Specific heat of vibrating bosons
I am trying to understand a solved exercise. It is about calculating the energy of vibrating bosons using the density of states, which is defined as follows:
$$g(omega) = frac{9N}{omega^3_D} omega$$
Afterwards, the specific heat $c_v$ is calculated ($E = frac{partial E}{partial T}$).
After giving some context, I will go to the point.
$$E = int_{0}^{infty} frac{hbar omega g(omega)}{e^{beta hbar omega} - 1} domega = frac{9N}{omega^3_D} int_{0}^{omega_D} frac{hbar omega^3}{e^{beta hbar omega} - 1} domega $$
On this integral, I have to use the following change of variables:
$$x = beta hbar omega$$
$$T_D = frac{omega_D hbar}{K_B}$$
After applying the first one I got:
$$E =frac{9N}{omega^3_D beta^4 hbar^3} int_{0}^{frac{omega_D}{beta hbar}} frac{ x^3}{e^{x} - 1} dx$$
Where:
$$beta = frac{1}{K_BT}$$
Once at this point I applied the second change of variables and derived E with respect to T (Temperature). My issue is that the solved problem skips this calculation and gives directly the $c_v$:
$$ c_v = 9K_BN (frac{T}{T_D})^3 int_{0}^{frac{T_D}{T}} frac{ x^4 e^x}{(e^{x} - 1)^2} dx$$
But I am not getting this outcome. May you help me out?
calculus integration mathematical-physics change-of-variable
asked Dec 4 '18 at 9:46
JD_PMJD_PM
10410
|
show 3 more comments
I am trying to understand a solved exercise. It is about calculating the energy of vibrating bosons using the density of states, which is defined as follows:
$$g(omega) = frac{9N}{omega^3_D} omega$$
Afterwards, the specific heat $c_v$ is calculated ($E = frac{partial E}{partial T}$).
After giving some context, I will go to the point.
$$E = int_{0}^{infty} frac{hbar omega g(omega)}{e^{beta hbar omega} - 1} domega = frac{9N}{omega^3_D} int_{0}^{omega_D} frac{hbar omega^3}{e^{beta hbar omega} - 1} domega $$
On this integral, I have to use the following change of variables:
$$x = beta hbar omega$$
$$T_D = frac{omega_D hbar}{K_B}$$
After applying the first one I got:
$$E =frac{9N}{omega^3_D beta^4 hbar^3} int_{0}^{frac{omega_D}{beta hbar}} frac{ x^3}{e^{x} - 1} dx$$
Where:
$$beta = frac{1}{K_BT}$$
Once at this point I applied the second change of variables and derived E with respect to T (Temperature). My issue is that the solved problem skips this calculation and gives directly the $c_v$:
$$ c_v = 9K_BN (frac{T}{T_D})^3 int_{0}^{frac{T_D}{T}} frac{ x^4 e^x}{(e^{x} - 1)^2} dx$$
But I am not getting this outcome. May you help me out?
calculus integration mathematical-physics change-of-variable
asked Dec 4 '18 at 9:46
JD_PMJD_PM
10410
1
@Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
– JD_PM
Dec 4 '18 at 10:00
1
Did you change the upper limit at first substitution?
– Nosrati
Dec 4 '18 at 10:06
1
Well, I don't see the $T$ variable anywhere in the $E$ formula. So, I don't know how to perform a calculation which does not result zero.
– Gustavo
Dec 4 '18 at 10:09
1
Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
– Nosrati
Dec 4 '18 at 11:30
1
$$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
– Nosrati
Dec 4 '18 at 11:35
|
show 3 more comments
I am trying to understand a solved exercise. It is about calculating the energy of vibrating bosons using the density of states, which is defined as follows:
$$g(omega) = frac{9N}{omega^3_D} omega$$
Afterwards, the specific heat $c_v$ is calculated ($E = frac{partial E}{partial T}$).
After giving some context, I will go to the point.
$$E = int_{0}^{infty} frac{hbar omega g(omega)}{e^{beta hbar omega} - 1} domega = frac{9N}{omega^3_D} int_{0}^{omega_D} frac{hbar omega^3}{e^{beta hbar omega} - 1} domega $$
On this integral, I have to use the following change of variables:
$$x = beta hbar omega$$
$$T_D = frac{omega_D hbar}{K_B}$$
After applying the first one I got:
$$E =frac{9N}{omega^3_D beta^4 hbar^3} int_{0}^{frac{omega_D}{beta hbar}} frac{ x^3}{e^{x} - 1} dx$$
Where:
$$beta = frac{1}{K_BT}$$
Once at this point I applied the second change of variables and derived E with respect to T (Temperature). My issue is that the solved problem skips this calculation and gives directly the $c_v$:
$$ c_v = 9K_BN (frac{T}{T_D})^3 int_{0}^{frac{T_D}{T}} frac{ x^4 e^x}{(e^{x} - 1)^2} dx$$
But I am not getting this outcome. May you help me out?
calculus integration mathematical-physics change-of-variable
asked Dec 4 '18 at 9:46
JD_PMJD_PM
10410
I am trying to understand a solved exercise. It is about calculating the energy of vibrating bosons using the density of states, which is defined as follows:
$$g(omega) = frac{9N}{omega^3_D} omega$$
Afterwards, the specific heat $c_v$ is calculated ($E = frac{partial E}{partial T}$).
After giving some context, I will go to the point.
$$E = int_{0}^{infty} frac{hbar omega g(omega)}{e^{beta hbar omega} - 1} domega = frac{9N}{omega^3_D} int_{0}^{omega_D} frac{hbar omega^3}{e^{beta hbar omega} - 1} domega $$
On this integral, I have to use the following change of variables:
$$x = beta hbar omega$$
$$T_D = frac{omega_D hbar}{K_B}$$
After applying the first one I got:
$$E =frac{9N}{omega^3_D beta^4 hbar^3} int_{0}^{frac{omega_D}{beta hbar}} frac{ x^3}{e^{x} - 1} dx$$
Where:
$$beta = frac{1}{K_BT}$$
Once at this point I applied the second change of variables and derived E with respect to T (Temperature). My issue is that the solved problem skips this calculation and gives directly the $c_v$:
$$ c_v = 9K_BN (frac{T}{T_D})^3 int_{0}^{frac{T_D}{T}} frac{ x^4 e^x}{(e^{x} - 1)^2} dx$$
But I am not getting this outcome. May you help me out?
calculus integration mathematical-physics change-of-variable
calculus integration mathematical-physics change-of-variable
asked Dec 4 '18 at 9:46
JD_PMJD_PM
10410
asked Dec 4 '18 at 9:46
JD_PMJD_PM
10410
edited Dec 4 '18 at 10:42
JD_PM
asked Dec 4 '18 at 9:46
JD_PMJD_PM
10410
asked Dec 4 '18 at 9:46
JD_PMJD_PM
10410
asked Dec 4 '18 at 9:46
JD_PMJD_PM
10410
10410
1
@Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
– JD_PM
Dec 4 '18 at 10:00
1
Did you change the upper limit at first substitution?
– Nosrati
Dec 4 '18 at 10:06
1
Well, I don't see the $T$ variable anywhere in the $E$ formula. So, I don't know how to perform a calculation which does not result zero.
– Gustavo
Dec 4 '18 at 10:09
1
Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
– Nosrati
Dec 4 '18 at 11:30
1
$$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
– Nosrati
Dec 4 '18 at 11:35
|
show 3 more comments
1
@Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
– JD_PM
Dec 4 '18 at 10:00
1
Did you change the upper limit at first substitution?
– Nosrati
Dec 4 '18 at 10:06
1
Well, I don't see the $T$ variable anywhere in the $E$ formula. So, I don't know how to perform a calculation which does not result zero.
– Gustavo
Dec 4 '18 at 10:09
1
Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
– Nosrati
Dec 4 '18 at 11:30
1
$$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
– Nosrati
Dec 4 '18 at 11:35
1
1
@Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
– JD_PM
Dec 4 '18 at 10:00
@Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
– JD_PM
Dec 4 '18 at 10:00
1
1
Did you change the upper limit at first substitution?
– Nosrati
Dec 4 '18 at 10:06
Did you change the upper limit at first substitution?
– Nosrati
Dec 4 '18 at 10:06
1
1
Well, I don't see the $T$ variable anywhere in the $E$ formula. So, I don't know how to perform a calculation which does not result zero.
– Gustavo
Dec 4 '18 at 10:09
Well, I don't see the $T$ variable anywhere in the $E$ formula. So, I don't know how to perform a calculation which does not result zero.
– Gustavo
Dec 4 '18 at 10:09
1
1
Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
– Nosrati
Dec 4 '18 at 11:30
Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
– Nosrati
Dec 4 '18 at 11:30
1
1
$$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
– Nosrati
Dec 4 '18 at 11:35
$$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
– Nosrati
Dec 4 '18 at 11:35
|
show 3 more comments
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1
@Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
– JD_PM
Dec 4 '18 at 10:00
1
Did you change the upper limit at first substitution?
– Nosrati
Dec 4 '18 at 10:06
1
Well, I don't see the $T$ variable anywhere in the $E$ formula. So, I don't know how to perform a calculation which does not result zero.
– Gustavo
Dec 4 '18 at 10:09
1
Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
– Nosrati
Dec 4 '18 at 11:30
1
$$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
– Nosrati
Dec 4 '18 at 11:35