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Do electron spin like a ball?

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Do electron spin like a ball? you have heard that electron have spin and magnitude of that spin is

$\dfrac{1}{2}$ . well it's actually

$\dfrac{1}{2}\hbar$ we just take

$\hbar = 1$ and we all know that

$\hbar$ is reduced plank constant. but is electron really spinning like ball and having angular momentum of

$\dfrac{1}{2}\hbar$ ? hmmm... what is velocity of the electron then? let's try to calculate that , let's assume electron is solid sphere and it's radius is

$r_e = 2.81×10^{-15}$ meters. (check this https://en.m.wikipedia.org/wiki/Classical_electron_radius) from high-school physics you can calculate moment of inertia of the solid sphere. and it's turn out to be

$I = \dfrac{2}{5}mr^2$ . and angular velocity

$\omega = \dfrac{v}{r}$ so we have all the equipment. now we know that angular momentum

$L = I\omega$ also

$L = \dfrac{1}{2}\hbar$ so we have

$I\omega = \dfrac{1}{2}\hbar$ putting Value of

$I$ and

$\omega$ $\dfrac{2}{5}mr^{2}\dfrac{v}{r} = \dfrac{1}{2}\hba...

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Do electron spin like a ball?

Do electron spin like a ball? you have heard that electron have spin and magnitude of that spin is

$\dfrac{1}{2}$ . well it's actually

$\dfrac{1}{2}\hbar$ we just take

$\hbar = 1$ and we all know that

$\hbar$ is reduced plank constant. but is electron really spinning like ball and having angular momentum of

$\dfrac{1}{2}\hbar$ ? hmmm... what is velocity of the electron then? let's try to calculate that , let's assume electron is solid sphere and it's radius is

$I = \dfrac{2}{5}mr^2$ . and angular velocity

$\omega = \dfrac{v}{r}$ so we have all the equipment. now we know that angular momentum

$L = I\omega$ also

$L = \dfrac{1}{2}\hbar$ so we have

$I\omega = \dfrac{1}{2}\hbar$ putting Value of

$I$ and

$\omega$ $\dfrac{2}{5}mr^{2}\dfrac{v}{r} = \dfrac{1}{2}\hba...

Why does time run slower near a black hole? Time dilation General relativity is Einstein's Theory of gravity, in General relativity the proper time of the observer for Schwarzschild black hole is given by

$d\tau ^2 = ( 1 - \frac{2GM}{r})dt^2 + ( 1 - \frac{2GM}{r})^{-1} dr^2 + r^2 d\Omega ^2$ Where

$d \Omega ^2 = d \theta ^2 + \sin^{2}\theta d\phi^2$ Here

$d\tau$ is proper time of an observer that means it's time what observer's clock reads Now consider that our observer is not having any motion in space that is

$dr = d\phi = d\theta = 0$ so our equation now becomes

$d\tau^2 = ( 1 - \frac{2GM}{r})dt^2$

$d\tau = (1 - \frac{2GM}{r})dt ...(1)$ But what does dt means in eq(1)? if your observer is far away from the gravitating object then we can take

$r \approx \infty$ then

$d\tau = dt$ , so

$dt$ is time experienced by observer who is really far away from the object or dt is time experienced by observer who is in flat spacetime. if observer is ou...

What is commutator of position and momentum operator?

What is commutator of

$\hat{x}$ and

$\hat{p}$ ? commutator of two operator

$\hat{A}$ and

$\hat{B}$ is defined as

$[ \hat{A}, \hat{B} ] \equiv \hat{A}\hat{B} - \hat{B}\hat{A}$ In quantum mechanics to even find the minimum uncertainty between two operator we need the commutator of two , so it's important to find such quantity. here our position and momentum are also operators and we need to find the

$[ \hat{x}, \hat{p} ]$ . so how will we find it?? as the name suggests they are operator and they operate on the functions (function of the Hilbert space). for position operator we just multiply function by

$x$ and for momentum operator we differentiate the function by

$x$ and then multiple by

$-i\hbar$ . ( Note we are working only in 1 dimentions) generally working with operators we use test function

$\phi(x)$ . this

$\phi(x)$ is arbitrary function. so let's try to find

$[ \hat{x}, \hat{p} ]\phi(x)$ by using defination of commutator we can write ...

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Do electron spin like a ball?

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Do electron spin like a ball?

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