Looking to understand time dilation

[Dale]

Dabx to Ebx and Dbax to Eax, where x is a, b, and c. You did two of the six possible calculations and got ratios of 2 for each of them, I'd like you to do the other four, please.

Sure. However, since the Lorentz transform is linear you automatically know that the ratio will always be 2. But here are the results:

Daba = (2.5cn,1.5cn,0,0)
Eba = (1.25cn,.75cn,0,0)
Dbaa = (2cn,0,0,0)
Eaa = (cn,0,0,0)


Dabb = (2cn,0,0,0)
Ebb = (cn,0,0,0)
Dbab = (2.5cn,-1.5cn,0,0)
Eab = (1.25cn,-.75cn,0,0)


Dabc = (2.12cn,.71cn,0,0)
Ebc = (1.06cn,.35cn,0,0)
Dbac = (2.12cn,-.71cn,0,0)
Eac = (1.06cn,-.35cn,0,0)

 

[ghwellsjr]

Sure. However, since the Lorentz transform is linear you automatically know that the ratio will always be 2.

Of course the Lorentz Transform has to get the correct answer of 2 no matter what frame you do the calculation from because this ratio is 2 and has nothing to do with Special Relativity. I just didn't want Grimble to think that since you originally only calculated the ratio that A sees from the frame in which A is at rest and the ratio that B sees from the frame in which B is at rest, that a conclusion could be drawn that these ratios were somehow associated with rest frames.

This ratio, as I pointed out in post #323 is the Relative Doppler factor (or its reciprocal depending on whether the clocks are advancing or retreating) and has nothing to do with Special Relativity. It is a real physical measurement made by observer/clocks in constant relative motion (with prior knowledge that they have identical clocks, of course). One of the points of Special Relativity is that you can use any inertial frame of reference and it will calculate that observer/clocks make the same measurements. It better, or it wouldn't match reality.

 

[Grimble]

But there is no sense in which we can say that any of the three clocks continue to be synchronized after the one monent when they are co-located. Only clocks that are at rest with respect to each other can be synchronized and remain synchronized.

Let us assign the space-time coordinates (0,0,0,0) to the event A and B pass C.

The interval between that event and A emitting the first flash of light, first 'tick' of the clock will be given by S² = (x²+y²+z²-c²t²)

Where S is the space-time interval
x is the displacement in proper length units = 0.333 light seconds
y and z = 0, c=1, t =1 second proper time.

which gives us S² = (0.333² + 0 + O - 1) = -0.889 and
s = 0.943

For B the displacement will be -0.333 giving the same interval of 0.943 and in which the displacement and time difference are the same as for A.

So by my reckoning the emission of light from A and B are synchronous, everything is symmetrical.

And if that is true for the first pulse of light from each clock it will also be true for the second pulses and all subsequent pairs of pulses.

 

[Dale]

The interval between that event and A emitting the first flash of light, first 'tick' of the clock will be given by S² = (x²+y²+z²-c²t²)

Where S is the space-time interval
x is the displacement in proper length units = 0.333 light seconds
y and z = 0, c=1, t =1 second proper time.

Careful here. x is the coordinate displacement and t is the coordinate time. Then s^2 is the square of the interval so that sqrt(-s^2/c^2) is the proper time for timelike intervals and sqrt(s^2) is the proper length for spacelike intervals.

That said, you are correct that the time coordinate in frame C is the same for Eac and Ebc. But they are not synchronized with coordinate time in frame C.

 

[Grimble]

Careful here. x is the coordinate displacement and t is the coordinate time. Then s^2 is the square of the interval so that sqrt(-s^2/c^2) is the proper time for timelike intervals and sqrt(s^2) is the proper length for spacelike intervals.

That said, you are correct that the time coordinate in frame C is the same for Eac and Ebc. But they are not synchronized with coordinate time in frame C.

So are you agreeing that the emission event for A is synchronous with the emission event at B? I didn't say it would be synchronised with coordinate time in C, yet is it not the coordinate time in frame C for the emission events at A and B that is equal?

I do agree that the emission from A as observed by B and that of B observed by A that will be subject to Time dilation - (and length contraction).

Yet the light emitted from those events will not be affected, will it? As light travels at c and this is not affected by the speed of the source. So the two pulses of light will travel the same distance at c.

Consider too that after 100 seconds each clock will have emitted 100 pulses of light. It is the arrival of these pulses of light that will be observed by the other observers who by subtracting the transmission time will count 100 pulses emitted in 100seconds. At the same rate by each clock.
There is not and cannot logically be a two for one ratio between the emission and receipt of the pulses.
The only way that we could have a two second coordinate time between the emission of the said pulses is if two seconds coordinate time have the same duration as one second proper time, which is what the two clocks are keeping.

 

[Dale]

So are you agreeing that the emission event for A is synchronous with the emission event at B?

In frame C, yes, in any other frame, no. That is what the relativity of simultaneity means.

Yet the light emitted from those events will not be affected, will it? As light travels at c and this is not affected by the speed of the source. So the two pulses of light will travel the same distance at c.

Correct. Dacx=Dbcx (where x is anyone of a,b,c). The distances traveled by the light pulses is only equal in frame c. In other frames the distances traveled are different, and because of the relativity of simultaneity the times of emission are also different, such that they arrive at C at the same event.

Consider too that after 100 seconds each clock will have emitted 100 pulses of light.

No, after 100 seconds in any given frame only the clock at rest in that frame will have emitted 100 pulses of light. For example, in A's frame:
Eaa = (cn,0,0,0)
Eba = (1.25cn,.75cn,0,0)
so after 100 seconds there will be 100 pulses emitted from clock A but only 80 emitted from clock B.

It is the arrival of these pulses of light that will be observed by the other observers who by subtracting the transmission time will count 100 pulses emitted in 100seconds. At the same rate by each clock.

No, even after accounting for the delay of light you will still find that fewer than 100 pulses were emitted.

There is not and cannot logically be a two for one ratio between the emission and receipt of the pulses.

Sure there can.

The only way that we could have a two second coordinate time between the emission of the said pulses is if two seconds coordinate time have the same duration as one second proper time, which is what the two clocks are keeping.

The coordinate time between the emissions is 1.25 seconds (for Eba), not 2 seconds. The 2 seconds is the coordinate time between the detections (for Dbaa)

 

[Grimble]

l3sm506608gc

No, after 100 seconds in any given frame only the clock at rest in that frame will have emitted 100 pulses of light. For example, in A's frame:
Eaa = (cn,0,0,0)
Eba = (1.25cn,.75cn,0,0)
so after 100 seconds there will be 100 pulses emitted from clock A but only 80 emitted from clock B.

I am still finding this way of expressing things a little confusing - what does the 'cn' mean?

No, even after accounting for the delay of light you will still find that fewer than 100 pulses were emitted.

And yes, of course, from one frame the time in another will be dilated. BUT In each frame, as observed from within that frame, 100 pulses will be emitted in 100 seconds. And as they frames that are not moving, according to that observer they are proper seconds (I am not sure whether this is significant or not)

And light from the pulse from each frame will arrive at C simultaneously. (And this has nothing to do with relative simultaneity as they are simultaneous at the same place, i.e. together they comprise a single event.)
And if a mirror was in place at C then the light from one frame would pass in the same time, the same distance and arrive at the other frame simultaneously with that frame's light reflected back from C, so each has to observe the other frame emitting 100 pulses in 100 seconds.

If one frame only observed 80 flashes in 100 seconds, what happens to the other 20 flashes that we know it emitted in 100 seconds in its own frame?

And so, regardless of how you calculate time passing what happens doesn't change, it is how it is observed that changes.

The coordinate time between the emissions is 1.25 seconds (for Eba), not 2 seconds. The 2 seconds is the coordinate time between the detections (for Dbaa)

I'm sorry but can you explain that? How do we have a greater time between detections than between emissions. Or are you adding in the transmission times again?

 

[yuiop]

I'm sorry but can you explain that? How do we have a greater time between detections than between emissions. Or are you adding in the transmission times again?

In the above diagram the green line represents a person firing projectiles (blue lines) at rate of one per second at a fleeing person represented by the red line. The red guy receives the projectiles at a rate of one every two seconds because he is going away from the emitter. The transmission time for the projectiles between being emitted and being received is steadily increasing from 1 to 5 seconds and yet the victim continues to receive them at a constant rate of one every 2 seconds. Therefore the rate of reception is independent of the transmission time and can differ from the emitter rate. This is a simple geometrical property that is the basis of classic Doppler shift and in this example we are not even considering relativistic velocities or time dilation / length contraction etc.

 

[Dale]

I am still finding this way of expressing things a little confusing - what does the 'cn' mean?

The c is the speed of light and n is simply an arbitrary integer indicating which pulse. For example, for the 5th pulse you would set n=5.

And yes, of course, from one frame the time in another will be dilated. BUT In each frame, as observed from within that frame, 100 pulses will be emitted in 100 seconds.

Only by an emitter which is at rest in that frame.

If one frame only observed 80 flashes in 100 seconds, what happens to the other 20 flashes that we know it emitted in 100 seconds in its own frame?

Nothing has happened to them. They simply haven't been emitted yet in the other frames. That is what the relativity of simultaneity implies.

I'm sorry but can you explain that? How do we have a greater time between detections than between emissions. Or are you adding in the transmission times again?

Yes, of course, by definition the time of detection includes the time of emission plus the time of transmission. If you negelct the transmission time then by definition you have the emission, not the detection.

 

[Grimble]

The c is the speed of light and n is simply an arbitrary integer indicating which pulse. For example, for the 5th pulse you would set n=5.

Thank you, that makes sense.

Only by an emitter which is at rest in that frame.

But those two Frames are in every way the same (apart from their relative velocity), their clocks are synchronised and identical operating in identical conditions. viz.The Principle of Relativity – The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.

So if that 1st postulate is adhered to they must be emitting pulses simultaneously.

* But you will no doubt say simultaneous to which observer?

And I would answer to Space time itself.

The emission of each 'pair' of simultaneous pulses are equal space time intervals from the origin event where/when A and B pass C.

Not only do their Space time intervals have to be equal but those space time intervals comprise a space element and a time element and in each case the space intervals are equal, as are the time elements.
So if the spacetime intervals are equal and the time elements of them are equal how can they not be simultaneous.

These events observed by A or B however will not be silultaneous, distances and time swill differ BECAUSE of where they are observed from.

 

[Dale]

But those two Frames are in every way the same (apart from their relative velocity), their clocks are synchronised and identical operating in identical conditions. viz.The Principle of Relativity – The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.

So if that 1st postulate is adhered to they must be emitting pulses simultaneously.

No, simultaneity is not a law of physics. It is, in fact, merely an arbitrary human-made convention.

Do you believe that if two events occur at the same x coordinate in one frame that they must therefore occur at the same x coordinate in all other frames? If not, then what would make you believe that is the case with the t coordinate?

The emission of each 'pair' of simultaneous pulses are equal space time intervals from the origin event where/when A and B pass C.

Yes, in all frames.

those space time intervals comprise a space element and a time element and in each case the space intervals are equal, as are the time elements.

Yes in frame C, but not in any other frame.

So if the spacetime intervals are equal and the time elements of them are equal how can they not be simultaneous.

They are simultaneous in frame C, but not in any other frame.

These events observed by A or B however will not be silultaneous, distances and time swill differ BECAUSE of where they are observed from.

Yes. And not only are the detection events not simultaneous, but the emission events are also not simultaneous in A and B, as shown above.

 

[Grimble]

No, simultaneity is not a law of physics. It is, in fact, merely an arbitrary human-made convention.

I have to agree with you there
It depends on how one defines the convention and how it is measured.
So let us say that if we take the difference in proper time from a single event A in space time and a second event B and compare that with the difference in proper time from A and a third event C.
If the proper time difference AB is equal to AC then, in space-time, is B not simultaneous with C?
Surely as A,B and C are all individual events in Space time there relationship is fixed, wherever it is measured from, inertially in spacetime.

Do you believe that if two events occur at the same x coordinate in one frame that they must therefore occur at the same x coordinate in all other frames? If not, then what would make you believe that is the case with the t coordinate?

No, and I do not.

Yes, in all frames.

Yes in frame C, but not in any other frame.[/quote]
Because from other frames the measurements will be coordinate measurements? And by the application of the LT equations they can be converted to proper measurements? And as proper measurements they will then be equal? (I am asking here, not telling)

Yes. And not only are the detection events not simultaneous, but the emission events are also not simultaneous in A and B, as shown above.

Yes that is just what I am saying they are simultaneous when measured from an impartial frame of reference yet from nowhere else.

Spacetime is a way of measuring the relationship of events.

If two intervals from a common event in space time are not only equal but identical then not only are their Space-time intervals equal but their space elements are equal as are their time elements.

As I understand it Space-time is described by a four dimensional coordinate system. With three space-like coordinates and one time-like coordinate.

Everything in time and space may be plotted with reference to those coordinates. By any inertial observer. The only things that can be different are the designated origin (0,0,0,0) that is used and which inertial frames of reference are at rest and which are moving relative to that designated origin.

All measurements made by an observer at rest within said inertial FoR are proper measurements: made by clocks at rest (proper times) and between the world lines of such clocks at rest (proper lengths/distances).

All measurements of moving objects, made by taking the measurements made within the moving object's own FoR (in which it is at rest) and transforming them using LT, will be coordinate measurements.

(I have a mind that sees things in pictures/ diagrams/ images/ in multiple dimensions and so I am trying to find a starting point from which I can show where I find things that don't seem to fit)

Thank you one and all for your efforts.

 

[Dale]

let us say that if we take the difference in proper time from a single event A in space time and a second event B and compare that with the difference in proper time from A and a third event C.
If the proper time difference AB is equal to AC then, in space-time, is B not simultaneous with C?

Not in general, no. B is simultaneous with C in a given frame iff the coordinate time difference AB is equal to the coordinate time difference AC.

Surely as A,B and C are all individual events in Space time there relationship is fixed, wherever it is measured from, inertially in spacetime.

I am not sure what you mean here by "relationship". The spacetime intervals AB, AC, and BC are indeed fixed, but of course the coordinate differences depend on the coordinate system.

And by the application of the LT equations they can be converted to proper measurements? And as proper measurements they will then be equal? (I am asking here, not telling)

You don't need the LT in order to transform the measurements into proper measurements. All you need is the formula for the spacetime interval:

ds² = -c²dt² + dx² + dy² + dz²

The spacetime interval squared, ds², is invariant, meaning that it is the same in all reference frames. The Lorentz transform is the transform which preserves the spacetime interval.

If two intervals from a common event in space time are not only equal but identical then not only are their Space-time intervals equal but their space elements are equal as are their time elements.

I don't know what you mean by "not only equal but identical".

As I understand it Space-time is described by a four dimensional coordinate system. With three space-like coordinates and one time-like coordinate.

Yes.

Everything in time and space may be plotted with reference to those coordinates. By any inertial observer. The only things that can be different are the designated origin (0,0,0,0) that is used and which inertial frames of reference are at rest and which are moving relative to that designated origin.

In spacetime a point particle is represented by a line, called a worldline, which indicates the particle's position at each moment of time. If the particle is inertial then the worldline is straight. If two particles are at rest wrt each other then their worldlines are parallel to each other.

The origin is an event, which is a point rather than a line in spacetime. It makes no sense to say that something is moving relative to the origin any more than it does to say that a given line is parallel to a point.

(I have a mind that sees things in pictures/ diagrams/ images/ in multiple dimensions and so I am trying to find a starting point from which I can show where I find things that don't seem to fit)

Me too. I highly recommend going through the exercise of making a few spacetime diagrams where you carefully plot the coordinate lines of a moving frame on the same diagram so that you can see how the coordinates differ. I also recommend learning about four-vectors, they are much more suited to a geometrically-oriented mind.

 

[ghwellsjr]

If two intervals from a common event in space time are not only equal but identical then not only are their Space-time intervals equal but their space elements are equal as are their time elements.

I don't know what you mean by "not only equal but identical".

Dalespam, it looks to me like Grimble is thinking that since the spacetime interval between two events is the same for two observers at rest in two different frames of reference, then this means that the space "distances" are equal in those two FoRs and the time "intervals" are equal in those two FoRs. He doesn't understand that it is only the "spacetime intervals" that are equal which are a combination of both the space and time "intervals". I know you have been repeating this over and over again, but he's still stuck on this point, I believe.

 

[Grimble]

Not in general, no. B is simultaneous with C in a given frame if the coordinate time difference AB is equal to the coordinate time difference AC.

But that was not what I was specifying was it? I am trying to relate to the concept of Space-Time which is a way of describing what is. Coordinate times and distances are measurements taken from different Reference Frames: different reference frames are, surely, how that same single reality is measured by different observers?[/quote]

I am not sure what you mean here by "relationship". The spacetime intervals AB, AC, and BC are indeed fixed, but of course the coordinate differences depend on the coordinate system.

I mean the relationship between events as defined by the space-time intervals between them

You don't need the LT in order to transform the measurements into proper measurements. All you need is the formula for the spacetime interval:

ds² = -c²dt² + dx² + dy² + dz²

The spacetime interval squared, ds², is invariant, meaning that it is the same in all reference frames. The Lorentz transform is the transform which preserves the spacetime interval.

I'm sorry but how does one convert between coordinate and proper measurements using the space time interval?
I understood (or thought that I did) that;

proper measure/coordinate measure= gamma

I don't know what you mean by "not only equal but identical".

Well to me 'equal' compares the totals, whereas identical compares the individual terms; i.e. the time vales are equal and the distance values are equal.

In spacetime a point particle is represented by a line, called a worldline, which indicates the particle's position at each moment of time. If the particle is inertial then the worldline is straight. If two particles are at rest wrt each other then their worldlines are parallel to each other.

Yes, as I said "Everything in time and space may be plotted with reference to those coordinates "
But this next bit has me confused:

The origin is an event, which is a point rather than a line in spacetime. It makes no sense to say that something is moving relative to the origin any more than it does to say that a given line is parallel to a point.

But if the origin (0,0,0,0) defines an event, a point in space at a specific time, then the world line of that point in space is a straight line – agreed.

Yet any other reference frame must also have an origin that will be in the form (ct, x, y, z) and which will either have a constant Space-time interval or a changing one. If it is constant then that Reference frame is at rest relative to the space-time origin and if the space-time interval is changing then that reference frame is moving with respect to the defined space-time origin.

Now it seems to me that within our 4 dimensional coordinate system any inertial frame of reference may be considered to have a clocks at rest with respect to it. Those clocks will have straight worldlines that are parallel one to the other. The world line of the one at the origin of that frame of reference will describe the time coordinate of our 4D set of coordinates. The world lines of all the clocks will measure proper time and the distances between their parallel world lines will therefore measure proper distances.

Me too. I highly recommend going through the exercise of making a few spacetime diagrams where you carefully plot the coordinate lines of a moving frame on the same diagram so that you can see how the coordinates differ. I also recommend learning about four-vectors, they are much more suited to a geometrically-oriented mind.

Space diagrams I am happy with but I will look into 'four-vectors'.

 

[Dale]

But that was not what I was specifying was it? I am trying to relate to the concept of Space-Time which is a way of describing what is. Coordinate times and distances are measurements taken from different Reference Frames: different reference frames are, surely, how that same single reality is measured by different observers?

Yes, although usually you would say something more along the lines of "different observer's perspective" rather than "measured by different observers". This is because what is actually measured by an observer is affected by the finite speed of light and the reference frame is what the observer calculates after accounting for the finite speed of light.

I'm sorry but how does one convert between coordinate and proper measurements using the space time interval?
I understood (or thought that I did) that;

proper measure/coordinate measure= gamma

The formula you just posted is only correct in certain particular circumstances. The one I posted is more general and also it easily generalizes further to non-inertial coordinates and curved spacetimes.

In the formula
ds² = -c²dt² + dx² + dy² + dz²

the dt, dx, dy, and dz are all coordinate times and distances in some inertial frame while the ds is the frame-invariant spacetime interval. If ds² > 0 then the interval is called "spacelike" and ds is the proper distance. If ds² < 0 then the interval is called "timelike" and dτ = sqrt(-ds²/c²) is the proper time. If ds² = 0 then the interval is called "lightlike" or "null" and represents the path of a pulse of light.

The value of this formula is that all reference frames can use their own coordinate values and come up with the same value for ds². This formula also let's you easily see where the formula you posted comes from:

dτ² = -ds²/c² = dt² - dx²/c² - dy²/c² - dz²/c²
dτ²/dt² = 1 - (dx²/dt² - dy²/dt² - dz²/dt²)/c²
dτ²/dt² = 1 - v²/c²
dτ/dt = 1/γ

identical compares the individual terms; i.e. the time vales are equal and the distance values are equal.

Then yes, clearly "if two intervals from a common event in space time are not only equal but 'the time vales are equal and the distance values are equal' then not only are their Space-time intervals equal but their space elements are equal as are their time elements".

Yes, as I said "Everything in time and space may be plotted with reference to those coordinates "
But this next bit has me confused:
But if the origin (0,0,0,0) defines an event, a point in space at a specific time, then the world line of that point in space is a straight line – agreed.

Yet any other reference frame must also have an origin that will be in the form (ct, x, y, z) and which will either have a constant Space-time interval or a changing one. If it is constant then that Reference frame is at rest relative to the space-time origin and if the space-time interval is changing then that reference frame is moving with respect to the defined space-time origin.

I think you are confusing the origin of spacetime, the event (0,0,0,0), with the coordinate time axis, the worldline (ct,0,0,0). Objects may be resting or moving wrt the time axis (parallel worldlines), but not wrt the origin (lines cannot be parallel or not parallel to a point).

Now it seems to me that within our 4 dimensional coordinate system any inertial frame of reference may be considered to have a clocks at rest with respect to it. Those clocks will have straight worldlines that are parallel one to the other. The world line of the one at the origin of that frame of reference will describe the time coordinate of our 4D set of coordinates.

Yes, exactly. That clock defines the time axis of the coordinate system, (ct,0,0,0).

 

[grav-universe]

It depends on how one defines the convention and how it is measured.
So let us say that if we take the difference in proper time from a single event A in space time and a second event B and compare that with the difference in proper time from A and a third event C.
If the proper time difference AB is equal to AC then, in space-time, is B not simultaneous with C?
Surely as A,B and C are all individual events in Space time there relationship is fixed, wherever it is measured from, inertially in spacetime.

Hi, Grimble. It looks like you are still getting stuck on this simultaneity thing. The difference in times between the events that occur at A and B, and at A and C, depend upon which frame is measuring the time of those events and the simultaneity convention by which their clocks are set. As Dalespam mentioned, simultaneity conventions are man-made, meaning there is no absolute concept of simultaneity.

For instance, let's say that events A and B are both measured from within the same frame and it is determined that they occurred simultaneously. What does that mean? It simply means that the clocks that an observer carries at A and another observer carries at B read the same when the events occurred, right? Now what if we were to set observer B's clock forward 5 minutes? Then according to the new simultaneity convention, event B now occurs five minutes after event A, not simultaneously at all anymore. That's what simultaneity is, it is man-made and depends upon the convention we have declared for how each frame should set their clocks.

The Einstein simultaneity convention gives a convenient way to set clocks by using light since all frames measure the same two way time in any direction for light to travel away and back over the same distance, between points 1 and 2 say, so a clock at the other end of that distance at point 2 can simply be set such that the time for the light to travel from point 1 to point 2 is the same as that from point 2 to point 1. After the simultaneity has been set accordingly for each frame, Relativity applies as measurements are then made between frames for events. This is what applies to space-time, not just space and not just time, since a simple re-adjustments in the clocks of a frame will cause that frame's observers to measure lengths in other frames differently as well, for example.

You can establish an altogether different simultaneity convention if you wish, but then you must first determine what it should be and by what procedure it would be physically established, then work out the details for the simultaneity issues between frames while still acknowledging the results of experiments that confirm Relativity.

 

[Grimble]

Yes, although usually you would say something more along the lines of "different observer's perspective" rather than "measured by different observers". This is because what is actually measured by an observer is affected by the finite speed of light and the reference frame is what the observer calculates after accounting for the finite speed of light.

Thank you, that makes sense, I will try and remember that in the future.

The formula you just posted is only correct in certain particular circumstances. The one I posted is more general and also it easily generalizes further to non-inertial coordinates and curved spacetimes.

In the formula
ds² = -c²dt² + dx² + dy² + dz²

the dt, dx, dy, and dz are all coordinate times and distances in some inertial frame while the ds is the frame-invariant spacetime interval. If ds² > 0 then the interval is called "spacelike" and ds is the proper distance. If ds² < 0 then the interval is called "timelike" and dτ = sqrt(-ds²/c²) is the proper time. If ds² = 0 then the interval is called "lightlike" or "null" and represents the path of a pulse of light.

The value of this formula is that all reference frames can use their own coordinate values and come up with the same value for ds². This formula also let's you easily see where the formula you posted comes from:

dτ² = -ds²/c² = dt² - dx²/c² - dy²/c² - dz²/c²
dτ²/dt² = 1 - (dx²/dt² - dy²/dt² - dz²/dt²)/c²
dτ²/dt² = 1 - v²/c²
dτ/dt = 1/γ

Thank you

I think you are confusing the origin of spacetime, the event (0,0,0,0), with the coordinate time axis, the worldline (ct,0,0,0). Objects may be resting or moving wrt the time axis (parallel worldlines), but not wrt the origin (lines cannot be parallel or not parallel to a point).

Oh yes, I see that now, I was becoming a little muddled there; talking about the path of a point in space at one particular point in time!
What I should have said was that the world line of the origin of any other inertial frame of reference would be parallel if that FoR was at rest with the first one.

Yes, exactly. That clock defines the time axis of the coordinate system, (ct,0,0,0).

And that time shown by that clock will be proper time? and the distance between any two clocks at rest will be a proper distance?
Which leads me to the conclusion that Proper time and Proper lengths are the units of an inertial FoR's coordinates? Of any inertial FoR?

 

[Grimble]

Hi, Grimble. It looks like you are still getting stuck on this simultaneity thing.

Yes I suppose that is true. It seems to me that something, that I see as very simple, is very hard to refer to without everyone becoming confused.

Are not all conventions, by definition, man-made?

To me simultaneity is is a straight forward concept – that two things happen at the same time.

The difficulty that is immediately seized upon is 'how does one define “at the same time”'.

The obvious way to do that, is to see how those two events are plotted onto a coordinate system that has time as an axis. Such as Space-time...

Then if two events have the same time coordinate, in one FoR, they must be simultaneous? For that is what having the same time coordinate means? Surely.

Now we know that the three space coordinates may be aligned any way we like in Space time, but what about the time axis? How is that aligned? Is it aligned as such?

Coordinate time depends on the observer's perspective, but what of proper-time, i.e. the time of space-time itself? Every observer can translate their coordinate time into proper time but is that into a common 'proper-time' that all agree upon or does it still vary according to the observer's perspective?

After all Space-time is not fixed is it? It is only a set of coordinates and the Space-time coordinates that are at rest with respect to One FoR will be moving with respect to another FoR.

 

[grav-universe]

Most of the statements you made are pretty much as you have said, but I will elaborate on these.

To me simultaneity is is a straight forward concept – that two things happen at the same time.

Just to be clear, if two events occur simultaneously according to a frame, then they happen at the same time according to that frame's coordinate system, yes, but simultaneity issues in general actually refer to differences in simultaneity, where events that occur simultaneously according to the coordinate system of one frame do not necessarily occur simultaneously according to the coordinate system of another frame, depending upon the simultaneity convention that is set for each frame.

For instance, if we have two observers in a frame that are separated by some distance, and they both then instantly and simultaneously accelerate to the same speed in the same direction and they leave their clocks alone, then if the original frame measures events simultaneously, the two observers in their new frame will still measure them as simultaneous also, but the two observers will also measure the speed of light anisotropically. But if the two observers re-synchronize their clocks to measure the same speed of light in every direction in accordance with the Einstein simultaneity convention, then they will no longer measure the events as simultaneous as the original frame does, because in order to synchronize in this way to measure an isotropic speed of light, the front observer's clock must be set back somewhat or the rear observer's clock set forward.

Now we know that the three space coordinates may be aligned any way we like in Space time, but what about the time axis? How is that aligned? Is it aligned as such?

Yes. We start by placing clocks at all points within a frame, all stationary to each other within the frame. Now let's say we want to re-synchronize the clocks along the x axis. We can begin at the origin and add, say, one second on the clocks per meter along the positive x direction, so add one second at one meter, two seconds at two meters, and so on, and subtract one second per meter in the negative x direction. Observers within the frame will consider the clocks to be synchronized regardless of how they are set since the readings upon clocks for events can only be measured directly by clocks that coincide in the same place, just as long as the synchronization method is linear along any direction so that objects that travel inertially will still be measured as such.

Coordinate time depends on the observer's perspective, but what of proper-time, i.e. the time of space-time itself? Every observer can translate their coordinate time into proper time but is that into a common 'proper-time' that all agree upon or does it still vary according to the observer's perspective?

Only one observer in a particular frame can measure the proper time between two events, since that observer's clock must coincide with both events to measure the time of the events directly. In other words, the proper time can only be measured directly by a single clock. However, since all of the observer's clocks in the same frame as that observer, no matter how they have actually been synchronized, are considered by observers within that frame to be in perfect synch with each other, then all observers within the frame with the observer that measures the proper time of the two events will agree upon the time of both events also, each occurring in the same place as that particular observer, so with zero spatial difference as measured by that frame. All other frames, however, moving at some relative speed to the frame in which the proper time is measured, will measure some distance between the events due to the relative speed. They will also measure a different difference in time between the events, depending upon how their clocks are synchronized. If both frame's clocks have been synchronized according to the Einstein simultaneity convention, then the difference in times and distances measured by each frame will be related by the Minkowski metric.

 

[Dale]

And that time shown by that clock will be proper time?

Yes. All ideal clocks measure proper time along their worldline.

Which leads me to the conclusion that Proper time and Proper lengths are the units of an inertial FoR's coordinates? Of any inertial FoR?

I don't know what you mean by "units of an inertial FoR's coordinates". You could certainly use a clock that gives units of seconds in a FoR using years, as long as you convert properly. Do you somehow think that proper time and coordinate time are measured in different units?

 

[Dale]

Then if two events have the same time coordinate, in one FoR, they must be simultaneous? For that is what having the same time coordinate means? Surely.

They must be simultaneous in that FoR. They will not be simultaneous in most other FoR's.

Coordinate time depends on the observer's perspective, but what of proper-time, i.e. the time of space-time itself? Every observer can translate their coordinate time into proper time but is that into a common 'proper-time' that all agree upon or does it still vary according to the observer's perspective?

There is not a common proper time, per se. Proper time is defined along a particular path in spacetime (i.e. along a worldline). It is geometrically equivalent to the arc length along that path, and just like arc length is invariant under rotations, so also the proper time is invariant under Lorentz transforms.

Consider two hikers that are looking at a map and determining how far they have to hike to reach their campsite. One hiker uses a GPS system so he uses celestial north and determines that he needs to hike 4 miles north and 3 miles east. The other hiker uses a compass so he uses magnetic north and determines that he needs to hike 3 miles north and 4 miles east. But both hikers agree that the distance they will hike is 5 miles. The GPS hiker will determine that the compass hiker's "north-meter" is "going slow", but that doesn't mean that they are using different units, just that they are talking about different directions when they each say "north".

 

[Grimble]

Most of the statements you made are pretty much as you have said, but I will elaborate on these.

Just to be clear, if two events occur simultaneously according to a frame, then they happen at the same time according to that frame's coordinate system, yes, but simultaneity issues in general actually refer to differences in simultaneity, where events that occur simultaneously according to the coordinate system of one frame do not necessarily occur simultaneously according to the coordinate system of another frame, depending upon the simultaneity convention that is set for each frame.

I must admit to being a little puzzled, perhaps just by the way you have expressed this but what do you mean by a 'simultaneity convention'?
To me, a set of coordinates are a means of relating points to different axes.
If we have a time axis then simultaneity relates to points having the same time coordinate.
It depends on where they are not how they are labelled.

For instance, if we have two observers in a frame that are separated by some distance, and they both then instantly and simultaneously accelerate to the same speed in the same direction and they leave their clocks alone, then if the original frame measures events simultaneously, the two observers in their new frame will still measure them as simultaneous also, but the two observers will also measure the speed of light anisotropically. But if the two observers re-synchronize their clocks to measure the same speed of light in every direction in accordance with the Einstein simultaneity convention, then they will no longer measure the events as simultaneous as the original frame does, because in order to synchronize in this way to measure an isotropic speed of light, the front observer's clock must be set back somewhat or the rear observer's clock set forward..

I'm sorry but can you expand upon this?
.

Yes. We start by placing clocks at all points within a frame, all stationary to each other within the frame. Now let's say we want to re-synchronize the clocks along the x axis.

Why? what can that achieve?
Synchronising one's clocks, as per Einstein to a common time, is a requirement for reading the coordinates of different points.
Synchronising to any scheme is only going to confuse things.

We can begin at the origin and add, say, one second on the clocks per meter along the positive x direction, so add one second at one meter, two seconds at two meters, and so on, and subtract one second per meter in the negative x direction. Observers within the frame will consider the clocks to be synchronized regardless of how they are set since the readings upon clocks for events can only be measured directly by clocks that coincide in the same place, just as long as the synchronization method is linear along any direction so that objects that travel inertially will still be measured as such.

You have lost me here as to what you are trying to elucidate.

Only one observer in a particular frame can measure the proper time between two events, since that observer's clock must coincide with both events to measure the time of the events directly. In other words, the proper time can only be measured directly by a single clock. However, since all of the observer's clocks in the same frame as that observer, no matter how they have actually been synchronized, are considered by observers within that frame to be in perfect synch with each other, then all observers within the frame with the observer that measures the proper time of the two events will agree upon the time of both events also, each occurring in the same place as that particular observer, so with zero spatial difference as measured by that frame. All other frames, however, moving at some relative speed to the frame in which the proper time is measured, will measure some distance between the events due to the relative speed. They will also measure a different difference in time between the events, depending upon how their clocks are synchronized. If both frame's clocks have been synchronized according to the Einstein simultaneity convention, then the difference in times and distances measured by each frame will be related by the Minkowski metric.

It seems to me after further thought that time must be isotropic and homogeneous.
I can imagine this like seeing the images from a CTI scan but being able to see them not as images of a single time but as moving images where one could scan backwards and forwards in time.

 

[grav-universe]

I must admit to being a little puzzled, perhaps just by the way you have expressed this but what do you mean by a 'simultaneity convention'?
To me, a set of coordinates are a means of relating points to different axes.
If we have a time axis then simultaneity relates to points having the same time coordinate.
It depends on where they are not how they are labelled.

I'm sorry but can you expand upon this?
.
Why? what can that achieve?

It doesn't achieve anything one way or the way, only depending upon what we decide is the best way to synchronize clocks. That is all a simultaneity convention is after all, a method by which to synchronize clocks, or a way to determine what a frame's clock settings will be in relation to each other.

Synchronising one's clocks, as per Einstein to a common time, is a requirement for reading the coordinates of different points.
Synchronising to any scheme is only going to confuse things.

You have lost me here as to what you are trying to elucidate.


It seems to me after further thought that time must be isotropic and homogeneous.
I can imagine this like seeing the images from a CTI scan but being able to see them not as images of a single time but as moving images where one could scan backwards and forwards in time.

There is no absolute way that clocks must be synchronized, that is the point. For instance, repeating the earlier scenario in a slightly different way, let's say that clocks in frame A are synchronized according to the Einstein simultaneity convention, so that observers in that frame measure light to travel at c in all directions. Now let's say that two observers with some distance between them simultaneously accelerate to a speed v relative to frame A. If the two observers leave their clocks alone, they will no longer measure the speed of light to be the same in each direction between themselves, but since they have both accelerated simultaneously in the same way to the same speed v, their clocks will read the same according to observers in frame A, and any events that frame A says occur simultaneously, the two observers in their new frame will say the events occur simultaneously as well.

But now let's re-synchronize the clocks of the two observers in their new frame by applying the Einstein simultaneity convention, meaning they will synchronize their clocks so that light will now be measured at the same speed between themselves. In order to do that, the front observer must set her clock back by d v / c^2 or the back observer must set his clock forward by d v / c^2. Now the two observers will measure the same speed of light in either direction, but since one of the observers has changed the setting on their clock, frame A says that the back observer's clock is now set ahead of the front observer's clock by d v / c^2, so events that occur simultaneously according to frame A cannot occur simultaneously to the two observers anymore since their clocks read differently.

 

[Grimble]

Yes. All ideal clocks measure proper time along their worldline.

I don't know what you mean by "units of an inertial FoR's coordinates". You could certainly use a clock that gives units of seconds in a FoR using years, as long as you convert properly. Do you somehow think that proper time and coordinate time are measured in different units?

Well they are, aren't they?

In post #346 you explained (with my highlighting):

In the formula
ds² = -c²dt² + dx² + dy² + dz²

the dt, dx, dy, and dz are all coordinate times and distances in some inertial frame while the ds is the frame-invariant spacetime interval. If ds² > 0 then the interval is called "spacelike" and ds is the proper distance. If ds² < 0 then the interval is called "timelike" and dτ = sqrt(-ds²/c²) is the proper time. If ds² = 0 then the interval is called "lightlike" or "null" and represents the path of a pulse of light.

The value of this formula is that all reference frames can use their own coordinate values and come up with the same value for ds². This formula also let's you easily see where the formula you posted comes from:

dτ² = -ds²/c² = dt² - dx²/c² - dy²/c² - dz²/c²
dτ²/dt² = 1 - (dx²/dt² - dy²/dt² - dz²/dt²)/c²
dτ²/dt² = 1 - v²/c²
dτ/dt = 1/γ

Where dτ/dt = (by your definitions above) proper time/coordinate time = 1/γ
i.e. dτ = dt/γ which is surely a conversion between different unit sizes? As in feet = yards/3?

They must be simultaneous in that FoR. They will not be simultaneous in most other FoR's.

I agree

There is not a common proper time, per se. Proper time is defined along a particular path in spacetime (i.e. along a worldline). It is geometrically equivalent to the arc length along that path, and just like arc length is invariant under rotations, so also the proper time is invariant under Lorentz transforms.

Yet if we examine ANY inertial FoR , we agreed that the ideal clock, at rest at the origin of that FoR has a worldline that defines the time axis of that FoR's coordinate system viz:

Now it seems to me that within our 4 dimensional coordinate system any inertial frame of reference may be considered to have a clocks at rest with respect to it. Those clocks will have straight worldlines that are parallel one to the other. The world line of the one at the origin of that frame of reference will describe the time coordinate of our 4D set of coordinates.

Yes, exactly. That clock defines the time axis of the coordinate system, (ct,0,0,0).

Therefore I conclude that ANY inertial FoR has a time axis that is defined by an ideal clock's worldline and that the time shown by that clock, the time of the time axis, is proper time; and according to Einstein's first postulate viz:

“The Principle of Relativity – The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.”

that proper-time will be the same for ANY IFoR.

Consider two hikers that are looking at a map and determining how far they have to hike to reach their campsite. One hiker uses a GPS system so he uses celestial north and determines that he needs to hike 4 miles north and 3 miles east. The other hiker uses a compass so he uses magnetic north and determines that he needs to hike 3 miles north and 4 miles east. But both hikers agree that the distance they will hike is 5 miles. The GPS hiker will determine that the compass hiker's "north-meter" is "going slow", but that doesn't mean that they are using different units, just that they are talking about different directions when they each say "north".

But is that not the equivalent of having two IFoR's at rest with one another but with their space coordinates just having a different alignment? Each will measure the same unit size, it is only when the observer is moving with respect to the observed that measurements are transformed according to the LT equations. transformed between different unit sizes.

 

[Dale]

Where dτ/dt = (by your definitions above) proper time/coordinate time = 1/γ
i.e. dτ = dt/γ which is surely a conversion between different unit sizes? As in feet = yards/3?

In the trigonometric expression r = x/cos(θ) would you say that it is a conversion between different unit sizes? The expressions are of the same form. What about the diameter and the circumference of a circle, or the length and width of a rectangle, are those conversions between different unit sizes? They also have the same form.

If you would agree with me that they are not conversions between different unit sizes, then can you identify what is the distinction between this class of expressions and unit conversions? After all, as you pointed out, they do have the same mathematical form, so what distinguishes them?

But is that not the equivalent of having two IFoR's at rest with one another but with their space coordinates just having a different alignment?

Yes, exactly, it is just the time axes having a different alignment instead of the space axes. A boost is geometrically equivalent to a rotation, rotations preserve the Euclidean distance and boosts preserve the Minkowski interval.

 

[Grimble]

In the trigonometric expression r = x/cos(θ) would you say that it is a conversion between different unit sizes? The expressions are of the same form. What about the diameter and the circumference of a circle, or the length and width of a rectangle, are those conversions between different unit sizes? They also have the same form.

If you would agree with me that they are not conversions between different unit sizes, then can you identify what is the distinction between this class of expressions and unit conversions? After all, as you pointed out, they do have the same mathematical form, so what distinguishes them?

The fact that they refer to quantities of the same kind: lengths, times etc.

The fact that they may be written as ratios e.g. τ : t

indeed I understand it may be described as *a dimensionless*quotient*of the two terms.

Yes, exactly, it is just the time axes having a different alignment instead of the space axes. A boost is geometrically equivalent to a rotation, rotations preserve the Euclidean distance and boosts preserve the Minkowski interval.

But how can the time axes have alignment?

As I said in post #353 viz:

“It seems to me after further thought that time must be isotropic and homogeneous.”

And if it is then how can it have any alignment? Does it not just exist at every point in space?

 

[Dale]

The fact that they refer to quantities of the same kind: lengths, times etc.

The fact that they may be written as ratios e.g. τ : t

indeed I understand it may be described as *a dimensionless*quotient*of the two terms.

That is not what distinguishes the above examples from a unit conversion. In both cases you can write it as a dimensionless quotient:
yard/foot = 3
circumference/diameter = 3.1416...
adjacent/hypotenuse = cos

All of the above are dimensionless quotients of the two terms. The difference is that in the case of the unit conversion you are measuring the same thing using two different units (yard or foot), but in the geometric ratios you are measuring different things (circumference or diameter) using the same unit.

When you are talking about time dilation you need to realize that clocks measure proper time, and proper time is the length of the worldline of the clock. So, if the clocks are not traveling along the same worldline then they are measuring different things. Thus the time dilation factor is a geometric factor (like cos), rather than a unit conversion factor.

But how can the time axes have alignment?

Draw a spacetime diagram showing two sets of coordinates. The answer should be clear.

 

[Grimble]

That is not what distinguishes the above examples from a unit conversion. In both cases you can write it as a dimensionless quotient:
yard/foot = 3
circumference/diameter = 3.1416...
adjacent/hypotenuse = cos

All of the above are dimensionless quotients of the two terms. The difference is that in the case of the unit conversion you are measuring the same thing using two different units (yard or foot), but in the geometric ratios you are measuring different things (circumference or diameter) using the same unit.

But those are not quantities of the same kind as I specified.

if you had said circumference/circumference or hypotenuse/hypotenuse you would be comparing like with like!

When you are talking about time dilation you need to realize that clocks measure proper time, and proper time is the length of the worldline of the clock. So, if the clocks are not traveling along the same worldline then they are measuring different things. Thus the time dilation factor is a geometric factor (like cos), rather than a unit conversion factor.

Draw a spacetime diagram showing two sets of coordinates. The answer should be clear.

I'm sorry I can't comment on that unless you specify which clocks you are referring to.

Proper time is the scale of time that is measured by an observer adjacent to an ideal clock.
Coordinate time is the scale of time measured by an observer remote from the same clock.
The remote observer uses the time measured by the adjacent observer (proper time) and, using the LT equations, transforms that proper time into coordinate time. (transforming between two different scales of measurement)

The only other way of taking the measurement remotely would be for the observer to use her rulers and clocks to measure a moving object - which would be very difficult.

 

[Dale]

if you had said circumference/circumference or hypotenuse/hypotenuse you would be comparing like with like!

Yes, exactly. And since you are measuring the same thing the ratio will be 1 (dimensionless). Similarly, if two clocks measure the same worldline the ratio will be 1.

I'm sorry I can't comment on that unless you specify which clocks you are referring to.

You have had plenty of time to do the spacetime diagram I suggested earlier. How is that coming. Have you finished it yet?