Question about full Lorentz transformation

[Sagittarius A-Star]

But, if the student has yet to learn or understand this distinction, then it can't resolve that confusion.

Therefore, I wrote:

The clock is at rest in the unprimed frame at $x=0$.
With equation (1) follows:
$\require{color} 0=\gamma(x'+vt') \ \ \ \ \ (3)$

... in analogy to ...

Or, you could write $t = \gamma t' \ (x'=0)$

 

[PeterDonis]

In $t' = \gamma{t}$, variable t is the "proper" time between two events measured in S by clock C1 at rest in the same frame, whereas t' is the "coordinate" time measured in frame S' with respect to which clock C1 is moving.
In $t = \gamma{t'}$, variable t' is the "proper" time between two events measured in S' by clock C2 at rest in the same frame, whereas t is the "coordinate" time measured in frame S with respect to which clock C2 is moving.

Your descriptions here are misleading for the OP. Both $t$ and $t'$ are coordinate times, in different frames. It is true that, for a clock at rest in a given frame, coordinate time in that frame is the same as proper time for that clock for events on the clock's worldline. But given the confusion the OP has about the fact that the Lorentz transformation transforms coordinates, I think it is important to be clear that $t$ and $t'$ are both coordinates.

 

[FactChecker]

I agree that they're measuring different things, but maybe that's because observers in different reference frames will measure different things?

For each equation, one of the IRFs just has one clock, A, which is stationary in that IRF. The other IRF is recording its own IRF times at two different locations as the clock A moves from one location to the other.
When you compare the two equations, remember that the situation is different for each equation. The IRF with one clock stationary, A, and the one with times at two locations are swapped.

 

[Chenkel]

I agree with this.I do not agree with this. The first equation works when $x=0$. The second equation works when $x'=0$.

There is no event recorder in the equations.I don't think that this change in terminology is helpful. You may as well call it an observer like everyone else.

So with $x=0$ we find with the Lorentz transformation that $t' = \gamma{t}$ and $x' = -\gamma{vt}$ which means $x' = -vt'$

In this case the unprimed clock (centered at x=0) moves with respect to the primed frame at velocity $-v$ on the primed frame's x axis and the unprimed clock ticks slowly relative to the primed frame's clock. (And the primed frame clock ticks normally)

With $x'=0$ we find with the inverse Lorentz transformation that $t = \gamma{t'}$ and $x = \gamma{vt'}$ which means $x = vt$

In this case the primed clock moves with respect to the unprimed frame along the unprimed frame's x axis at velocity $v$ and the primed clock ticks slowly relative to the unprimed clock (And the unprimed clock ticks normally)

 

[Chenkel]

For each equation, one of the IRFs just has one clock, A, which is stationary in that IRF. The other IRF is recording its own IRF times at two different locations as the clock A moves from one location to the other.
When you compare the two equations, remember that the situation is different for each equation. The IRF with one clock stationary, A, and the one with times at two locations are swapped.

I think I might see what you are saying but it's a little over my head.

 

[Dale]

So with $x=0$ we find with the Lorentz transformation that $t' = \gamma{t}$ and $x' = -\gamma{vt}$ which means $x' = -vt'$

In this case the unprimed clock (centered at x=0) moves with respect to the primed frame at velocity $-v$ on the primed frame's x axis and the unprimed clock ticks slowly relative to the primed frame's clock. (And the primed frame clock ticks normally)

With $x'=0$ we find with the inverse Lorentz transformation that $t = \gamma{t'}$ and $x = \gamma{vt'}$ which means $x = vt$

In this case the primed clock moves with respect to the unprimed frame along the unprimed frame's x axis at velocity $v$ and the primed clock ticks slowly relative to the unprimed clock (And the unprimed clock ticks normally)

Yes.

Note, where you say “we find with the inverse Lorentz transformation”, you can also find the same with the Lorentz transform, it is just a little more algebra. Similarly where you say “we find with the Lorentz transformation”, we can find the same with the inverse transform.

 

[PeroK]

So with $x=0$ we find with the Lorentz transformation that $t' = \gamma{t}$ and $x' = -\gamma{vt}$ which means $x' = -vt'$

In this case the unprimed clock (centered at x=0) moves with respect to the primed frame at velocity $-v$ on the primed frame's x axis and the unprimed clock ticks slowly relative to the primed frame's clock. (And the primed frame clock ticks normally)

With $x'=0$ we find with the inverse Lorentz transformation that $t = \gamma{t'}$ and $x = \gamma{vt'}$ which means $x = vt$

In this case the primed clock moves with respect to the unprimed frame along the unprimed frame's x axis at velocity $v$ and the primed clock ticks slowly relative to the unprimed clock (And the unprimed clock ticks normally)

The only thing I would add is that all these coordinate transformations are valid whether or not there are any physical clocks involved.

 

[Nugatory]

In the Lorentz transformation (non inverse) is the observer at rest relative to the primed frame?

There are no observers in the Lorentz transformations. There are points in spacetime, which we call “events”; you can think of these as dots on a spacetime diagram.

Different frames assign different x and t coordinates to the same event. The Lorentz transformations relate the x and t coordinates assigned by one frame, which we call the “unprimed frame, to the coordinates assigned by another frame, which we call the “primed frame”, when the primed frame is moving at speed v relative to the unprimed frame.

 

[Chenkel]

You should think of an "observer" as an inertial reference frame that can record the position of events in its spacetime coordinate system, no matter where or when they occur.
This is so important for keeping SR simple that Bernard Schutz describes it immediately on page 3 of "A First Course in General Relativity":

"1.2 Definition of an inertial observer in SR
It is important to realize that an 'observer' is in fact a huge information-gathering system, not simply one person with binoculars. In fact, we shall remove the human element entirely from our definition and say that an inertial observer is simply a spacetime coordinate system, which makes an observation simply by recording the location (x, y, z) and the time (t) of any event. ...."

I recommend reading the first chapter of his book because, IMO, it addresses many misconceptions that people have about SR.

Inertial frames don't require an observer definition but I like imagining an observer at rest with respect to an IRF recording events.

 

[PeroK]

Inertial frames don't require an observer definition but I like imagining an observer at rest with respect to an IRF recording events.

You need an array of infinitely many local observers, each at a known location and all with synchronized clocks. Each local observer is responsible for noting the time of any event at their location. The full picture is put together when all the data is consolidated.

That said, the theoretical modelling of spacetime and natural events, in general, must take place without this framework of local observers. For cosmology, we have only observations from Earth telescopes or space telescopes in near-Earth orbits.

 

[Nugatory]

Inertial frames don't require an observer definition but I like imagining an observer at rest with respect to an IRF recording events.

Given the difficulty you are having understanding this stuff, you might want to consider the possibility that the way you’re imagining what’s going on is part of the problem….

“An observer at rest with respect to an IRF recording events” can only record the events that happen at one place, where the observer is: these are the events with x-coordinate equal to zero, forming a vertical line through the origin.

It is possible to think in terms of observers recording events, but you need more observers. Taylor and Wheeler (“Spacetime Physics”) take this approach: we imagine that the entire universe is full of observers stationed one meter apart in a grid, at rest relative to one another, and carrying synchronized clocks. Whenever something happens, the one at that spot makes a note on a slip of paper, something like “I am the observer at x=10,y=0,z=0. This spaceship passed right where I am when my clock read 12:55 PM”. After our thought experiment is done we gather all these slips of paper and use them to reconstruct what happened according to the frame in which they were all at rest. We’ll call this the unprimed frame.

Now suppose that we have another such flock of observers, also all at rest relative to one another and carrying synchronized clocks but all moving relative to our first group. We’ll call the frame in which they are at rest the primed frame.

The Lorentz transformations let me calculate, from all the notes written by the unprimed frame observers, what the various primed-frame observers will have written down. The inverse transformations go the other way, they let me calculate the unprimed-frame results from the primed-frame ones.

 

[FactChecker]

Inertial frames don't require an observer definition but I like imagining an observer at rest with respect to an IRF recording events.

Only if you want to get confused. Including the time required to transfer all information to a single location unnecessarily complicates things. Suppose that you were in New York and wanted to know the exact time an airplane landed in Los Angeles. You could call them later and ask. Or would you insist that some mechanism be set up to signal you at light speed and then subtract out that signal travel time. The first approach clearly requires that the clocks in LA are exactly synchronized with yours. That is where the "relativity of simultaneity" comes in, which is fundamental to SR. But we assume that any IRF has an exactly synchronized time coordinate.

 

[Ibix]

Inertial frames don't require an observer definition but I like imagining an observer at rest with respect to an IRF recording events.

The problem with that is that a frame is defined everywhere across spacetime, while an observer has a single location. What an observer records is what that observer experiences, which are only things at their location - anything that happens elsewhere they only see after the fact, when the light from events reaches them.

An inertial reference frame needs an infinite army of observers if they are to report their experience in a way consistent with the frame coordinates.

 

[Sagittarius A-Star]

Inertial frames don't require an observer definition but I like imagining an observer at rest with respect to an IRF recording events.

Also Morin recommends, before introducing the LT, to remove the observers.

Lattice of clocks and meter sticks
In everything we’ve done so far, we’ve taken the route of having observers sitting in various frames, making various measurements. But as mentioned above, this can cause some ambiguity, because you might think that the time when light reaches the observer is important, whereas what we are generally concerned with is the time when something actually happens.

A way to avoid this ambiguity is to remove the observers and then imagine filling up space with a large rigid lattice of meter sticks and synchronized clocks.

Source (page XI-18):
https://scholar.harvard.edu/files/david-morin/files/cmchap11.pdf

via:
https://scholar.harvard.edu/david-morin/classical-mechanics

 

[Dale]

Inertial frames don't require an observer definition but I like imagining an observer at rest with respect to an IRF recording events.

That is fine. There is nothing wrong with defining observers and plenty of people have that same preference. As long as you know that it is a matter of personal preference. And this …

But don't we need to an observer to record events in the reference frame?

is not necessary.

As a corollary, if you choose to use observers then it is up to you to be explicit about what frame each observer is at rest in. They are not implicit in the math, so you must make them explicit in the description (if you choose to use them). In other words there is a little additional care required on your part if you want to do this. You need to be careful and explicit in your description of any observers.

 

[Chenkel]

Given the difficulty you are having understanding this stuff, you might want to consider the possibility that the way you’re imagining what’s going on is part of the problem….

“An observer at rest with respect to an IRF recording events” can only record the events that happen at one place, where the observer is: these are the events with x-coordinate equal to zero, forming a vertical line through the origin.

It is possible to think in terms of observers recording events, but you need more observers. Taylor and Wheeler (“Spacetime Physics”) take this approach: we imagine that the entire universe is full of observers stationed one meter apart in a grid, at rest relative to one another, and carrying synchronized clocks. Whenever something happens, the one at that spot makes a note on a slip of paper, something like “I am the observer at x=10,y=0,z=0. This spaceship passed right where I am when my clock read 12:55 PM”. After our thought experiment is done we gather all these slips of paper and use them to reconstruct what happened according to the frame in which they were all at rest. We’ll call this the unprimed frame.

Now suppose that we have another such flock of observers, also all at rest relative to one another and carrying synchronized clocks but all moving relative to our first group. We’ll call the frame in which they are at rest the primed frame.

The Lorentz transformations let me calculate, from all the notes written by the unprimed frame observers, what the various primed-frame observers will have written down. The inverse transformations go the other way, they let me calculate the unprimed-frame results from the primed-frame ones.

So if the frames are moving with a relative velocity of 86.6 percent the speed of light (gamma factor of 2) then an event at x=1 y=0, and z=0, and t=20 seconds will be recorded in the primed frame at t' = 2*(20 -.866*1) = 38.268 and x'= 2*(1-.866*20) = -32.64 light seconds

The event of recording an event at x'=-32.64, y' = 0, z' = 0, and t' = 38.268 seconds in the primed frame will happen in the unprimed frame at t=2(38.268 + .866*(-32.64)) = 20.00352
And x=2(-32.64 + .866*38.268) = 1.000176

We got roughly the same value for t and x as when we started so this makes some sense to me.

 

[Nugatory]

We got roughly the same value for t and x as when we started so this makes some sense to me.

yep, that’s how it should work (I haven’t checked your calculations though).

When you’re setting up examples, try choosing $v=\frac{3}{5}c$ or $\frac{4}{5}c$ so $\gamma=\frac{5}{4}$ or $\frac{5}{3}$ - the arithmetic will be a lot easier and you’ll find that the inverse comes out exact.

 

[Chenkel]

yep, that’s how it should work (I haven’t checked your calculations though).

When you’re setting up examples, try choosing $v=\frac{3}{5}c$ or $\frac{4}{5}c$ so $\gamma=\frac{5}{4}$ or $\frac{5}{3}$ - the arithmetic will be a lot easier and you’ll find that the inverse comes out exact.

I just plugged the Lorentz transformation into the inverse Lorentz transformation algebraicly and found that after simplifying x = x and t = t so the Lorentz transformation and the inverse Lorentz transformation must be true for an event in the primed frame and a corresponding event in the unprimed frame for both Lorentz transformation and inverse Lorentz transformation.

 

[Chenkel]

I was testing my intuition about space time intervals.

So if you have an event where a juggler is juggling balls at 1 meter round trip intervals and a ball returns to his hand every a second then the spacetime interval is 1 meter minus 1 second times the speed of light.

If the reference frame the juggler is juggling in is moving very fast relative to earth then the distance between balls leaving the hand and coming back will increase but the spacetime interval will be invariant across inertial reference frames, so the ball will travel a longer distance but the time between the ball leaving the hand and coming back will also increase when viewed relative to someone on earth to maintain invariance of space time intervals.

What do you guys think about this?

 

[Nugatory]

I just plugged the Lorentz transformation into the inverse Lorentz transformation algebraicly and found that after simplifying x = x and t = t so the Lorentz transformation and the inverse Lorentz transformation must be true for an event in the primed frame and a corresponding event in the unprimed frame for both Lorentz transformation and inverse Lorentz transformation.

You got it!

Now, if you want your next challenge, try deriving the time dilation formula $\Delta T' = \Delta T/\gamma$ for when a spaceship leaves earth travelling at speed $v$ - when one year has passed on earth, how much time has passed on the spaceship?
There are two relevant events:
1) Spaceship passes earth, x=0, t=0, x'=0, t'=0.
2) The spaceship is where it is after $\Delta T$ years have passed on earth: $x=v\Delta T$, $t=\Delta T$. This will require calculating the x' and t' coordinates of that event.

Get past that challenge and we'll have another one for you involving the symmetry of time dilation.

 

[Chenkel]

You got it!

Now, if you want your next challenge, try deriving the time dilation formula $\Delta T' = \Delta T/\gamma$ for when a spaceship leaves earth travelling at speed $v$ - when one year has passed on earth, how much time has passed on the spaceship?
There are two relevant events:
1) Spaceship passes earth, x=0, t=0, x'=0, t'=0.
2) The spaceship is where it is after $\Delta T$ years have passed on earth: $x=v\Delta T$, $t=\Delta T$. This will require calculating the x' and t' coordinates of that event.

Get past that challenge and we'll have another one for you involving the symmetry of time dilation.

I'll work on this one, thanks!

 

[robphy]

Now, if you want your next challenge

Bonus points: next, draw a spacetime diagram of the situation!

 

[Chenkel]

I was testing my intuition about space time intervals.

So if you have an event where a juggler is juggling balls at 1 meter round trip intervals and a ball returns to his hand every a second then the spacetime interval is 1 meter minus 1 second times the speed of light.

If the reference frame the juggler is juggling in is moving very fast relative to earth then the distance between balls leaving the hand and coming back will increase but the spacetime interval will be invariant across inertial reference frames, so the ball will travel a longer distance but the time between the ball leaving the hand and coming back will decrease when viewed relative to someone on earth.

What do you guys think about this?

Maybe I should have said the time interval between the juggling events relative to someone on earth increases but the light travel time between the events also increases to keep the spacetime interval invariant.

Because the distance between the ball leaving the hand and coming back increases the time interval must also increase because the spacetime interval is event distance minus light travel time in meters.

 

[Chenkel]

You got it!

Now, if you want your next challenge, try deriving the time dilation formula $\Delta T' = \Delta T/\gamma$ for when a spaceship leaves earth travelling at speed $v$ - when one year has passed on earth, how much time has passed on the spaceship?
There are two relevant events:
1) Spaceship passes earth, x=0, t=0, x'=0, t'=0.
2) The spaceship is where it is after $\Delta T$ years have passed on earth: $x=v\Delta T$, $t=\Delta T$. This will require calculating the x' and t' coordinates of that event.

Get past that challenge and we'll have another one for you involving the symmetry of time dilation.

These are my equations:

$t' = \gamma(t - vx/c^2)$

$x=vt$

So

$t' = \gamma{(t - \frac {v^2{t}} {c^2})}$

$t' = \gamma{t(1 - \frac {v^2} {c^2})}$

$t' = t\sqrt{(1 - \frac {v^2} {c^2})}$

$t' = \frac t \gamma$

 

[jbriggs444]

So if you have an event where a juggler is juggling balls at 1 meter round trip intervals and a ball returns to his hand every a second then the spacetime interval is 1 meter minus 1 second times the speed of light.

You have lost me. We are using an inertial frame where the juggler is moving at 1 meter per second and the ball takes 1 second to rise and fall back down?

So in our chosen frame, the displacement from toss event to catch event is 1 meter in space and 1 second in time?

Then the squared separation is ${(\Delta x)}^2 - {(c \Delta t)}^2$ which is approximately minus one squared light-second. Because it is negative, it is a time-like interval. Within measurement accuracy, we might simply say that the [unsquared] interval is "one second".

The key point being that you square before doing the subtraction.

If one repeated the calculation using a different frame, the full Lorentz transform could be used to obtain new coordinates and new coordinate deltas. The computed interval would again be approximately one second regardless of the chosen frame.

 

[Chenkel]

You have lost me. We are using an inertial frame where the juggler is moving at 1 meter per second and the ball takes 1 second to rise and fall back down?

So in our chosen frame, the displacement from toss event to catch event is 1 meter in space and 1 second in time?

Then the squared separation is ${(\Delta x)}^2 - {(c \Delta t)}^2$ which is approximately minus one squared light-second. Because it is negative, it is a time-like interval. Within measurement accuracy, we might simply say that the [unsquared] interval is "one second".

The key point being that you square before doing the subtraction.

If one repeated the calculation using a different frame, the full Lorentz transform could be used to obtain new coordinates and new coordinate deltas. The computed interval would again be approximately one second regardless of the chosen frame.

The juggler is in a spaceship and is at rest with respect to the spaceship and a ball leaves the hand and comes back to the hand in 1 meter roundtrip distance and 1 second of travel time so 1 second of light travel time relative to the rest frame of the spaceship. This means 1 - 300000000 * 1 second = -299999999 meters is the spacetime interval in meters. (I rounded the speed of light to 300000000 meters per second for simplicity.)

If the spaceship is moving really fast then the balls will go a longer distance relative to a rest frame on earth, but the spacetime interval will be the same (-299999999), so if we let D be the distance traveled by the ball relative to earth then $D - \Delta t * c = -299999999$ if D is 225000000 (the distance an object travels at 75 percent the speed of light in one second) then we have $225000000 - \Delta t * 300000000 = -299999999$ so $-300000000\Delta t = -524999999$ so $\Delta t = 524999999/300000000 = 1.75 seconds$

So an event on the spaceship of a ball leaving the hand and coming back will be 1 second on the spaceship but it will be 1.75 seconds for the ball to leave the hand and come back relative to someone on earth if the spaceship is going 75 percent the speed of light.

 

[jbriggs444]

The juggler is in a spaceship and is at rest with respect to the spaceship and a ball leaves the hand and comes back to the hand in 1 meter roundtrip distance and 1 second of travel time so 1 second of light travel time relative to the rest frame of the spaceship. This means 1 - 300000000 * 1 second = -299999999 meters is the spacetime interval in meters. (I rounded the speed of light to 300000000 meters per second for simplicity.)

Assuming that we are using the flat space of special relativity, the ball is traversing a parabolic free fall trajectory. A curved trajectory. Not a straight line trajectory. Not a pair of straight line trajectories.

The invariant length of this time-like trajectory will be the proper time recorded by a wrist-watch attached to the ball. One could obtain that time by integrating the space-time interval over small segments of the curved path.

Subtracting 1 meter from 300000000 meters strikes me as a nonsense calculation for this purpose. If we had a straight-line trajectory covering 1 meter of coordinate distance over 1 second of coordinate time then we could compute interval $t$ as:$$t = \frac {\sqrt{300000000^2-1^2}} {300000000} = 0.99999999999999999444444444444444 \text{ sec}$$

 

[Chenkel]

Assuming that we are using the flat space of special relativity, the ball is traversing a parabolic free fall trajectory. A curved trajectory. Not a straight line trajectory. Not a pair of straight line trajectories.

The invariant length of this time-like trajectory will be the proper time recorded by a wrist-watch attached to the ball. One could obtain that time by integrating the space-time interval over small segments of the curved path.

That's an interesting idea to attach a clock to the ball.

In my calculations I made some simplification, I also included some gravity on the spaceship (how can you juggle without gravity?)

 

[Ibix]

The juggler is in a spaceship and is at rest with respect to the spaceship and a ball leaves the hand and comes back to the hand in 1 meter roundtrip distance and 1 second of travel time so 1 second of light travel time relative to the rest frame of the spaceship. This means 1 - 300000000 * 1 second = -299999999 meters is the spacetime interval in meters. (I rounded the speed of light to 300000000 meters per second for simplicity.)

No, this isn't right. The ball started and ended in the same place, so $\Delta x=0$. Thus the spacetime interval between throw and catch is $\Delta s^2=\Delta x^2-c^2\Delta t^2=-c^2×1\ \mathrm{s^2}$. So you forgot the squares, and you have misapplied the formula anyway.

If you want to calculate the spacetime interval along the path followed by the ball (rather than the "straight line" distance calculated above) you can do so, but there's an integral involved.

 

[jbriggs444]

That's an interesting idea to attach a clock to the ball.

In my calculations I made some simplification, I also included some gravity on the spaceship (how can you juggle without gravity?)

I assumed a gravitational force and, hence, a parabolic trajectory.

 

[Chenkel]

No, this isn't right. The ball started and ended in the same place, so $\Delta x=0$. Thus the spacetime interval between throw and catch is $\Delta s^2=\Delta x^2-c^2\Delta t^2=-c^2×1\ \mathrm{s^2}$. So you forgot the squares, and you have misapplied the formula anyway.

If you want to calculate the spacetime interval along the path followed by the ball (rather than the "straight line" distance calculated above) you can do so, but there's an integral involved.

Thanks for pointing that out, I did forget the square.

Also interesting I would need an integral, I was trying to simplify the problem by approximating the path with a straight line.

 

[Ibix]

Thanks for pointing that out, I did forget the square.

Also interesting I would need an integral, I was trying to simplify the problem by approximating the path with a straight line.

But it's two straight lines, one upwards and one downwards. If you actually simplify it to a no gravity situation where you bounce a ball off the ceiling then (idealising the collision as instantaneous) then you don't need an integral - you can compute $\Delta s$ for each leg and add them up (the integral is just the limit of this process for curved paths, with infinitesimal straight lines segments). Note that you need to take a square root because you need to add the two $\Delta s$ values, not the $\Delta s^2$ values, so in this case you are better off working with the $\Delta s^2 =c^2\Delta t^2-\Delta x^2$ convention - otherwise you have to compute $\sqrt{|\Delta s^2|}$.

 

[Chenkel]

But it's two straight lines, one upwards and one downwards. If you actually simplify it to a no gravity situation where you bounce a ball off the ceiling then (idealising the collision as instantaneous) then you don't need an integral - you can compute $\Delta s$ for each leg and add them up (the integral is just the limit of this process for curved paths, with infinitesimal straight lines segments). Note that you need to take a square root because you need to add the two $\Delta s$ values, not the $\Delta s^2$ values, so in this case you are better off working with the $\Delta s^2 =c^2\Delta t^2-\Delta x^2$ convention - otherwise you have to compute $\sqrt{|\Delta s^2|}$.

I thought to get the spacetime interval you need to find the square of the distance between two events minus the square of distance light travels between the two events occurring.

 

[jbriggs444]

I thought to get the spacetime interval you need to find the square of the distance between two events minus the square of distance light travels between the two events occurring.

The "distance light travels" is not exactly relevant. The $c$ enters into the formula as a conversion factor between the chosen units for time and space. Light is not travelling anywhere.

You start with two events. You express them each in coordinate form using your chosen inertial frame. E.g. as $(x_1, y_1, z_1, t_1)$ and $(x_2, y_2, z_2, t_2)$.

You subtract the two coordinate tuples yielding $(\Delta x, \Delta y, \Delta z, \Delta t)$

You compute $${\Delta x}^2 + {\Delta y}^2 + {\Delta z}^2 - {(c \Delta t)}^2$$That is the "squared separation".

If the result is positive, you have a spacelike interval. Take the square root and you have the length of the interval in your chosen units of distance.

If the result is negative, you have a timelike interval. Invert it and take the square root. You have $c$ times the duration of the interval. Divide by $c$ and you can extract a value for time in your chosen units of time.

If the result is zero, you have a lightlike or "null" interval.

This assumes a normal coordinate system on a flat space-time with a (-+++) metric signature. You can save yourself some paper and ink if you use units where $c = 1$.

 

[Chenkel]

If the result is negative, you have a timelike interval. Invert it and take the square root. You have c times the duration of the interval. Divide by c and you can extract a value for time in your chosen units of time.

If dx^2 + dy^2 + dz^2 - (c*dt)^2 is negative and I can invert it and take the square root? What do you mean by invert?