Specific heat of vibrating bosons












0












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
    $endgroup$
    – JD_PM
    Dec 4 '18 at 10:00






  • 1




    $begingroup$
    Did you change the upper limit at first substitution?
    $endgroup$
    – Nosrati
    Dec 4 '18 at 10:06






  • 1




    $begingroup$
    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.
    $endgroup$
    – Gustavo
    Dec 4 '18 at 10:09






  • 1




    $begingroup$
    Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
    $endgroup$
    – Nosrati
    Dec 4 '18 at 11:30






  • 1




    $begingroup$
    $$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
    $endgroup$
    – Nosrati
    Dec 4 '18 at 11:35
















0












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
    $endgroup$
    – JD_PM
    Dec 4 '18 at 10:00






  • 1




    $begingroup$
    Did you change the upper limit at first substitution?
    $endgroup$
    – Nosrati
    Dec 4 '18 at 10:06






  • 1




    $begingroup$
    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.
    $endgroup$
    – Gustavo
    Dec 4 '18 at 10:09






  • 1




    $begingroup$
    Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
    $endgroup$
    – Nosrati
    Dec 4 '18 at 11:30






  • 1




    $begingroup$
    $$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
    $endgroup$
    – Nosrati
    Dec 4 '18 at 11:35














0












0








0





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 10:42







JD_PM

















asked Dec 4 '18 at 9:46









JD_PMJD_PM

10410




10410








  • 1




    $begingroup$
    @Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
    $endgroup$
    – JD_PM
    Dec 4 '18 at 10:00






  • 1




    $begingroup$
    Did you change the upper limit at first substitution?
    $endgroup$
    – Nosrati
    Dec 4 '18 at 10:06






  • 1




    $begingroup$
    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.
    $endgroup$
    – Gustavo
    Dec 4 '18 at 10:09






  • 1




    $begingroup$
    Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
    $endgroup$
    – Nosrati
    Dec 4 '18 at 11:30






  • 1




    $begingroup$
    $$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
    $endgroup$
    – Nosrati
    Dec 4 '18 at 11:35














  • 1




    $begingroup$
    @Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
    $endgroup$
    – JD_PM
    Dec 4 '18 at 10:00






  • 1




    $begingroup$
    Did you change the upper limit at first substitution?
    $endgroup$
    – Nosrati
    Dec 4 '18 at 10:06






  • 1




    $begingroup$
    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.
    $endgroup$
    – Gustavo
    Dec 4 '18 at 10:09






  • 1




    $begingroup$
    Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
    $endgroup$
    – Nosrati
    Dec 4 '18 at 11:30






  • 1




    $begingroup$
    $$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
    $endgroup$
    – Nosrati
    Dec 4 '18 at 11:35








1




1




$begingroup$
@Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
$endgroup$
– JD_PM
Dec 4 '18 at 10:00




$begingroup$
@Gustavo Note that my issue is with the second change of variables and derivative (i.e I am stuck with maths).
$endgroup$
– JD_PM
Dec 4 '18 at 10:00




1




1




$begingroup$
Did you change the upper limit at first substitution?
$endgroup$
– Nosrati
Dec 4 '18 at 10:06




$begingroup$
Did you change the upper limit at first substitution?
$endgroup$
– Nosrati
Dec 4 '18 at 10:06




1




1




$begingroup$
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.
$endgroup$
– Gustavo
Dec 4 '18 at 10:09




$begingroup$
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.
$endgroup$
– Gustavo
Dec 4 '18 at 10:09




1




1




$begingroup$
Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
$endgroup$
– Nosrati
Dec 4 '18 at 11:30




$begingroup$
Use $dfrac{partial E}{partial T}=dfrac{partial E}{partial beta}dfrac{partialbeta}{partial T}$
$endgroup$
– Nosrati
Dec 4 '18 at 11:30




1




1




$begingroup$
$$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
$endgroup$
– Nosrati
Dec 4 '18 at 11:35




$begingroup$
$$dfrac{partial E}{partial T}= frac{-9N}{omega^3_D} int_{0}^{omega_D} frac{hbar^2 omega^4e^{beta hbar omega}}{(e^{beta hbar omega} - 1)^2} domegatimesdfrac{-1}{K_BT^2}$$ now let $beta hbar omega=x$ and simplify.
$endgroup$
– Nosrati
Dec 4 '18 at 11:35










0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025355%2fspecific-heat-of-vibrating-bosons%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3025355%2fspecific-heat-of-vibrating-bosons%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei