# What is the hypervolume?

1. Mar 26, 2007

### MeJennifer

What is the hypervolume of a hypercube in a Minkowski space?

2. Mar 26, 2007

### Hurkyl

Staff Emeritus
4-Volume = Duration * Length * Width * Height.

3. Mar 26, 2007

### MeJennifer

I understand that that is the case for a Euclidean space.
But I fail to understand how you conclude that that also is the case for Minkowski space.

4. Mar 26, 2007

### Hurkyl

Staff Emeritus
4-volume is the integral of a 4-form, so it must be given by

$$V = \iiiint f(t, x, y, z) \, dt \, dx \, dy \, dz$$

The symmetry of Minkowski space would force f(t, x, y, z) to be a constant function. All that's left is to determine the constant.

It would be reasonable enough to declare by fiat that the constant is 1, but a short google search turns up that there is a canonical choice of volume form, by setting f to be $\sqrt{\left| \det g \right|}$, where g is the metric tensor. Since we (presumably) chose (t, x, y, z)-coordinates to be orthonormal (a.k.a. an inertial coordinate chart), det g = -1, and f(t, x, y, z) = 1.

(p.s. wow, LaTeX does have a quadruple integral symbol! I would have expected it to stop at 3)

Last edited: Mar 26, 2007
5. Mar 26, 2007

### quasar987

If I remember correctly, $\sqrt{\left| \det g \right|}$ is the jacobian of the lorentz transformations.

6. Mar 26, 2007

### robphy

Technically speaking, I think the notion of a "tensor density" arises here.
But I think Hurkyl's response is correct.

7. Mar 26, 2007

### MeJennifer

How about the volume of a unit 4-sphere and the 4-volume of a unit 4-ball in Minkowski space?

These questions seem so basic, surely I am not the first person who asks such questions.

Anybody who can provide some numbers?

8. Mar 26, 2007

### Hurkyl

Staff Emeritus
To be honest, I really dislike the notion of a tensor density. I much prefer thinking about the differential 4-form
f(t, x, y, z) dt dx dy dz​
which is an honest-to-goodness tensor, rather than treating f(t, x, y, z) as a geometric entity in its own right.

Just to make sure we're on the same page -- the unit 4-sphere is not the set of all points a unit (Minkowski) distance away from the origin. That object is... well, in Minkowski 2-space it would be a hyperbola. I'm not sure what it's called in Minkowski 4-space.

The 4-volume of the unit ball is a straightforward quadruple integral. It's the same calculation as for the 4-sphere in Euclidean 4-space.

It's far too late for me to go searching for what the right notion of 3-volume would be.

9. Mar 27, 2007

### MeJennifer

Well Hurkyl you seem to be much better in visualizing what a sphere is in Minkowski space, I already have enough trouble visualizing Euclidean 4-space let alone being able to visualize a sphere in Minkowski space, but whatever you want to call it, that is what I am asking for.

So all that I am asking for is the volume of the set of all points a unit distance away from the origin and the 4-volume of the set of all points from the origin up to a unit distance away from the origin.

So I am looking for two numbers, anybody who can tell me what they are?

Last edited: Mar 27, 2007
10. Mar 27, 2007

### quasar987

I don't think the unit sphere has a finite volume. With metric signature -+++, the integral is

$$\iiiint_{\mathcal{D}}dxdydzdt$$

where

$$\mathcal{D}=\{(x,y,z,t)\in\mathcal{R}^4:-t^2+x^2+y^2+z^2\leq 1\}=\{(x,y,z,t)\in\mathcal{R}^4:x^2+y^2+z^2\leq 1+t^2\}$$

So given a t, we integrate the volume of the 2-sphere of radius 1+t². And t goes from -infinity to +infinity.

11. Mar 27, 2007

### MeJennifer

I think you are right.

12. Mar 27, 2007

### robphy

In 1+1 Minkowski space, you can calculate the area swept by a radius vector with tip on the unit hyperbola as $$A=\frac{1}{2}r^2\theta$$, where $$\theta$$ is the intercepted Minkowski-angle (rapidity). Since the rapidity ranges from $$(-\infty, \infty)$$, the area is infinite.

In n+1 Minkowski space, the analogous hypersurface is called the hyperboloid (asymptotic to the light cone).