The h Bar

So, the metric is not on the 4D space, but on the space of 2 points and a curve between them. Not really, I never learned TeX

though we typically use \tau for the variable that parameterises the curve

5:06 PM

@JohnRennie Sigh...

Only in physicist GR

And you use the standard convention that tau or whatever goes from 0 to 1?

It's a physics site

But the chat is general.

Wald, Hawking-Ellis and others just use $t$

@JohnRennie and @barrycarter is a mathematician

Or I claim to be one anyway.

5:07 PM

@barrycarter Sometimes.

I may have to retract that claim if I sound any stupider.

@barrycarter you're enough of an ass that I believe you

Oh, if you'd put the word mathematician into that, I would've starred it.

OK, so this metric takes a curve in 4D space and assigns it a number?

NOooooo

NOOOOOOOOOOOOO

OOOOOOOOOOOOOOOOOOOOO

This would go faster if you let me explain ...

5:08 PM

KHAAAAAAAAAAAAAN

@JohnRennie go for it

Please be precise.

You can both try.

@JohnRennie You seemed hesitant earlier, I would prefer your help, only so I don't have to be nice to Ocelot

I believe fundamentally that everything can be reduced to first principles.

you're just mad an engineer knows more geometry than you

Is what I'm saying not first principles?

Ocelot, no, you're fine, but I'm letting @JohnRennie know

The cotangent bundle is very intuitive

5:10 PM

IF you want to both continue, that's good too.

@JohnRennie you go, I have to go grab my stuff from the office.

Spacetime is a manifold that looks locally like R^4 ...

I'm with you so far.

So to uniquely identify points in the manifold we need four numbers

Not very precise.

Oh god

5:12 PM

I'm with you @JohnRennie

@JohnRennie that's not correct but OK

@0celo7 I'm seeing if I can give you apoplexy

Not smart enough to know what that means

@JohnRennie Keep going. Ocelot, keep heckling.

Bernard Meurer

@DanielSank I'm perverting your octopus passion

ACuriousMind

5:13 PM

@0celo7 A stroke. An apoplexy is a stroke.

@barrycarter I'm doing the mathematics god's work

Now I choose a coordinate system. This can be any coordinate system e.g. t,x,y,z or t,r,\theta,\phi or the Kruskal-Szekeres coordinates or anything - the details don't matter.

I don't understand the last one, but sure.

Ok let me choose t,x,y,z. Suppose I consider an infinitesimal displacement in my coordinates (dt,dx,dy,dz) ...

OK...

5:15 PM

Then the metric allows me to calculate the length (the norm) of this four-vector.

OK, so you are assigning a number to every (t,x,y,z) coordinate?

So far all I'm doing is calculating the length of an infinitesimal four-vector

Or do you mean the (t,x,y,z) to (t+dt,x+dx,y+dy,z+dz) vector?

Got it. And I assume it's not the standard one.

>standard one??

sqrt(dt^2 + dx^2 + dy^2 + dz^2)

Since you said it was locally R^4

5:17 PM

Yes, yes, yes, you're nearly there!!! The key thing about relativity is that the signature is (1,3) i.e. one of the squares inside the root has a negative sign.

The 't'?

It's the only one that's quasi distinct

In special relativity the norm is sqrt(-dt^2 + dx^2 + dy^2 + dz^2)

OK, I'd call it something other than a norm, but I'm with you.

But this only applies for small values of dx, dy, dz, dt, right?

Yes

OK, keep going.

5:19 PM

But consider some curve that goes from point A to point B

Oh, I think I see it now.

Because of that -dt^2 factor, when you integrate the length between two points, a straight line isn't necessarily the shortest distance?

(back in 2m)

We can split up that curve into a sum of infinitesimal vectors, calculate their length using the metric and integrate along the curve to add them all up.

@barrycarter Yes, yes, yes, yes. And that's the key point. There is a class of curves called null curves that have a length of zero, and those are the curves that light rays follow.

Bernard Meurer

@dmckee I summon you

Back.. reading.

@JohnRennie In theory, you could solve this problem using calculus of variations, but, in this simple case, you should be able to get a closed form solution.

@barrycarter yes, to calculate a geodesic you can either use calculus of variations or you can start with the geodesic equation and grind away. Both approaches work.

5:24 PM

@JohnRennie But there's no simple closed form?

Well it depends on the metric and the curve. Generally no, there is no closed form.

OK, but we have the metric and the curve is just any curve connecting the start and end points?

Or is there more that I'm missing?

With simple metrics or high symmetry there may be a closed form. For example in SR the geodesics are obviously straight lines.

Not obvious to me, but ok.

You've jumped ahead a bit. So far all I've explained is how to calculate the length along a curve between two points.

5:26 PM

Yes, and what a null curve is. Keep going.

Now, hold onto your hat because this is the key point ...

Suppose I use my coordinates to calculate ds^2 ...

Yes...

(are there curves with negative length, I ponder... must be)

And some other observer uses any, I repeat any, other coordinates and also calculates ds^2 ...

I see this coming, but go ahead.

We will get the same value. ds^2 is an invariant. All observers in all coordinate systems will agree on its value.

5:29 PM

thinking....

So you're saying if two observers compute sqrt(dx^2+dy^2+dz^2-dt^2) for a given point, regardless of where and when they see it, they will get the same value?

> for a given point???

A given delta

(dt, dx, dy, dz) is a vector not a point

That's what I meant :)

So if one observer says the something moved one way in space and time and another says it moved in another way, that ds coordinate remains the same.

ds^2.
ds^2 may be negative meaning there is no real root.

5:33 PM

@JohnRennie Uhhh

Fair enough. OK, let me see if I can conjure some examples.

Oh my God he's back

@JohnRennie Had to grab Shankar and my notebook from the office

Observer A: the object is moving along the x axis at velocity v. Thus ds^2 = (v*dt)^2 - dt^2 = dt^2*(v-1). Good so far?

@barrycarter if you'll bear with me I'll give you a very simpe explanation of time dilation

5:34 PM

Currently working on electron loss of heavy ions in targets

@JohnRennie OK, continue.

@barrycarter Yes, that's the example I was going to use.

@barrycarter did John tell you how to define "vector"?

And observer B would see the object moving at a different velocity in the (1,1,1) xyz direction?

Ocelot, OK, I do know what a vector is, I think.

Start in my rest frame. In my rest frame I am stationary so in a time dt I will move by (dt,0,0,0)

5:36 PM

OK, so in the object's rest frame, ds^2 = -dt^2 ?

Yes

With you so far.

Now suppose you are moving at velocity v relative to me

Actually make that -v for convenience

@barrycarter what is it

And I choose to define my x axis in that direction?

5:37 PM

on a manifold

Ocelot, a member of R^n

@barrycarter on a manifold?

how does that work

@barrycarter yes, for convenience we'll assume motion is along the x axis

Ocelot, I think I know, but give me a few more seconds.

@barrycarter: if you get distracted arguing with 0celo7 this isn't going to work

5:38 PM

@JohnRennie :o

So, in my frame, the object is (dt, -v*dt, 0, 0)?

:(

Fine, I'll leave

Ocelot, chill, I'll get to you in a few seconds I said.

@barrycarter just know that this is all unrigorous

it's something a civil engineer would do :(

@JohnRennie Continue, and I'll let Ocelot heckle when we're done.

Ocelot, could you please assign t' = t + 5 minutes?

5:39 PM

@barrycarter let's call your time coordinate t and mine \tau to avoid confusion.

So in my frame ds^2 = -d\tau^2

So, in moving guy's frame... ok... -dt^2 + (-v*dt)^2 = dt^2 (1-v) ?

(2m break)

And in your frame ds^2 = -dt^2 + (vdt)^2 or ds^2 = - (1 - v^2)dt^2

@barrycarter @ me when you're done

Ocelot, will do if I can figure out how.

@JohnRennie OK, good, I almost got there with a mistake

So, I can figure out v

Since ds^2 must be the same in both cases we can equate the two expressions to get:
d\tau^2 = (1-v^2)dt^2

5:43 PM

Good.

And do d\tau and dt cancel out?

Well, yes, so we then get the standard gamma thing.

d\tau is the time I measured, and dt is the time you measured, and:

d\tau/dt = sqrt(1 - v^2)

Does that last equation look familiar?

That's time dilation, right?

Yes. sqrt(1-v^2) is just 1/gamma

With the usual convention that c=1

OK, I think I get this. The invariance of ds^2 can probably yield the Lorentz contraction or some variant of ti tooo.

c=1

@barrycarter Exactly. The Lorentz transforations are the linear transformations that preserve ds^2

There's bound to be a derivation on the site somewhere

5:46 PM

The Lorentz contraction (not transform) should be interesting, since I will have to integrate.

Q: How do I derive the Lorentz contraction from the invariant interval?

The Lorentz contraction is what I calculate in:

5

Reviewing some basic special relativity, and I stumbled upon this problem: From the definition of the proper time: $$c^2d\tau^2=c^2dt^2-dx^2$$ I was able to derive the time dilation formula by using $x=vt$: $$c^2d\tau^2=c^2dt^2-v^2dt^2=c^2dt^2\left(1-\frac{v^2}{c^2}\right)\rightarrow d\tau = dt\...

Ahhhhhh!

And that should now make sense

They're still skipping steps, but I can follow. Is there more? I'd like to give Ocelot a chance to tear you down :)

But be careful about Lorentz contraction because it is not really the same as time dilation even though it appears superficially similar.

5:48 PM

@JohnRennie No, you showed me earlier, that there's really no such thing, it's more of a existence persistence thing.

If a meter stick flashed on for a fraction of a second and then off, it would appear longer.

The point abut time dilation is that we measure it by comparing clocks.

In the twin paradox we synchronise clocks then speed off and compare clocks on our return

The clocks are different because the lengths of the curves we have followed are different

Where those lengths are calculated with the metric

@JohnRennie Not to rush you, but I do want to hear what Ocelot has to say, especially if he's going to ream you for what you've said so far.

Cool @0celo7 we're ready for you now :-)

@0celo7 Ocelot, ocelot, ocelot

I hope he didn't kill himself... before making his argument.

I love explaining relativity to an interested audience :-)

Even though I've probably butchered it

5:52 PM

@JohnRennie Your explanation was fantastic... but it's not the standard one for SR. And, until @0celo7 explains why not, it seems fairly simple.

To make the argument rigorous is somewhat involved, and that's what 0celo7 can do and I can't.

But, my butchery aside, this is the way to understand relativity. GR is exactly the same except that the metric is different.

Oh crap, that wasn't GR?

Want me to explain time dilation by a black hole?

No, I thought I'd just learned GR and SR both.

Or does the metric involve gravity too?

@barrycarter you did, but we used the SR metric, the Minkowksi metric, in our calculation

5:54 PM

OK, and the GR metric involves gravity, which is 0 in the SR metric?

Or more than that?

GR splits into two bits:

the easy bit is taking a metric and doing calculations with it

the hard bit is calculating what the metric should be

@JohnRennie No.

Factually incorrect.

Ocelot, tear him apart.

Oh no, it's back

Ocelot, start at the beginning, please :)

5:55 PM

In GR you don't have global coordinates

No, we summoned him, we should've expected that.

As long as we stand outside the circle, we're safe.

IN THE BEGINNING THERE WAS A TOPOLOGICAL HAUSDORFF SPACE

You mean a Hausdorff topological space, but sure.

Sure

But seriously, tell me what you find wrong with @JohnRennie's argument.

5:58 PM

It's too long

You said a vector on an n-manifold could not be regarded as a member of R^(n-1)

It's not wrong, just not very mathematical.

@barrycarter It can be, but that doesn't make sense

How do you define a vector at $p\in M$?

OK, a manifold is like a spherical surface in 3D, right?

@barrycarter oh, that's defining manifolds via embedding in $\mathbb{R}^n$

which can be done, but it's not very elegant

the standard way is to define them intrinsically