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| https://prezi.com/jgjtnwr7mgz4/life-cycle-of-a-star-white-dwarf/ |
(a) Express the kinetic energy of a particle of mass m in terms of its momentum p instead of the usual notation using its speed v.
To begin, we can consider the standard expression of kinetic energy: \[ K = \frac{1}{2} m v^2 \] To convert this to being in terms of momentum, we can use the following expression for momentum: \[ p = mv \] Solving this for v, we get: \[ v = \frac{p}{m} \] Now, we can plug this value into v for our kinetic energy expression: \[ K = \frac{p^2}{2m} \] (b) What is the relationship between the total kinetic energy of the electrons that are supplying the pressure in a white dwarf, and the total gravitational energy of the WD?
To relate the total kinetic energy with potential energy, we can apply the virial theorem: \[ K = - \frac{1}{2} U \] From part (a), we know that \[ K = \frac{p^2}{2m} \]. Since we want to know the total kinetic energy for all electrons, we also want to include the term, n, for number of electrons, giving us the following expression: \[ K_{total} = n \times \frac{p^2}{2m} \] Next, we can apply the equation for potential energy: \[ U = - \frac{GM^2}{R} \] Substituting these into the virial theorem, we get: \[ n \times \frac{p^2}{2m} = \frac{GM^2}{R} \]
(c) According to the Heisenberg uncertainty Principle, one cannot know both the momentum and position of an election such that ∆p∆x > h/4π . Use this to express the relationship between the kinetic energy of electrons and their number density ne (Hint: what is the relationship between an object’s kinetic energy and its momentum? From here, assume p « ∆p and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume ~ (∆x)^3 .)
To begin, we can derive an expression for number density. Since the problem has given that the volume occupied by an electron is \( V = { \Delta x }^3 \), we can assume the number density to be approximately : \[ n_e = \frac{ 1}{ { \Delta x }^3 } \] From part (a), we related Kinetic energy to momentum and mass. For this problem, we need to solve for momentum, so we can rearrange our expression from part (a) to do so: \[ K = \frac{p^2}{2m} \] \[ p = \sqrt{2 \times m \times K} \] Since we now have terms that involve \( \Delta p \) and \( \Delta x \), we can start plugging into the uncertainty principle expression. This expression is: \[ \Delta p \Delta x > \frac{h}{4 \pi } \] We can begin by removing constants from this expression, which are \( 4 \pi \) and h. This gives us the following expression: \[ \Delta p \Delta x > 1 \] \[ \Delta p > \frac{ 1}{ \Delta x} \] Now, from our number density expression, we can achieve the following relationship: \[ \Delta p > { n_e }^{ \frac{1}{3} } \] \[{ \Delta p}^3 > n_e \] We can then insert our expression for momentum in terms of Kinetic energy: \[ { \sqrt{ 2 \times m \times K } }^3 > n_e \] \[ ( 2 \times m \times K )^ { \frac{3}{2} } > n_e \] Again, we can remove the constants from this expression, giving us: \[ K^ { \frac{3}{2}} > n_e \] \[ K > {n_e}^{ \frac{2}{3}} \] \[ K \sim {n_e}^{ \frac{2}{3}} \]
(d) Substitute back into your Virial energy statement. What is the relationship between ne and the mass M and radius R of a WD?
As we've seen, the Virial Theorem states that: \[ K \sim \frac{1}{2} U \] Earlier in part (b), we expanded this expression to: \[ n_e \times K = \frac{GM^2}{R} \] We can now apply our expression for K from part (c): \[ n \times {n_e}^{ \frac{2}{3} } = \frac{ G M^2}{R} \] We can reduce this by canceling out constants: \[ n \times {n_e}^{ \frac{2}{3} } \sim \frac{ M^2}{R} \] To get the mass of a white dwarf, we can use the number density of protons, and multiply it by the mass of a proton and the overall volume. Additionally, since the number of protons and electrons should be equal, we can use the following expression: \[ n \sim M \] \[{n_e}^{ \frac{2}{3} } \sim \frac{ M}{R} \] \[ {n_e} \sim ({ \frac{M}{R}})^ { \frac{3}{2}} \]
(e) Now, aggressively yet carefully drop constants, and relate the mass and radius of a WD.
Given that the number density is equal to the number of electrons divided by \( R^3 \) we have the following expression: \[ n_e \sim \frac{n}{R^3} \] Additionally, from part (d), we showed that \( n \sim M \), which gives: \[ n_e \sim \frac{M}{R^3} \] Inserting this into our expression from part (d), we get: \[ \frac{M}{R^3} \sim ( \frac{M}{R} )^ {\frac{3}{2} } \] \[ M \sim \frac{1}{R^3} \]
(f) What would happen to the radius of a white dwarf if you add mass to it?
Our expression in part (e) shows that ass mass is added to a White Dwarf the radius will, in fact, decrease! This makes sense, since the density will increase, and the gravitational pull will draw the volume of the WD inwards.
Acknowledgements: Team EE
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