4 ms·

Every experiment so far has detected gravitational waves a tiny bit before they detected light based evidence. Consistent with the light being slowed by the ref

Every experiment so far has detected gravitational waves a tiny bit before they detected light based evidence. Consistent with the light being slowed by the refraction in that very small amount of matter that exists in the interstellar medium and gravity passing through that dust more or less unaffected. Of course going faster than light which is being slowed by absorption and re-emission isn't the same as breaking the speed of light since light itself is going slower than the speed of light in this case. So yes you're right that it isn't exactly the same arrival time but we're not talking about curvature differences here, we're talking about physical interactions that the light undergoes that gravity doesn't.

piyh 5y ago
We need to replace faster than light with faster than gravity
- spsoto 5y ago
  I don’t have a physics background but I’ve always seen “c” as the speed of causality. The light happens to go at that speed in the absence of gravitational disturbances. Gravity and others fields should also move at this maximum speed. That said, I’m still trying to come to terms with the fact that breaking this speed limit just means that causality would be potentially broken. Isn’t that just something we axiomatically believed based on experience and we just haven’t observed otherwise?
  - Twisol 5y ago
    My (mostly layperson's) understanding is that our laws of physics demand that causality would be broken; it's not taken as an axiom. Because of how the three dimensions of space and one dimension of time are put together, you can think of there being a balance or trade between motion in space and motion in time. If you aren't moving in space, you're moving through time at the maximum possible "rate". The more rapidly you move through space, the slower you move through time. This trade bottoms out at "c", at which point you're not moving through time at all. (Since motion is impossible without time passing, "c" itself is unachievable; you can only approach it asymptotically. Something about massless particles makes "motion" not a thing in the first place, I think, meaning they can actually propagate at exactly "c" as seen by an observer.) You can visualize this as a dial on an X-Y graph which starts out pointing in the Y direction, and as you speed up, it turns more toward the X direction. When you're pointing completely in the X direction, you're moving "at the speed of light", purely in space and not at all through time. If you turn the dial even further, you're trading some of that speed back for motion in time... but in the opposite direction. Of course, this is all super-handwavey; most importantly, velocity has to be measured relative to an observer, so all of this about rates has to be anchored relative to an observer. (But this is also precisely why massless particles propagate at the same rate regardless of observer -- insert timey-wimey Doctor Who reference.) Greg Egan has a lovely trilogy, Orthogonal, set in a universe where space and time don't have this trade (formally, the sign on the time variable in some critical equation is flipped to match the spatial dimensions). He has some great material on the exact physics of such a world. [0] [0] https://www.gregegan.net/ORTHOGONAL/00/PM.html https://www.gregegan.net/ORTHOGONAL/00/PM.html
- evanb 5y ago
  Light can take any speed---when it is traveling through a medium. Capital-T 'The' speed of light is the speed of light through a vacuum, and equal to 'The' speed of gravity.
tsegratis 5y ago
May I ask since gravity obeys curvature like light, why do we see gravity from black holes, when it is the curvature that stops us seeing light? Maybe the gravity emanates from outside the event horizon, but then why would it pull us inside? Thanks
- AnotherGoodName 5y ago
  >We study static, spherically symmetric black hole solutions of the Einstein equations with a positive cosmological constant and a conformally coupled self interacting scalar field. Exact solutions for this model found by Mart´ınez, Troncoso, and Zanelli, (MTZ) >The final conclusion of our analysis is that there appear to be no physically acceptable stable solutions of the MTZ system https://arxiv.org/pdf/0710.1735.pdf https://arxiv.org/pdf/0710.1735.pdf Basically it's a huge hole in black hole theory right now. It should be made clear though that both gravity is self interacting and black holes do exist. It's just when you get down to specifics it's a case of "we don't know how to make this work".
  - tsegratis 5y ago
    Thanks for the reply I assume punching a hole in spacetime, punches an equivalent hole in maths aswell Let me ask a few simpler questions first, my main question is at the end This punched hole might be like measuring angles with a differential. When the difference between the measured points hits zero, the other end of the equation hits infinity and the angle becomes meaningless So would a true vertical curvature in spacetime equivalently require an infinite amount of mass? They say that at the event horizon the deformation is so strong that from a black hole all paths lead inwards. But isn't gravity commutative? A.k.a. coming from inside, vertical curvature is reached. But if the curvature is vertical, then presumably there is also no way into a black hole? --- So main question; could we just say that vertical curvature is impossible, and black holes are simply extreeeeme curvature to the extent that a 1.7second difference between light waves and gravitational waves over 130million years is enough to stop light escaping, but not gravity? Is that solution too simple, what am i missing? Thanks
    - 5y ago

piyh 5y ago

We need to replace faster than light with faster than gravity

spsoto 5y ago
I don’t have a physics background but I’ve always seen “c” as the speed of causality. The light happens to go at that speed in the absence of gravitational disturbances. Gravity and others fields should also move at this maximum speed. That said, I’m still trying to come to terms with the fact that breaking this speed limit just means that causality would be potentially broken. Isn’t that just something we axiomatically believed based on experience and we just haven’t observed otherwise?
- Twisol 5y ago
  My (mostly layperson's) understanding is that our laws of physics demand that causality would be broken; it's not taken as an axiom. Because of how the three dimensions of space and one dimension of time are put together, you can think of there being a balance or trade between motion in space and motion in time. If you aren't moving in space, you're moving through time at the maximum possible "rate". The more rapidly you move through space, the slower you move through time. This trade bottoms out at "c", at which point you're not moving through time at all. (Since motion is impossible without time passing, "c" itself is unachievable; you can only approach it asymptotically. Something about massless particles makes "motion" not a thing in the first place, I think, meaning they can actually propagate at exactly "c" as seen by an observer.) You can visualize this as a dial on an X-Y graph which starts out pointing in the Y direction, and as you speed up, it turns more toward the X direction. When you're pointing completely in the X direction, you're moving "at the speed of light", purely in space and not at all through time. If you turn the dial even further, you're trading some of that speed back for motion in time... but in the opposite direction. Of course, this is all super-handwavey; most importantly, velocity has to be measured relative to an observer, so all of this about rates has to be anchored relative to an observer. (But this is also precisely why massless particles propagate at the same rate regardless of observer -- insert timey-wimey Doctor Who reference.) Greg Egan has a lovely trilogy, Orthogonal, set in a universe where space and time don't have this trade (formally, the sign on the time variable in some critical equation is flipped to match the spatial dimensions). He has some great material on the exact physics of such a world. [0] [0] https://www.gregegan.net/ORTHOGONAL/00/PM.html https://www.gregegan.net/ORTHOGONAL/00/PM.html
evanb 5y ago
Light can take any speed---when it is traveling through a medium. Capital-T 'The' speed of light is the speed of light through a vacuum, and equal to 'The' speed of gravity.

tsegratis 5y ago

May I ask since gravity obeys curvature like light, why do we see gravity from black holes, when it is the curvature that stops us seeing light? Maybe the gravity emanates from outside the event horizon, but then why would it pull us inside? Thanks

AnotherGoodName 5y ago
>We study static, spherically symmetric black hole solutions of the Einstein equations with a positive cosmological constant and a conformally coupled self interacting scalar field. Exact solutions for this model found by Mart´ınez, Troncoso, and Zanelli, (MTZ) >The final conclusion of our analysis is that there appear to be no physically acceptable stable solutions of the MTZ system https://arxiv.org/pdf/0710.1735.pdf https://arxiv.org/pdf/0710.1735.pdf Basically it's a huge hole in black hole theory right now. It should be made clear though that both gravity is self interacting and black holes do exist. It's just when you get down to specifics it's a case of "we don't know how to make this work".
- tsegratis 5y ago
  Thanks for the reply I assume punching a hole in spacetime, punches an equivalent hole in maths aswell Let me ask a few simpler questions first, my main question is at the end This punched hole might be like measuring angles with a differential. When the difference between the measured points hits zero, the other end of the equation hits infinity and the angle becomes meaningless So would a true vertical curvature in spacetime equivalently require an infinite amount of mass? They say that at the event horizon the deformation is so strong that from a black hole all paths lead inwards. But isn't gravity commutative? A.k.a. coming from inside, vertical curvature is reached. But if the curvature is vertical, then presumably there is also no way into a black hole? --- So main question; could we just say that vertical curvature is impossible, and black holes are simply extreeeeme curvature to the extent that a 1.7second difference between light waves and gravitational waves over 130million years is enough to stop light escaping, but not gravity? Is that solution too simple, what am i missing? Thanks
  - 5y ago