Minkowski spacetime interval

In the Minkowski space time equation in one dimensional space , ds^2 = dx^2 - (ct)^2, what is the value to use for x and t, and what does the space time interval ds represent? For example, if Alpha Centauri is 4 light years away, what values are. used for x and t, based on speed I guess, and what is the meaning of ds? What is it value when the observer is stationary, traveling at half the speed of light, and at the speed of light?

The equation is the equivalent of Pythagoras' Theorem, but correct for Minkowski spacetime instead of Euclidean space. Each term on the right is a difference of coordinates (three space and one time, although your version has suppressed two spatial dimensions) in some orthogonal coordinate system, and ##\Delta s^2## is the squared distance between the two events defined by the coordinates. Like in Euclidean geometry, that distance is the same whatever coordinate system you use to label the points.

Your questions about Alpha Centauri don't make sense. You can ask for the interval between Earth, now, and Alpha Centauri now (or in one hour, or next week, or whatever). In that case, ##\Delta x## is 4ly and ##\Delta t## is zero (or one hour, or one week, or whatever).

Both the ##\Delta x## and ##\Delta t## will depend on which frame you choose to use. The resulting ##\Delta s^2## will not.

I’ll just add that the form ##ds^2 = dx^2 - dt^2## (units with c=1) represents the infinitesimal version of the Pythagoras’ theorem equivalent in Minkowski space. With the deltas we typically mean finite coordinate differences and ##\Delta s^2## being the straight line spacetime interval. This way one can compute the invariant spacetime distance ##\int ds## for any path.

For me personally, taking the Minkowski space equivalent of Pythagoras’ theorem is usually where I start my SR class.

Thanks for responses but I don't get it, because I guess I asked the wrong question. I want to know what is the spacetime interval dependent upon what the are the variables for different values of x and t when witnessing the same event by different observers. Also if c = 1, then t must be in length units? I need some numbers.. What happens when ds is imaginary?.

PhanthomJay said:

I want to know what is the spacetime interval dependent upon what the are the variables for different values of x and t when witnessing the same event by different observers

The spacetime interval is not something for a single event. It is a measure related to the difference between two events. As stated in #2, it is the Minkowski space equivalent of the Pythagorean theorem.

It is unclear to me what you mean by “what are the variables”. Both values of t and x will differ between frames, the spacetime interval is invariant.

PhanthomJay said:

Also if c = 1, then t must be in length units?

Indeed. It is simply a way of emphasizing the geometrical nature of spacetime using a set of units where it is not obfuscated by multiples of c.

Just as c = 299792458 m/s is a choice in SI units, you can use a different set of units with fewer base dimensions such that length and time have the same physical dimension. It is just a convenient choice of units. If you don’t like it, think of it as using years and lightyears such that c = 1 ly/year.

PhanthomJay said:

What happens when ds is imaginary?.

It is not used that way. If ##ds^2 > 0## then* ##\sqrt{ds^2}## represents a spatial distance. If ##ds^2 < 0## then ##\sqrt{-ds^2}## represents a (proper) time interval.

* This is assuming the definition ##ds^2 = dx^2 - dt^2##. It is common to use the definition ##ds^2 = dt^2 - dx^2## instead and then the sign changes (always check what convention the text you are reading is using!). There are pros and cons with both.

PhanthomJay said:

I want to know what is the spacetime interval dependent upon what the are the variables for different values of x and t when witnessing the same event by different observers.

As Orodruin says, it's a "distance" between two events, not something relating to one event.

Pick a coordinate system in which Earth is at rest at the origin, so ##x_E=0##, and Alpha Centauri is at rest at 4ly, so ##x_\alpha=+4\mathrm{ly}##. Then ##\Delta x=x_\alpha-x_E=4\mathrm{ly}##. Let's say you wanted to know the interval between an event on Earth at time zero, ##t_1=0##, and another at Alpha Centauri five years later, ##t_2=5\mathrm{y}##. Then ##\Delta t=t_2-t_1=5\mathrm{y}##. Plug those numbers in, noting that in these units ##c=1\mathrm{ly/y}## and you get ##\Delta s^2=16-25=-9\mathrm{ly^2}##.

Given your choice of sign convention, negative ##\Delta s^2## tells you that things travelling below the speed of light can get from one event to the other, so something that happens at ##t=5## on Alpha Centauri could be affected by something that happened on Earth at ##t=0##. Zero would mean that ##\Delta x=\pm c\Delta t##, and only light could get there. Positive would mean not even light could cross the gap and the events cannot have a causal relationship (they might be caused by some common event further back in time, but one cannot affect the other). As Orodruin notes, the sign convention is by no means universal, so check what you are reading carefully.

If you want to know the coordinates that would be assigned by a frame moving at some velocity ##v## in the Earth-to-Centauri direction, apply the Lorentz transforms to the coordinates of each event. Recalculate ##\Delta s^2## and you will get the same value. It's an exercise in simple algebra to plug the Lorentz transforms into the formula and prove that this is always true.

Rather than me asking a nonsensical question, can you please ask a sample question regarding the spacetime interval between 2 events and give numerical results or inputs for x, t, and s. Thanks.

PhanthomJay said:

Thanks for responses but I don't get it, because I guess I asked the wrong question. I want to know what is the spacetime interval dependent upon what the are the variables for different values of x and t when witnessing the same event by different observers. Also if c = 1, then t must be in length units? I need some numbers.. What happens when ds is imaginary?.

You need to consult a textbook on Special Relativity. Chapter 1 of Morin is available here:

https://scholar.harvard.edu/files/david-morin/files/cmchap11.pdf

It might be worth studying that text from the beginning. The Minkowski spacetime interval is covered in section 11.6.

PhanthomJay said:

Rather than me asking a nonsensical question, can you please ask a sample question regarding the spacetime interval between 2 events and give numerical results or inputs for x, t, and s. Thanks.

See #6.

Ibix said:

As Orodruin says, it's a "distance" between two events, not something relating to one event.

Pick a coordinate system in which Earth is at rest at the origin, so ##x_E=0##, and Alpha Centauri is at rest at 4ly, so ##x_\alpha=+4\mathrm{ly}##. Then ##\Delta x=x_\alpha-x_E=4\mathrm{ly}##. Let's say you wanted to know the interval between an event on Earth at time zero, ##t_1=0##, and another at Alpha Centauri five years later, ##t_2=5\mathrm{y}##. Then ##\Delta t=t_2-t_1=5\mathrm{y}##. Plug those numbers in, noting that in these units ##c=1\mathrm{ly/y}## and you get ##\Delta s^2=16-25=-9\mathrm{ly}##.

Given your choice of sign convention, negative ##\Delta s^2## tells you that things travelling below the speed of light can get from one event to the other, so something that happens at ##t=5## on Alpha Centauri could be affected by something that happened on Earth at ##t=0##. Zero would mean that ##\Delta x=\pm c\Delta t##, and only light could get there. Positive would mean not even light could cross the gap and the events cannot have a causal relationship (they might be caused by some common event further back in time, but one cannot affect the other). As Orodruin notes, the sign convention is by no means universal, so check what you are reading carefully.

If you want to know the coordinates that would be assigned by a frame moving at some velocity ##v## in the Earth-to-Centauri direction, apply the Lorentz transforms to the coordinates of each event. Recalculate ##\Delta s^2## and you will get the same value. It's an exercise in simple algebra to plug the Lorentz transforms into the formula and prove that this is always true.

That is most helpful, thank you for the example.

Ibix said:

Plug those numbers in, noting that in these units ##c=1\mathrm{ly/y}## and you get ##\Delta s^2=16-25=-9\mathrm{ly}##.

ly² …

(Sorry, couldn’t help myself …)

Orodruin said:

ly² …

I always thought units were over-rated.

Thanks - corrected above.

Ibix said:

I always though units were over-rated.

Thanks - corrected above.

I guess also time is overrated (you're missing a t).

PAllen said:

I guess also time is overrated (you're missing a t).

Maybe it's just late here but I'm not seeing that one - can you be more specific?

Edit: never mind, I got the joke and have corrected the typo. But I had to hink about it.

PhanthomJay said:

Thanks for responses but I don't get it, because I guess I asked the wrong question. I want to know what is the spacetime interval dependent upon what the are the variables for different values of x and t when witnessing the same event by different observers. Also if c = 1, then t must be in length units? I need some numbers.. What happens when ds is imaginary?.

The spacetime interval is a sort of “distance” in spacetime. I put “distance” in quotes because there are some important differences between ordinary spatial distances, but there are also a lot of conceptual similarities.

The most important similarity is that both distance and the spacetime interval are measures of a path.

In spacetime the points are called events and are both a location and a time. So the top of Mt. Everest at noon today ##(today, Everest)## is a different event than the top of Mt. Everest at noon tomorrow ##(tomorrow,Everest)##, and it is also a different event than the top of Mt. Fuji at noon today ##(today, Fuji)##.

So one example path through spacetime the path sitting on the top of Mt. Everest from today to tomorrow. The length of this path would be measured with a clock.

Another example path would be the spatial path from Mt. Everest to Mt. Fuji, all at exactly noon today. No one object could travel this path because you would have to go faster than the speed of light. But you could measure the length of the path with a ruler stretched from Mt. Everest to Mt. Fuji.

When ##ds^2<0## that just means the interval is measured with a clock. And when ##ds^2>0## that means the interval is measured with a ruler.

Dale said:

Another example path would be the spatial path from Mt. Everest to Mt. Fuji, all at exactly noon today. No one object could travel this path because you would have to go faster than the speed of light. But you could measure the length of the path with a ruler stretched from Mt. Everest to Mt. Fuji.

I believe we have to be careful in defining the lenght between those two events. We need to establish a simultaneity convention and then, as @Dale pointed out, you can measure the lengh juxtaposing rulers one after the other.

cianfa72 said:

I believe we have to be careful in defining the lenght between those two events.

If you use a ruler to measure the distance (as in @Dale's post) and choose ##\mathrm{diag}(-1,1,1,1)## as the metric tensor (as in the OP) then your simultaneity convention is already chosen. It's orthogonal to the worldlines of the markings on the ruler.

PhanthomJay said:

Thanks for responses but I don't get it, because I guess I asked the wrong question.

There are no "wrong" questions ...

PhanthomJay said:

Also if c = 1, then t must be in length units?

Yes, and vice versa... lenght in "time" units

to Alpha Centauri = 4 light years
a foot = one light nano second
to the Moon = approximately one light second
to the Sun = 8 light minutes
to the planet Neptune = 4.2 light hours

PhanthomJay said:

I need some numbers.. What happens when ds is imaginary?.

The distance between Earth "now" and Alpha Centauri "now" ##\Delta t=0## is 4 light years.

If you leave a washing machine :-) in the same place for 4 years ##\Delta x=0## , it will pass through space-time the same interval of 4 years.

However, if the light starts from Earth now and reaches Alpha Centauri in 4 years, it will cross the zero (null) distance. Means ##0^2=4^2-4^2##

If an event on Alpha Centauri happens in, for example, 6 years, then it is theoretically possible to reach it. (We say that it is inside the light cone)

If some event on Alpha Centauri happens in, for example, 2 years, then it is impossible to reach it.

There are two conventions:
either ( the same thing has several names space-time interval ##\Delta s##, proper length ##\Delta L##, proper distance ##\Delta \sigma##)
##\Delta s^2=\Delta x^2 - \Delta t^2##

or ( the same thing has two names space-time interval ##\Delta s##, proper time ##\Delta \tau##)
##\Delta s^2=\Delta t^2 - \Delta x^2##

It's a matter of personal choice. (Only the plus and minus signs will swap places in formulas and results)

If you choose the first one, then you will get negative (imaginary) numbers for events inside the light cone and positive (real) numbers outside.

If you choose second convention, then the spatial distances will be negative(imaginary) and the events inside the light cone, in the future or the past, will be positive (real) numbers.

PhanthomJay said:

In the Minkowski space time equation in one dimensional space , ds^2 = dx^2 - (ct)^2, what is the value to use for x and t, and what does the space time interval ds represent?

Imagine that you want to make it to an event that takes place on Alpha Centauri in 5 years.

(important question)
At what time should you set your clock to wake you up from hibernation?

You enter your spaceship :-) ... and set the speed to 4/5 = 0.8 c and the ship's computer prints the following data:

destination: Alpha Centauri
speed: 0.8 c
gamma contraction: 0.6
distance: 2.4 light years
travel time: 3 years

How?
Lorentz factor
## \gamma=\sqrt{1-0.8^2}=\sqrt{1-0.64}=\sqrt{0.36}=0.6##

Length contraction ...
##L=0.6 \cdot 4 ly=2.4 ly ##

Time dilation
##t'=0.6 \cdot 5 yrs =3 yrs##

If we multiply the right side of the equation ##\Delta s^2=\Delta x^2 - \Delta t^2 ## by ##1=\frac{\Delta t^2}{\Delta t^2}## we get:

##\Delta s^2=\Delta x^2\frac{\Delta t^2}{\Delta t^2} - \Delta t^2\frac{\Delta t^2}{\Delta t^2 } ##

##\Delta s^2=\left(\frac{\Delta x^2}{\Delta t^2} - \frac{\Delta t^2}{\Delta t^2} \right) \Delta t^ 2##

##\Delta s^2=\left(v^2 - 1 \right) \Delta t^2##

##\Delta s=\sqrt{\left(v^2 - 1 \right) \Delta t^2}##

##\Delta s=\gamma i \cdot \Delta t##

The space-time interval ##\Delta s## is in this case the same as the travel time ##\Delta t'=3 yrs##
##\Delta s## is the time at what traveler should set his clock to wake him up from hibernation?

PhanthomJay said:

For example, if Alpha Centauri is 4 light years away, what values are. used for x and t, based on speed I guess, and what is the meaning of ds?

For each event, the spatial position (for example, Alpha Centauri = 4 ly) and time (for example, 10 years ago, now, in 5 years, ...) should be determined.

Bosko said:

There are no "wrong" questions ...

There certainly are. They can be predicated on faulty assumptions or simply be the wrong question to ask based on what it is you actually want to know.

Orodruin said:

There certainly are. They can be predicated on faulty assumptions or simply be the wrong question to ask based on what it is you actually want to know.

They are certainly the first group of questions :-) haha ... until one is able to asks a proper question.

PhanthomJay said:

what does the space time interval ds represent?

It is defined as follows:

Definition. Two events E and F in spacetime determine a physical quantity ∆s, the spacetime interval between the events:

• If an inertial clock can move between E and F, define ∆s to be the time between E and F as measured by the clock. Call ∆s the proper time between the events and say that the events are timelike separated.
• If light can move between E and F, define ∆s = 0. Say that the events are lightlike separated.
• If neither light nor clocks can move between E and F then, as we shall see, a rigid rod can have its ends simultaneously at E and F. (Simultaneously in the sense that light flashes emitted at E and F reach the center of the rod simultaneously, or equivalently, that E and F are simultaneous in the rest frame of the rod.) Define |∆s| to be the rest length of this rod. (The reason for the absolute value will be clear later.) Call |∆s| the proper distance between the events and say that the events are spacelike separated.

Source:
http://www.faculty.luther.edu/~macdonal/Interval/Interval.html