I What is the Term for ds in Spacetime Interval Equation?

[nomadreid]

In the usual relativistic equation, ds² = (cdt)² - dx² - dy ² -dz² or dx² + dy ² + dz² - (cdt)², depending on the convention of your choice, and ds² is called the spacetime interval between the corresponding events, the square being used to avoid nasty ambiguities and irritating imaginary numbers. Fine. But if one wants to refer to ds, is there any standard term for it besides the clumsy "the square root of the spacetime interval" or "that which when squared gives the spacetime interval"?
Thanks.

 

[PeroK]

In the usual relativistic equation, ds² = (cdt)² - dx² - dy ² -dz² or dx² + dy ² + dz² - (cdt)², depending on the convention of your choice, and ds² is called the spacetime interval between the corresponding events, the square being used to avoid nasty ambiguities and irritating imaginary numbers. Fine. But if one wants to refer to ds, is there any standard term for it besides the clumsy "the square root of the spacetime interval" or "that which when squared gives the spacetime interval"?
Thanks.

The (infinitesimal) spacetime interval is the four-vector $(c dt, dx, dy, dz)$. That's essentially what $ds$ is, although you should really use a vector notation, such as $\textbf{ds}$.

$ds^2 = \textbf{ds} \cdot \textbf{ds}$

So, $ds^2$ is really the inner product of the spacetime interval with itself. And, in a different context, this is the definition of the (in this case flat) spacetime metric.

The quantity $\sqrt{\pm ds^2}$ is the proper time or proper length of the spacetime interval, depending on whether $ds^2$ is positive or negative. If it's $0$ you have a null spacetime interval.

 

[nomadreid]

Ah, thank you, PeroK. The explanation clears up the confusion. This confusion was caused by the Wikipedia site https://en.wikipedia.org/wiki/Spacetime#Spacetime_interval which calls the scalar quantity ds² the spacetime interval, but your pointing out that the spacetime interval is the vector ds, and the scalar quantity the inner product ds⋅ds and so forth, makes sense.

 

[Andrew Mason]

In the usual relativistic equation, ds² = (cdt)² - dx² - dy ² -dz² or dx² + dy ² + dz² - (cdt)², depending on the convention of your choice, and ds² is called the spacetime interval between the corresponding events, the square being used to avoid nasty ambiguities and irritating imaginary numbers. Fine. But if one wants to refer to ds, is there any standard term for it besides the clumsy "the square root of the spacetime interval" or "that which when squared gives the spacetime interval"?
Thanks.

Generally, s² is a useful not because of what it represents physically but because it is a metric associated with two events that is the same in all reference frames (and because its sign tells us whether the interval is timelike or spacelike). It has physical significance only in certain frames: If they took place at the same location in a reference frame, s represents the time interval between the events (multiplied by the factor c); if the two events took place at the same time but at different locations in a particular reference frame, s = $\sqrt{s^2}$ is the distance between the two events in that frame multiplied by the imaginary number i. For this reason, we refer to s² rather than s as the metric for the space-time interval.

AM

 

[nomadreid]

Thanks, Andrew Mason. That is a good elaboration of the last sentence of PeroK's last sentence when adopting the convention [+,-,-,-].

 

[robphy]

For $ds^2$, I prefer “[spacetime] square-interval”.
For the 4-vector $d\vec s$ (or $d\tilde s$, if I need to use $d\vec V$ for 3-vectors), I prefer “[spacetime] displacement vector”.
Both may have to be prefixed by infinitesimal, if needed for clarity.

 

[nomadreid]

Thanks, robphy. Interesting suggestions. Are they standard (or at least one of the standards)?

 

[robphy]

I haven’t surveyed the literature to see if my terms are standard. But one of my motivations is asking what their Euclidean or Galilean analogues would sound like.

In addition, my possibly annoying use of the hyphen suggests that it’s a single term “square-interval”, not to be mistakenly separated into an adjective “square” modifying an object noun “interval”.

 

[nomadreid]

Thanks, robphy. If your terms are acceptable and understood, then since there seems to be no set term, yours are also good.

 

[romsofia]

I'll add a lot more to this discussion, but I'll keep it without time for now, and i'll put it in spherical form. After typing this all out, i recognize this might not be what you were looking for, and more so just wanted to know what to call this vector, but I'm not sure so might as well post.

Let $ds^2=dr^2+r^2d\theta^2+r^2\sin^2\theta^2 d\phi^2$ then, as stated above, you can recognize some vector $d\vec{r} = dr\hat{r}+rd\theta \hat{\theta} + r\sin\theta d\phi \hat{\phi}$ In other words, $d\vec{r}$ is a vector-valued one-form! Which is a great introduction into the wonderful world of differential forms.

From here, you can recognize your basis as $\omega = \{dr, rd\theta, r\sin\theta d\phi \} = dr \wedge rd\theta \wedge r\sin\theta d\phi$

So, what's the big deal? By putting things in this format, it's a lot easier to find your connections, curvature, etc which I will do for you so you can recognize how to do this in practice (also, I've been lazy this summer so, i might as well refresh myself).

This post would be way too long if I had to type out motivation for the formulas, but you can ask questions if you're unsure where the formulas come from!

Let $d\hat{e_i} = a^j_i \hat{e_j}$ where $a^j_i$ is called a "connection" and the derivative in this post is called an "exterior derivative". Let's type out how our basis vectors change (Recognize that 1 refers to the first coordinate in my basis, 2 as second, 3 as third)
$d\hat{e_1} = a^1_1 \hat{e_1} + a^2_1 \hat{e_2} + a^3_1 \hat{e_3} = 0 +a^2_1 \hat{\theta} + a^3_1 \hat{\phi}$. Now, from the levi-cievta condition, we recognize that $a^j_j = 0$
$d\hat{e_2} = a^1_2 \hat{e_1} + a^2_2 \hat{e_2} + a^3_2 \hat{e_3} = a^1_2 \hat{r} + 0 + a^3_2 \hat{\phi}$
$d\hat{e_3} = a^1_3 \hat{e_1} + a^2_3 \hat{e_2} + a^3_3 \hat{e_3} = a^1_3 \hat{r} + a^2_3 \hat{\theta} + 0$

Now, we must use our levi-cievta connection to complete how to solve for the connection. The Levi-cievta connection has two properties which are metric compatibility (If you're interested in these, I believe they come from your structure equations, but someone feel free to correct me!), which looks like $a^j_i +a^i_j = 0$ or $a^j_i = -a^i_j$ in orthongonal basis, but i believe in a general basis, this looks like $a^j_i+a^i_j = dg_{ij}$ and torsion free which looks like $d{b^i} + a^i_j \wedge b^j = 0$ where $b^i$ is your ACTUAL components of your basis! That is, you have to recognize that $b^2 = rd\theta$. Since i decided to work in orthogonal basis, I have to take the whole thing. I think if you decide to not take the r in this, you'd be in a "coordinate basis" (once again, more math-y people, feel free to correct me).

Great, now using these two we can finally see how our basis changes from point to point, which will lead us to curvature!
So, going through the equation we get:
$0 = d(b^1) + a^1_1 \wedge b^1 + a^1_2 \wedge b^2 + a^1_3 \wedge b^3 = 0 + 0 + a^1_2 \wedge rd\theta + a^1_3 \wedge r\sin\theta d\phi$ So here you need to know a little more about wedge products, and I'll supply the fact that if you have two of the same one forms wedged together, the equal zero. That is, $dr \wedge dr = d\theta \wedge d\theta = d\phi \wedge d\phi = 0$ Which we must utilize in order to solve the equations we will have ahead! The other one is a property of the exterior derivative, which is: the second exterior derivative is equal to 0. Which is why $d(b^1) = d(dr) = 0$ in this case.

So just from my first one, I know that my $a^1_2$ MUST contain a $d\theta$ term, otherwise I can't satisfy my torsion free condition. Like wise, $a^1_3$ MUST contain a $d\phi$. Once again, a natural question may arise if you haven't seen wedge products before! What about the $r\sin\theta$ term before the $d\phi$?

Well, wedge products can be thought of as multiplication, so we can just bypass them by just wedging together the one forms. For example, if $a^1_3 = d\phi$ we'd have $d\phi \wedge r\sin\theta d\phi = r\sin\theta d\phi \wedge d\phi = 0$ Or, if that doesn't satisfy you, you can also use the property of wedge products that switching the order of the placements will give you a minus sign (this is known as an "anti-symmetric" property). So, using this, you can see that $d\phi \wedge r\sin\theta d\phi = - r\sin\theta d\phi \wedge d\phi = 0$ We will be using the anti-symmetric property on the other two torsion free equations!

$0 = d(b^2) + a^2_1 \wedge b^1 + a^2_2 \wedge b^2 + a^2_3 \wedge b^3 = d(rd\theta) + a^2_1 \wedge dr + 0 + a^2_3 \wedge r\sin\theta d\phi = dr \wedge d\theta +a^2_1 \wedge dr + a^2_3 + \wedge r\sin\theta d\phi = dr \wedge d\theta - dr \wedge a^2_1 + a^2_3 + \wedge r\sin\theta d\phi = 0$

Now, let's pause here! Where did that $dr \wedge rd\theta$ term come from? It came from the product rule with derivatives (yes, exterior derivatives still follow this rule!) $d(rd\theta) = dr \wedge d\theta + rd^2\theta = dr \wedge d\theta + 0$ and the $- dr \wedge a^2_1$ came from the anti-symmetric property of wedge products, and I used it because I want to get rid of that first term! I have to recognize that $a^2_1 = d\theta$ WHICH IS GREAT! Recall from my first torsion free equation, I recognized that $a^1_2$ HAD to have $d\theta$ in it! Even better, this is basically just the metric compatibility condition! That is, if $a^1_2 = -a^2_1$, or in our case, $a^2_1 = d\theta$ therefore $a^1_2 = -d\theta$

The only thing left, is we must recognize that $a^2_3$ HAS to have a term $d\phi$ in it to satisfy the torsion free condition.

Finally, we have: $0 = d(b^3) + a^3_1 \wedge b^1 + a^3_2 \wedge b^2 + a^3_3 \wedge b^3 = d(r\sin\theta d\phi) + a^3_1 \wedge dr + a^3_2 \wedge rd\theta + 0 = dr \wedge \sin\theta d\phi + r\cos\theta d\theta \wedge d\phi - dr \wedge a^3_1 - rd\theta \wedge a^3_2 = 0$

Once again, we used product rule for derivatives, and used the anti symmetric property of wedge products. Here, we recognize that $a^3_1 = \sin\theta d\phi$ where we NEED all the stuff out front because in order to completely get rid of the first term, we need all of those (You can see this by imagining if I factored out a dr \wedge d\phi from the equation, what would you have left?) and then let's try to get $a^3_2$.

If recall from my 2nd torsion free equation that $a^2_3$ had to have something with $d\phi$ in it. So, let's see if it was just $d\phi$.

I'd get $rcos\theta d\theta \wedge d\phi - rd\theta \wedge d\phi$ if I factor out a $d\theta \wedge d\phi$ term, I'll be left with $d\theta \wedge d\phi(r\cos\theta - r)$ Well, that is obviously not 0! So I know that I have to have a $\cos\theta$ term in my $a^3_2$. That is, $a^3_2 = \cos\theta d\phi$ which would make $a^2_3 = -\cos\theta d\phi$Now, you can plug these all in, and check if you get zero. So finally, we can see how our basis changes!

That is:
$d\hat{r} = d\theta \hat{\theta} + \sin\theta d\phi \hat{\phi}$
$d\hat{\theta} = -d\theta \hat{r} + \cos\theta d\phi \hat{\phi}$
$d\hat{\phi} = -\sin\theta d\phi \hat{r} -\cos\theta d\phi \hat{\theta}$

Curvature is then found with the equation: $\Omega^i_j = d(a^i_j) + a^i_k \wedge a^k_j$ (This is called the curvature 2-form, and it also has the property $\Omega^i_j = -\Omega^j_1$) This post is long enough, so you can try this out on your own!

And last, but not least, with that vector valued one form $d\vec{r}$ you can calculate your geodesics! First recognize that a geodesic is when we have no acceleration, that is $\vec{a} = 0$. How do we get this from $d\vec{r}$?

Once again, long enough post, but the way you do it is: Start with $d\vec{r}$ then divide by $dt$. To make this one explicit, I'll do the first step so you can see what i mean by "divide".

Let $d\vec{r} = dr\hat{r}+rd\theta \hat{\theta} + r\sin\theta d\phi \hat{\phi}$ If I want to find velocity, that is just the change of my position with respect to time, or "dividing" by dt. So, let's do that $\vec{v} = \frac{d\vec{r}}{dt} = \dot{r}\hat{r}+r\dot{\theta} \hat{\theta} + r\sin\theta \dot{\phi} \hat{\phi}$ Where the dots over the variables are derivatives with respect to time as usual!

From here, find $d\vec{v}$ then "divide" by dt, then you have $\vec{a}$ now set it equal to zero, and solve your differential equations!

So, long post is done! Sorry if this was more than you wanted, but there is so much that can be done with that $d\vec{r}$! The math doesn't change by adding another dimension, but it just gets longer. Which is why I did a 3D one!

Sorry for any typos.

 

[nomadreid]

Thanks, romsofia. Although this was more than I asked for, it is not superfluous. There is a lot in your post that I must slowly digest, and perhaps I will come back to you later to ask questions on specific points, if this is OK. In the meantime, a question concerning post #2 of PeroK (I am slow to react)

The quantity ñds2\sqrt{\pm ds^2} is the proper time or proper length of the spacetime interval, depending on whether ds2ds^2 is positive or negative. If it's 00 you have a null spacetime interval.

First, I thought that whether ±ds^2 is positive or negative depended not only on the separation of the events (i.e., whether (ct)² is greater or less than the sum of the squares of the spatial coordinate changes) but also upon which convention one used [+, -, -, -] or [-, + ,+, +], so it would seem to say that whether ±ds^2 corresponds to a proper time or a proper distance would depend on which convention one selected? That sounds strange, so I presume I am missing something here. Secondly, if the same quantity [hence the same units] can be either time or length, I am presuming that one is setting c=1? Furthermore, in Post #4, Andrew Mason states that such a correspondence only works in the special cases -- or did I misunderstand that?

 

[PeroK]

Thanks, romsofia. Although this was more than I asked for, it is not superfluous. There is a lot in your post that I must slowly digest, and perhaps I will come back to you later to ask questions on specific points, if this is OK. In the meantime, a question concerning post #2 of PeroK (I am slow to react)

First, I thought that whether ±ds^2 is positive or negative depended not only on the separation of the events (i.e., whether (ct)² is greater or less than the sum of the squares of the spatial coordinate changes) but also upon which convention one used [+, -, -, -] or [-, + ,+, +], so it would seem to say that whether ±ds^2 corresponds to a proper time or a proper distance would depend on which convention one selected? That sounds strange, so I presume I am missing something here. Secondly, if the same quantity [hence the same units] can be either time or length, I am presuming that one is setting c=1? Furthermore, in Post #4, Andrew Mason states that such a correspondence only works in the special cases -- or did I misunderstand that?

Whether it is proper time or proper length depends on the sign and the convention. In any case, Timelike means that the time component is greater than the spatial component. Or, alternatively, that there is a reference frame in which the spatial component is zero. Spacelike means the opposite, and there is reference frame where the time component is zero.

If you multiply a time by a speed (of light) you get a length. You can either have $(t, x, y, z)$ or $(ct, x, y, z)$ or, use units where $c =1$.

But, you have to be clear which one you are using.

 

[nomadreid]

Thanks very much, PeroK. That makes sense.

 

[SiennaTheGr8]

If you multiply a time by a speed (of light) you get a length.

Maybe I'm being pedantic, but I think it's better to say: "If you multiply a time by the speed of light, you get that same time expressed in length-units" (assuming the time was originally expressed in time-units).

 

[nomadreid]

Thanks, SiennaTheGr8. You are of course correct; that is probably what PeroK meant by

But, you have to be clear which one you are using.

 

[Andrew Mason]

First, I thought that whether ±ds^2 is positive or negative depended not only on the separation of the events (i.e., whether (ct)² is greater or less than the sum of the squares of the spatial coordinate changes) but also upon which convention one used [+, -, -, -] or [-, + ,+, +], so it would seem to say that whether ±ds^2 corresponds to a proper time or a proper distance would depend on which convention one selected? That sounds strange, so I presume I am missing something here.

What the sign signifies depends on the convention. If you use $s^2 = (ct)^2 - x^2 - y^2 - z^2$ as the space-time interval, then the interval between the events will be timelike if positive and spacelike if negative. It is the other way around if you use the convention $s^2 = x^2 + y^2 + z^2 -(ct)^2$.

Furthermore, in Post #4, Andrew Mason states that such a correspondence only works in the special cases -- or did I misunderstand that?

What I said was that $s = \sqrt{s^2}$ corresponds to something physical only in certain reference frames where the interval is either purely time or purely distance. Otherwise, s² is just a useful metric (because it is has the same value in all reference frames).

AM

 

[nomadreid]

Thanks, Andrew Mason.

What the sign signifies depends on the convention.

Yes, I should have inserted the obvious "respectively"; that will teach me to post before breakfast (in my time zone).

What I said was

Ah, sorry, I mangled your statement in my post. I understand, and I stand corrected.
Thanks for the corrections.