Re: (ds)^2, some questions about modelling spacetime

[Brent Meeker]

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

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

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[Alan Grayson]

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end. he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. With c multiplied by coordinate time along time axis for a particle spatially at rest, isn't this tantamount to a definition with the intended result that spacetime velocity is c? You refer to time dilation, but this definition seems unrelated to that concept. The key question is how a physical clock measures something other than coordinate time when moving along some arbitrary path? AG 

[Alan Grayson]

On Friday, October 31, 2025 at 11:36:14 PM UTC-6 Alan Grayson wrote:

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end. he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. With c multiplied by coordinate time along time axis for a particle spatially at rest, isn't this tantamount to a definition with the intended result that spacetime velocity is c? You refer to time dilation, but this definition seems unrelated to that concept. The key question is how a physical clock measures something other than coordinate time when moving along some arbitrary path? AG 

Since c is multiplied by coordinate time, the distance between coordinate time units is the distance light travels in a vaccum during one second. From this, we get that bodies at rest in spacetime are traveling at light speed, even though relativity assumes material bodies cannot travel at light speed. If this makes sense, I'd like to know how. TY, AG

[Brent Meeker]

On 10/31/2025 10:36 PM, Alan Grayson wrote:

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end. he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. 

I don't understand that.  If when you're stationary in some coordinate system you can adjust you clock to sync with the stationary clocks and so long as you remain stationary you would have d(tau)/dt=1.  But once you started to move relative that coord system one-second-per-second in your frame is not one-second-per-second in the stationary frame.  Proper time is one-second-per-second along your path thru spacetime.  In general it will be shorter as measured from the stationary frame, but remember inertial motion is strictly relative so you will see the stationary clocks as running slow.

With c multiplied by coordinate time along time axis for a particle spatially at rest

That would define a forward light cone for that particle.  Why would you do that?

, isn't this tantamount to a definition with the intended result that spacetime velocity is c? 

Along a path at velocity c the proper time is zero.

You refer to time dilation, but this definition seems unrelated to that concept. The key question is how a physical clock measures something other than coordinate time when moving along some arbitrary path? 

Every  clock measures only its proper time.  Coordinate time in general is just labels.   But for convenience it is usually chosen to be the the time kept by some clock(s) that are taken to define a stationary frame (so they're stationary relative to one another).  So any clock moving relative to that frame will measure proper time different from that frame.

Brent

AG 

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

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[Alan Grayson]

On Saturday, November 1, 2025 at 5:15:36 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 10:36 PM, Alan Grayson wrote:

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end, he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. 

 

I don't understand that. 

It's pretty straightforward. If you're at rest in some frame in spacetime, you're moving along the time axis only. Along that axis are coordinate labels, but since you've multiplied these lables by c, you're left with distances (as on spatial axis), and the distance separation of two adjacent coordinate unit times, has a distance which light traverses in one second of proper time. IOW, along the time axis, proper and coordinate time are identical. Thus, d(tau)/dt=1. When motion is not strictly along time axis, that is, when you're not at rest, coordinate and proper time no longer coincide, no longer have equal values. The non trivial existential question is why a clock which measures only proper time, "knows" to adjust its rate when moving along some arbitrary path in spacetime?  AG

 

If when you'r.e stationary in some coordinate system you can adjust you clock to sync with the stationary clocks and so long as you remain stationary you would have d(tau)/dt=1.  But once you started to move relative that coord system one-second-per-second in your frame is not one-second-per-second in the stationary frame.  Proper time is one-second-per-second along your path thru spacetime.  In general it will be shorter as measured from the stationary frame, but remember inertial motion is strictly relative so you will see the stationary clocks as running slow.

With c multiplied by coordinate time along time axis for a particle spatially at rest

That would define a forward light cone for that particle.  Why would you do that?

, isn't this tantamount to a definition with the intended result that spacetime velocity is c? 

Along a path at velocity c the proper time is zero.

No. When at rest, proper time increments, and is indistinguishable from coordinate time. If this weren't the case, it would be impossible to make the unintelligible conclusion that all motion in spacetime is c. AG 

[Alan Grayson]

On Saturday, November 1, 2025 at 9:07:12 PM UTC-6 Alan Grayson wrote:

On Saturday, November 1, 2025 at 5:15:36 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 10:36 PM, Alan Grayson wrote:

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end, he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. 

 

I don't understand that. 

It's pretty straightforward. If you're at rest in some frame in spacetime, you're moving along the time axis only. Along that axis are coordinate labels, but since you've multiplied these lables by c, you're left with distances (as on spatial axis), and the distance separation of two adjacent coordinate unit times, has a distance which light traverses in one second of proper time. IOW, along the time axis, proper and coordinate time are identical. Thus, d(tau)/dt=1. When motion is not strictly along time axis, that is, when you're not at rest, coordinate and proper time no longer coincide, no longer have equal values. The non trivial existential question is why a clock which measures only proper time, "knows" to adjust its rate when moving along some arbitrary path in spacetime?  AG

 

If when you'r.e stationary in some coordinate system you can adjust you clock to sync with the stationary clocks and so long as you remain stationary you would have d(tau)/dt=1.  But once you started to move relative that coord system one-second-per-second in your frame is not one-second-per-second in the stationary frame.  Proper time is one-second-per-second along your path thru spacetime.  In general it will be shorter as measured from the stationary frame, but remember inertial motion is strictly relative so you will see the stationary clocks as running slow.

With c multiplied by coordinate time along time axis for a particle spatially at rest

That would define a forward light cone for that particle.  Why would you do that?

, isn't this tantamount to a definition with the intended result that spacetime velocity is c? 

Along a path at velocity c the proper time is zero.

No. When at rest, proper time increments, and is indistinguishable from coordinate time. If this weren't the case, it would be impossible to make the unintelligible conclusion that all motion in spacetime is c. AG 

One can always plot distance versus time, but in this case we must convert time to units of distance because we want to calculate (ds)^2 and have consistent units (of length). Toward the end of the video the author does exactly that, using the Pythagorean formula, but I don't see how he gets the negative sign preceding the spatial differrentials. AG 

[Brent Meeker]

On 11/3/2025 9:20 AM, Alan Grayson wrote:

On Saturday, November 1, 2025 at 9:07:12 PM UTC-6 Alan Grayson wrote:

On Saturday, November 1, 2025 at 5:15:36 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 10:36 PM, Alan Grayson wrote:

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end, he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. 

 

I don't understand that. 

It's pretty straightforward. If you're at rest in some frame in spacetime, you're moving along the time axis only. Along that axis are coordinate labels, but since you've multiplied these lables by c, you're left with distances (as on spatial axis), and the distance separation of two adjacent coordinate unit times, has a distance which light traverses in one second of proper time. IOW, along the time axis, proper and coordinate time are identical. Thus, d(tau)/dt=1. 

OK, you've used proper (clock) time to mark the intervals of coordinate time.

When motion is not strictly along time axis, that is, when you're not at rest, coordinate and proper time no longer coincide, no longer have equal values. The non trivial existential question is why a clock which measures only proper time, "knows" to adjust its rate when moving along some arbitrary path in spacetime?  AG

It doesn't "adjust its rate".  The clock continues to measure proper time along the new spacetime direction.  But because the new direction is not parallel to the old one the intervals don't match the intervals of the clock that remained on the stationary worldline.  Motion is only relative.  So each clock sees the other as running slow because they judge the other clock to not be going in the futureward direction.

Brent

 

If when you'r.e stationary in some coordinate system you can adjust you clock to sync with the stationary clocks and so long as you remain stationary you would have d(tau)/dt=1.  But once you started to move relative that coord system one-second-per-second in your frame is not one-second-per-second in the stationary frame.  Proper time is one-second-per-second along your path thru spacetime.  In general it will be shorter as measured from the stationary frame, but remember inertial motion is strictly relative so you will see the stationary clocks as running slow.

With c multiplied by coordinate time along time axis for a particle spatially at rest

That would define a forward light cone for that particle.  Why would you do that?

, isn't this tantamount to a definition with the intended result that spacetime velocity is c? 

Along a path at velocity c the proper time is zero.

No. When at rest, proper time increments, and is indistinguishable from coordinate time. If this weren't the case, it would be impossible to make the unintelligible conclusion that all motion in spacetime is c. AG 

One can always plot distance versus time, but in this case we must convert time to units of distance because we want to calculate (ds)^2 and have consistent units (of length). Toward the end of the video the author does exactly that, using the Pythagorean formula, but I don't see how he gets the negative sign preceding the spatial differrentials. AG 

You refer to time dilation, but this definition seems unrelated to that concept. The key question is how a physical clock measures something other than coordinate time when moving along some arbitrary path? 

Every  clock measures only its proper time.  Coordinate time in general is just labels.   But for convenience it is usually chosen to be the the time kept by some clock(s) that are taken to define a stationary frame (so they're stationary relative to one another).  So any clock moving relative to that frame will measure proper time different from that frame.

Brent

AG 

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

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[Alan Grayson]

On Tuesday, November 4, 2025 at 1:35:06 AM UTC-7 Brent Meeker wrote:

On 11/3/2025 9:20 AM, Alan Grayson wrote:

On Saturday, November 1, 2025 at 9:07:12 PM UTC-6 Alan Grayson wrote:

On Saturday, November 1, 2025 at 5:15:36 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 10:36 PM, Alan Grayson wrote:

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end, he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. 

 

I don't understand that. 

It's pretty straightforward. If you're at rest in some frame in spacetime, you're moving along the time axis only. Along that axis are coordinate labels, but since you've multiplied these lables by c, you're left with distances (as on spatial axis), and the distance separation of two adjacent coordinate unit times, has a distance which light traverses in one second of proper time. IOW, along the time axis, proper and coordinate time are identical. Thus, d(tau)/dt=1. 

OK, you've used proper (clock) time to mark the intervals of coordinate time.

When motion is not strictly along time axis, that is, when you're not at rest, coordinate and proper time no longer coincide, no longer have equal values. The non trivial existential question is why a clock which measures only proper time, "knows" to adjust its rate when moving along some arbitrary path in spacetime?  AG

It doesn't "adjust its rate".  The clock continues to measure proper time along the new spacetime direction.  But because the new direction is not parallel to the old one the intervals don't match the intervals of the clock that remained on the stationary worldline.  Motion is only relative.  So each clock sees the other as running slow because they judge the other clock to not be going in the futureward direction.

Brent

I don't see any daylight between "adjusting its rate" and "judging" how another clock is moving. That aside, you seem to be affirming the TP. AG 

[Alan Grayson]

On Tuesday, November 4, 2025 at 11:38:20 AM UTC-7 Alan Grayson wrote:

On Tuesday, November 4, 2025 at 1:35:06 AM UTC-7 Brent Meeker wrote:

On 11/3/2025 9:20 AM, Alan Grayson wrote:

On Saturday, November 1, 2025 at 9:07:12 PM UTC-6 Alan Grayson wrote:

On Saturday, November 1, 2025 at 5:15:36 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 10:36 PM, Alan Grayson wrote:

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end, he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. 

 

I don't understand that. 

It's pretty straightforward. If you're at rest in some frame in spacetime, you're moving along the time axis only. Along that axis are coordinate labels, but since you've multiplied these lables by c, you're left with distances (as on spatial axis), and the distance separation of two adjacent coordinate unit times, has a distance which light traverses in one second of proper time. IOW, along the time axis, proper and coordinate time are identical. Thus, d(tau)/dt=1. 

OK, you've used proper (clock) time to mark the intervals of coordinate time.

I didn't do it. The community of physicists did it. How can a test particle at rest move at light speed? Makes no sense AFAICT. AG 

[Brent Meeker]

On 11/6/2025 2:49 PM, Alan Grayson wrote:

On Tuesday, November 4, 2025 at 11:38:20 AM UTC-7 Alan Grayson wrote:

On Tuesday, November 4, 2025 at 1:35:06 AM UTC-7 Brent Meeker wrote:

On 11/3/2025 9:20 AM, Alan Grayson wrote:

On Saturday, November 1, 2025 at 9:07:12 PM UTC-6 Alan Grayson wrote:

On Saturday, November 1, 2025 at 5:15:36 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 10:36 PM, Alan Grayson wrote:

On Friday, October 31, 2025 at 4:15:29 PM UTC-6 Brent Meeker wrote:

On 10/31/2025 6:17 AM, Alan Grayson wrote:

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.

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" is it moving, making that derivative non-zero. AG 

You're muddling things.  For a clock moving inertially in flat spacetime, the coordinate times are arbitrary up to a linear transformation.  So d(tau)/dt=const.  not necessarily 1.  And the constant depends on the speed (time dilation).  So the coordinate speed depends on the choice of coordinate time, i.e. relativity of motion.

Brent

In the video toward the end, he claims d(tau)/dt=1, so every 1 sec increment in coordinate time is set to 1 sec increment in proper time. 

 

I don't understand that. 

It's pretty straightforward. If you're at rest in some frame in spacetime, you're moving along the time axis only. Along that axis are coordinate labels, but since you've multiplied these lables by c, you're left with distances (as on spatial axis), and the distance separation of two adjacent coordinate unit times, has a distance which light traverses in one second of proper time. IOW, along the time axis, proper and coordinate time are identical. Thus, d(tau)/dt=1. 

OK, you've used proper (clock) time to mark the intervals of coordinate time.

I didn't do it. The community of physicists did it. How can a test particle at rest move at light speed? Makes no sense AFAICT. AG 

When motion is not strictly along time axis, that is, when you're not at rest, coordinate and proper time no longer coincide, no longer have equal values. The non trivial existential question is why a clock which measures only proper time, "knows" to adjust its rate when moving along some arbitrary path in spacetime?  AG

It doesn't "adjust its rate".  The clock continues to measure proper time along the new spacetime direction.  But because the new direction is not parallel to the old one the intervals don't match the intervals of the clock that remained on the stationary worldline.  Motion is only relative.  So each clock sees the other as running slow because they judge the other clock to not be going in the futureward direction.

Brent

I don't see any daylight between "adjusting its rate" and "judging" how another clock is moving. That aside, you seem to be affirming the TP. AG 

"Adjusting its rate" would imply that there was some absolute motion that would tell it how to adjust.  Judging how some other clock, that's moving relative to you, keeps time can depend on relative motion and doesn't imply any absolute.

Brent

[Alan Grayson]

When the ds vector is inclined wrt the time axis, all the test particle has to "know" is how far its velocity differs from rest in spacetime, which we know is relative. This has nothing to do with absolute anything. But I see you ignore the more important issue; namely, how can a test particle at rest, move at lightspeed in spacetime, if SR shows it can never reach that speed. And if this is mistaken, it implies that your alleged solutions to the TP is on very shaky ground, because it assumes what has been demonstrated is impossible. AG

[Brent Meeker]

On 11/7/2025 12:05 AM, Alan Grayson wrote:

I don't see any daylight between "adjusting its rate" and "judging" how another clock is moving. That aside, you seem to be affirming the TP. AG 

"Adjusting its rate" would imply that there was some absolute motion that would tell it how to adjust. 

When the ds vector is inclined wrt the time axis, all the test particle has to "know" is how far its velocity differs from rest in spacetime, which we know is relative. 

"differs from rest" implies there's some "at rest".  The particle in motion has many different velocities relative to many other objects.  I can't have many different time dilations at once.

This has nothing to do with absolute anything. But I see you ignore the more important issue; namely, how can a test particle at rest, move at lightspeed in spacetime, 

No need to answer "How?" since it can't.

if SR shows it can never reach that speed. And if this is mistaken, it implies that your alleged solutions to the TP is on very shaky ground, because it assumes what has been demonstrated is impossible. AG

You keep referring to "solution to the Twin Paradox".  What exactly is the problem which you think needs a solution.  The fact that two twins who travel different paths between the same two events in spacetime measure different path lengths doesn't seem like a problem to me.  And it has nothing to do with light speed.

Brent

[Alan Grayson]

On Friday, November 7, 2025 at 2:49:12 PM UTC-7 Brent Meeker wrote:

On 11/7/2025 12:05 AM, Alan Grayson wrote:

I don't see any daylight between "adjusting its rate" and "judging" how another clock is moving. That aside, you seem to be affirming the TP. AG 

"Adjusting its rate" would imply that there was some absolute motion that would tell it how to adjust. 

When the ds vector is inclined wrt the time axis, all the test particle has to "know" is how far its velocity differs from rest in spacetime, which we know is relative. 

"differs from rest" implies there's some "at rest".  The particle in motion has many different velocities relative to many other objects.  I can't have many different time dilations at once.

This has nothing to do with absolute anything. But I see you ignore the more important issue; namely, how can a test particle at rest, move at lightspeed in spacetime, 

No need to answer "How?" since it can't.

Never heard of this???  https://www.reddit.com/r/AskPhysics/comments/1fmpay3/everything_travels_through_spacetime_at_the_speed/ AG

[Brent Meeker]

On 11/7/2025 10:01 PM, Alan Grayson wrote:

On Friday, November 7, 2025 at 2:49:12 PM UTC-7 Brent Meeker wrote:

On 11/7/2025 12:05 AM, Alan Grayson wrote:

I don't see any daylight between "adjusting its rate" and "judging" how another clock is moving. That aside, you seem to be affirming the TP. AG 

"Adjusting its rate" would imply that there was some absolute motion that would tell it how to adjust. 

When the ds vector is inclined wrt the time axis, all the test particle has to "know" is how far its velocity differs from rest in spacetime, which we know is relative. 

"differs from rest" implies there's some "at rest".  The particle in motion has many different velocities relative to many other objects.  I can't have many different time dilations at once.

This has nothing to do with absolute anything. But I see you ignore the more important issue; namely, how can a test particle at rest, move at lightspeed in spacetime, 

No need to answer "How?" since it can't.

Never heard of this???  https://www.reddit.com/r/AskPhysics/comments/1fmpay3/everything_travels_through_spacetime_at_the_speed/ AG

Well then why did you ask "how can a test particle at rest, move at lightspeed in spacetime,"
 Brent

if SR shows it can never reach that speed. And if this is mistaken, it implies that your alleged solutions to the TP is on very shaky ground, because it assumes what has been demonstrated is impossible. AG

You keep referring to "solution to the Twin Paradox".  What exactly is the problem which you think needs a solution.  The fact that two twins who travel different paths between the same two events in spacetime measure different path lengths doesn't seem like a problem to me.  And it has nothing to do with light speed.

Brent

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[Alan Grayson]

On Saturday, November 8, 2025 at 5:12:49 PM UTC-7 Brent Meeker wrote:

On 11/7/2025 10:01 PM, Alan Grayson wrote:

On Friday, November 7, 2025 at 2:49:12 PM UTC-7 Brent Meeker wrote:

On 11/7/2025 12:05 AM, Alan Grayson wrote:

I don't see any daylight between "adjusting its rate" and "judging" how another clock is moving. That aside, you seem to be affirming the TP. AG 

"Adjusting its rate" would imply that there was some absolute motion that would tell it how to adjust. 

When the ds vector is inclined wrt the time axis, all the test particle has to "know" is how far its velocity differs from rest in spacetime, which we know is relative. 

"differs from rest" implies there's some "at rest".  The particle in motion has many different velocities relative to many other objects.  I can't have many different time dilations at once.

This has nothing to do with absolute anything. But I see you ignore the more important issue; namely, how can a test particle at rest, move at lightspeed in spacetime, 

No need to answer "How?" since it can't.

Never heard of this???  https://www.reddit.com/r/AskPhysics/comments/1fmpay3/everything_travels_through_spacetime_at_the_speed/ AG

Well then why did you ask "how can a test particle at rest, move at lightspeed in spacetime,"
 Brent

First you wrote it can't. Now, apparently, you write it can. Doesn't give me much confidence either way. But since SR says it can't, I have to go with that unless you have a contrary argument. AG