அனைவருக்கும் வணக்கம். இந்த காணொளி இயற்பியல் மாணவர்களுக்கு உருவாக்க பட்டது. இதன் கணிதம் சற்று கடினமாக இருக்கலாம். Entanglement பற்றி அனைவருக்கும் புரியும் படியாக பொதுவான காணொளி ஏற்கனவே என்னுடைய சேனலில் பதிவில் உள்ளது. தயவு செய்து அந்த காணொளியை கானாதவார்கள் அதை காணலாம். அதில் கணிதம் எதுவும் இருக்காது. உங்களின் அன்பிற்கும் ஆதரவிர்க்கும் என்றும் நன்றிகள். நட்புடன் சாம்
Please clarify sir ....A and B nu entanglement panna Electron ah na A ah madurai layum b Chennai layum irukku...yennakku ippa A yenna spin la irukkunu theriyathu until i measure ...na ippa A ah measure panni patha upspin ..now we know b is down spin...but ithula yennakku yenna use ...ippa yennakku theriyum A upspin nu... na ippa A ah down spin ah mathuna Chennai ah irukkura b downspin la iruntha upspin ah maruma... instant ah ..apti electron spin ah namma matha mudiyuma..apti spin direction ah mathuna thana nammaku use... illana atha measure mattum panni yenna panna porom..unga video la yeppayum electron ah first time measure pannuratha mattum solluringa ..next measure pannathku aprama atha electron state ah namma mathuna enna agum nu nenga sollala ...na Google panni patha electron spin direction ah yepti change pannalamnu intha method la result ah vanthuchu 1.Applying an external magnetic field can influence the spin of electrons. 2.Spin-Orbit Coupling 3.Electron Spin Resonance (ESR) 4.Optical Pumping Ithum method pathiyum sollunga...
@@ScienceWithSam sorry sir..now i understand.. "Upon measuring the state of one of the entangled particles, the entangled state collapses, and both particles take on definite states. For example, if you measure the first particle and find it to be in the "up" state, the second particle instantaneously takes on the "down" state (or whatever state is correlated).After the measurement, the particles are no longer in an entangled superposition; they have definite, individual states." Na measure pannalum collapse agathunu ninaithen..but in this explanation -- ""if i measure the first particle and find it to be in the "up" state, the second particle instantaneously takes on the "down" state (or whatever state is correlated).After the measurement, the particles are no longer in an entangled superposition; they have definite, individual states."----- just imagine after the measurement the particles not collapse still in entangle..now i change the position of second particle down to up ..the first particle instantaneously takes on the "up" state..now the common statement that "entanglement is faster than the speed of light."...but once I measure entanglement is collapse...but why people use this context "entanglement is faster than the speed of light."
Sir I bring a new thought about energy in physics ever I wish to share my thought experiment to you that one time I experienced that ,when I run. the energy is inversly proportional to velocity by this concept I derive the equation for my doubt , let Derivation, E inversly proportional to v which is represented as (1/v) I really experience that if wave carries velocity or velocity carries wave so it is expressed in the form of( lambda × velocity ) so it make in the order of ,. E is inversely proportional to lambda times of velocity which is denoted by (1/lambda × v) Which equal this equation I use the constant k therefore, the eq is written as E = k/( lambda × v) Now I want the value of k so I use the dimension formula in this equation E = ML²/T² (which dimensionally equal), lambda = L (dimensionally equal), v = L/T (dimensionally equal) Therefore I form this eq in the order of .,, ML²/T²= K / (L×L/T) ML²/T² = K/(L²/T) ML²/T²× (L²/T) = K ML⁴/T³ = K Therefore in energy concept there is two constant h - Planck constant another one c - speed of light as a constant now analysis the dimension formula for these constants are (c² = L²/T²), and the (h is ML²/T) now product this constant we got , h×c²=(ML²/T) ×(L²/T²)=ML⁴/T³=K Therefore the product of Planck constant and c² is K k=hc² which it is read as dhanush constant therefore the equation is written as "E = hc²/(lambda × v)" Now simplify this equation we got E =hc²/(v/f)v Where as the formula of* lambda =(v/f)* Therefore E =hc²f/v²..... Therefore my equation states that , If an man runs ,.. the energy of the man is inversely proportional to his squared velocity while directly proportional to his frequency where I use mine constant which equal this equation hc² When I deeply derive my equation it tells a Scooby answer to me that is...., this equation is directly equal to Einstein 's energy E = mc² Proof , E = hc²f/v²......my thought of energy E=mc².....Einstein thought of energy E = hc²v/(lambda)v² as per the formula of * f = v/lambda* v is cancelling in both numerator and denominator we got ,. E = hc²/ (lambda )×v E =hc²/(h/mv)v as per the de Broglie equation *lambda =h /mv* E = hc²/(hv/mv) Therefore v is canceling I got E = hc²/ h /m m is reciprocal as per mathemical rule then I got E = hc²m/h now canceling h. I got Einstein's energy mc² E = mc² ....... Energy statement Really Einstein is great..... what you think about my theorical thought please comment your answer
அனைவருக்கும் வணக்கம். இந்த காணொளி இயற்பியல் மாணவர்களுக்கு உருவாக்க பட்டது. இதன் கணிதம் சற்று கடினமாக இருக்கலாம். Entanglement பற்றி அனைவருக்கும் புரியும் படியாக பொதுவான காணொளி ஏற்கனவே என்னுடைய சேனலில் பதிவில் உள்ளது. தயவு செய்து அந்த காணொளியை கானாதவார்கள் அதை காணலாம். அதில் கணிதம் எதுவும் இருக்காது.
உங்களின் அன்பிற்கும் ஆதரவிர்க்கும் என்றும் நன்றிகள்.
நட்புடன்
சாம்
Great effort bro
So nice❤❤❤❤❤❤❤❤❤❤❤❤
What a channel ❤
Thanks for the video sir
Quantum mechanics continue pannunga brother.
Bro tell about Richard Feynman interesting facts
Prof. Sam and I are like an entangled pair, I can instantly grasp complex theories before he even explains them.
Please explain entanglement process in one video. How to make two electrons as entangled pair?
Sure. I am planning
whether Entanglement applicable only for electrons or photons too?
Only for sub atomic particles
ok thanks@@ScienceWithSam
Boss neenga physics professors'ku Quantum Mechanics solli thanthuttu irukinga. yengala madhiri normal audience ku Quantum Mechanics solli thaanga...🙏🙏
Please clarify sir ....A and B nu entanglement panna Electron ah na A ah madurai layum b Chennai layum irukku...yennakku ippa A yenna spin la irukkunu theriyathu until i measure ...na ippa A ah measure panni patha upspin ..now we know b is down spin...but ithula yennakku yenna use ...ippa yennakku theriyum A upspin nu... na ippa A ah down spin ah mathuna Chennai ah irukkura b downspin la iruntha upspin ah maruma... instant ah ..apti electron spin ah namma matha mudiyuma..apti spin direction ah mathuna thana nammaku use... illana atha measure mattum panni yenna panna porom..unga video la yeppayum electron ah first time measure pannuratha mattum solluringa ..next measure pannathku aprama atha electron state ah namma mathuna enna agum nu nenga sollala ...na Google panni patha electron spin direction ah yepti change pannalamnu intha method la result ah vanthuchu
1.Applying an external magnetic field can influence the spin of electrons.
2.Spin-Orbit Coupling
3.Electron Spin Resonance (ESR)
4.Optical Pumping
Ithum method pathiyum sollunga...
Watch my video on Quantum Cryptography to know the application
@@ScienceWithSam sorry sir..now i understand..
"Upon measuring the state of one of the entangled particles, the entangled state collapses, and both particles take on definite states. For example, if you measure the first particle and find it to be in the "up" state, the second particle instantaneously takes on the "down" state (or whatever state is correlated).After the measurement, the particles are no longer in an entangled superposition; they have definite, individual states."
Na measure pannalum collapse agathunu ninaithen..but in this explanation -- ""if i measure the first particle and find it to be in the "up" state, the second particle instantaneously takes on the "down" state (or whatever state is correlated).After the measurement, the particles are no longer in an entangled superposition; they have definite, individual states."----- just imagine after the measurement the particles not collapse still in entangle..now i change the position of second particle down to up ..the first particle instantaneously takes on the "up" state..now the common statement that "entanglement is faster than the speed of light."...but once I measure entanglement is collapse...but why people use this context "entanglement is faster than the speed of light."
Sir I bring a new thought about energy in physics ever I wish to share my thought experiment to you that one time I experienced that ,when I run. the energy is inversly proportional to velocity by this concept I derive the equation for my doubt , let
Derivation,
E inversly proportional to v which is represented as (1/v)
I really experience that if wave carries velocity or velocity carries wave so it is expressed in the form of( lambda × velocity ) so it make in the order of ,.
E is inversely proportional to lambda times of velocity which is denoted by (1/lambda × v)
Which equal this equation I use the constant k therefore, the eq is written as
E = k/( lambda × v)
Now I want the value of k so I use the dimension formula in this equation
E = ML²/T² (which dimensionally equal), lambda = L (dimensionally equal), v = L/T (dimensionally equal)
Therefore I form this eq in the order of .,,
ML²/T²= K / (L×L/T)
ML²/T² = K/(L²/T)
ML²/T²× (L²/T) = K
ML⁴/T³ = K
Therefore in energy concept there is two constant h - Planck constant another one c - speed of light as a constant now analysis the dimension formula for these constants are (c² = L²/T²),
and the (h is ML²/T) now product this constant we got ,
h×c²=(ML²/T) ×(L²/T²)=ML⁴/T³=K
Therefore the product of Planck constant and c² is K
k=hc² which it is read as dhanush constant therefore the equation is written as
"E = hc²/(lambda × v)"
Now simplify this equation we got
E =hc²/(v/f)v
Where as the formula of* lambda =(v/f)*
Therefore E =hc²f/v².....
Therefore my equation states that , If an man runs ,.. the energy of the man is inversely proportional to his squared velocity while directly proportional to his frequency where I
use mine constant which equal this equation hc²
When I deeply derive my equation it tells a Scooby answer to me that is...., this equation is directly equal to
Einstein 's energy E = mc²
Proof ,
E = hc²f/v²......my thought of energy
E=mc².....Einstein thought of energy
E = hc²v/(lambda)v²
as per the formula of * f = v/lambda*
v is cancelling in both numerator and denominator we got ,.
E = hc²/ (lambda )×v
E =hc²/(h/mv)v
as per the de Broglie equation
*lambda =h /mv*
E = hc²/(hv/mv)
Therefore v is canceling I got
E = hc²/ h /m
m is reciprocal as per mathemical rule then I got
E = hc²m/h
now canceling h. I got Einstein's energy mc²
E = mc² ....... Energy statement
Really Einstein is great..... what you think about my theorical thought please comment your answer
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Super sir 👌