Recent content by jag

Thank you @Orodruin . I learned a lot from this exercise; fixing my misconceptions. I don't have much formal math and physics education, but I always wanted to learn more in-depth than reading popular science books. Hence, I embarked on a journey to self-learn math and physics. Thank you again...

Sorry, I know I need to type the equations in Latex. It just got too long when I expanded the sums. Hence, I captured my workings in the image below. Unfortunately, the picture quality seems to decrease when I uploaded. You could zoom in. Please kindly let me know if I'm wrong. Thank you.

This is my fourth attempt: ##R = R^{\alpha}_{\, \alpha}## The left-hand side of the equation remains the same in the next steps. ##g^{\alpha \beta}R_{\beta \alpha} ## ##g^{\alpha \beta}R_{\alpha \beta} ## ##g^{\alpha \beta}R^\lambda_{\, \alpha \lambda \beta}## => using Ricci tensor definition...

Thank you for your feedback and the reading you gave me. It was very helpful! This is my third attempt: ##R = R^{\alpha}_{\, \alpha}## The left-hand side of the equation remains the same in the next steps. ##g^{\alpha \beta}R_{\beta \alpha} ## ##g^{\alpha \beta}R_{\alpha \beta} ## ##g^{\alpha...

Thank you for your response @Orodruin . This is my second attempt: ##R = R^{\alpha}_{\, \alpha}## The left-hand side of the equation remains the same in the next steps. ##\frac {R^{\alpha}_{\, \alpha} g_{\beta \alpha}} {g_{\beta \alpha}}## ##\frac {R_{\beta \alpha}} {g_{\beta \alpha}}##...

Hi, I'm self-learning some physics topics and came across an exercise to derive the relationship between Ricci scalar and Gauss curvature in 2-surface, ##R=2K##, where ##K \equiv \frac {R_{1212}} {g}##; given the Ricci tensor ##R_{\alpha \beta} \equiv R^\lambda_{\, \alpha \lambda \beta}## and...

Hi All, it is super clear for me now. Thank you very much for your help! As a self-learner of physics, this forum has been really helpful for me! :smile:

For question number (1), I multiplied the Lorentz transformation matrix for each axis and I get the result $$\Lambda_{\nu}^{\mu} = \begin{pmatrix} 1 & v^1 & v^2 & v^3 \\ v^1 & 1 & v^1v^2 & v^1v^3 \\ v^2 & 0 & 1 & v^2v^3 \\ v^3 & 0 & 0 & 1 \\ \end{pmatrix}$$ My assumption is ##v_iv_j = 0## and...

@anuttarasammyak Sorry, I meant to write multiplying in the post.

1. Show that the infinitesimal boost by ##v^j## along the ##x^j##-axis is given by the Lorentz transformation $$\Lambda_{\nu}^{\mu} = \begin{pmatrix} 1 & v^1 & v^2 & v^3 \\ v^1 & 1 & 0 & 0 \\ v^2 & 0 & 1 & 0 \\ v^3 & 0 & 0 & 1 \\ \end{pmatrix}$$ Attempted solution I know that for x-axis...

Hey All, thank you very much for the explanation. I took some time to digest all the information you gave me but it is super clear now. I successfully derived the Lorentz force! :smile:

@TSny Thank you for the answer. It is clear to me why the spatial derivatives doesn't act on ##\dot x##. One question, for the term (v⋅∇)⋅A, does the spatial derivative also doesn't act on ##A##?

@Steve4Physics Thank you for the reply. I had some thoughts similar to you but I am not sure whether it is correct. Another thought I was having is whether the terms are just "canceled" because of some vector identities but I'm not sure on this one too. Looking forward to some clarification from...

I know that ##∇(A⋅v)=(A⋅∇)⋅v+(v⋅∇)⋅A+v×(∇×A)+A×(∇×v)## The third term ##v×(∇×A)## simplifies to ##v×B##. I'm just now sure how to "get rid" of the other terms. I tried checking for some vector identities but couldn't make any headways. Any guidance?