# Question to build intuition for GR

Tags:
1. Dec 28, 2014

### danieron

I am having trouble connecting some of the differential geometry in gr to what is actually measurable in the real world.

As far as i understand we can measure physical quantities in terms of coordinates $x^\mu$ on the tangent space of a 4-dim Pseudo-Riemannian manifold M. So lets say we are in the tangent space at some point p on M and observe a free falling particle on a geodesic passing trough that point and we measure some acceleration $\ddot{x^\mu}$. Since it is in free fall the geodesic equation must be satisfied
$(\nabla_{\dot x} \dot{x})^\mu = \ddot{x^\mu}+\Gamma^{\mu}_{\sigma\delta}\dot{x}^\sigma\dot{x}^\delta = 0$.
Is it correct to say that we see the coordinate acceleration $\ddot x^\mu$ because the velocity vector is trying to compensate for the change of the basis vectors of $T_{p}M$ (or equivalently change of the metric ), by changing it's coordinates with respect to these in order to keep the intrinsic length of the vector fixed (hence covariant derivative of it equals zero)?

Would that mean that some hyper dimensional being, that could see our 4-dim space-time as embedded in some R^n, would see the basis vectors of the tangent spaces (in an appropriate chart) change in size just as i can envision them changing from point to point on the 2-sphere. So he would basically see a tiny human with a ruler in his hand measuring stuff that is actually of constant size with a ruler of variable length (coordinates of ruler stay the same but basis vector changes) and ascribing the change to said quantities rather than to his ruler changing?

If so I could finally understand the redshift of distant galaxies due to expansion of universe in the sense that the wavelength is some extrinsic quantity that stays constant, but the basis vectors are shrinking everywhere making our rulers/coordinates smaller/worth less, if seen from the outside, hence leading us to measure an increased wavelength.

So again to summarize you have something that wants to stay constant, but space-time is contracting/expanding beneath it's feet, so to us it has to look like it is changing basically because we change with space-time.
This way of explaining things makes sense to me but i've never been really able to read this out of a textbook (maybe I don't read carefully enough hehe).

Could you please point out any flaws in my point of view?

2. Dec 28, 2014

### Staff: Mentor

Hi danieron, welcome to PF!
I think that is fine to say. Of course, the English statement is much less exact than the mathematical statement, so in the end it is the mathematical one that is important and any English statement that you understand as being consistent with the math will work.

There are lots of embedding theorems about embedding our 3+1 D curved spacetime into higher dimensional flat spacetimes, but they are never necessary from a practical standpoint since the presence or absence of those higher dimensions doesn't affect anything here. However, in most embedding theorems the higher-dimensional flat space is considered to be isometric to the lower-dimensional curved space.

Consider the example of a 2-sphere embedded in a flat 3-space. Measurements within the sphere could be done with perfectly acceptable rulers, meaning that a path measured to be 1 m long in the sphere would also be measured to be 1 m long in the flat space. The thing is that unit changes in lattitude and longitude would differ from unit changes in length across the manifold, as measured by rulers in either the sphere or in the embedding space.

3. Dec 29, 2014

### Matterwave

I feel like you are focusing too much on length. The parallel transport not only has to preserve length, but also preserves "direction" in a sense (as best as can be done on a curved manifold) hence why it's called "parallel transport". What the components are doing is they are changing in order to compensate for the fact that the basis coordinate vectors are not parallel transported along the curve we are considering. The lengths of the basis vectors need not be changing in order for you to obtain Christoffel symbol terms.

4. Dec 29, 2014

### stevendaryl

Staff Emeritus
I think a good way to get a feel for geodesics in curved spacetime is to look at the simpler example of constant-velocity nonrelativistic motion in curvilinear coordinates in Euclidean space.

In a flat 2D space, let $\vec{V}$ be a velocity vector. The criterion that the velocity is constant is just:

$\frac{d \vec{V}}{dt} = 0$

In terms of Cartesian coordinates, we can write: $\vec{V} = V^x e_x+ V^y e_y$, where $e_x$ and $e_y$ are basis vectors in the x-direction and y-direction, respectively, and where $V^x = \frac{dx}{dt}$ and $V^y = \frac{dy}{dt}$. The criterion that $\vec{V}$ is constant becomes:

$\frac{d V^x}{dt} e_x + V^x (\frac{d}{dt} e_x) + \frac{d V^y}{dt} e_y + V^y (\frac{d}{dt} e_y) = 0$

Since $e_x$ and $e_y$ are constant vectors, we can rewrite this vector equation as two equations for components:

$\frac{d V^x}{dt} = 0$

$\frac{d V^y}{dt} = 0$

That's pretty obvious, of course. But now, let's switch to polar coordinates: $r$ and $\theta$ where $x= r cos(\theta)$ and $y=r sin(\theta)$. We can define velocity components $V^r = \frac{dr}{dt}$ and $V^\theta = \frac{d\theta}{dt}$.

In terms of polar coordinates, constant velocity motion looks like this:

$\frac{d V^r}{dt} e_r + V^r (\frac{d}{dt} e_r) + \frac{d V^\theta}{dt} e_\theta + V^\theta (\frac{d}{dt} e_\theta) = 0$

In order for $\vec{V}$ to have the same value in either coordinate system, the basis vectors for polar coordinates $e_r$ and $e_\theta$ must relate to $e_x$ and $e_y$ through:

$e_\theta = -r sin(\theta) e_x + r cos(\theta) e_y$

$e_r = cos(\theta) e_x + sin(\theta) e_y$

These basis vectors are NOT constant:

$\frac{d}{dt} e_\theta = - \frac{dr}{dt} sin(\theta) e_x - \frac{d\theta}{dt} r cos(\theta) e_x + \frac{dr}{dt} cos(\theta) e_y - \frac{d\theta}{dt} r sin(\theta) e_y$
$= V^r \frac{1}{r} e_\theta - V^\theta r e_r$

$\frac{d}{dt} e_r = - \frac{d\theta}{dt} sin(\theta) e_x + \frac{d\theta}{dt} cos(\theta) e_y$
$= V^\theta \frac{1}{r} e_\theta$

So in terms of components, $\frac{d\vec{V}}{dt} = 0$ becomes, in polar coordinates:

$\frac{d V^r}{dt} - V^\theta V^\theta r = 0$

$V^r V^\theta \frac{1}{r} + \frac{d V^\theta}{dt} + V^\theta V^r \frac{1}{r} = 0$

We can combine these into a single equation with dummy coordinates:

$\frac{d}{dt} V^i + \sum_{j k} \Gamma^i_{jk} V^j V^k = 0$

where $\Gamma^r_{\theta \theta} = r$, $\Gamma^\theta_{r \theta} = \Gamma^\theta_{\theta r} = \frac{1}{r}$, and all the rest are zero.

5. Dec 29, 2014

### PAllen

I have a few questions related to the physics of the OP. If an object is in free fall, any effectively co-located instrument that is also in free fall (even with a different velocity) will measure the given object as having no acceleration. So I am stuck on the first sentence of what exactly this measured acceleration is?

Similarly, if one imagines putting Cartesian coordinates on the tangent manifold (which is considered flat), the connection components vanish, and the acceleration for a geodesic will be zero in these coordinates (of the tangent manifold) at the tangent point.

To observe acceleration for a free fall object, you either need to have a non-free fall instrument, or a family of instruments implementing extended coordinates originating from a non-colocated free fall instrument.

6. Dec 29, 2014

### danieron

First of all thank you everyone for answering!

Yes you are right, in fact I meant a fixed reference frame on some point on the geodesic.
I understand the tangent space as the best linear approximation to the manifold endowed with a constant metric (hence tangent space is flat) but I am not really sure that the Christoffel symbols have to vanish. I guess you mean Riemann normal coordinates which I do not understand from a mathematical point of view but I guess from a physical standpoint mean that your coordinate system is moving along on a geodesic. In particular it is said that Riemann normal coordinates are such that geodesics look like straight lines as can bee seen by the geodesic equation if the gammas vanish, right?

I don't really have a problem with this. What I do have a problem with is the logical order of things.

I can move around on the manifold by sitting still in my room or go to the moon, but as I set myself up to do an experiment and define some coordinate system I am effectively setting myself up to do measurements in the tangent space at that point. Therefore my coordinate system is not globally valid and the distances that I measure are related to proper distances through the metric right? So if I now go to another tangent space and use the same ruler as before, which i have defined to be 1m, (hence, have lets say coordinate 1 along x axis) this ruler is going to correspond to a different proper length because the metric has changed right?
What is completely weirding me out at the moment is that I would expect the wavelength of light to have some intrinsic proper length, but which then is measured in different points on the manifold with a ruler which to me always looks the same, but which actually varies in proper length.
The problem is now the following.
I take a measurement of lets say the peak wavelength of the CMBR at some point in space and time. Then I wait and take it again . So i took two measurements in two different tangent spaces with two metrics which are different because the universe is intrinsically expanding. But if the metric is becoming bigger it means that my rulers and clocks are becoming worth more proper length and I would need less multiples of them to fit the wavelength i see, but less length measured by me means I see blueshift.

I also see I am missing something because of how some sources treat the case of redshift, this time due to the light escaping a gravitational field.
What they do is say: I have an observer at $\vec x_1$ which measures a coordinate time $\Delta t_1$ elapsing between adjacent crests in the wave. The proper time elapsed must then be $(\Delta \tau_1)^2=g_{00}(\vec x_1)(\Delta t)^2$. Then you write the same equation for an observer at $\vec x_2$ and what they do next is say $\Delta t_1=\Delta t_2$ hence getting $\frac{\nu_2}{\nu_1}=\frac{\Delta \tau_1}{\Delta \tau_2}=\sqrt{\frac{g_{00}(\vec x_1)}{g_{00}(\vec x_2)}}$. But that is the exact opposite of what I would have done as to me it would have made sense to set the proper times equal to each other.
To me $\nu_1=\frac{1}{\Delta t_1}$ makes sense, while $\nu_1=\frac{1}{\Delta \tau_1}$ does not.
By the way another thing I don't understand is the notation $g_{00}(\vec x_1)$ and $g_{00}(\vec x_2)$ as if the two different values for the metric were assigned to two different points in one and the same tangent space. I would understand $g_{00}(p_i)$ were the p_i are points on the manifold.

Anyone can see where the hell I must be terribly wrong with my way of seeing things?

7. Dec 29, 2014

### stevendaryl

Staff Emeritus
Let's suppose that
1. The first crest passes $\vec{x_1}$ at time $t_1$.
2. That crest passes $\vec{x_2}$ at time $t_1'$.
3. The second crest passes $\vec{x_1}$ at time $t_2$
4. The second crest passes $\vec{x_2}$ at time $t_2'$
If it takes a certain amount of time, $\delta t_{12}$ for light to pass from $\vec{x_1}$ to $\vec{x_2}$, then:

• $t_1' = t_1 + \delta t_{12}$
• $t_2' = t_2 + \delta t_{12}$
So subtracting gives:
$t_2' - t_1' = t_2 - t_1$

So the coordinate time between crests is the same at both locations. But since proper time is not the same as coordinate time (and the conversion factor is location-dependent), the proper time between crests is different in the two locations.

8. Dec 29, 2014

### Staff: Mentor

This doesn't happen in GR because the Levi Civta connection is metric compatible. It therefore preserves all lengths and angles, even when parallel transported along non geodesic paths.

9. Dec 29, 2014

### pervect

Staff Emeritus
I find it helpful to consider the situation of the surface of the globe to illustrate how to handle curvature. What we have then is a curved 2d manifold that we have some intuition for.

When you set yourself up to do some experiment you've got a choice of coordinates to use. I suspect you may be thinking of using local Fermi normal coordinates.

[change]
But rather than go on about this as length, I'm going to trim the heck out of my previous post, and just say: you can h ave globally valid coordinate systems as seen in our example of the Earth. So you don't have to obsess too much about the behavior of Fermi Normal coordinates (assuming that's what you are thinking of) not being global. It's related to that specific choice of coordinates, when you think in a broader context you realize you can have global coordinaes, and that in fact a lot of the machinery of GR was developed so you could do physics in whatever coordinates you needed to make them global.

I don't think so. Consider you have a 100 meter long arc on the globe. It doesn't matter what coordinates you use, the length of the arc is 100 meters. If you define a different tangent space, the arc is still 100 meters. Distances do not change when you change coordinates, or tangent spaces. The distances are physical, the coordinate description of them varies, but the distances do not vary.

What varies are the coordinates, not the distances.

Example: If you go to a higher latitude, 1 degree of longitude represents different lengths. But a meter is still a meter on the equator or at/near the north pole. The meter doesn't change it's length ever, it's physical. What changes is the amount of distance a certain change in longitude represents.

Well, enough for now - hope this helps.

Last edited: Dec 29, 2014
10. Dec 31, 2014

### danieron

That's true I had not noticed that metric compatibility has that meaning! Thank you very much for pointing that out!
Ok let's see if I understand things better now.
Let's say I have a small cannon in my room. I set up a coordinates system and now contemplate the space-time distance tau a ball travels in 1s of my time, when i shoot the ball from $(0,0,0,0)$ to $(1s ,x^1,x^2,x^3)$, in my coordinate system. This distance, which incidentally corresponds to the time a clock moving with the ball would measure , is going to be given by $(\Delta \tau)^2=g_{ij}x^ix^j$ where the g_ij's depend on where I am right? Now I parallel transport the cannon and my meters and clocks I used to define my coordinate system to another far away place. If I now repeat the same experiment there I might not get the same spatial coordinates $(x^1,x^2,x^3)$ because the gravitational pull on the ball has changed (making it for example travel less distance in that 1s). But this change in coordinates is exactly balanced by the fact that the metric changes as well depending on the gravitational pull at my location, and hence the previous equation spits out the same $(\Delta\tau)^2$! Can I also say that my coordinates changed because parallel transport doesn't necessarily preserve coordinates? (I think that thats a property of parallel transport...)

But what if instead of that experiment I just give you fixed coordinates $(x^0 ,x^1,x^2,x^3)$ in both locations and ask for the corresponding space-time interval? Then the metric still changes but the coordinates don't compensate for that. Is it that the coordinates even though numerically the same just do not correspond to the same physical thing at the two locations? (In the previous example demanding the coordinates to stay the same would mean I would have to change to a more powerful cannon to compensate for gravity….)

11. Dec 31, 2014

### Staff: Mentor

This is an odd experiment. You are using a coordinate dependent criterion, t=1s in your coordinates, to determine the end of the experiment.

Technically that would be $(\Delta \tau)^2=\int g_{ij}dx^i dx^j$, but I think you realized that already.

I cannot think of how to analyze this scenario. I am not sure that the result would not change since the result seems to depend on the coordinates and the coordinates may have changed. Not because the physics depends on the coordinates but because the point that you are choosing as "the end" does.

If, instead of a coordinate-dependent end, you chose some physical event as "the end" then I think what you are saying is true.

As mentioned above, the interval is calculated by $(\Delta \tau)^2=\int g_{ij}dx^i dx^j$. So you need the metric in order to calculate it. If you do not have the metric then you cannot calculate the interval.

Last edited: Dec 31, 2014
12. Jan 1, 2015

### danieron

@DaleSpam Sigh I just don't get it then…
I am not a priori against what you said but I don't get what the coordinates system $(t,x^0,x^1,x^2,x^3)$ is supposed to describe. It cannot be coordinates on the tangent space because there are two different values for the metric at two locations $\vec x_1$ and $\vec x_2$. Moreover in my head the conversion formula between proper time and coordinate time $(d\tau)^2=g_{ij}dx^idx^j$ only applies between a fixed observer on a tangent space who measures a change in coordinates $dx^i$, and a moving observer, who travels along with whatever object has changed it's coordinates by the $dx^i$, and who's clock is going to measure $d\tau$.

I will attach a snapshot of this problem I found in a script, and if someone takes me by hand and tells me what there corresponds to what in the mathematics of curved manifolds and then what it means physically I would be very thankful. In particular what that aforementioned coordinate system is and who it is that would physically measure the time $\Delta t$ if it's not the time measured by the the observers fixed at $\vec x$. (as seems to be mentioned there).

#### Attached Files:

• ###### gr.png
File size:
95.5 KB
Views:
67
13. Jan 1, 2015

### Staff: Mentor

Sorry about that. My favorite introductory reference is this one:
http://preposterousuniverse.com/grnotes/
especially Ch 2.

Coordinates are not on the tangent space. Coordinates are on the manifold itself. Coordinates map events in the manifold to points in R4.

I am not sure what you mean by "locations $\vec x_1$ and $\vec x_2$". Points on a manifold are not vectors, so I don't understand the notation.

14. Jan 1, 2015

### Staff: Mentor

It depends on what you mean by "my coordinate system". Since you appear to be saying that 1s of coordinate time corresponds to 1s of time elapsed on your clock, then the coordinates you are using must be local inertial coordinates in which you are at rest (assuming you are in free fall), which only cover a small patch of spacetime around a given event (in this case, the event of you firing the cannon). In these coordinates, the metric is always the same (the Minkowski metric $\eta_{\mu \nu}$ ), and if we assume that the cannonball is in free fall once it's fired and that it's always fired with the same velocity relative to you, then the coordinates it reaches after 1s will always be the same, regardles of where in spacetime you run the experiment. Differences in gravity won't affect this because they are all "factored out" by the use of a local inertial frame in which you are at rest.

If, OTOH, you are using some sort of global coordinates (such as Schwarzschild coordinates in the spacetime around a gravitating mass), then you can't assume that 1s of coordinate time is the same as 1s of "your" time (time elapsed on your clock). That means the conditions of the experiment are not specified well enough, because you can't measure the cannon ball's position after 1s of coordinate time unless you know the relationship between your clock's time and coordinate time, since you have to use your clock to measure time. See further comments below.

The example you give is using global coordinates in a stationary spacetime (which means the metric does not depend on coordinate time). No observer at any finite $\vec{x}$ (this notation, btw, is not often used because, as DaleSpam pointed out, points on a manifold are not vectors, so thinking of spatial positions as vector displacements from some spatial origin is not a good idea) measures coordinate time in these coordinates; the coordinate time is just a bookkeeping device. Observers fixed at a given spatial position $\vec{x}$ measure their own proper time; however, a stationary spacetime has the property that there is a function $\phi(\vec{x})$ which gives the ratio of proper time to coordinate time for an observer fixed at $\vec{x}$. The example you give is simplly illustrating that if you send a light signal between two spatial positions $\vec{x}_1$ and $\vec{x}_2$ which have different values of $\phi$, the light will be redshifted or blueshifted, because, since the lapse of coordinate time $t$ between two wave crests or troughs is the same, the lapse of proper time for the two observers fixed at $\vec{x}_1$ and $\vec{x}_2$ must be different.

Last edited: Jan 1, 2015
15. Jan 1, 2015

### stevendaryl

Staff Emeritus
No. Coordinates are labels on points in space (or spacetime, in the case of relativity). A coordinate system assigns four numbers: $x^0, x^1, x^2, x^3$ to each event. A coordinate system defines "simultaneity" of two events: if they have the same value for $x^0$, then they are at the same "time". ($t$ is used interchangeably with $x^0$, usually).

No. It doesn't have to do with "observers". If you have a clock that travels at constant velocity from the point with coordinates
$(x,y,z,t)$ to the point $(x+\delta x, y+\delta y, z+\delta z, t+\delta t)$, then the elapsed time on the clock changes by the amount

$\delta \tau = \sqrt{g_{ij} \delta x^i \delta x^j}$

It doesn't have anything to do with observers, unless you're considering the clock to be an observer.

16. Jan 2, 2015

### pervect

Staff Emeritus
OK, lets stop right there.

You can transport a tensor along a curve. However, cannons (and meters and clocks) aren't tensors. So I'm not sure what you are doing. My best guess at an attempt of what you could legitimately do (which probably doesn't resemble what you think you are doing) is this:

You have a vector that represents the initial velocity of some cannonball. The vector is a tensor, so you can parallel transport it. You need to specify some particular curve you're going to parallel transport it on, though.

The cannonball follows a geodesic. You parallel transport the initial velocity vector along some curve, then that paralllel transported vector defines a second geodesic "parallel" to the first.

You could possibly apply the geodesic equation here, but I'm not going to go into great detail about that unless there is some reason to beleive that was your intent, which I don't see as likely at the moment. I could be wrong cause I don't understand your intent.

If you have a vector that represents the final velocity of some cannonball, you could also parallel transport that vector along a different curve than the curve you used to parallel transport the initial velocity. It couldn't possibly be the same curve if the initial and final points are not the same. So I don't see how you can make the statements you make, you specify a couple of curves, you parallel transport a couple of velocity vectors, but there's no guarantee that after you do so that these vectors have any relation or lie on the same geodesic.

17. Jan 2, 2015

### pervect

Staff Emeritus
There wouldn't be any "t", t would usually be the $x^0$. You'd have $(x^0,x^1,x^2,x^3)$

Correct, as others have mentioned the $x^i$ are on the manifold.

You might want to recall the definition of a manifold. The short version says that charts exist that map points in the manfiold into R^n via some collection of charts (which I'm not going to get into the details of).

$(x^0, x^1, x^2, x^3)$ is a point in R^4 . So do get from this to the manifold, you really need to specify not only the manifold but also the chart. Given this, everythign else follows from the fundamental defintion of a manifold (and a chart).

The somewhat good news is that maps do exist from the tangent space to the manifold, such as the exponential map. I wrote something on this, but I think it'd be better left until asked for, it is more technical than the above and it's not clear it's relevant.

You mention later a snapshot of something. I haen't seen any snapshots, I remain cheerfully optimistic that it would help to see it to understand your question.