Mixing
Physical meaning of potential in heat equation
$begingroup$
I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows :
Let $Omega$ be a bounded region of the space and $T>0$ a fixed time. In $Omega_T=(0,T)times Omega$ we consider the following equation
$$u_t =alphaDelta u -a(x)u,$$
$$u(0,x)=f(x),$$
where $f$ is the initial condition, $a$ a bounded potential, $alpha>0$ is a constant, and $Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
thermodynamics potential diffusion heat-conduction
edited Jan 6 at 23:23
Qmechanic♦
103k121851177
asked Jan 6 at 12:57
S. ChoS. Cho
1536
$endgroup$
add a comment |
$begingroup$
I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows :
Let $Omega$ be a bounded region of the space and $T>0$ a fixed time. In $Omega_T=(0,T)times Omega$ we consider the following equation
$$u_t =alphaDelta u -a(x)u,$$
$$u(0,x)=f(x),$$
where $f$ is the initial condition, $a$ a bounded potential, $alpha>0$ is a constant, and $Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
thermodynamics potential diffusion heat-conduction
edited Jan 6 at 23:23
Qmechanic♦
103k121851177
asked Jan 6 at 12:57
S. ChoS. Cho
1536
$endgroup$
$begingroup$
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
$endgroup$
– N. Steinle
Jan 6 at 15:15
$begingroup$
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
$endgroup$
– S. Cho
Jan 6 at 16:16
$begingroup$
Your title asks about a potential? What potential are you talking about?
$endgroup$
– Drew
Jan 7 at 0:34
$begingroup$
We call the coefficient $a$ a potential
$endgroup$
– S. Cho
Jan 7 at 11:09
add a comment |
$begingroup$
I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows :
Let $Omega$ be a bounded region of the space and $T>0$ a fixed time. In $Omega_T=(0,T)times Omega$ we consider the following equation
$$u_t =alphaDelta u -a(x)u,$$
$$u(0,x)=f(x),$$
where $f$ is the initial condition, $a$ a bounded potential, $alpha>0$ is a constant, and $Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
thermodynamics potential diffusion heat-conduction
edited Jan 6 at 23:23
Qmechanic♦
103k121851177
asked Jan 6 at 12:57
S. ChoS. Cho
1536
$endgroup$
I'm working on the mathematical theory of parabolic equations. The prototype of such equations is heat equation given as follows :
Let $Omega$ be a bounded region of the space and $T>0$ a fixed time. In $Omega_T=(0,T)times Omega$ we consider the following equation
$$u_t =alphaDelta u -a(x)u,$$
$$u(0,x)=f(x),$$
where $f$ is the initial condition, $a$ a bounded potential, $alpha>0$ is a constant, and $Delta$ is the Laplacian. I'm wondering to know the physical meaning of the coefficient $a$ (and may be $alpha$) and it's role in the heat process? Any reference or suggestion would be helpful.
thermodynamics potential diffusion heat-conduction
thermodynamics potential diffusion heat-conduction
edited Jan 6 at 23:23
Qmechanic♦
103k121851177
asked Jan 6 at 12:57
S. ChoS. Cho
1536
edited Jan 6 at 23:23
Qmechanic♦
103k121851177
asked Jan 6 at 12:57
S. ChoS. Cho
1536
edited Jan 6 at 23:23
Qmechanic♦
103k121851177
edited Jan 6 at 23:23
Qmechanic♦
103k121851177
edited Jan 6 at 23:23
Qmechanic♦
103k121851177
103k121851177
asked Jan 6 at 12:57
S. ChoS. Cho
1536
asked Jan 6 at 12:57
S. ChoS. Cho
1536
asked Jan 6 at 12:57
S. ChoS. Cho
1536
1536
$begingroup$
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
$endgroup$
– N. Steinle
Jan 6 at 15:15
$begingroup$
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
$endgroup$
– S. Cho
Jan 6 at 16:16
$begingroup$
Your title asks about a potential? What potential are you talking about?
$endgroup$
– Drew
Jan 7 at 0:34
$begingroup$
We call the coefficient $a$ a potential
$endgroup$
– S. Cho
Jan 7 at 11:09
add a comment |
$begingroup$
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
$endgroup$
– N. Steinle
Jan 6 at 15:15
$begingroup$
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
$endgroup$
– S. Cho
Jan 6 at 16:16
$begingroup$
Your title asks about a potential? What potential are you talking about?
$endgroup$
– Drew
Jan 7 at 0:34
$begingroup$
We call the coefficient $a$ a potential
$endgroup$
– S. Cho
Jan 7 at 11:09
$begingroup$
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
$endgroup$
– N. Steinle
Jan 6 at 15:15
$begingroup$
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
$endgroup$
– N. Steinle
Jan 6 at 15:15
$begingroup$
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
$endgroup$
– S. Cho
Jan 6 at 16:16
$begingroup$
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
$endgroup$
– S. Cho
Jan 6 at 16:16
$begingroup$
Your title asks about a potential? What potential are you talking about?
$endgroup$
– Drew
Jan 7 at 0:34
$begingroup$
Your title asks about a potential? What potential are you talking about?
$endgroup$
– Drew
Jan 7 at 0:34
$begingroup$
We call the coefficient $a$ a potential
$endgroup$
– S. Cho
Jan 7 at 11:09
$begingroup$
We call the coefficient $a$ a potential
$endgroup$
– S. Cho
Jan 7 at 11:09
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
answered Jan 6 at 16:34
ChemomechanicsChemomechanics
4,77131023
$endgroup$
$begingroup$
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
$endgroup$
– S. Cho
Jan 7 at 11:28
1
$begingroup$
What is $partial_nu$ in this case?
$endgroup$
– Chemomechanics
Jan 7 at 19:16
$begingroup$
$partial_nu y = nabla y cdot nu$ is the normal derivative of $y$ where $nu$ is the unit outer normal vector on $Gamma$.
$endgroup$
– S. Cho
Jan 7 at 22:08
1
$begingroup$
This would seem to correspond to conduction within the system with thermal diffusivity $beta$, lateral convective heat transfer to the environment as mediated by a coefficient $crho b(x)$, and lateral conductive heat transfer to the environment as mediated by a thermal conductivity $(-crhoalpha)$.
$endgroup$
– Chemomechanics
Jan 8 at 18:31
add a comment |
$begingroup$
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
answered Jan 6 at 13:24
N. SteinleN. Steinle
1,528117
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1
$begingroup$
@S.Cho My pleasure! Which wiki article, specifically?
$endgroup$
– N. Steinle
Jan 6 at 14:30
1
$begingroup$
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
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– Chester Miller
Jan 6 at 14:44
1
$begingroup$
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
$endgroup$
– t t t t
Jan 6 at 14:49
1
$begingroup$
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
$endgroup$
– N. Steinle
Jan 6 at 15:20
1
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Oops. You're right. I didn't notice that.
$endgroup$
– Chester Miller
Jan 6 at 18:08
|
show 5 more comments
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The proper physical name of your equation is diffusion equation with a source term. The equation can be rearranged to continuity equation - $u_t-alpha u_{xx} = Q$. For $Q=0$ the time dependent solution can be shown to have time independent norm, which is manifestation of local mass conservation law. Source term means that particles can be created and destroyed locally, according to $Q(x)$ variation.
Continuity equation is a restatement of Gauss law - at given infinitesimal volume, the change in the number of particles in the volume is exactly equal to the number of particles crossed the surface of it in/out of this volume.
You may gain some physical intuition exploring compressible aspect of the Navier-Stockes equation. The compressability is exactly the violation of continuity equation.
Closed form solution is given here. This wierd document seems related, however didn't find any peer reviewed articles.
answered Jan 9 at 6:36
AlexanderAlexander
1,058724
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add a comment |
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In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
answered Jan 6 at 13:13
Vincent FraticelliVincent Fraticelli
5004
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add a comment |
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The equation describes the flow of heat in presence of sources or sinks. The first term on the right hand side is the normal diffusion term. The second term can be thought of as a source or sink term. For more details see: https://www.math.ubc.ca/~peirce/M257_316_2012_Lecture_19.pdf
answered Jan 8 at 5:47
noisyoscillatornoisyoscillator
318
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
answered Jan 6 at 16:34
ChemomechanicsChemomechanics
4,77131023
$endgroup$
$begingroup$
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
$endgroup$
– S. Cho
Jan 7 at 11:28
1
$begingroup$
What is $partial_nu$ in this case?
$endgroup$
– Chemomechanics
Jan 7 at 19:16
$begingroup$
$partial_nu y = nabla y cdot nu$ is the normal derivative of $y$ where $nu$ is the unit outer normal vector on $Gamma$.
$endgroup$
– S. Cho
Jan 7 at 22:08
1
$begingroup$
This would seem to correspond to conduction within the system with thermal diffusivity $beta$, lateral convective heat transfer to the environment as mediated by a coefficient $crho b(x)$, and lateral conductive heat transfer to the environment as mediated by a thermal conductivity $(-crhoalpha)$.
$endgroup$
– Chemomechanics
Jan 8 at 18:31
add a comment |
$begingroup$
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
answered Jan 6 at 16:34
ChemomechanicsChemomechanics
4,77131023
$endgroup$
$begingroup$
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
$endgroup$
– S. Cho
Jan 7 at 11:28
1
$begingroup$
What is $partial_nu$ in this case?
$endgroup$
– Chemomechanics
Jan 7 at 19:16
$begingroup$
$partial_nu y = nabla y cdot nu$ is the normal derivative of $y$ where $nu$ is the unit outer normal vector on $Gamma$.
$endgroup$
– S. Cho
Jan 7 at 22:08
1
$begingroup$
This would seem to correspond to conduction within the system with thermal diffusivity $beta$, lateral convective heat transfer to the environment as mediated by a coefficient $crho b(x)$, and lateral conductive heat transfer to the environment as mediated by a thermal conductivity $(-crhoalpha)$.
$endgroup$
– Chemomechanics
Jan 8 at 18:31
add a comment |
$begingroup$
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
answered Jan 6 at 16:34
ChemomechanicsChemomechanics
4,77131023
$endgroup$
As discussed here under "Incorporating lateral heat transfer" (disclaimer: my site), if you're considering a 1-D transient heat transfer problem as suggested by the variables $x$ and $t$, then the equation $$kDelta T(x,t)-h(x)T(x,t)=crhodot T(x,t)$$
represents axial conduction with thermal conductivity $k$, linear lateral heat dissipation (through conduction, convection, and/or slight radiation) with spatially dependent coefficient $h(x)$, and energy storage with specific heat capacity $c$ and density $rho$. $T(x,t)$ is the temperature excursion from some ambient value.
The qualification of "slight" radiation is to ensure linearity of $T(x,t)$ in that term. For convection, $h$ is simply a convection coefficient. As discussed in the link, $h$ might also represent lateral conduction to an adjacent temperature sink (for a suspended microfabricated beam, say).
If we change variables from $T(x,t)$ to $u$ and divide by $crho$, then we have
$$alpha Delta u-left(frac{h(x)}{crho}right)u=u_t,$$
with $alpha$ being the thermal diffusivity, which matches your equation and indicates that $a(x)$ corresponds to a spatially varying lateral heat coefficient divided by the specific heat capacity and the density. This is the physical interpretation of that parameter for this type of system (I solve the equation here).
answered Jan 6 at 16:34
ChemomechanicsChemomechanics
4,77131023
answered Jan 6 at 16:34
ChemomechanicsChemomechanics
4,77131023
answered Jan 6 at 16:34
ChemomechanicsChemomechanics
4,77131023
answered Jan 6 at 16:34
ChemomechanicsChemomechanics
4,77131023
4,77131023
$begingroup$
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
$endgroup$
– S. Cho
Jan 7 at 11:28
1
$begingroup$
What is $partial_nu$ in this case?
$endgroup$
– Chemomechanics
Jan 7 at 19:16
$begingroup$
$partial_nu y = nabla y cdot nu$ is the normal derivative of $y$ where $nu$ is the unit outer normal vector on $Gamma$.
$endgroup$
– S. Cho
Jan 7 at 22:08
1
$begingroup$
This would seem to correspond to conduction within the system with thermal diffusivity $beta$, lateral convective heat transfer to the environment as mediated by a coefficient $crho b(x)$, and lateral conductive heat transfer to the environment as mediated by a thermal conductivity $(-crhoalpha)$.
$endgroup$
– Chemomechanics
Jan 8 at 18:31
add a comment |
$begingroup$
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
$endgroup$
– S. Cho
Jan 7 at 11:28
1
$begingroup$
What is $partial_nu$ in this case?
$endgroup$
– Chemomechanics
Jan 7 at 19:16
$begingroup$
$partial_nu y = nabla y cdot nu$ is the normal derivative of $y$ where $nu$ is the unit outer normal vector on $Gamma$.
$endgroup$
– S. Cho
Jan 7 at 22:08
1
$begingroup$
This would seem to correspond to conduction within the system with thermal diffusivity $beta$, lateral convective heat transfer to the environment as mediated by a coefficient $crho b(x)$, and lateral conductive heat transfer to the environment as mediated by a thermal conductivity $(-crhoalpha)$.
$endgroup$
– Chemomechanics
Jan 8 at 18:31
$begingroup$
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
$endgroup$
– S. Cho
Jan 7 at 11:28
$begingroup$
Is it the same for the parameters appearing in the dynamic boundary conditions ? More precisely, if $Gamma=partial Omega$ is the boundary of the domain, the dynamic boundary conditions on $Gamma$ is $partial_t y_Gamma =beta Delta_Gamma y_Gamma +alpha partial_nu y -b(x) y_Gamma$, on $Gamma$. I mean the parameters $beta$ and $b$, ($alpha$ is the same as the first equation).
$endgroup$
– S. Cho
Jan 7 at 11:28
1
1
$begingroup$
What is $partial_nu$ in this case?
$endgroup$
– Chemomechanics
Jan 7 at 19:16
$begingroup$
What is $partial_nu$ in this case?
$endgroup$
– Chemomechanics
Jan 7 at 19:16
$begingroup$
$partial_nu y = nabla y cdot nu$ is the normal derivative of $y$ where $nu$ is the unit outer normal vector on $Gamma$.
$endgroup$
– S. Cho
Jan 7 at 22:08
$begingroup$
$partial_nu y = nabla y cdot nu$ is the normal derivative of $y$ where $nu$ is the unit outer normal vector on $Gamma$.
$endgroup$
– S. Cho
Jan 7 at 22:08
1
1
$begingroup$
This would seem to correspond to conduction within the system with thermal diffusivity $beta$, lateral convective heat transfer to the environment as mediated by a coefficient $crho b(x)$, and lateral conductive heat transfer to the environment as mediated by a thermal conductivity $(-crhoalpha)$.
$endgroup$
– Chemomechanics
Jan 8 at 18:31
$begingroup$
This would seem to correspond to conduction within the system with thermal diffusivity $beta$, lateral convective heat transfer to the environment as mediated by a coefficient $crho b(x)$, and lateral conductive heat transfer to the environment as mediated by a thermal conductivity $(-crhoalpha)$.
$endgroup$
– Chemomechanics
Jan 8 at 18:31
add a comment |
$begingroup$
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
answered Jan 6 at 13:24
N. SteinleN. Steinle
1,528117
$endgroup$
1
$begingroup$
@S.Cho My pleasure! Which wiki article, specifically?
$endgroup$
– N. Steinle
Jan 6 at 14:30
1
$begingroup$
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
$endgroup$
– Chester Miller
Jan 6 at 14:44
1
$begingroup$
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
$endgroup$
– t t t t
Jan 6 at 14:49
1
$begingroup$
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
$endgroup$
– N. Steinle
Jan 6 at 15:20
1
$begingroup$
Oops. You're right. I didn't notice that.
$endgroup$
– Chester Miller
Jan 6 at 18:08
|
show 5 more comments
$begingroup$
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
answered Jan 6 at 13:24
N. SteinleN. Steinle
1,528117
$endgroup$
1
$begingroup$
@S.Cho My pleasure! Which wiki article, specifically?
$endgroup$
– N. Steinle
Jan 6 at 14:30
1
$begingroup$
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
$endgroup$
– Chester Miller
Jan 6 at 14:44
1
$begingroup$
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
$endgroup$
– t t t t
Jan 6 at 14:49
1
$begingroup$
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
$endgroup$
– N. Steinle
Jan 6 at 15:20
1
$begingroup$
Oops. You're right. I didn't notice that.
$endgroup$
– Chester Miller
Jan 6 at 18:08
|
show 5 more comments
$begingroup$
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
answered Jan 6 at 13:24
N. SteinleN. Steinle
1,528117
$endgroup$
The heat equation, as you've written it, models the flow of energy via thermal conduction (heat) through some region with well defined boundary conditions. You have yet to provide the specifics of the boundary region, so my answer will remain general and vague.
The $alpha$ is the "diffusion coefficient" which is the isotropic form (diagonal terms only) of the diffusion tensor - alas the heat equation is a special case of the diffusion equation. So this coefficient tells us about how thermally diffuse the material that composes the region is (how diffusely distributed is the matter that the heat flows through?).
the physical meaning of the coefficient $a$
Sorry, at first I misread the equation (I thought it was merely a forcing term at first). And then secondly I mistook it for a convective term, but that's not correct since a convective term is typically proportional to $frac{partial u}{partial x}$ (see equation 27 here). I have found that the term, $a(x)u$, could represent an approximative radiative term that is position dependent (for small radiative losses), i.e. see the last equation here. In which case, $a$ determines the strength of the radiation emitted from the conductor as a function of position. This radiation term is only meaningful for temperature variations in the rod that are small compared to the temperature of the surroundings, and in the case of larger fluctuations one must use a $u^4$ dependence instead (in accordance with Stefan-Boltzmann Law).
Here is a very nice write-up for non-homogeneous heat problems.
answered Jan 6 at 13:24
N. SteinleN. Steinle
1,528117
edited Jan 7 at 18:17
answered Jan 6 at 13:24
N. SteinleN. Steinle
1,528117
answered Jan 6 at 13:24
N. SteinleN. Steinle
1,528117
answered Jan 6 at 13:24
N. SteinleN. Steinle
1,528117
1,528117
1
$begingroup$
@S.Cho My pleasure! Which wiki article, specifically?
$endgroup$
– N. Steinle
Jan 6 at 14:30
1
$begingroup$
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
$endgroup$
– Chester Miller
Jan 6 at 14:44
1
$begingroup$
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
$endgroup$
– t t t t
Jan 6 at 14:49
1
$begingroup$
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
$endgroup$
– N. Steinle
Jan 6 at 15:20
1
$begingroup$
Oops. You're right. I didn't notice that.
$endgroup$
– Chester Miller
Jan 6 at 18:08
|
show 5 more comments
1
$begingroup$
@S.Cho My pleasure! Which wiki article, specifically?
$endgroup$
– N. Steinle
Jan 6 at 14:30
1
$begingroup$
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
$endgroup$
– Chester Miller
Jan 6 at 14:44
1
$begingroup$
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
$endgroup$
– t t t t
Jan 6 at 14:49
1
$begingroup$
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
$endgroup$
– N. Steinle
Jan 6 at 15:20
1
$begingroup$
Oops. You're right. I didn't notice that.
$endgroup$
– Chester Miller
Jan 6 at 18:08
1
1
$begingroup$
@S.Cho My pleasure! Which wiki article, specifically?
$endgroup$
– N. Steinle
Jan 6 at 14:30
$begingroup$
@S.Cho My pleasure! Which wiki article, specifically?
$endgroup$
– N. Steinle
Jan 6 at 14:30
1
1
$begingroup$
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
$endgroup$
– Chester Miller
Jan 6 at 14:44
$begingroup$
In a heat transfer problem like this, the term involving a(x) is not a sink. It represents the convective transport of heat by fluid flow in the x direction.
$endgroup$
– Chester Miller
Jan 6 at 14:44
1
1
$begingroup$
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
$endgroup$
– t t t t
Jan 6 at 14:49
$begingroup$
@S.Cho That's very strange. Radiative losses are usually of the form $u^4$. Unless you make the approximation that u does not vary appreciably, then I guess one can use a linear approximation...
$endgroup$
– t t t t
Jan 6 at 14:49
1
1
$begingroup$
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
$endgroup$
– N. Steinle
Jan 6 at 15:20
$begingroup$
Interesting. Indeed it's not a sink. I don't think it's a convective term, since convective terms are typically proportional to $u_{x}$. I suspect it is a weak radiation term as @tttt suggests
$endgroup$
– N. Steinle
Jan 6 at 15:20
1
1
$begingroup$
Oops. You're right. I didn't notice that.
$endgroup$
– Chester Miller
Jan 6 at 18:08
$begingroup$
Oops. You're right. I didn't notice that.
$endgroup$
– Chester Miller
Jan 6 at 18:08
|
show 5 more comments
$begingroup$
The proper physical name of your equation is diffusion equation with a source term. The equation can be rearranged to continuity equation - $u_t-alpha u_{xx} = Q$. For $Q=0$ the time dependent solution can be shown to have time independent norm, which is manifestation of local mass conservation law. Source term means that particles can be created and destroyed locally, according to $Q(x)$ variation.
Continuity equation is a restatement of Gauss law - at given infinitesimal volume, the change in the number of particles in the volume is exactly equal to the number of particles crossed the surface of it in/out of this volume.
You may gain some physical intuition exploring compressible aspect of the Navier-Stockes equation. The compressability is exactly the violation of continuity equation.
Closed form solution is given here. This wierd document seems related, however didn't find any peer reviewed articles.
answered Jan 9 at 6:36
AlexanderAlexander
1,058724
$endgroup$
add a comment |
$begingroup$
The proper physical name of your equation is diffusion equation with a source term. The equation can be rearranged to continuity equation - $u_t-alpha u_{xx} = Q$. For $Q=0$ the time dependent solution can be shown to have time independent norm, which is manifestation of local mass conservation law. Source term means that particles can be created and destroyed locally, according to $Q(x)$ variation.
Continuity equation is a restatement of Gauss law - at given infinitesimal volume, the change in the number of particles in the volume is exactly equal to the number of particles crossed the surface of it in/out of this volume.
You may gain some physical intuition exploring compressible aspect of the Navier-Stockes equation. The compressability is exactly the violation of continuity equation.
Closed form solution is given here. This wierd document seems related, however didn't find any peer reviewed articles.
answered Jan 9 at 6:36
AlexanderAlexander
1,058724
$endgroup$
add a comment |
$begingroup$
The proper physical name of your equation is diffusion equation with a source term. The equation can be rearranged to continuity equation - $u_t-alpha u_{xx} = Q$. For $Q=0$ the time dependent solution can be shown to have time independent norm, which is manifestation of local mass conservation law. Source term means that particles can be created and destroyed locally, according to $Q(x)$ variation.
Continuity equation is a restatement of Gauss law - at given infinitesimal volume, the change in the number of particles in the volume is exactly equal to the number of particles crossed the surface of it in/out of this volume.
You may gain some physical intuition exploring compressible aspect of the Navier-Stockes equation. The compressability is exactly the violation of continuity equation.
Closed form solution is given here. This wierd document seems related, however didn't find any peer reviewed articles.
answered Jan 9 at 6:36
AlexanderAlexander
1,058724
$endgroup$
The proper physical name of your equation is diffusion equation with a source term. The equation can be rearranged to continuity equation - $u_t-alpha u_{xx} = Q$. For $Q=0$ the time dependent solution can be shown to have time independent norm, which is manifestation of local mass conservation law. Source term means that particles can be created and destroyed locally, according to $Q(x)$ variation.
Continuity equation is a restatement of Gauss law - at given infinitesimal volume, the change in the number of particles in the volume is exactly equal to the number of particles crossed the surface of it in/out of this volume.
You may gain some physical intuition exploring compressible aspect of the Navier-Stockes equation. The compressability is exactly the violation of continuity equation.
Closed form solution is given here. This wierd document seems related, however didn't find any peer reviewed articles.
answered Jan 9 at 6:36
AlexanderAlexander
1,058724
answered Jan 9 at 6:36
AlexanderAlexander
1,058724
answered Jan 9 at 6:36
AlexanderAlexander
1,058724
answered Jan 9 at 6:36
AlexanderAlexander
1,058724
1,058724
add a comment |
add a comment |
$begingroup$
In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
answered Jan 6 at 13:13
Vincent FraticelliVincent Fraticelli
5004
$endgroup$
add a comment |
$begingroup$
In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
answered Jan 6 at 13:13
Vincent FraticelliVincent Fraticelli
5004
$endgroup$
add a comment |
$begingroup$
In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
answered Jan 6 at 13:13
Vincent FraticelliVincent Fraticelli
5004
$endgroup$
In the study of the thermal transfer within a heat sink, we have a term of the form a * (T-Text) which correspond to the conducto convective exchanges between the heat sink and the air surrounding it: it is proportional to the temperature difference in accordance with Newton's law. The first term, in alpha, corresponds to the thermal conduction within the material. The thermal current density is proportional to the first spatial derivative of the temperature and the variation of this density leads to the second derivative (Laplacian). The alpha in the équation is the thermal diffusivity (conductivity/mu*C)
answered Jan 6 at 13:13
Vincent FraticelliVincent Fraticelli
5004
edited Jan 6 at 13:19
answered Jan 6 at 13:13
Vincent FraticelliVincent Fraticelli
5004
answered Jan 6 at 13:13
Vincent FraticelliVincent Fraticelli
5004
answered Jan 6 at 13:13
Vincent FraticelliVincent Fraticelli
5004
5004
add a comment |
add a comment |
$begingroup$
The equation describes the flow of heat in presence of sources or sinks. The first term on the right hand side is the normal diffusion term. The second term can be thought of as a source or sink term. For more details see: https://www.math.ubc.ca/~peirce/M257_316_2012_Lecture_19.pdf
answered Jan 8 at 5:47
noisyoscillatornoisyoscillator
318
$endgroup$
add a comment |
$begingroup$
The equation describes the flow of heat in presence of sources or sinks. The first term on the right hand side is the normal diffusion term. The second term can be thought of as a source or sink term. For more details see: https://www.math.ubc.ca/~peirce/M257_316_2012_Lecture_19.pdf
answered Jan 8 at 5:47
noisyoscillatornoisyoscillator
318
$endgroup$
add a comment |
$begingroup$
The equation describes the flow of heat in presence of sources or sinks. The first term on the right hand side is the normal diffusion term. The second term can be thought of as a source or sink term. For more details see: https://www.math.ubc.ca/~peirce/M257_316_2012_Lecture_19.pdf
answered Jan 8 at 5:47
noisyoscillatornoisyoscillator
318
$endgroup$
The equation describes the flow of heat in presence of sources or sinks. The first term on the right hand side is the normal diffusion term. The second term can be thought of as a source or sink term. For more details see: https://www.math.ubc.ca/~peirce/M257_316_2012_Lecture_19.pdf
answered Jan 8 at 5:47
noisyoscillatornoisyoscillator
318
answered Jan 8 at 5:47
noisyoscillatornoisyoscillator
318
answered Jan 8 at 5:47
noisyoscillatornoisyoscillator
318
answered Jan 8 at 5:47
noisyoscillatornoisyoscillator
318
318
add a comment |
add a comment |
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$begingroup$
Could you tell us where you're studying this from? Are there assumptions made about the form of the equations (i.e. small radiation losses, etc...)?
$endgroup$
– N. Steinle
Jan 6 at 15:15
$begingroup$
From a mathematical point of vue, we use this form in abstract way without precising the physical statement of the equation.
$endgroup$
– S. Cho
Jan 6 at 16:16
$begingroup$
Your title asks about a potential? What potential are you talking about?
$endgroup$
– Drew
Jan 7 at 0:34
$begingroup$
We call the coefficient $a$ a potential
$endgroup$
– S. Cho
Jan 7 at 11:09