My professors are so bad at explaining... They always skip the explanation... But you sir... Bravo... You make even a plebian like me understand perfectly!!
that comes from the fact that the product was defined diffrently. There are two diffrent conventions, and the one prefered by Physicists is ai* • aj= delta(ij)• 2pi, which you can check on wikipedia. On wikipedia reciprocal basis vectors are described as bi
This is purely a question of convention. The convention used by me is preferred by crystallographers where ai*aj=delta_ij. The convention often used by physicists is ai*aj*=2pi delta_ij. One can use either convention.
Dear Professor, I have a simple question: The Kroneker's delta is defined based on the cubic system? I don't get 1 from dot product of the vector a and the vector a* in parallelopiped and in any other lattice systrm not having the angle of 90 degree between the basis vectors in real space.
Kronecker's delta is just another name for the unit matrix. In the unit matrix only the diagonal terms are unity and all off-diagonal terms are zero. Similarly for kronecker's delta delta_11=delta_22=delta_33=1 and delta_12=delta_21=delta_13=delta_31=delta_23=delta_32=0. Thus ai. a*j=delta_ij is just a short way of writing nine dot products a1.a*1=1; a1.a*2=0; a1.a*3=0; a2.a*1=0; a2.a*2=1; a2.a*3=0; a3.a*1=0; a3.a*2=1; a3.a*3=1; The above relations are true for all systems and not only for cubic system. Note that we are not requiring a1.a2=0 which is true for orthogonal system (cubic, tetragonal and orthorhombic) but not for a general parallelopiped. We are requiring a1.a*2=0 where a*2 is a basis vector of reciprocal lattice and is not the same as a2. So by the relation a1.a*2=0 we are requiring that the second basis vector (a*2) of the reciprocal basis is orthogonal to the first basis vector (a1) of the real lattice. Hope this clarifies.
No comparison.... Greatest teacher ever seen on the subject... Highest regards Respected Sir 🙏🙏🙏
My professors are so bad at explaining... They always skip the explanation... But you sir... Bravo... You make even a plebian like me understand perfectly!!
I am a big fan of your videos. Thank you do much sir please uploading more videos. I wish I could have a teacher like you 😔
thank you for your great videos on the subject.
You handwriting is beautiful 😍
Thank you so much!👍
question please why in my texxtbook it says a1*= 2*pi a2xa3/V ? i mean from where that 2 pi came?
same question.
that comes from the fact that the product was defined diffrently. There are two diffrent conventions, and the one prefered by Physicists is ai* • aj= delta(ij)• 2pi, which you can check on wikipedia. On wikipedia reciprocal basis vectors are described as bi
Seriously. Wow!
Shouldn't the product of ai* and aj vectors be delta(ij) times 2pi?
This is purely a question of convention. The convention used by me is preferred by crystallographers where ai*aj=delta_ij. The convention often used by physicists is ai*aj*=2pi delta_ij. One can use either convention.
@@rajeshprasadlectures thank you a lot, now everything is clear
Thank you
Dear Professor,
I have a simple question:
The Kroneker's delta is defined based on the cubic system?
I don't get 1 from dot product of the vector a and the vector a* in parallelopiped and in any other lattice systrm not having the angle of 90 degree between the basis vectors in real space.
Kronecker's delta is just another name for the unit matrix. In the unit matrix only the diagonal terms are unity and all off-diagonal terms are zero. Similarly for kronecker's delta delta_11=delta_22=delta_33=1 and delta_12=delta_21=delta_13=delta_31=delta_23=delta_32=0. Thus ai. a*j=delta_ij is just a short way of writing nine dot products
a1.a*1=1; a1.a*2=0; a1.a*3=0;
a2.a*1=0; a2.a*2=1; a2.a*3=0;
a3.a*1=0; a3.a*2=1; a3.a*3=1;
The above relations are true for all systems and not only for cubic system. Note that we are not requiring a1.a2=0 which is true for orthogonal system (cubic, tetragonal and orthorhombic) but not for a general parallelopiped. We are requiring a1.a*2=0 where a*2 is a basis vector of reciprocal lattice and is not the same as a2. So by the relation a1.a*2=0 we are requiring that the second basis vector (a*2) of the reciprocal basis is orthogonal to the first basis vector (a1) of the real lattice.
Hope this clarifies.
Bite me ... literally can't understand anything
Sorry indeed. I fail.
Sir you are truly awakening the teachers inside us....keep up the EXCELLENT work Sir