# What is the hypervolume of a 4D tetrahedron ($5$-cell)?

Here's how this question arose in my mind:

• area of a triangle: $\frac{1}{2} \cdot b \cdot h$

• volume of a tetrahedron: $\frac{1}{3} \cdot A \cdot h$

So the 2D object has $\frac{1}{2}$ in the formula, the 3D object has $\frac{1}{3}$ in its formula...does the 4D object have $\frac{1}{4}$ in its formula? And if so, why is there a linear progression instead of something exponential?

The concrete question is, what is the hypervolume of a $4$-dimensional tetrahedron?, aka $5$-cell or pentachoron.

(PS, i saw 2D and 3D as tags, but not 4D or hyper dimensional tag. Don't know what else to tag.)

• The technical name is "pentachoron". Commented Aug 12, 2016 at 23:57

Indeed, the $n$-simplex = "$n$-dimensional tetrahedron" (and indeed, every $n$-dimensional cone, of which the $n$-simplex is just a special case) has that factor $1/n$. This is easily seen by the fact that the volume of the $n$-dimensional tetrahedron is \begin{aligned} V_n &= \int_0^h V_{n-1}(z)\,\mathrm dz\\ &= \int_0^h V_{n-1}(0)\left(\frac{h-z}{h}\right)^{n-1}\,\mathrm dz && \text{substitute $w=h-z$}\\ &= \frac{V_{n-1}}{h^{n-1}}\int_0^hw^{n-1}\,\mathrm dw\\ &= \frac{V_{n-1}}{h^{n-1}}\frac{h^n}{n}\\ &= \frac{V_{n-1}h}{n} \end{aligned} Here $V_n$ is the volume of the $n$-dimensional cone, $V_{n-1}(z)$ is the $(n-1)$-dimensional volume ("area") of the "horizontal" slice at height $z$ above the base, $V_{n-1}=V_{n-1}(0)$ is the $(n-1)$-dimensional volume of the base, and $h$ is the height of the cone.

In the second line, I've used the fact that all slices of the pace are scaled versions of the base, with a scaling factor $(h-z)/h$, and volumes scale with the power of the dimension.

• So it basically comes from the fact that integral of x^n puts n in the denominator? That's not as complex an underpinning as I feared. Thanks. Commented Aug 12, 2016 at 8:42
• That is exactly right. Commented Aug 12, 2016 at 11:19

The $$\frac{1}{n}$$ term was understood well before calculus. Archimedes found the volume of a pyramid by slicing it into thin triangles and summing their areas and throwing away the small terms, prefiguring calculus by 2000 years. In general, it was well known that: $$\sum^N_{i=0} i^n \sim \frac{N^{n+1}}{n+1}$$ The area under the curve $$x^n$$ was shown geometrically to be equal to $$\frac{x^{n+1}}{n+1}$$ for positive integer values of $$n$$ a few decades before calculus.

However, the concept of higher dimensional geometry only came some 100 years after calculus.

Since I couldn't find it anywhere else and I ended up on this page when I searched for it, it might be useful to others to put the actual values for the (hyper)volumes of the first few regular n-polytopes with $$n+1$$ vertices here. The tetrahedron is $$V_3$$ and the 5-cell/pentachoron is $$V_4$$.

The height can be calculated using the centroid of the previous shape and Pythagorus' Theorem giving the following recurrence relation: $${h_n}^2 = 1 - {\left({\frac{n-1}{n}}\right)}^2{h_{n-1}}^2$$ with $$h_1=1$$

$$V_1 = 1$$

$$V_2 =\frac{1}{2}\times V_1 \times h_2 = \frac{1}{2}\times 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$$

$$V_3 =\frac{1}{3}\times V_2 \times h_3 = \frac{1}{3}\times \frac{\sqrt{3}}{4} \times \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2}}{12}$$

$$V_4 =\frac{1}{4}\times V_3 \times h_4 = \frac{1}{4}\times \frac{\sqrt{2}}{12} \times \frac{\sqrt{5}}{\sqrt{8}} = \frac{\sqrt{5}}{96}$$

$$V_5 = \frac{1}{5}\times V_4 \times h_5 = \frac{1}{5}\times \frac{\sqrt{5}}{96} \times \frac{\sqrt{3}}{\sqrt{5}} = \frac{\sqrt{3}}{480}$$

Update: I found a source with more information on regular polytopes, there is a polytope wiki at: https://polytope.miraheze.org/wiki