What is the simplest way of getting the solid angle $Omega_d$ in a space of $d$ dimensions?
It is known that the solid angle in a flat space of $d$ dimensions ($d = 2 n$ or $d = 2 n + 1$) is given by these formulae:
begin{align}tag{1}
Omega_{2 n} &= frac{1}{(n - 1)!} , 2 pi^n, qquad
&Omega_{2 n + 1} &= frac{2^{2 n} , n!}{(2 n)!} , 2 pi^n.
end{align}
For examples: $Omega_1 = 2$, $Omega_2 = 2 pi$, $Omega_3 = 4 pi$, $Omega_4 = 2 pi^2$. In some papers, it is described as the volume $mathrm{Vol}(mathbb{S}^{d−1})$ of the $d - 1$ unit-sphere (is there a difference?).
Instead of calculating a volume of some object (the unit sphere), suppose an observer is in free fall in a $D$ dimensional spacetime, so space appears to be flat locally (space around the observer has $d = D - 1$ dimensions). The observer wants to calculate the solid angle all around himslef, by looking in every orientations. He should get (1). What would be the simplest way in deriving these expressions?
integration geometry volume
asked Dec 28 '18 at 16:48
ChamCham
5210
migrated from physics.stackexchange.com Dec 28 '18 at 19:05
This question came from our site for active researchers, academics and students of physics.
add a comment |
It is known that the solid angle in a flat space of $d$ dimensions ($d = 2 n$ or $d = 2 n + 1$) is given by these formulae:
begin{align}tag{1}
Omega_{2 n} &= frac{1}{(n - 1)!} , 2 pi^n, qquad
&Omega_{2 n + 1} &= frac{2^{2 n} , n!}{(2 n)!} , 2 pi^n.
end{align}
For examples: $Omega_1 = 2$, $Omega_2 = 2 pi$, $Omega_3 = 4 pi$, $Omega_4 = 2 pi^2$. In some papers, it is described as the volume $mathrm{Vol}(mathbb{S}^{d−1})$ of the $d - 1$ unit-sphere (is there a difference?).
Instead of calculating a volume of some object (the unit sphere), suppose an observer is in free fall in a $D$ dimensional spacetime, so space appears to be flat locally (space around the observer has $d = D - 1$ dimensions). The observer wants to calculate the solid angle all around himslef, by looking in every orientations. He should get (1). What would be the simplest way in deriving these expressions?
integration geometry volume
asked Dec 28 '18 at 16:48
ChamCham
5210
migrated from physics.stackexchange.com Dec 28 '18 at 19:05
This question came from our site for active researchers, academics and students of physics.
Suggestion: take the gamma function expression, change it to a recursive form (i.e. $Omega_d=frac{sqrt{pi}}{d/2-1}Omega_{d-1}$, and think about how to prove that recursive form. Then, use either $Omega_1=2$ or $Omega_2 = 2pi$ as the root of the hierarchy.
– Sean Lake
Dec 28 '18 at 17:25
@SeanE.Lake, but then what is the relation with the solid angle? The problem is to find the solid angle in $d$ space, in the simplest way, from local considerations only.
– Cham
Dec 28 '18 at 17:39
1
Would Mathematics be a better home for this question?
– Qmechanic
Dec 28 '18 at 18:15
@Qmechanic, probably you're right. But since geometry is physics...
– Cham
Dec 28 '18 at 19:00
add a comment |
It is known that the solid angle in a flat space of $d$ dimensions ($d = 2 n$ or $d = 2 n + 1$) is given by these formulae:
begin{align}tag{1}
Omega_{2 n} &= frac{1}{(n - 1)!} , 2 pi^n, qquad
&Omega_{2 n + 1} &= frac{2^{2 n} , n!}{(2 n)!} , 2 pi^n.
end{align}
For examples: $Omega_1 = 2$, $Omega_2 = 2 pi$, $Omega_3 = 4 pi$, $Omega_4 = 2 pi^2$. In some papers, it is described as the volume $mathrm{Vol}(mathbb{S}^{d−1})$ of the $d - 1$ unit-sphere (is there a difference?).
Instead of calculating a volume of some object (the unit sphere), suppose an observer is in free fall in a $D$ dimensional spacetime, so space appears to be flat locally (space around the observer has $d = D - 1$ dimensions). The observer wants to calculate the solid angle all around himslef, by looking in every orientations. He should get (1). What would be the simplest way in deriving these expressions?
integration geometry volume
asked Dec 28 '18 at 16:48
ChamCham
5210
It is known that the solid angle in a flat space of $d$ dimensions ($d = 2 n$ or $d = 2 n + 1$) is given by these formulae:
begin{align}tag{1}
Omega_{2 n} &= frac{1}{(n - 1)!} , 2 pi^n, qquad
&Omega_{2 n + 1} &= frac{2^{2 n} , n!}{(2 n)!} , 2 pi^n.
end{align}
For examples: $Omega_1 = 2$, $Omega_2 = 2 pi$, $Omega_3 = 4 pi$, $Omega_4 = 2 pi^2$. In some papers, it is described as the volume $mathrm{Vol}(mathbb{S}^{d−1})$ of the $d - 1$ unit-sphere (is there a difference?).
Instead of calculating a volume of some object (the unit sphere), suppose an observer is in free fall in a $D$ dimensional spacetime, so space appears to be flat locally (space around the observer has $d = D - 1$ dimensions). The observer wants to calculate the solid angle all around himslef, by looking in every orientations. He should get (1). What would be the simplest way in deriving these expressions?
integration geometry volume
integration geometry volume
asked Dec 28 '18 at 16:48
ChamCham
5210
asked Dec 28 '18 at 16:48
ChamCham
5210
asked Dec 28 '18 at 16:48
ChamCham
5210
asked Dec 28 '18 at 16:48
ChamCham
5210
asked Dec 28 '18 at 16:48
ChamCham
5210
5210
migrated from physics.stackexchange.com Dec 28 '18 at 19:05
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Dec 28 '18 at 19:05
This question came from our site for active researchers, academics and students of physics.
Suggestion: take the gamma function expression, change it to a recursive form (i.e. $Omega_d=frac{sqrt{pi}}{d/2-1}Omega_{d-1}$, and think about how to prove that recursive form. Then, use either $Omega_1=2$ or $Omega_2 = 2pi$ as the root of the hierarchy.
– Sean Lake
Dec 28 '18 at 17:25
@SeanE.Lake, but then what is the relation with the solid angle? The problem is to find the solid angle in $d$ space, in the simplest way, from local considerations only.
– Cham
Dec 28 '18 at 17:39
1
Would Mathematics be a better home for this question?
– Qmechanic
Dec 28 '18 at 18:15
@Qmechanic, probably you're right. But since geometry is physics...
– Cham
Dec 28 '18 at 19:00
add a comment |
Suggestion: take the gamma function expression, change it to a recursive form (i.e. $Omega_d=frac{sqrt{pi}}{d/2-1}Omega_{d-1}$, and think about how to prove that recursive form. Then, use either $Omega_1=2$ or $Omega_2 = 2pi$ as the root of the hierarchy.
– Sean Lake
Dec 28 '18 at 17:25
@SeanE.Lake, but then what is the relation with the solid angle? The problem is to find the solid angle in $d$ space, in the simplest way, from local considerations only.
– Cham
Dec 28 '18 at 17:39
1
Would Mathematics be a better home for this question?
– Qmechanic
Dec 28 '18 at 18:15
@Qmechanic, probably you're right. But since geometry is physics...
– Cham
Dec 28 '18 at 19:00
Suggestion: take the gamma function expression, change it to a recursive form (i.e. $Omega_d=frac{sqrt{pi}}{d/2-1}Omega_{d-1}$, and think about how to prove that recursive form. Then, use either $Omega_1=2$ or $Omega_2 = 2pi$ as the root of the hierarchy.
– Sean Lake
Dec 28 '18 at 17:25
Suggestion: take the gamma function expression, change it to a recursive form (i.e. $Omega_d=frac{sqrt{pi}}{d/2-1}Omega_{d-1}$, and think about how to prove that recursive form. Then, use either $Omega_1=2$ or $Omega_2 = 2pi$ as the root of the hierarchy.
– Sean Lake
Dec 28 '18 at 17:25
@SeanE.Lake, but then what is the relation with the solid angle? The problem is to find the solid angle in $d$ space, in the simplest way, from local considerations only.
– Cham
Dec 28 '18 at 17:39
@SeanE.Lake, but then what is the relation with the solid angle? The problem is to find the solid angle in $d$ space, in the simplest way, from local considerations only.
– Cham
Dec 28 '18 at 17:39
1
1
Would Mathematics be a better home for this question?
– Qmechanic
Dec 28 '18 at 18:15
Would Mathematics be a better home for this question?
– Qmechanic
Dec 28 '18 at 18:15
@Qmechanic, probably you're right. But since geometry is physics...
– Cham
Dec 28 '18 at 19:00
@Qmechanic, probably you're right. But since geometry is physics...
– Cham
Dec 28 '18 at 19:00
add a comment |
1 Answer
1
active
oldest
votes
Let $mathbf{x}=(x_1,...,x_N)$ denote an $N$-tuple of real variables, and
$$
requivsqrt{x_1^2+cdots+x_N^2}.
$$
Define
$$
Omega(N) = frac{f(N)}{g(N)}
tag{1}
$$
with
$$
f(N)equiv int d^N x exp(-mathbf{x}^2)
tag{2}
$$
and
$$
g(N)equiv int_0^infty dr r^{N-1} exp(-r^2).
tag{3}
$$
The definition (1) implies that $Omega(N)$ is the desired quantity (with $N$ denoted $d$ in the OP). The integral $f(N)$ is
$$
f(N)equiv left(int dx e^{-x^2}right)^N = pi^{N/2}.
tag{4}
$$
To evaluate the integral $g(N)$, first consider odd $N$. Then $(N-1)/2$ is an integer, so we can use
begin{align}
g(N)
&= left(-frac{d}{da}right)^{(N-1)/2}
left.int_0^infty dr e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-1)/2}
sqrt{frac{pi}{4a}}right|_{a=1}.
tag{5}
end{align}
When $N$ is even, the quantity $(N-2)/2$ is an integer, so we can use
begin{align}
g(N)
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
int_0^infty dr r,e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
frac{1}{2a}right|_{a=1}.
tag{6}
end{align}
answered Dec 28 '18 at 18:43
Dan YandDan Yand
1467
This is what I believed to be the simplest way. I think it's the proper answer to my question.
– Cham
Dec 28 '18 at 18:59
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $mathbf{x}=(x_1,...,x_N)$ denote an $N$-tuple of real variables, and
$$
requivsqrt{x_1^2+cdots+x_N^2}.
$$
Define
$$
Omega(N) = frac{f(N)}{g(N)}
tag{1}
$$
with
$$
f(N)equiv int d^N x exp(-mathbf{x}^2)
tag{2}
$$
and
$$
g(N)equiv int_0^infty dr r^{N-1} exp(-r^2).
tag{3}
$$
The definition (1) implies that $Omega(N)$ is the desired quantity (with $N$ denoted $d$ in the OP). The integral $f(N)$ is
$$
f(N)equiv left(int dx e^{-x^2}right)^N = pi^{N/2}.
tag{4}
$$
To evaluate the integral $g(N)$, first consider odd $N$. Then $(N-1)/2$ is an integer, so we can use
begin{align}
g(N)
&= left(-frac{d}{da}right)^{(N-1)/2}
left.int_0^infty dr e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-1)/2}
sqrt{frac{pi}{4a}}right|_{a=1}.
tag{5}
end{align}
When $N$ is even, the quantity $(N-2)/2$ is an integer, so we can use
begin{align}
g(N)
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
int_0^infty dr r,e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
frac{1}{2a}right|_{a=1}.
tag{6}
end{align}
answered Dec 28 '18 at 18:43
Dan YandDan Yand
1467
This is what I believed to be the simplest way. I think it's the proper answer to my question.
– Cham
Dec 28 '18 at 18:59
add a comment |
Let $mathbf{x}=(x_1,...,x_N)$ denote an $N$-tuple of real variables, and
$$
requivsqrt{x_1^2+cdots+x_N^2}.
$$
Define
$$
Omega(N) = frac{f(N)}{g(N)}
tag{1}
$$
with
$$
f(N)equiv int d^N x exp(-mathbf{x}^2)
tag{2}
$$
and
$$
g(N)equiv int_0^infty dr r^{N-1} exp(-r^2).
tag{3}
$$
The definition (1) implies that $Omega(N)$ is the desired quantity (with $N$ denoted $d$ in the OP). The integral $f(N)$ is
$$
f(N)equiv left(int dx e^{-x^2}right)^N = pi^{N/2}.
tag{4}
$$
To evaluate the integral $g(N)$, first consider odd $N$. Then $(N-1)/2$ is an integer, so we can use
begin{align}
g(N)
&= left(-frac{d}{da}right)^{(N-1)/2}
left.int_0^infty dr e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-1)/2}
sqrt{frac{pi}{4a}}right|_{a=1}.
tag{5}
end{align}
When $N$ is even, the quantity $(N-2)/2$ is an integer, so we can use
begin{align}
g(N)
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
int_0^infty dr r,e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
frac{1}{2a}right|_{a=1}.
tag{6}
end{align}
answered Dec 28 '18 at 18:43
Dan YandDan Yand
1467
This is what I believed to be the simplest way. I think it's the proper answer to my question.
– Cham
Dec 28 '18 at 18:59
add a comment |
Let $mathbf{x}=(x_1,...,x_N)$ denote an $N$-tuple of real variables, and
$$
requivsqrt{x_1^2+cdots+x_N^2}.
$$
Define
$$
Omega(N) = frac{f(N)}{g(N)}
tag{1}
$$
with
$$
f(N)equiv int d^N x exp(-mathbf{x}^2)
tag{2}
$$
and
$$
g(N)equiv int_0^infty dr r^{N-1} exp(-r^2).
tag{3}
$$
The definition (1) implies that $Omega(N)$ is the desired quantity (with $N$ denoted $d$ in the OP). The integral $f(N)$ is
$$
f(N)equiv left(int dx e^{-x^2}right)^N = pi^{N/2}.
tag{4}
$$
To evaluate the integral $g(N)$, first consider odd $N$. Then $(N-1)/2$ is an integer, so we can use
begin{align}
g(N)
&= left(-frac{d}{da}right)^{(N-1)/2}
left.int_0^infty dr e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-1)/2}
sqrt{frac{pi}{4a}}right|_{a=1}.
tag{5}
end{align}
When $N$ is even, the quantity $(N-2)/2$ is an integer, so we can use
begin{align}
g(N)
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
int_0^infty dr r,e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
frac{1}{2a}right|_{a=1}.
tag{6}
end{align}
answered Dec 28 '18 at 18:43
Dan YandDan Yand
1467
Let $mathbf{x}=(x_1,...,x_N)$ denote an $N$-tuple of real variables, and
$$
requivsqrt{x_1^2+cdots+x_N^2}.
$$
Define
$$
Omega(N) = frac{f(N)}{g(N)}
tag{1}
$$
with
$$
f(N)equiv int d^N x exp(-mathbf{x}^2)
tag{2}
$$
and
$$
g(N)equiv int_0^infty dr r^{N-1} exp(-r^2).
tag{3}
$$
The definition (1) implies that $Omega(N)$ is the desired quantity (with $N$ denoted $d$ in the OP). The integral $f(N)$ is
$$
f(N)equiv left(int dx e^{-x^2}right)^N = pi^{N/2}.
tag{4}
$$
To evaluate the integral $g(N)$, first consider odd $N$. Then $(N-1)/2$ is an integer, so we can use
begin{align}
g(N)
&= left(-frac{d}{da}right)^{(N-1)/2}
left.int_0^infty dr e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-1)/2}
sqrt{frac{pi}{4a}}right|_{a=1}.
tag{5}
end{align}
When $N$ is even, the quantity $(N-2)/2$ is an integer, so we can use
begin{align}
g(N)
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
int_0^infty dr r,e^{-ar^2}right|_{a=1}
\
&=
left.left(-frac{d}{da}right)^{(N-2)/2}
frac{1}{2a}right|_{a=1}.
tag{6}
end{align}
answered Dec 28 '18 at 18:43
Dan YandDan Yand
1467
answered Dec 28 '18 at 18:43
Dan YandDan Yand
1467
answered Dec 28 '18 at 18:43
Dan YandDan Yand
1467
answered Dec 28 '18 at 18:43
Dan YandDan Yand
1467
1467
This is what I believed to be the simplest way. I think it's the proper answer to my question.
– Cham
Dec 28 '18 at 18:59
add a comment |
This is what I believed to be the simplest way. I think it's the proper answer to my question.
– Cham
Dec 28 '18 at 18:59
This is what I believed to be the simplest way. I think it's the proper answer to my question.
– Cham
Dec 28 '18 at 18:59
This is what I believed to be the simplest way. I think it's the proper answer to my question.
– Cham
Dec 28 '18 at 18:59
add a comment |
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Suggestion: take the gamma function expression, change it to a recursive form (i.e. $Omega_d=frac{sqrt{pi}}{d/2-1}Omega_{d-1}$, and think about how to prove that recursive form. Then, use either $Omega_1=2$ or $Omega_2 = 2pi$ as the root of the hierarchy.
– Sean Lake
Dec 28 '18 at 17:25
@SeanE.Lake, but then what is the relation with the solid angle? The problem is to find the solid angle in $d$ space, in the simplest way, from local considerations only.
– Cham
Dec 28 '18 at 17:39
1
Would Mathematics be a better home for this question?
– Qmechanic
Dec 28 '18 at 18:15
@Qmechanic, probably you're right. But since geometry is physics...
– Cham
Dec 28 '18 at 19:00