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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 4 months |
| seen | May 30 at 9:19 | |
| stats | profile views | 180 |
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May 22 |
revised |
Path Integrals Page Peskin deleted 30 characters in body; edited title |
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May 9 |
comment |
Contracting the Riemann tensor issues, p540 hobson it is not but i have done it now anyway |
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May 2 |
comment |
Singularities in Schwarzchild space-time yeah thx, so would it be r=0 geometric and then r=2GM co-oridante |
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May 1 |
comment |
Singularities in Schwarzchild space-time Is the singulatiry when the coeficient goes to $\infty$ so at r=0 and at infinity? |
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May 1 |
asked | Singularities in Schwarzchild space-time |
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Apr 28 |
revised |
Discretization of action in path integral added 2 characters in body |
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Apr 28 |
comment |
Discretization of action in path integral I thinks so, does this mean that $k\Delta t=t$?, Sorry i didnt put this in but it is actually $S=T-V dt$ not $dx$ |
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Apr 28 |
comment |
Discretization of action in path integral Im sorry I don't follow from maths line 2 to 3 why has the denominator disappeared? |
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Apr 27 |
asked | Discretization of action in path integral |
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Apr 19 |
revised |
Poles bit in a propagator deleted 478 characters in body |
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Apr 19 |
revised |
Path Integrals Page Peskin deleted 610 characters in body |
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Apr 19 |
accepted | Poles bit in a propagator |
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Apr 19 |
comment |
Poles bit in a propagator Thanks I get it now |
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Apr 19 |
comment |
Poles bit in a propagator Sorry I am getting muddled with conventions firstly is $E_{p}^{2}=m^{2}+p^{2}_{i}$, also shouldn't the exponentials have a $p$ not an $E_{p}$, I don't really see why there is a $p^{2}-m^{2}$, sorry for being awkwrd |
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Apr 19 |
comment |
Poles bit in a propagator sorry I am still a bit confused, I dont see how you did the last integral. I realise that p can be split into $p^{0}$ and its spatial terms but i stil cant see how to do the last integral |
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Apr 19 |
asked | Poles bit in a propagator |
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Apr 18 |
asked | Path Integrals Page Peskin |
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Apr 11 |
asked | What does Planck/WMAP/COBE actually measure when studying the CMB? |
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Apr 8 |
awarded | Disciplined |
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Apr 8 |
comment |
Contracting the Riemann tensor issues, p540 hobson surely though the $g^{\alpha \rho}$ takes away this index? If not thewn is $g_{\sigma\alpha}g^{\alpha rho} \Gamma^{\sigma}_{\mbox{ }\mu \rho}=\Gamma^{\rho}_{\mbox{ }\mu \rho}$ |
