Holy Golindukha (Mary) of Persia — July 12 / July 25 (Julian Calendar)

On this day, July 12 according to the Julian Calendar (July 25 civil calendar), the Georgian Orthodox Church commemorates Holy Golindukha (Mary) of Persia.

Holy Golindukha (in holy baptism, Mary) lived in Persia during the reign of Khosrow I. She was the wife of the chief magus (high priest) of the Persian kingdom. Gifted with a keen and searching mind, Golindukha felt the emptiness of the pagan Zoroastrian religion and pondered deeply the meaning of true faith.

By the grace of God, she learned about the Christian faith. Through prayer and the light of divine grace, she came to believe in Christ with her whole heart. Despite the enormous danger this posed to her life in a Zoroastrian court, she presented herself to a Christian bishop and received holy baptism, taking the name Mary.

When her conversion became known, she was seized and cast into prison. For seven years she endured imprisonment and severe torments with remarkable steadfastness and joy, sustained by heavenly visions and the comfort of the Holy Spirit.

After seven years, she was miraculously released from prison. She spent the remaining years of her life in holy ascetic labor, healing the sick through prayer and bringing many to the Christian faith.

Holy Golindukha reposed in the Lord around the year 591. The Georgian Orthodox Church especially venerates her, as her life and witness are closely connected to the Christianization of the Caucasus region. She is commemorated on July 12 according to the Julian Calendar.