(ds)^2, and questions about modelling spacetime

7 views
Skip to first unread message

Alan Grayson

unread,
Oct 31, 2025, 9:35:09 AM (11 days ago) Oct 31
to Everything List
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. 

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

Alan Grayson

unread,
Oct 31, 2025, 10:13:05 AM (11 days ago) Oct 31
to Everything List
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. 

Reply all
Reply to author
Forward
0 new messages