Recent content by PhyAmateur

What does it mean that a Killing vector and a total differential of a certain theory are related to bilinears? In other words, why would bilinears (e.g. of the forms ##<\gamma_0\epsilon, \gamma_5\gamma_{\mu}\epsilon>## and ##<\gamma_0\epsilon, \gamma_{\mu}\epsilon>## tell us anything about...

$$\epsilon^{\rho\sigma\mu\nu}\epsilon_{\mu\nu\rho'\sigma'}=-2(\delta^{\rho}_{\rho'}\delta^{\sigma}_{\sigma'}-\delta^{\rho}_{\sigma'}\delta^{\sigma}_{\rho'})$$ I hope this is what you mean as I am new to those terminologies and to differential geometry in general. Please bear with me @Orodruin .

I will give it a shot though I feel I am mistaken: Maybe, this would be more like: ##2\epsilon^{\mu\nu\rho\sigma}\tilde{G}_{\rho\sigma}=G^{\mu\nu}##? Is this by any chance correct? @Orodruin

Yes, that is what I am having troubles in. To take the epsilon to the other side where the index placement is giving me a hard time.

Do you mean that ##\tilde{G}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}G_{\rho\sigma}##?

I have an equation that says $$C_1\partial_{\mu}G^{\mu\nu}+C_2\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}\partial_{\mu}G_{\rho\sigma}=0$$ If I want to get rid of the ##\epsilon^{\mu\nu\rho\sigma}## in the second term, I know I must multiply the equation by some other ##\epsilon## with different set...

Ok thank you, but about the last thing I mentioned about the complex function in my previous comment? I would appreciate if you reread my previous comment again? Does the complex function leap over the linear Hodge dual with no problem the same as if it were a real function?

Oh great, I understand now, then for a linear map ##\star## and if ##c## is a complex function then it will not affect the complex function upon leaping it over the ##\star## That is to say, is the following correct if the c is a complex function: $$\star(c\omega)=c\star\omega$$ Thank you...

What do you mean here by conjugate-linear map? Let me be more specific with my question, do you mean if "c" is a real number or real function then $$ \star(c\omega)=(c\star \omega)$$ If so, then what will happen if c were a complex function instead? Note: I am trying to using a single "$" for...

Hodge star in R^3.

If we have an equation that looks like $$H=Y$$ and we want to multiply H by either $$ReM_{IJ}$$ or $$ImM_{IJ}$$ where $$M_{IJ}$$ is a complex matrix. But the thing is that $$Y=\star(...)$$ where $$\star$$ is hodge star and (...) is set of complex functions and other numerical stuff, my question...

So yes, considering that I used this notation istead of $\nabla^2$ is my final answer correct or at least is the procedure correct that I considered all terms must be set equal to zero?

$$\nabla^2$$ is the normal laplacian of (x,y,z). t is chosen like this such that dt is a 1 form.

If we have,$$A=d[(\bar{\alpha}-\alpha)(dt+\lambda)]$$ where $$\alpha$$ is a complex function and $$\lambda$$ is a 1-form. t here represents the time coordinate. If we want to calculate $$d\star A=0$$ where $$\star$$ is hodge star, we get if I did my calculations correctly...

Are you telling me that Ward Takashi holds for the amplitude and not necessarily for each of the Feynman diagrams whose sum is the amplitude? @vanhees71