Ask Science
Ask a science question, get a science answer.
Community Rules
Rule 1: Be respectful and inclusive.
Treat others with respect, and maintain a positive atmosphere.
Rule 2: No harassment, hate speech, bigotry, or trolling.
Avoid any form of harassment, hate speech, bigotry, or offensive behavior.
Rule 3: Engage in constructive discussions.
Contribute to meaningful and constructive discussions that enhance scientific understanding.
Rule 4: No AI-generated answers.
Strictly prohibit the use of AI-generated answers. Providing answers generated by AI systems is not allowed and may result in a ban.
Rule 5: Follow guidelines and moderators' instructions.
Adhere to community guidelines and comply with instructions given by moderators.
Rule 6: Use appropriate language and tone.
Communicate using suitable language and maintain a professional and respectful tone.
Rule 7: Report violations.
Report any violations of the community rules to the moderators for appropriate action.
Rule 8: Foster a continuous learning environment.
Encourage a continuous learning environment where members can share knowledge and engage in scientific discussions.
Rule 9: Source required for answers.
Provide credible sources for answers. Failure to include a source may result in the removal of the answer to ensure information reliability.
By adhering to these rules, we create a welcoming and informative environment where science-related questions receive accurate and credible answers. Thank you for your cooperation in making the Ask Science community a valuable resource for scientific knowledge.
We retain the discretion to modify the rules as we deem necessary.
view the rest of the comments
Really? What is it hypothesized that they decay into?
I looked it up, after 6.6 x 10e28 years or so they are theorised to decay into neutrinos and photons.
Huh, interesting. So would charge not be conserved in that process? Neither neutrinos nor photons are charged.
Charge conservation would indeed be violated, which is why this decay is not expected. Dave is mistaken: the half-life they're referring to is an experimental lower-bound, not a actual expected value.
Thanks, that makes more sense.