Which Equation for the Space-Time Interval Should You Use?

[Thepolak]

I have two questions about the space-time interval question.

I'm doing a little research about space time, and about space time intervals, but I'm not sure which equation to take. Some sources say that its s^2 = x^2 - c^2 t^2, other say its s^2 =c^2 t^2 - x^2. So which one do I take?

The other question is why do we subtract?

 

[bcrowell]

The choice of +- or -+ signs is arbitrary. Different people use one or the other.

One way of understanding why there's a subtraction is that s=0 for a light ray, and the speed of light is the same for all observers.

 

[Thepolak]

One way of understanding why there's a subtraction is that s=0 for a light ray, and the speed of light is the same for all observers.

I still don't understand.

 

[Mentz114]

I still don't understand.

if you write ds² = c²dt²-dx² then if ds=0 ( which it is for light ) thenc²dt² = dx²

c² = (dx /dt)²

 

[Phrak]

To add to the above, for spacetime intervals in units of time, it is customary to use the symbol Δtau, Δτ, and the interval is positive in units of time for the world line of massive objects. If the interval squared is positive in time, it is negative in space, and if positive in space, negative in time which can be confusing to anyone new at it.

Also, to be picky, your equations are not space time intervals in general, but displacements from the origin. You need to put a Δ in front of your variables, or change them to differences like x₂- x₁ as Mentz hinted at by using differentials.

To finally get the interval, Δτ, in special relativity, in units of time,

\Delta\tau = \sqrt{(\Delta t)^2 - \left( \frac{\Delta x}{c} \right)^2}}

For space-like separated events,

\Delta s = \sqrt{(\Delta x)^2 - (c \Delta t)^2}

 

[bcrowell]

Thepolak, we're going to have a hard time helping you unless you explain more about what you've been reading, what's confusing you, etc.

 

[Thepolak]

Thepolak, we're going to have a hard time helping you unless you explain more about what you've been reading, what's confusing you, etc.

I have to talk about space time intervals, and thus explaining why do we have to subtract in the equation.

 

[homology]

Bernard Schutz's book "A First Course in General Relativity" has a couple chapters on special relativity at the beginning which are good.

If you have a light ray then in one frame with coordinates x,y,z,t you have (\Delta x)^2+(\Delta y)^2+(\Delta z)^2-(c\Delta t)^2=0, nothing fancy there, the distance the photon travels is just the usual euclidean metric...BUT... all observers agree on this fact, that is, if you define \Delta s^2=(\Delta x)^2+(\Delta y)^2+(\Delta z)^2-(c\Delta t)^2 then everyone agrees when its zero.

Then you use this (we're still just playing with a postulate here) to deduce that in fact \Delta s^2 is an invariant, not just when its zero. Schutz's derivation is straightforward but a bit long to repeat here.

But it is, as mentioned above, the fact that we all agree on the speed of light. Oh and it doesn't matter which sign convention you use pick what you like...I like +--- in my metric tensor (for SR).

 

[Thepolak]

Bernard Schutz's book "A First Course in General Relativity" has a couple chapters on special relativity at the beginning which are good.

If you have a light ray then in one frame with coordinates x,y,z,t you have (\Delta x)^2+(\Delta y)^2+(\Delta z)^2-(c\Delta t)^2=0, nothing fancy there, the distance the photon travels is just the usual euclidean metric...BUT... all observers agree on this fact, that is, if you define \Delta s^2=(\Delta x)^2+(\Delta y)^2+(\Delta z)^2-(c\Delta t)^2 then everyone agrees when its zero.

Then you use this (we're still just playing with a postulate here) to deduce that in fact \Delta s^2 is an invariant, not just when its zero. Schutz's derivation is straightforward but a bit long to repeat here.

But it is, as mentioned above, the fact that we all agree on the speed of light. Oh and it doesn't matter which sign convention you use pick what you like...I like +--- in my metric tensor (for SR).

Thanks, this helps a lot!