Recent content by AdvaitDhingra

I think Python determines types at runtime unlike other languages like C++.

Ahh ok. I think I get it now. Thanks

The ## \bar{x^1} ## grid squares are twice as big as the ## x^1 ## grid squares so the ## \bar{V^1} ## component should be half as much to compensate for that (contravarient), right? Okay. I found online that you get a transformation matrix as an answer but how do I use that to get the...

Okay, now I think I understand this. I just have one question. Above I wrote ## \bar{x}^1 = 2x^1, \bar{x}^2 = 2x^2 ## Now if I use the formula for calculating the components of the vector ## V(1, 2) ## (which uses ## x^1 ## and ## x^2 ## coordinates) in ## \bar{x}^1 ## and ## \bar{x}^2 ##...

Thank you so much this really cleared things up.

Ahh ok. Thank you.

Let's say I have coordinate system ## x^n = (x^1, x^2) ## and another ## x'^m = (2x^1, 2x^2) ## . I know this isn't the best example. Let's say I have a vector ## V=(1,2) ## Now according to the formula: ## V'^m = /frac{/partial x'^m}{/partial x^n} V^n ## Now if I'm looking for the first...

Okay, thanks

@PeroK just out of curiosity, what is this specific formula called? I'd like to learn more and I couldn't find exactly what I was looking for.

Okay, thanks a lot for your help on this. I bought a textbook on QFT and this is one of the things I need to know. I'm 16 so buying the textbook was a bit ambitious looking back but I'm still going to give it a shot. Thanks for the link btw. Have a nice day :D

@PeroK ok, so after scratching my head for a couple of hours I think I get it now. Is this correct? Suppose we have a coordinate system ## x^n = (x^1, x^2, x^3) ## and another ## \bar{x^m} = (x^1 * sin(x^2) * cos(x^3), x^1 * sin(x^2) * sin(x^3), x^1 * cos(x^3)) ## . As you can see, ##...

I'm trying to learn this using the internet, so sadly no textbook. But thanks for your explanations, I think I have a better understanding now.

Okay. Sorry, but this is all very new to me. So let's assume that I have a vector ##V## in the ## x^m ## frame of reference and it's components are: ## \left( 1, 2, 3 \right) ## With the basis vector ## \hat{i} , \hat{j} ,\hat{k} ## one can write ## V = 1 * \hat{i}, 2 * \hat{j}, 3 * \hat{k}...

Ah ok. But if I'm trying to convert a Vector from one reference frame to another, how can I know the component of the vector in the new reference frame. Isnt that what I'm calculating? And what is the difference between V^m and x^m ?

Hello, I have a question regarding the contravarient transformation of vectors. So the formula: V'n = dx'n / dxm Vm So in words, the nth basis vector in the ' frame of reference over the mth (where m is the summation term) basis vector in the original frame of reference times the mth...