I'm just looking closely at this "oblate spheroid hyper-cylinder" and its features. It is instructive to me to find a shape from which I can identify lengths which correspond to kinematic variables in addition to the understanding gained from their presence in equations. This shape, at least for me, maps out all the fundamentals, encapsuled into one quantum, one pill to comprehend.
However I had to innovate to define hypervolume with the imaginary unit. So is this valid, as it seems to make the parcle an invariant that I gave the value of i ( pi / 6 ) D^4? I am proposing that this is the definition of bullet hypervolume.

4. Apr 21, 2015

### Staff: Mentor

I don't understand. An "event" is just a point in spacetime. Any point on a worldline is an event; and a world tube is just a family of worldlines that are contiguous. What stops us from distinguishing any of those points?

No. As above, "events" are points in spacetime. "Particles" are described in relativity by their worldlines (or world tubes if they are extended objects). A worldline is a continuous sequence of events, not a single event.

I think you need to rethink your question in the light of the above; you appear to have an incorrect understanding of what an event is and how objects are modeled in spacetime.

5. Apr 22, 2015

### Helios

What your saying is obvious. I have not said anything in contradiction. How can a world tube be a family of worldlines? By family, you imply a multiplicity. A world line is one dimensional. The multiplicity of one dimension is two dimensional. A world tube is four dimensional. By modeling of objects do you mean these objects in spacetime have hypervolume and this can be calculated and you can tell me this?

6. Apr 22, 2015

### Ibix

A set of world lines can form an object of any dimensionality higher than one. Think of un-cooked spaghetti. You can lay the pieces out side-by-side on a table and form a 2d sheet, or you can hold a bunch in your fist and form a 3d cylinder. The difference between spaghetti and world lines is that world lines are truly one dimensional, so you have infinitely many of them in any finite higher dimensional region.

You can calculate the hyper-volume of a finitely long world tube (for example, a particle that comes into existence at some time and goes out of existence at another). The answer isn't simple in general because space-time does not have a simple geometry.

If you are talking about the hyper-volume of a particle "now", this is zero, since it is the intersection between the world tube and a 3d slice of space-time, so is only 3d itself. I don't think this is obviously a problem, not least because "now" for an extended object is not a unisersally agreed concept.

7. Apr 22, 2015

### Helios

So I think you imply that there is a special case in which space-time does have a simple geometry and one can calculate the hyper-volume of a finitely long world tube simply. Can you give an example?

8. Apr 22, 2015

### Ibix

Noting that I've only learned a little about integration in manifolds, my understanding is as follows. I'd wait for the nod from @PeterDonis before accepting my answer as correct.

In the case of an object of constant volume, V, which lives for a time $\Delta t$, never accelerates and is far enough away from any source of gravity that space-time can be considered flat, the hyper volume of that world tube is pretty much what you'd expect: $Vc\Delta t$.

I'm not sure what you expect that to tell you.

9. Apr 22, 2015

### Staff: Mentor

This is correct for the case of flat spacetime and an object moving inertially. The only thing that might need clarification is that the volume $V$ is the proper volume of the object, i.e., the volume in its rest frame.

10. Apr 22, 2015

### Helios

I would not expect this. I expect the volume to be reduced by the Lorentz contraction factor. PeterDonis claims VcΔt Vc\Delta t is correct which omits the Lorentz contraction factor. Can Ibex or PeterDonis show that my understanding is incorrect as to "how objects are modeled in spacetime" ( as asserted by PeterDonis ) or how I calculate the hypervolume of a world tube section?

11. Apr 22, 2015

### Staff: Mentor

See my clarification in post #9. $V$ is the volume of the object in its rest frame, and similarly $V c \Delta t$ is the hypervolume of the object's world tube over a time interval $\Delta t$ in its rest frame.

If you transform to a different frame, in which the object is moving, you have to make a decision: do you want to look at the same hypervolume as before (i.e., the same geometric object), or do you want to look at a different hypervolume, defined the same way as before, but relative to the new frame? The original hypervolume, in the object's rest frame, is bounded by two surfaces of constant time in that frame, separated by a time interval $\Delta t$. If you compute the hypervolume of that exact geometric object, the world tube bounded by those two surfaces, in any frame, you will get the same answer; you must, because the hypervolume of a fixed geometric object is an invariant, the same in every frame. (But in the new frame, in which the object is moving, the two bounding surfaces at the start and end of the hypervolume will no longer be surfaces of constant time, because of relativity of simultaneity.)

If, however, you look at the hypervolume bounded by surfaces of constant time in the new frame (in which the object is moving), that is a different geometric object, and computing its hypervolume will give you a different answer. The Lorentz factor for the object in the new frame, in which it moves at some velocity $v$, will indeed occur in the formula for this different answer.