On Friday, October 31, 2025 at 7:35:09 AM UTC-6 Alan Grayson wrote:
*Re 2): Note that for a body at rest, coordinate and proper time are identical. Hence, d(tau)/dt = 1, where t is coordinate time and tau is proper time. But this is not true for a body not at rest. How does a physical clock "know" it is moving, making that derivative non-zero. AG * On Friday, October 31, 2025 at 2:40:07 AM UTC-6 Alan Grayson wrote: 1) For a body at rest, we multiply clock time, aka proper time, and/or coordinate time by some velocity, so its units become spatial. But why multiply by c? Is this procedure really a *definition* to get a velocity of c in spacetime? 2) Proper time and coordinate time are not equal along some arbitrary path in spacetime. How does a clock "know" it isn't reading coordinate time, but something else called proper time? Alternatively, what principle can we apply to put proper time on a logically necessary footing? 3) When moving along some arbitrary path in spacetime, the Pythagorean theorem holds; that is, (ds)^2 = (ct)^2 + (dx)^2. So how do we get a negative sign preceding the spatial differentials? Here I'm referring to a YouTube video whose link I will post later. *Link to video; https://www.youtube.com/watch?v=TcOLyqfA5k8&t=120s* 4) If (ds)^2 is an *invariant *under SR, does this hold only for the LT, but is it true for any linear transformation, as well as non-linear transformations? AG -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion visit https://groups.google.com/d/msgid/everything-list/ab69c960-6c45-4553-b7f7-d928de91989an%40googlegroups.com.
