勤行









曖昧さ回避
 この項目では、仏教用語のお勤めについて説明しています。「勤め」の美化語については「おつとめ」をご覧ください。


勤行（ごんぎょう）とは勤め励むこと。仏教用語としては、仏教における実践徳目である波羅蜜のひとつ精進波羅蜜（しょうじんはらみつ。この場合は仏道修行に勤め励むこと）と同一視され、寺院や自宅の仏壇の前で時を定めて行う読誦・礼拝などの儀式をいう。お勤め（おつとめ）ともいう。


勤行には、平時の日課として行われる日常勤行のほか、彼岸会、盂蘭盆会などの年中行事に行われるもの、故人の追善の法要として行われるものがある。


その意義については、仏教徒である事を自ら証するため、経典に説かれる善業（ぜんごう）のひとつとしての実践などと説かれる。また追善法要として行うものは、善業を行った功徳を、故人の冥福を祈って回向するものである。また仏恩・師恩に対する報恩報謝、仏徳讃嘆の為に行うという意義も説かれる。



関連項目


勤行の作法は宗旨・宗派によって違う。また、地方によっても違う場合もある。詳しくは下記関連項目を参照されたい。



  • 勤行 (天台宗)

  • 勤行 (真言宗)

  • 智山勤行式

  • 勤行 (浄土宗)

  • 勤行 (浄土真宗本願寺派)

  • 勤行 (真宗大谷派)

  • 勤行 (臨済宗)

  • 勤行 (曹洞宗)

  • 勤行 (日蓮宗)

  • 勤行 (日蓮正宗)





















































Popular posts from this blog

Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

Image
28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
Read more

Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

Image
up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
Read more

A Topological Invariant for $pi_3(U(n))$

Image
3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
Read more