# Confused on how to calculate invariant interval.

1. May 13, 2013

### ash64449

Hello friends,

I have now began to read a new book called space-time physics by Edwin.F.Taylor.

In the first chapter(parable of surveyors),First it talks about invariant interval and it says the equation is:

$\sqrt{ (ct)^2 - x^2 }$

In Wikipedia,i saw a different answer for the same invariant interval:

$\sqrt{ r^2 - (ct)^2 }$

Anyway i think $x$ in first one is same as $r$ in second one.

So what is actually happening here?

2. May 13, 2013

### Staff: Mentor

It's just a difference in sign convention. The definition in the Taylor book is using what's called a "timelike convention", where the sign of the $ct$ term is positive. Note that the quantity under the square root will only be positive if $ct > x$, i.e., if the interval is timelike; if $ct < x$, the quantity under the square root is negative and there is no real root.

The Wikipedia definition is using what's called a "spacelike convention", where the sign of the $r$ term (or the $x$ term, yes, they're supposed to be the same thing) is positive. Note that the quantity under the square root in this case will be positive only if $r > ct$, i.e., if the interval is spacelike.

3. May 13, 2013

### Mentz114

They the same thing. We can write the Minkowski metric two ways, and still get the same physics,

-ds2 = -c2dt2 + dr2 or ds2 = c2dt2 - dr2.

These metrics have 'signature' (-+++) and (+---) respectively.

4. May 14, 2013

### ash64449

I think i have understood your post.

The difference between those two equation is only in "sign convention".

So,In the Taylor book,they used Timelike convention.

In Wikipedia,they used Spacelike convention.

From your post,i understood why conventions are used.
To calculate Spacelike and Timelike intervels.

5. May 14, 2013

### ghwellsjr

Don't you remember asking about the invariant interval in another thread that you started just four days ago? Even if you didn't remember, it's always a good idea whenever you have a question to do a search. You could have hit the Search this Forum button and typed in "invariant interval" and there it would have found your previous thread as the first hit. You could have looked down to post #9 and read in the middle of Ibix'x very thorough post:

Also, whenever you reference something online, you should provide a link. The wikipedia article on this subject, I presume the one you are referring to, states:

Then further down in the article, they show it both ways, first the way the book shows it and then the way that you said the article shows it.

Was this the wikipedia article you saw or did you see another one?

6. May 14, 2013

### ash64449

And I think we can use this sort of conventions and get the invariant interval is because there are Three different intervals.

That is:

$s = \sqrt{(ct)^2 - x^2}$ $(1)$

$s =\sqrt { x^2 - (ct)^2 }$ $(2)$

This two equations are correct,that they measure invariant interval.

But the only difference is that they help us to measure different types of invariant interval.

The first equation helps us to calculate Time-like interval.

Second equation helps us to calculate Space-like interval.

But we cannot calculate Space-like interval using first equation and Time-like interval using second equation because square of negative number is imaginary.

And we can use both of those two equations to calculate Light-like interval because in light-like interval $x^2 = (ct)^2$, so there isn't any problem of square root of negative number.

Am I right?

7. May 14, 2013

### ash64449

Sorry what is this?

8. May 14, 2013

### ash64449

Yes. I do remember,But i didn't know what he said. And i haven't learned about this from that thread at all. That you can put variables the other way around.From there i only studied what is Space-like interval and what is Time-like interval etc..

This is the one. But i didn't know what it meant by this statement. And none of the posters said that you can change the variables and get different invariant intervals. I only studied what are different intervals..

9. May 14, 2013

### ash64449

And changing the sign changes the L.H.S too and i think that should not happen because interval cannot be negative.

I mean $s^2 = (ct)^2 - x^2$
$-s^2 = x^2 - (ct)^2$

Second one shows that interval is negative. That cannot happen because interval cannot be negative.

The only thing is that we can interchange the variables and have positive invariant interval because both explain different types of intervals.

10. May 14, 2013

### ghwellsjr

If you look after the equal sign, you will see four terms, the first one has an assumed + sign and the other three have - signs. This is the (+---) signature. If all the signs are changed then you get the (-+++) signature.

11. May 14, 2013

### Mentz114

Are you content with this now ?

12. May 14, 2013

### ash64449

Oh. that way!! I didn't include $y$ and $z$. Forgot about them!!

13. May 14, 2013

### ash64449

From my reply to George,you must have understood that i didn't include $y$ and $z$ axial coordinates and i quite forgot about them. And that is why i didn't understand your convention.

Now i understood that you said exactly the same what first poster said.

Anyway Thank you for the reply!

14. May 14, 2013

### ash64449

And i do understand why we omit $y$ and $z$ because we consider motion in only $x$ axis.

15. May 14, 2013

### Staff: Mentor

If you write the formulas with the square roots, yes. But it's more common to write the formula for the squared interval, as other posters have been doing. If you do that, you can represent any kind of interval with either sign convention: the timelike convention

$$s^2 = c^2 t^2 - x^2 - y^2 - z^2$$

or the spacelike convention

$$s^2 = - c^2 t^2 + x^2 + y^2 + z^2$$

Squared intervals can be positive, zero, or negative, so the different sign conventions just determine which sign goes with a timelike squared interval and which one goes with a spacelike squared interval.

16. May 14, 2013

### Fredrik

Staff Emeritus
I have never really been a fan of the "interval" terminology or the $s^2$ (or $\Delta s^2$) notation. The notation is pretty horrible, since it suggests that s is a real number.

@Ash64449: The definition doesn't define s (or $\Delta s$). It only defines $s^2$ (or $\Delta s^2$). You should think of $s^2$ as a single symbol that represents a real number. The $s$ doesn't have an independent meaning here. It's just a curved line that makes up about half the area of the symbol $s^2$.

Last edited: May 14, 2013
17. May 14, 2013

### robphy

A better notation would be $\vec s \cdot \vec s$, with the dot-product notation for the metric.

Similarly, the line element should be $d\vec s \cdot d\vec s$
and the Pythagorean theorem
$\vec c \cdot \vec c = \vec a \cdot \vec a + \vec b \cdot \vec b$ when $\vec c = \vec a + \vec b$ and $\vec a \cdot \vec b =0.$

18. May 14, 2013

### WannabeNewton

Unfortunately, the formal definition of $ds^{2}$ requires one to know what a tensor product is. Until then, as others have noted, just think of $\Delta s^{2}$ as nothing more than a symbol for the space-time interval, not something you should take the square root of.