# Hypervolume of a Hypercube in Minkowski Space

• MeJennifer
In summary: Its Minkowski volume would be infinite as well.In summary, the hypervolume of a hypercube in a Minkowski space is given by V = \iiiint f(t, x, y, z) \, dt \, dx \, dy \, dz, where f(t, x, y, z) = \sqrt{\left| \det g \right|} and g is the metric tensor. This is due to the symmetry of Minkowski space, which forces f(t, x, y, z) to be a constant function. However, the concept of a "tensor density" arises here, and it is important to consider the differential 4-form f(t, x, y, z) dt
MeJennifer
What is the hypervolume of a hypercube in a Minkowski space?

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

Hurkyl said:
4-Volume = Duration * Length * Width * Height.
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-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)

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

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

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?

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

Hurkyl said:
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.
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?

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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.

quasar987 said:
I don't think the unit sphere has a finite volume.
I think you are right.

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).

## 1. What is the hypervolume?

The hypervolume is a mathematical concept that measures the space occupied by a multi-dimensional object. It is a generalization of the concept of volume in three dimensions to higher dimensions.

## 2. How is the hypervolume calculated?

The hypervolume is calculated by multiplying the lengths of all the sides of the multi-dimensional object. For example, in two dimensions (a square), the hypervolume would be calculated by multiplying the length and width. In three dimensions (a cube), the hypervolume would be calculated by multiplying the length, width, and height.

## 3. What is the importance of the hypervolume in science?

The hypervolume has various applications in science, including in physics, biology, and computer science. It is used to measure the size and shape of complex objects, model and analyze high-dimensional data, and to understand the behavior of systems with multiple variables.

## 4. How does the concept of hypervolume relate to the concept of dimensionality?

The concept of hypervolume is closely related to the concept of dimensionality. As the number of dimensions increases, the hypervolume of an object also increases. Furthermore, the hypervolume of a multi-dimensional object can be thought of as the amount of space it occupies in that particular dimension.

## 5. Can the hypervolume of an object be negative?

No, the hypervolume of an object cannot be negative. It is a measure of the space occupied by an object, and space cannot have a negative value. In some cases, the hypervolume of an object can be zero if all the dimensions of the object have a length of zero.

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