Examples of alpha in the following topics:
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- In alpha decay an atomic nucleus emits an alpha particle and transforms into an atom with smaller mass (by four) and atomic number (by two).
- Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle that consists of two protons and two neutrons, as shown in .
- Alpha decay is the most common cluster decay because of the combined extremely high binding energy and relatively small mass of the helium-4 product nucleus (the alpha particle).
- Alpha decay typically occurs in the heaviest nuclides.
- Alpha decay is one type of radioactive decay.
-
- $\begin{aligned}
\cos(\alpha + \beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \\
\cos(\alpha - \beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta
\end{aligned}$
- $\begin{aligned}
\sin(\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\
\sin(\alpha - \beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta
\end{aligned}$
- $\displaystyle{
\begin{aligned}
\tan(\alpha + \beta) &= \frac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\
\tan(\alpha - \beta) &= \frac{ \tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}
\end{aligned}
}$
- Apply the formula $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:
- We can thus apply the formula for sine of the difference of two angles: $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$.
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- When $a$ is equal to one, $\alpha_1$ and $\alpha_2$ both equal one, and $\beta_1$ and $\beta_2$ are factors of the constant $c$ such that:
- When $a$ is not equal to one and not equal to zero, you can FOIL the above expression for the factored form of the quadratic to find that
$\alpha_1$ and $\alpha_2$
are factors of $a$ such that:
- In other words, the coefficient of the $x^2$ term is given by the product of the coefficients $\alpha_1$ and $\alpha_2$, and the coefficient of the $x$ term is given by the inner and outer parts of the FOIL process.
- In some cases, it will be impossible to factor the quadratic such that
$\alpha_1$ and $\alpha_2$ are integers.
- Such that $b = \alpha_1 \beta_2 + \alpha_2 \beta_1$.
-
- Many aldehydes and ketones undergo substitution reactions at an alpha carbon, as shown in the following diagram (alpha-carbon atoms are colored blue).
- If the alpha-carbon is a chiral center, as in the second example, the products of halogenation and isotopic exchange are racemic.
- First, these substitutions are limited to carbon atoms alpha to the carbonyl group.
- Cyclohexanone (the first ketone) has two alpha-carbons and four potential substitutions (the alpha-hydrogens).
- This is demonstrated convincingly by the third ketone, which is structurally similar to the second but has no alpha-hydrogen.
-
- It is often convenient to define $\alpha_\nu = \rho \kappa_\nu$ where $\kappa_\nu$ is the opacity of the material.
- The quantity $\alpha_\nu$ has both positive and negative contributions.
-
- Deriving the double-angle formula for sine begins with the sum formula that was introduced previously: $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$.
- If we let $\alpha = \beta = \theta$, then we have:
- They are useful for finding the trigonometric function of an angle $\theta$ which is half of a special angle $\alpha$ (in other words, $\displaystyle{\theta = \frac{\alpha}{2}}$).
- $\displaystyle{
\tan{\left(\frac{\alpha}{2}\right)} = \pm \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}
}$
- We can thus apply the half-angle formula with $\alpha = 30^{\circ}$:
-
- $\displaystyle J^\mu (x^\alpha) \equiv c \int \frac{d^3 {\bf p}}{E_{\bf p}} p^\mu f(x^\alpha, {\bf p}).$
- $\displaystyle J^0(x^\alpha) = \int \frac{d^3 {\bf p}}{E_{\bf p}} c p^0 f(x^\alpha, {\bf p}) = \int d^3 {\bf p} f(x^\alpha, {\bf p}) = n(x^\alpha) \\ {\bf J}(x^\alpha) = \int \frac{d^3 {\bf p}}{E_{\bf p}} c {\bf p} f(x^\alpha, {\bf p}) = \frac{1}{c} \int d^3 {\bf p} {\bf v} f(x^\alpha, {\bf p}) = \frac{\langle {\bf v} \rangle}{c} n(x^\alpha) $
-
- $\displaystyle I_\nu (z,\mu ) = S_\nu - \frac{\mu}{\alpha_\nu +\sigma_\nu} \frac{\partial I_\nu}{\partial z}.$
- $\displaystyle I_\nu^{(1)} (z,\mu) \approx B_\nu(T) - \frac{\mu}{\alpha_\nu +\sigma_\nu} \frac{\partial B_\nu}{\partial z}.$
- $\displaystyle F_\nu(z) = \int I_\nu^{(1)} \cos\theta d \Omega = -2\pi \frac{\partial B_\nu}{\partial z} \frac{1}{\alpha_\nu +\sigma_\nu} \int_{-1}^{+1} \mu^2 d \mu \\ \displaystyle = -\frac{4\pi}{3} \frac{1}{\alpha_\nu +\sigma_\nu} \frac{\partial B_\nu}{\partial z} = -\frac{4\pi}{3} \frac{1}{\alpha_\nu +\sigma_\nu} \frac{\partial B_\nu}{\partial T} \frac{\partial T}{\partial z}$
- $\displaystyle \frac{1}{\alpha_R} \equiv \frac{\int_0^\infty \left (\alpha_\nu +\sigma_\nu\right)^{-1} \frac{\partial B_\nu}{\partial T} d\nu}{\int_0^\infty \frac{\partial B_\nu}{\partial T} d\nu} = \frac{\pi}{4\sigma T^3} \int_0^\infty \left (\alpha_\nu +\sigma_\nu\right)^{-1} \frac{\partial B_\nu}{\partial T} d\nu$
- $\displaystyle F(z) = -\frac{16 \sigma T^3}{3\alpha_R} \rho \frac{\partial T}{\partial \Sigma}.= -\frac{16 \sigma T^3}{3\kappa_R} \frac{\partial T}{\partial \Sigma}
-
- $\displaystyle \frac{dI_\nu}{ds} = -\alpha_\nu \left (I_\nu - B_\nu \right ) - \sigma_\nu \left (I_\nu - J_\nu \right ) \\ \displaystyle = -\left (\alpha_\nu + \sigma_\nu \right ) \left ( I_\nu - S_\nu \right )$
- The net absorption coefficient is $\alpha_\nu+\sigma_\nu.$ On average a photon will travel a distance
- ${\cal L}_\nu = 4\pi \alpha_\nu B_\nu V, ~~~ (\tau_* \ll 1)$
- ${\cal L}_\nu = 4\pi \alpha_\nu B_\nu A l_* = 4\pi \sqrt{\epsilon_\nu} B_\nu A, ~~~ (\tau_* \gg 1)
-
- so the value of $\alpha $ must be less than or equal to unity.
- We can combine the $\alpha$-stress with the angular momentum transport equation to give
- $\displaystyle h^3 = \frac{1}{\alpha} \frac{\dot M}{4\pi \rho \Omega} \left [ 1 - \beta \left ( \frac{r_I}{r}\right )^{1/2} \right ], \frac{h^2}{r^2} = \frac{1}{2\alpha} \frac{v_r}{r \Omega} \left [ 1 - \beta \left ( \frac{r_I}{r} \right )^{1/2} \right ].$
- The disk gets thinner as the value of $\alpha$ increases and gets fatter as the infall velocity approaches the orbital velocity.
- The thickness increases with the accretion rate and decreases rapidly with $\alpha$.