The spacetime interval versus a chain of causative events

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Posted (edited)

Imagine a pair of fictional events  the first being the murder of Caesar by one of his enemies using a sword  and the second fictional event the murder of JFK by someone using the same sword some 2000 years  later.

The spacetime interval between the two events is calculated  by choosing the earth as a reference frame ,measuring the distance between Rome and Dallas ,choosing the units of time and applying the s^2=(ct)^2-r^2 formula.

Suppose instead we were to follow the sword across the centuries  and make a note of the time and places in which it was recorded as having been prior to it being used again in the fictional killing of JFK.

Could we calculate spacetime intervals at every stop on the way,add them all up and end up with a figure that was comparable to the first spacetime interval when we just used the two events?

What if the sword had disappeared down a hole in the earth and reappeared in N. Zealand for a time (or had been taken to the moon and back) ?

Would that have increased the"cumulative" spacetime interval  even though the beginning and the end of its journey were still the same?

Sorry if that sounds a bit weird or overly tortuous

Edited by geordief
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Posted (edited)
1 hour ago, geordief said:

s^2=(ct)^2-r^2

No, the correct method is to choose a spacetime coordinate system, establish the metric for that coordinate system, establish the spacetime trajectory of the sword in that coordinate system, then integrate the expression for ds along that trajectory. Bear in mind that the sword has been around the sun a couple of thousand times and around the earth hundreds of thousands of times if one chooses a heliocentric coordinate system. One could choose a geocentric coordinate system but then the metric will be more complicated. One could even create a "sword-centric" coordinate system in which the spacetime trajectory and integration are trivial at the expense of an even more complicated metric.

Edited by KJW
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The interval is the space-time distance between two events.
The path-history of the sword is its worldline.

Not necessarily the same.

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Posted (edited)
34 minutes ago, MigL said:

The interval is the space-time distance between two events.
The path-history of the sword is its worldline.

Not necessarily the same.

Are they related?Can they be the same under any  particular conditions?

If the worldline  of an object is integrated is the result given in the same units as  the spacetime interval?(which just considers beginning and end points of the worldline,if I have that right)

Edit :just saw @KJW 's response.Probably too late to answer him today...

Edited by geordief
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55 minutes ago, geordief said:

Are they related?Can they be the same under any  particular conditions?

Not related; except by the two end points/events.
One is a separation, the other, a path dependent length.

Reduce it to two dimensions to make it clearer.
Going around, for example,  a circle has the same start and end points so separation of the two points is zero; but the path length travelled is definitely non-zero ( for any R other than  R=0 ).

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Posted (edited)

One thing I like to say is that $ds$ is not an exact differential. If $ds$ were an exact differential, there would exist a function of the coordinates, $s(x, y, z, t)$, and the spacetime distance between any two points would simply be the difference of the two values of $s$ at the endpoints and independent of the path between them. But $ds$ is not an exact differential and the spacetime distance between two points does depend on the path between them. For distances between pairs of locations in three-dimensional space, it seems rather obvious that this depends on the path between them, so it's rather natural that the same applies in four-dimensional spacetime and that the notion of universal time does not exist.

Mathematically, if $ds$ were an exact differential:

$ds = \dfrac{\partial s}{\partial x} dx + \dfrac{\partial s}{\partial y} dy + \dfrac{\partial s}{\partial z} dz + \dfrac{\partial s}{\partial t} dt = \dfrac{\partial s}{\partial x^{\mu}} dx^{\mu}$

and:

$(ds)^2 = g_{\mu\nu} dx^{\mu} dx^{\nu} = \dfrac{\partial s}{\partial x^{\mu}} \dfrac{\partial s}{\partial x^{\nu}} dx^{\mu} dx^{\nu}$

$g_{\mu\nu} = \dfrac{\partial s}{\partial x^{\mu}} \dfrac{\partial s}{\partial x^{\nu}}$

which, as the tensor product of two vectors, is singular, in violation of the requirement that the metric tensor be invertible.

Edited by KJW
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1 hour ago, geordief said:

Are they related?Can they be the same under any  particular conditions?

Yes, a timelike spacetime interval is the square of the proper time measured by an inertial clock moving between the 2 events in flat spacetime.

If the sword remained at rest the whole time, and gravity was neglected, the interval would be the square of how much the sword aged between the two events.

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Posted (edited)
1 hour ago, KJW said:

One thing I like to say is that ds is not an exact differential. If ds were an exact differential, there would exist a function of the coordinates, s(x,y,z,t) , and the spacetime distance between any two points would simply be the difference of the two values of s at the endpoints and independent of the path between them. But ds is not an exact differential and the spacetime distance between two points does depend on the path between them. For distances between pairs of locations in three-dimensional space, it seems rather obvious that this depends on the path between them, so it's rather natural that the same applies in four-dimensional spacetime and that the notion of universal time does not exist.

Mathematically, if ds were an exact differential:

ds=sxdx+sydy+szdz+stdt=sxμdxμ

and:

(ds)2=gμνdxμdxν=sxμsxνdxμdxν

gμν=sxμsxν

which, as the tensor product of two vectors, is singular, in violation of the requirement that the metric tensor be invertible.

Just a side note if you have a metric with off diagonal terms such as the Kerr metric one can perform a blockwise inversion.  Strictly an FYI for others as I'm sure KJW already knows this detail just in case anyone thinks only orthogonal tensors can be metric tensors that isn't always the case.

Edited by Mordred
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Posted (edited)

Might it be that the quantative difference between the  overall spacetime interval and the integral of the causative chain of events is the  aging  and  spatial displacements of the sword?

If the sword was a smaller system  (and followed a geodesic?) would the difference be less?

Also is entropy another description of aging?

edit ,am I just repeating what @md65536 posted a little above

8 hours ago, md65536 said:

Yes, a timelike spacetime interval is the square of the proper time measured by an inertial clock moving between the 2 events in flat spacetime.

If the sword remained at rest the whole time, and gravity was neglected, the interval would be the square of how much the sword aged between the two events.

Edited by geordief
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The interval is simply a distance, in the space-time co-ordinate system, calculated using the Pythagorean Theorem.
Now you might ask "how is the temporal dimension converted to a distance".
It is multiplied by a 'conversion' factor c , and because it must be orthogonal to the three spatial dimensions, we make it imaginary.
IOW, the temporal distance is ict, such that when we square all the distances, using the Pythagorean Theorem to find the length of the hypotenuse ( aka distance between end points ), the temporal distance squared is always of the opposite sign as the spatial distances.

The path taken by the sword, on the other hand, could be any one of the infinite number of squiggly lines/curves between the two endpoints. The only constraint on them is that they have to lie within the light cone for massive objects like swords.

Don't overthink things.

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Posted (edited)
8 hours ago, MigL said:

It is multiplied by a 'conversion' factor c , and because it must be orthogonal to the three spatial dimensions, we make it imaginary.

To add to this, these are choices that are made in defining the interval, to make it useful and simple, not to make it somehow supremely meaningful. It's not even a distance, but a square (so that you can deal with negative squares, instead of imaginary times or distances).

However since it's made up of distances and times, you can use maths to convert it into something with the meaning you want. The proper time along an arbitrary world line should be the same as an integral of infinitesimal times measured in momentary rest frames at each event on the world line, so you could integrate the square root of infinitesimal spacetime intervals.

Edited by md65536
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Posted (edited)

10 hours ago, MigL said:

Now you might ask "how is the temporal dimension converted to a distance".
It is multiplied by a 'conversion' factor c , and because it must be orthogonal to the three spatial dimensions, we make it imaginary

Is that a bit of "back engineering"

Was it Minkowski who arranged the axes so that they all had the same units of spatial distance?

Was he just trying to convert time to spatial distance or did be have in mind (as I always thought was  implied) that ct was the distance traveled by light  in the units of time chosen by the timepiece?

If it was him,did he explain his reasoning at the time?(not that we would need to respect his intentions ,necessarily if the model speaks for itself-I have read that  Einstein didn't  approve at first)

Would the model work as well if c was replaced with  an arbitrary speed such as ,for no particular reason   the speed of the earth around the Sun ?

Edited by geordief
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8 hours ago, geordief said:

Would the model work as well if c was replaced with  an arbitrary speed such as ,for no particular reason   the speed of the earth around the Sun ?

No. Only c is invariant to Lorentz transformations. Bear in mind that c is derived from the electromagnetic constants in Maxwell's equations.

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I think there are many ways to see c (do not pronounce that sentence). Just to mention a few:

• speed of light in vacuum
• speed of gravitation
• the speed limit of every object with mass
• the only speed massless particles can have
• maximum speed of causality
• the conversion factor between time and space, so that they can be used on equal footing

Surely you could draw a spacetime diagram with another speed as conversion factor (you need it to get a distance, as you can only 'make distances' on paper). However for calculations you would still need c, and if you would like to derive formulas from the diagram, they would be awfully complicated. The Minkowski diagram, with c as conversion factor show everything much more directly.

1 hour ago, KJW said:

Bear in mind that c is derived from the electromagnetic constants in Maxwell's equations.

This might be a bit over-precise, but I would have written something like 'was historically (or originally) derived'. Logically seen, the values of e0 and µ0 'must adapt' to 'the only speed massless particles can have', not the other way round. Maxwell took 'the empirical way' based on Faraday's results, and so was the first to derive c, however without knowing how fundamental c is for the structure of spacetime. And that is why I like the last description of c best.

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On 4/8/2024 at 7:51 PM, geordief said:

Was it Minkowski who arranged the axes so that they all had the same units of spatial distance?

It's the constant speed of light that allows times to be expressed in terms of a distance that light travels, and distances to be expressed (and literally defined) by how far light travels in a given time.

Consider a beam of light from one event to another event. Those two events will be generally different distances apart for different moving observers, meaning different frames will have the same beam of light travel different distances, meaning that the time between the events must be different in different frames.

Draw a beam of light as a line. If you draw it in the ct dimension, that represents the time between two events at the same spatial location, like an abstract beam whose distance is measured only by time, or you could make it a real beam in the y direction. Now consider the same "beam" in a frame that is relatively moving in the x direction. It is orthogonal to ct, so if you draw this out, you end up with a right triangle. The hypotenuse can represent the same time as measured in another frame where the the start and end events are at different locations, ie. the distance between them is longer ie. the time is dilated. The relationship between the lengths of the edges of the triangle is given by the Pythagorean theorem.

The meaning of it comes from the maths. Maybe you could say it's a geometrical representation of the invariance of the speed of light.

For any timelike spacetime interval, there's always a rest frame where the spatial distance between the two events is zero, so you can always represent the interval with simple right triangles. I suspect if you want more meaning than that, it would be found in a more mathematical description.

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On 4/9/2024 at 12:48 PM, Eise said:

I think there are many ways to see c (do not pronounce that sentence). Just to mention a few:

• speed of light in vacuum
• speed of gravitation
• the speed limit of every object with mass
• the only speed massless particles can have
• maximum speed of causality
• the conversion factor between time and space, so that they can be used on equal footing

Surely you could draw a spacetime diagram with another speed as conversion factor (you need it to get a distance, as you can only 'make distances' on paper). However for calculations you would still need c, and if you would like to derive formulas from the diagram, they would be awfully complicated. The Minkowski diagram, with c as conversion factor show everything much more directly.

This might be a bit over-precise, but I would have written something like 'was historically (or originally) derived'. Logically seen, the values of e0 and µ0 'must adapt' to 'the only speed massless particles can have', not the other way round. Maxwell took 'the empirical way' based on Faraday's results, and so was the first to derive c, however without knowing how fundamental c is for the structure of spacetime. And that is why I like the last description of c best.

It has always struck me as rather surprising that empty space should have quantitatively measurable properties, such as  e0 and µ0.

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2 hours ago, exchemist said:

It has always struck me as rather surprising that empty space should have quantitatively measurable properties, such as  e0 and µ0.

Is there such a thing as "empty space"?

If "it"

is permeable  or permittive  then it has a property/properties  .

Can anything with properties be considered "empty"?

And if it is permeable/permittive  then does that mean it it "permeated"? -(with radiation?)

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Those coefficients define ease or effort for radiation to pass through that medium ( or lack thereof ).
For example, when we say the index of refraction of free space is 1, as opposed to glass at 1.5, is n=1 a property of space ?
Or is it a lack of that property.

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26 minutes ago, MigL said:

Those coefficients define ease or effort for radiation to pass through that medium ( or lack thereof ).
For example, when we say the index of refraction of free space is 1, as opposed to glass at 1.5, is n=1 a property of space ?
Or is it a lack of that property.

Is the "medium" of free space the radiation itself?

Those coefficients apply to em radiation: Are there forms of radiation that apply to other fields (as per the fields in QFT)?

Eg does the strong force radiate?

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I'm not well versed enough in QFT to know whether those fields have coefficients in the same way.
Or they may be just defined as 1, as is the index of refraction.

Keep in mind that permeability refers to the magnetic field and permittivity to the electric, and, since both have units, with a suitable choice they could both be set to 1.

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Posted (edited)

The other fields have different coupling constants. They don't apply the index of refraction as per the EM field. Though all fields are still subjective to spacetime curvature that's a different devil. As well as the speed limit.

Edited by Mordred
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On 4/7/2024 at 8:07 PM, geordief said:

Could we calculate spacetime intervals at every stop on the way,add them all up and end up with a figure that was comparable to the first spacetime interval when we just used the two events?

Yes, they'd be comparable. You're asking if a straight line (or a geodesic one) is a similar length to one with some minor curvature to it, and both are about 2000 years long, differing only in perhaps the 6th significant digit (a guess).

The length of the actual worldline would be a tiny bit shorter than the interval between the two events. This is essentially the gist of the twins scenario where one twin has a shorter worldline than the one with the straighter worldline.

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