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Vincent Thacker
Vincent Thacker
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It is correct to suspect that something more fundamental is occurring when the integral is written with the quantity $\partial q$$\mathrm{d}q$. ButBut consider the electric field created by a charge. HowHow can one represent the $\vec D$ fields created by a particular charge in a defined shell volume $4*pi*r^2\partial r$$4\pi r^2\mathrm{d}r$? ThatThat comes down to Maxwell's Equation $\vec \nabla \bullet \vec D=\rho$$\vec \nabla \cdot \vec D=\rho$ so that $$\large\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4*\pi*r^2\partial r \right)=q=\int_\text{Volume}\vec D \bullet \hat{n} \left(\partial \text{ Surface}\right)$$. $$\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4\pi r^2\mathrm{d}r \right) \\ =q=\int_\text{Volume}\vec D \cdot \hat{n} \left(\partial \text{ Surface}\right).$$

So where exactly is this charge located to create a radially-symmetric $\vec D$ field pointing outwards, and why is a charge density $\rho$ used in Maxwell's equation instead of only the charge $e_c$ for an electron or $e_p$ for a proton? FromFrom this question, one can try to understand $\vec \nabla \bullet \vec D=\rho$$\vec \nabla \cdot \vec D=\rho$ more critically. SometimesSometimes there is such a great number of individual charges that contribute to the field, that so-to-speak, the pennies count up to dollars, and the change does not appreciably affect the result because the answer is close enough using the charge density approximation. AtAt other times, the $\vec D$ field itself outside a certain radius may not change instantaneously with motion of an electron or proton within the radius. SoSo then for accounting-on-the-radial-shell $\vec r$, the quantity $\rho$ can be used to make a calculation as an equivalent charge-density on the shell, and not as an individual electron or proton quantity (since the proton or electron motion within the shell may not have enough time to alter the expected charge density $\vec D$ on the outside of the shell).

Applying Quantum Mechanicsquantum mechanics applies a probability density to finding the electron in a particular volume, and therefore also a most probable $\vec D$ field until the electron wave-function is actually measured at a point. So Quantum MechanicsSo quantum mechanics gives up on the point-view of the electron altogether when making probability calculations for important values such as the ionization energy or most probable location of finding the electron and the probability that it would be found there. InIn that sense, then the electron wave function is spread out across the volume of the universe, and only has an expected value of being measured within any restricted volume ($\partial\text{Volume}$) so it makes sense the quantity $\partial q=\int_\text{Volume} \rho \left(\partial \text{Volume}\right)$, where $\rho$ is taken from Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$$\vec \nabla \cdot \vec D =\rho$.

Appendix: Maxwell Equations Reference as Quoted

From Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$$\vec \nabla \cdot \vec D =\rho$

Maxwell's Equations Reference2

It is correct to suspect that something more fundamental is occurring when the integral is written with the quantity $\partial q$. But consider the electric field created by a charge. How can one represent the $\vec D$ fields created by a particular charge in a defined shell volume $4*pi*r^2\partial r$? That comes down to Maxwell's Equation $\vec \nabla \bullet \vec D=\rho$ so that $$\large\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4*\pi*r^2\partial r \right)=q=\int_\text{Volume}\vec D \bullet \hat{n} \left(\partial \text{ Surface}\right)$$.

So where exactly is this charge located to create a radially-symmetric $\vec D$ field pointing outwards, and why is a charge density $\rho$ used in Maxwell's equation instead of only the charge $e_c$ for an electron or $e_p$ for a proton? From this question, one can try to understand $\vec \nabla \bullet \vec D=\rho$ more critically. Sometimes there is such a great number of individual charges that contribute to the field, that so-to-speak, the pennies count up to dollars, and the change does not appreciably affect the result because the answer is close enough using the charge density approximation. At other times, the $\vec D$ field itself outside a certain radius may not change instantaneously with motion of an electron or proton within the radius. So then for accounting-on-the-radial-shell $\vec r$, the quantity $\rho$ can be used to make a calculation as an equivalent charge-density on the shell, and not as an individual electron or proton quantity (since the proton or electron motion within the shell may not have enough time to alter the expected charge density $\vec D$ on the outside of the shell).

Applying Quantum Mechanics applies a probability density to finding the electron in a particular volume, and therefore also a most probable $\vec D$ field until the electron wave-function is actually measured at a point. So Quantum Mechanics gives up on the point-view of the electron altogether when making probability calculations for important values such as the ionization energy or most probable location of finding the electron and the probability that it would be found there. In that sense, then the electron wave function is spread out across the volume of the universe, and only has an expected value of being measured within any restricted volume ($\partial\text{Volume}$) so it makes sense the quantity $\partial q=\int_\text{Volume} \rho \left(\partial \text{Volume}\right)$, where $\rho$ is taken from Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$.

Appendix: Maxwell Equations Reference as Quoted

From Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$

Maxwell's Equations Reference

It is correct to suspect that something more fundamental is occurring when the integral is written with the quantity $\mathrm{d}q$. But consider the electric field created by a charge. How can one represent the $\vec D$ fields created by a particular charge in a defined shell volume $4\pi r^2\mathrm{d}r$? That comes down to Maxwell's Equation $\vec \nabla \cdot \vec D=\rho$ so that $$\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4\pi r^2\mathrm{d}r \right) \\ =q=\int_\text{Volume}\vec D \cdot \hat{n} \left(\partial \text{ Surface}\right).$$

So where exactly is this charge located to create a radially-symmetric $\vec D$ field pointing outwards, and why is a charge density $\rho$ used in Maxwell's equation instead of only the charge $e_c$ for an electron or $e_p$ for a proton? From this question, one can try to understand $\vec \nabla \cdot \vec D=\rho$ more critically. Sometimes there is such a great number of individual charges that contribute to the field, that so-to-speak, the pennies count up to dollars, and the change does not appreciably affect the result because the answer is close enough using the charge density approximation. At other times, the $\vec D$ field itself outside a certain radius may not change instantaneously with motion of an electron or proton within the radius. So then for accounting-on-the-radial-shell $\vec r$, the quantity $\rho$ can be used to make a calculation as an equivalent charge-density on the shell, and not as an individual electron or proton quantity (since the proton or electron motion within the shell may not have enough time to alter the expected charge density $\vec D$ on the outside of the shell).

Applying quantum mechanics applies a probability density to finding the electron in a particular volume, and therefore also a most probable $\vec D$ field until the electron wave-function is actually measured at a point. So quantum mechanics gives up on the point-view of the electron altogether when making probability calculations for important values such as the ionization energy or most probable location of finding the electron and the probability that it would be found there. In that sense, then the electron wave function is spread out across the volume of the universe, and only has an expected value of being measured within any restricted volume ($\partial\text{Volume}$) so it makes sense the quantity $\partial q=\int_\text{Volume} \rho \left(\partial \text{Volume}\right)$, where $\rho$ is taken from Maxwell's Electromagnetic Equation $\vec \nabla \cdot \vec D =\rho$.

Appendix: Maxwell Equations Reference as Quoted

From Maxwell's Electromagnetic Equation $\vec \nabla \cdot \vec D =\rho$

2

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edit approved Jun 13 at 16:23
Sahil Muhammed
edit approved Jun 13 at 16:23
Sahil Muhammed
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It is correct to suspect that something more fundamental is occurring when the integral is written with the quantity $\partial q$. But consider the electric field created by a charge. How can one represent the $\vec D$ fields created by a particular charge in a defined shell volume $4*pi*r^2\partial r$? That comes down to Maxwell's Equation $\vec \nabla \bullet \vec D=\rho$ so that $\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4*\pi*r^2\partial r \right)=q=\\=\int_\text{Volume}\vec D \bullet \hat{n} \left(\partial \text{ Surface}\right)$$$\large\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4*\pi*r^2\partial r \right)=q=\int_\text{Volume}\vec D \bullet \hat{n} \left(\partial \text{ Surface}\right)$$.

So where exactly is this charge located to create a radially-symmetric $\vec D$ field pointing outwards, and why is a charge density $\rho$ used in Maxwell's equation instead of only the charge $e_c$ for an electron or $e_p$ for a proton? From this question, one can try to understand $\vec \nabla \bullet \vec D=\rho$ more critically. Sometimes there is such a great number of individual charges that contribute to the field, that so-to-speak, the pennies count up to dollars, and the change does not appreciably affect the result because the answer is close enough using the charge density approximation. At other times, the $\vec D$ field itself outside a certain radius may not change instantaneously with motion of an electron or proton within the radius. So then for accounting-on-the-radial-shell $\vec r$, the quantity $\rho$ can be used to make a calculation as an equivalent charge-density on the shell, and not as an individual electron or proton quantity (since the proton or electron motion within the shell may not have enough time to alter the expected charge density $\vec D$ on the outside of the shell).

Applying Quantum Mechanics applies a probability density to finding the electron in a particular volume, and therefore also a most probable $\vec D$ field until the electron wave-function is actually measured at a point. So Quantum Mechanics gives up on the point-view of the electron altogether when making probability calculations for important values such as the ionization energy or most probable location of finding the electron and the probability that it would be found there. In that sense, then the electron wave function is spread out across the volume of the universe, and only has an expected value of being measured within any restricted volume $\partial\text{ Volume}$($\partial\text{Volume}$) so it makes sense the quantity $\partial q=\int_\text{Volume} \rho \left(\partial \text{Volume}\right)$, where $\rho$ is taken from Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$.

Appendix: Maxwell Equations Reference as Quoted

From Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$

Maxwell's Equations Reference

It is correct to suspect that something more fundamental is occurring when the integral is written with the quantity $\partial q$. But consider the electric field created by a charge. How can one represent the $\vec D$ fields created by a particular charge in a defined shell volume $4*pi*r^2\partial r$? That comes down to Maxwell's Equation $\vec \nabla \bullet \vec D=\rho$ so that $\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4*\pi*r^2\partial r \right)=q=\\=\int_\text{Volume}\vec D \bullet \hat{n} \left(\partial \text{ Surface}\right)$.

So where exactly is this charge located to create a radially-symmetric $\vec D$ field pointing outwards, and why is a charge density $\rho$ used in Maxwell's equation instead of only the charge $e_c$ for an electron or $e_p$ for a proton? From this question, one can try to understand $\vec \nabla \bullet \vec D=\rho$ more critically. Sometimes there is such a great number of individual charges that contribute to the field, that so-to-speak, the pennies count up to dollars, and the change does not appreciably affect the result because the answer is close enough using the charge density approximation. At other times, the $\vec D$ field itself outside a certain radius may not change instantaneously with motion of an electron or proton within the radius. So then for accounting-on-the-radial-shell $\vec r$, the quantity $\rho$ can be used to make a calculation as an equivalent charge-density on the shell, and not as an individual electron or proton quantity (since the proton or electron motion within the shell may not have enough time to alter the expected charge density $\vec D$ on the outside of the shell).

Applying Quantum Mechanics applies a probability density to finding the electron in a particular volume, and therefore also a most probable $\vec D$ field until the electron wave-function is actually measured at a point. So Quantum Mechanics gives up on the point-view of the electron altogether when making probability calculations for important values such as the ionization energy or most probable location of finding the electron and the probability that it would be found there. In that sense, then the electron wave function is spread out across the volume of the universe, and only has an expected value of being measured within any restricted volume $\partial\text{ Volume}$ so it makes sense the quantity $\partial q=\int_\text{Volume} \rho \left(\partial \text{Volume}\right)$, where $\rho$ is taken from Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$.

Appendix: Maxwell Equations Reference as Quoted

From Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$

Maxwell's Equations Reference

It is correct to suspect that something more fundamental is occurring when the integral is written with the quantity $\partial q$. But consider the electric field created by a charge. How can one represent the $\vec D$ fields created by a particular charge in a defined shell volume $4*pi*r^2\partial r$? That comes down to Maxwell's Equation $\vec \nabla \bullet \vec D=\rho$ so that $$\large\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4*\pi*r^2\partial r \right)=q=\int_\text{Volume}\vec D \bullet \hat{n} \left(\partial \text{ Surface}\right)$$.

So where exactly is this charge located to create a radially-symmetric $\vec D$ field pointing outwards, and why is a charge density $\rho$ used in Maxwell's equation instead of only the charge $e_c$ for an electron or $e_p$ for a proton? From this question, one can try to understand $\vec \nabla \bullet \vec D=\rho$ more critically. Sometimes there is such a great number of individual charges that contribute to the field, that so-to-speak, the pennies count up to dollars, and the change does not appreciably affect the result because the answer is close enough using the charge density approximation. At other times, the $\vec D$ field itself outside a certain radius may not change instantaneously with motion of an electron or proton within the radius. So then for accounting-on-the-radial-shell $\vec r$, the quantity $\rho$ can be used to make a calculation as an equivalent charge-density on the shell, and not as an individual electron or proton quantity (since the proton or electron motion within the shell may not have enough time to alter the expected charge density $\vec D$ on the outside of the shell).

Applying Quantum Mechanics applies a probability density to finding the electron in a particular volume, and therefore also a most probable $\vec D$ field until the electron wave-function is actually measured at a point. So Quantum Mechanics gives up on the point-view of the electron altogether when making probability calculations for important values such as the ionization energy or most probable location of finding the electron and the probability that it would be found there. In that sense, then the electron wave function is spread out across the volume of the universe, and only has an expected value of being measured within any restricted volume ($\partial\text{Volume}$) so it makes sense the quantity $\partial q=\int_\text{Volume} \rho \left(\partial \text{Volume}\right)$, where $\rho$ is taken from Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$.

Appendix: Maxwell Equations Reference as Quoted

From Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$

Maxwell's Equations Reference

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Stephen Elliott
Stephen Elliott
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It is correct to suspect that something more fundamental is occurring when the integral is written with the quantity $\partial q$. But consider the electric field created by a charge. How can one represent the $\vec D$ fields created by a particular charge in a defined shell volume $4*pi*r^2\partial r$? That comes down to Maxwell's Equation $\vec \nabla \bullet \vec D=\rho$ so that $\int_\text{Inside Sphere Volume} \rho \left(\partial \text{Volume}\right)=\int_\text{Inside Sphere Volume} \rho \left(4*\pi*r^2\partial r \right)=q=\\=\int_\text{Volume}\vec D \bullet \hat{n} \left(\partial \text{ Surface}\right)$.

So where exactly is this charge located to create a radially-symmetric $\vec D$ field pointing outwards, and why is a charge density $\rho$ used in Maxwell's equation instead of only the charge $e_c$ for an electron or $e_p$ for a proton? From this question, one can try to understand $\vec \nabla \bullet \vec D=\rho$ more critically. Sometimes there is such a great number of individual charges that contribute to the field, that so-to-speak, the pennies count up to dollars, and the change does not appreciably affect the result because the answer is close enough using the charge density approximation. At other times, the $\vec D$ field itself outside a certain radius may not change instantaneously with motion of an electron or proton within the radius. So then for accounting-on-the-radial-shell $\vec r$, the quantity $\rho$ can be used to make a calculation as an equivalent charge-density on the shell, and not as an individual electron or proton quantity (since the proton or electron motion within the shell may not have enough time to alter the expected charge density $\vec D$ on the outside of the shell).

Applying Quantum Mechanics applies a probability density to finding the electron in a particular volume, and therefore also a most probable $\vec D$ field until the electron wave-function is actually measured at a point. So Quantum Mechanics gives up on the point-view of the electron altogether when making probability calculations for important values such as the ionization energy or most probable location of finding the electron and the probability that it would be found there. In that sense, then the electron wave function is spread out across the volume of the universe, and only has an expected value of being measured within any restricted volume $\partial\text{ Volume}$ so it makes sense the quantity $\partial q=\int_\text{Volume} \rho \left(\partial \text{Volume}\right)$, where $\rho$ is taken from Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$.

Appendix: Maxwell Equations Reference as Quoted

From Maxwell's Electromagnetic Equation $\vec \nabla \bullet \vec D =\rho$

Maxwell's Equations Reference