Rod shortening of General Relativity

[Pierre007080]

Hi Guys,
Is there a simple formula (like lorentz equations for SR) for rod shortening ALONG THE RADIAL (geodesic) that depends on gravitational potential. I realize that this is a theoretical case without rotation of the mass etc. If possible this formula should apply to a point of reference on the geodesic and ONLY apply to the single axis along the geodesic path (ie no rotation around the axis itself either)

 

[yuiop]

Hi Guys,
Is there a simple formula (like lorentz equations for SR) for rod shortening ALONG THE RADIAL (geodesic) that depends on gravitational potential. I realize that this is a theoretical case without rotation of the mass etc. If possible this formula should apply to a point of reference on the geodesic and ONLY apply to the single axis along the geodesic path (ie no rotation around the axis itself either)

The simplest equation is:

$$dL = dr/\sqrt{1-2m/r}$$

in units of G = c =1, where dL is the length of a short rod according to a local observer and dr is the length of the rod according to the Schwarzschild observer at infinity.

Over longer distances the ruler distance L as measured by laying short unit length rods ends to end from Schwarzschild radial coordinates r=a to r=b is obtained by integrating the above and is:

$$L = \sqrt{a(a-2m)} - \sqrt{b(b-2m)} + 2m \ln \left(\frac{\sqrt{a} + \sqrt{a-2m}}{\sqrt{b} + \sqrt{b-2m}} \right)$$

Is that what you are looking for?

 

[bcrowell]

Is there a simple formula (like lorentz equations for SR) for rod shortening ALONG THE RADIAL (geodesic) that depends on gravitational potential. I realize that this is a theoretical case without rotation of the mass etc. If possible this formula should apply to a point of reference on the geodesic and ONLY apply to the single axis along the geodesic path (ie no rotation around the axis itself either)

I think you need to clarify what you're asking for. When you say "radial," it sounds like you're assuming something specific, like the Schwarzschild spacetime, or some other spacetime with spherical symmetry.

 

[Pierre007080]

Hi yuiop,
Thanks for your response. This is exactly what I needed.
Does this short formula also have an inverse relationship that determines time dilation along that same radial, which is also dependent on the gravitational potential?

 

[yuiop]

Hi yuiop,
Thanks for your response. This is exactly what I needed.
Does this short formula also have an inverse relationship that determines time dilation along that same radial, which is also dependent on the gravitational potential?

Yes, the gravitational time dilation of a stationary clock at Schwarzschild radial coordinate r relative to the clock of an observer at infinity is:

$$dt' = dt \sqrt{1-2m/r}$$

If the clock at r is moving with velocity v relative to a stationary observer at r, the time dilation of the clock relative to the clock at infinity is:

$$dt' = dt \sqrt{1-2m/r} \sqrt{1-v^2/c^2}$$

An object that is initially at rest and dropped from infinity has velocity $\sqrt{(2m/r)}$ when it gets to r and its time dilation in this special case relative to the clock at infinity is:

[tex]dt' = dt(1-2m/r) [/itex]

 

[Pierre007080]

Thanks for the help yuiop. I appreciate your uncomplicated response.
Regards. Pierre

 

[Pierre007080]

Does this rod shortening formula apply to the wavelength of electromagnetic waves traveling along this same coordinate?

 

[yuiop]

Does this rod shortening formula apply to the wavelength of electromagnetic waves traveling along this same coordinate?

The relationship between the coordinate wavelength $\lambda$ of a lightwave traveling along a radial geodesic according to the Schwarzschild observer at infinity, relative to the wavelength $\lambda_r$ measured by a local observer at r is given by:

$$\lambda_r' = \frac{\lambda}{ \sqrt{1-2m/r}}$$

Note that the locally measured wavelength is longer (not shorter!) than the coordinate wavelength. This unintuitive result can be explained in very broad simplistic terms by thinking of it as the "length" of the wave being measured by a local observer is longer because the local observer is using a gravitationally length contracted measuring rod to measure the wavelength.

The equivalent relationship between the coordinate frequency of a lightwave traveling along a radial geodesic according to the Schwarzschild observer at infinity, relative to the frequency measured by a local observer at r is simply:

$$f_r' = \frac{f}{\sqrt{1-2m/r}}$$

The above equations are probably not what you would initially expect, but it should be noted that they are coordinate measurements related to local measurements and not the normally quoted redshift relationship between the emitted frequency $f_e$ of the wave measured by a stationary observer at $r_e$ relative to the observed frequency $f_o$ measured by a stationary receiver at $r_o$, which is given by:

$$\frac{f_o}{f_e} = \frac{\sqrt{1-2m/r_e}}{\sqrt{1-2m/r_o}}$$

which when the photon is emitted at infinity so that $r_e$ is infinite, the locally observed frequency of the receiver at coordinate r is the more familiar:

$$f_o = f_e \frac{1}{\sqrt{1-2m/r}}$$

Equivalently the wavelength relationship is given by:

$$\frac{\lambda_o}{\lambda_e} = \frac{\sqrt{1-2m/r_o}}{\sqrt{1-2m/r_e}}$$

which for the case of the photon being emitted at infinity reduces to:

$$\lambda_o = \lambda_e \sqrt{1-2m/r}$$

for the wavelength as measured by the local observer receiving the photon at r.

Finally according to the observer at infinity, the coordinate wavelength of a falling lightwave at r is:

$$\lambda = \lambda_e (1-2m/r)$$

and the coordinate frequency of the falling lightwave at r is:

$$f = f_e$$

so the Schwarzschild observer at infinity considers the coordinate frequency (and therefore energy) of the falling photon to remain constant.

In Newtonian mechanics the speed/ frequency/ wavelength relationship is given by $dr/dt = \lambda f = c$ but in the radial Schwarzschild case the coordinate relationship is $dr/dt = \lambda f = c(1-2m/r)$

 

[Pierre007080]

Yuiop, you are a STAR! Your answers are so well presented and concise. I am still struggling to internalise the basics, but there is a glimmer emerging.
So far you have clarified the rod, time wavelength and frequency ALONG the geodesic. If I may move to the next step and envisage a sphere at which the gravitational potential is equal at any point of its surface. This sphere would be at a "right angle" to the radial geodesic discussed so far. Can you say the geodesic radial is normal to the sphere surface?
Question:
1. Would the time be the same at any point of the surface because the gravitational surface is the same?
2. Would rods and wavelengths be unaffected at right angles to the radial geodesic along the surface of the sphere?
3. Would frequencies along this surface thus only be affected by the time dilation?

 

[yuiop]

If I may move to the next step and envisage a sphere at which the gravitational potential is equal at any point of its surface. This sphere would be at a "right angle" to the radial geodesic discussed so far. Can you say the geodesic radial is normal to the sphere surface?

I think that is fine.

Question:
1. Would the time be the same at any point of the surface because the gravitational surface is the same?

Yes.

2. Would rods and wavelengths be unaffected at right angles to the radial geodesic along the surface of the sphere?

Yes.

3. Would frequencies along this surface thus only be affected by the time dilation?

The coordinate horizontal speed of light is $dr/dt = \sqrt{(1-2m/r)}$ in contrast to the vertical speed of light $dr/dt = (1-2m/r)$. This means that the frequency measured by a local observer $f_r'$ of a horizontal lightwave at r, is related to the coordinate frequency $f$ according to the observer at infinity by:

$$f_r' = \frac{f}{\sqrt{1-2m/r}}$$

which is the same as the local/coordinate frequency ratio for a vertical lightwave.

So the short answer to question 3 is yes. :tongue:

 

[pervect]

The relationship between the coordinate wavelength $\lambda$ of a lightwave traveling along a radial geodesic according to the Schwarzschild observer at infinity, relative to the wavelength $\lambda_r$ measured by a local observer at r is given by:

$$\lambda_r' = \frac{\lambda}{ \sqrt{1-2m/r}}$$

Note that the locally measured wavelength is longer (not shorter!) than the coordinate wavelength. This unintuitive result can be explained in very broad simplistic terms by thinking of it as the "length" of the wave being measured by a local observer is longer because the local observer is using a gravitationally length contracted measuring rod to measure the wavelength.

Eeeep!

I assume that by "coordinate distance" you mean "coordinate difference". I think I owe it to the new readers here unfamiliar with GR to point out that in general, coordinate differences aren't generally regarded as distances in most textbooks or professional papers (at least that I've seen), that coordinate differences aren't something you measure with a ruler, that the choice of a particular coordinate system is purely a matter of convention - and there are a lot of other choices than the Schwarzschild metric for a massive body (isotropic coordiantes, for one), and that the coordinate difference depends on this choice and thus is not physically meaningful in and of itself.

The short version of my position is that if you happen to be using polar coordinates, and you have one point at r=1, theta =0, and another point at r=2, theta=90, you don't generally say the "distance between the points is 90 degrees". And it makes just as little sense to do the same thing in other generalized coordinates.

It's also worth noting that on Einstein's elevator, the standard treatment doesn't have anything that even resembles "gravitational length dilation", it's perfectly feasible to have all the distance-related metric coefficients unity. Not so with the time metric coefficients for the elevator. This is a pretty strong clue that "length dilation" - to the extend the term makes sense, the fact that a coordinate difference has to be scaled in order to be turned into a distance - has everything to do with the particular choice of coordinates and nothing to do with "gravity".

 

[bcrowell]

I would echo pervect's doubts. I don't think Pierre007080's original question was posed in a meaningful way.

 

[Passionflower]

I assume that by "coordinate distance" you mean "coordinate difference". I think I owe it to the new readers here unfamiliar with GR to point out that in general, coordinate differences aren't generally regarded as distances in most textbooks or professional papers (at least that I've seen), that coordinate differences aren't something you measure with a ruler, that the choice of a particular coordinate system is purely a matter of convention - and there are a lot of other choices than the Schwarzschild metric for a massive body (isotropic coordiantes, for one), and that the coordinate difference depends on this choice and thus is not physically meaningful in and of itself.

I just looked at some fairly recent relativity books that use the term 'coordinate distance':

Relativity, Gravitation, and Cosmology - Cheng (Oxford, 2005)
Relativity - Special, General and Cosmological - Rindler (Oxford, 2006)
General Relativity - Hobson, Efstathiou, Lasenby (Cambridge, 2006)
Einstein’s General Theory of Relativity - Grøn, Hervik (Springer, 2007)
Introduction to 3+1 Numerical Relativity - Alcubierre (Oxford, 2008)
A First Course in General Relativity (2nd Edition) - Schutz (Cambridge, 2009)

The term 'coordinate difference' seems to avoid a potential confusion. So in that sense it is better than using the term 'coordinate distance' but clearly 'coordinate distance' is used in many books.

Could you point out some textbooks where the term 'coordinate difference' is used?

 

[Pierre007080]

Hi Yuiop.
Thanks for the manner in which you have dealt with my questions. I think your answers enlightened my simple understanding. Sadly, I'm afraid that this thread has been taken over by "experts" that will not allow my further questions ... because you need a PhD in physics to "present the question in a meaningful way".
Regards.
Pierre

 

[pervect]

Schutz uses the term "coordinate distance" about 3x, and "coordinate difference" about 2x. Undertaking a more deliberate search, so far I haven't found anyone else who uses either term (I don't have the other references on your list).

But you're right, there is precedent (though I haven't seen the other examples you mention) for usage of the term "coordinate distance".

MTW doesn't use either term as near as I can tell (though it does discuss the topic of distance, which is in the index). MTW's index entry under "distance" points to the page on the metric, which page explains that the metric is used to calculate the distance between nearby events.

Wald, I don't have in a searchable format, just paper, but there's not any mention of distance in the index. I believe Wald prefers to deal entirely with length - he defines the length of a curve, which for his purpose is sufficient. (And avoids arguments :-) ).

Caroll's lecture notes don't seem to use either "coordinate distance" or "coordinate difference", though Caroll does discuss distance.

I don't have the other texts you mention.

However, when Schutz (the only source I've seen that does use coordinate distance) uses it, it appears to be mostly to caution people that it's not measurable. Something that I'd like to reinforce.

To wit:

" If we were to work in a TT-coordinate system, as in the previous paragraph, then
the fact that the particles remain at rest in the coordinates means that the components of ξ￿ would remain constant; although correct, this would not be a helpful result since we
have not associated the components of ξ￿ in TT-coordinates with the result of any measurement. Instead we shall work in a coordinate system closely associated with measurements, the local inertial frame at the point of the first geodesic where ξ￿ originates. In this frame, coordinate distances are proper distances, as long as we can neglect quadratic terms in the coordinates."

"But what measure of distance is suitable over vast cosmological separations? Not coordinate distance, which would be unmeasurable."

The third usage isn't particularly related:

"Then when the photons reach our coordinate distance from the emitting object"

 

[pervect]

Hi Yuiop.
Thanks for the manner in which you have dealt with my questions. I think your answers enlightened my simple understanding. Sadly, I'm afraid that this thread has been taken over by "experts" that will not allow my further questions ... because you need a PhD in physics to "present the question in a meaningful way".
Regards.
Pierre

I strongly suspect that you are going off on a wrong tangent, Pierre, and taking an answer that agrees with and reinforces your preconceptions of how you think things out to work over the way that they actually do work. Though I suppose it's all moot if we've scared you away.

 

[Pierre007080]

I strongly suspect that you are going off on a wrong tangent, Pierre, and taking an answer that agrees with and reinforces your preconceptions of how you think things out to work over the way that they actually do work. Though I suppose it's all moot if we've scared you away.

Hi Pervect,
Thanks for your comment ... I apologise for "going off at a tangent". Please remember that the truth of how things actually work is not only valid in terms of your precise terminology. Envisage trying to explain something to a child. You try not to lie to him, but tell the truth that is determined by his capabilities. This will then maintain his interest and when he is grown up you still have him as a student. There is no way that I as the person asking the question on a certain thread can sort out the technical relevance of what you guys are discussing now ... is there no way you can keep me in the conversation in stead of going over my head. This is the art of teaching! I again apologise for my rudeness!

 

[yuiop]

Eeeep!

I assume that by "coordinate distance" you mean "coordinate difference".

I would agree that "coordinate difference" is a better term than "coordinate distance". I should perhaps also have it clearer that the formula I gave for the relationship between the coordinate and local wavelength of a vertical lightwave applies to short distances or wavelengths and for longer distances the integrated distances have to be used, but I did mention this in my first post about lengths and distances in general.

One way to measure the wavelength of a lightwave and keep the the measurement local is to measure the local frequency and then calculate the implied wavelength assuming the relation wavelength = c/frequency using a single local clock at r. The coordinate observer at infinity in regular Schwarzschild polar coordinates will note that the clock of the stationary observer at r is running slow by a time dilation factor of sqrt(1-2m/r) and therefore estimates the coordinate wavelength to be shorter than the locally measured wavelength by a factor of sqrt(1-2m/r).

The short version of my position is that if you happen to be using polar coordinates, and you have one point at r=1, theta =0, and another point at r=2, theta=90, you don't generally say the "distance between the points is 90 degrees". And it makes just as little sense to do the same thing in other generalized coordinates.

I think what Pierre meant was "horizontal distance" in the infinitesimal limit for a short vector that is at right angles to where the radial geodesic intersects the shell sphere surface and tangential to the surface of the sphere.

Perhaps you will agree that the following is true?

The circumference of ring with Schwarzschild radius r as measured by a local observer r is 2*Pi*r whether the local observer measures the circumference using a tape measure or using radar measurements. This circumference will agree with circumference according to the Schwarzschild observer at infinity. There is a one to one relationship for horizontal distance measurements. At every infinitesimal point along the circumference the surface is orthogonal (or at right angles) to a radial geodesic intersecting the ring at that point. I think is what Pierre meant loosely by distances that are "at right angles" to vertical radial geodesics and I think we could also call these distances horizontal distances. If Pierre actually meant at right angles over extended distances from a fixed radial geodesic then hopefully he will clarify that.

This is a pretty strong clue that "length dilation" - to the extend the term makes sense, the fact that a coordinate difference has to be scaled in order to be turned into a distance - has everything to do with the particular choice of coordinates and nothing to do with "gravity".

Well, I think this is interesting subject in itself and I will start a new thread on the subject in the near future. I hope you will agree that "gravity" has an effect on lengths and distances that makes distances measured in curved spacetime different to lengths and distances measured in flat space, in a way that has more physical significance than just a choice of coordinates. In other words, intrinsic curvature of curved spacetime is not just something that can be transformed away.

 

[Passionflower]

Well, I think this is interesting subject in itself and I will start a new thread on the subject in the near future. I hope you will agree that "gravity" has an effect on lengths and distances that makes distances measured in curved spacetime different to lengths and distances measured in flat space, in a way that has more physical significance than just a choice of coordinates. In other words, intrinsic curvature of curved spacetime is not just something that can be transformed away.

Question for you yuiop, when you are talking about lengths changing under gravity do you mean:

- a change in ruler distance caused by tidal effects.
- a change in radar distance.
- proper lengths (e.g. as measured by a local co-moving observer) v.s. Lorentz adjusted lengths.​

Or perhaps something else?

 

[Pierre007080]

I would agree that "coordinate difference" is a better term than "coordinate distance". I should perhaps also have it clearer that the formula I gave for the relationship between the coordinate and local wavelength of a vertical lightwave applies to short distances or wavelengths and for longer distances the integrated distances have to be used, but I did mention this in my first post about lengths and distances in general.

Yes, thanks Yuiop, I am talking about short distances and understand that the distance changes along the axes in GR.
May I comment on the "coordinate distance vs coordinate difference". It is my understanding that the free falling observer is the "coordinate observer" at point r even though he is accelerating down the geodesic radial. At that instant under discussion his coordinate system has its zero point of that axis along the radial at point r. The "difference" minus zero then becomes the "distance". Does this not qualify the use of "distance" under the circumstances?

 

[Pierre007080]

I think what Pierre meant was "horizontal distance" in the infinitesimal limit for a short vector that is at right angles to where the radial geodesic intersects the shell sphere surface and tangential to the surface of the sphere.

Perhaps you will agree that the following is true?

The circumference of ring with Schwarzschild radius r as measured by a local observer r is 2*Pi*r whether the local observer measures the circumference using a tape measure or using radar measurements. This circumference will agree with circumference according to the Schwarzschild observer at infinity. There is a one to one relationship for horizontal distance measurements. At every infinitesimal point along the circumference the surface is orthogonal (or at right angles) to a radial geodesic intersecting the ring at that point. I think is what Pierre meant loosely by distances that are "at right angles" to vertical radial geodesics and I think we could also call these distances horizontal distances. If Pierre actually meant at right angles over extended distances from a fixed radial geodesic then hopefully he will clarify that.

Yes. I think you have described what I tried to ask. I think that the radial must be "perpendicular" to the surface of the sphere.

Question:
1.I did intend to mean only short "horisontal" distances, but if I understand you correctly then the distance measurement along the axis along the surface remains untransformed all the way around the circle of the surface of the sphere?
2. As discussed earlier, the time will also remain untransformed along this "horisontal" axis?

 

[yuiop]

Yes. I think you have described what I tried to ask. I think that the radial must be "perpendicular" to the surface of the sphere.

Question:
1.I did intend to mean only short "horisontal" distances, but if I understand you correctly then the distance measurement along the axis along the surface remains untransformed all the way around the circle of the surface of the sphere?

Yes, the Schwarzschild metric has spherical symmetry so a great circle on the surface of a shell with uniform radius r, will have the same circumference as any other great circle chosen at random that lies of the same shell. The important thing here is that horizontal is defined as a curve that has constant radial coordinate r at every point on the curve. If on the other hand we chose a given radial geodesic and drew a straight line that remained at right angles to the radial line over extended distances then the relationship is non trivial as the points along the tangential line will have different radial coordinates.

It might also help to consider a short rigid rod that is calibrated to be one metre long when very far from the gravitational source. When transported down to the shell and orientated horizontally, its coordinate difference according to the observer at infinity is still one metre. By contrast, if the same rod was orientated vertically at the shell surface, then its coordinate difference would be approximately 1 metre * sqrt(1-2m/r) = 0.7071 metres. The circumference of the great circle on the shell, using a series of such short horizontal rulers laid end to end, would be 2*pi*r according to the local observer at rest on the shell in agreement with what the observer at infinity considers the circumference of the great circle on the shell to be. If the shell happened to have the same radius as the photon sphere, then a shell observer could send a photon all the way around the shell and time how long it takes for the the photon to circumnavigate. The circumference implied by the circumnavigation time of the photon as timed by the shell observer, agrees with the circumference according to the observer at infinity.

2. As discussed earlier, the time will also remain untransformed along this "horisontal" axis?

The rate that clocks run at, will be the same anywhere on the shell surface, so in that sense they are "untransformed" because all clocks with equal radial coordinate distances will run at the same rate. They are transformed relative to the observer at infinity, but by the same equal amount, wherever they are on the shell. For Schwarzschild coordinates, poles and an equator are defined to map out the coordinates, but the choice of where the initial pole is located is arbitrary due to the symmetry. In a rotating metric such as the Kerr metric, the choice is not arbitrary.

May I comment on the "coordinate distance vs coordinate difference". It is my understanding that the free falling observer is the "coordinate observer" at point r even though he is accelerating down the geodesic radial. At that instant under discussion his coordinate system has its zero point of that axis along the radial at point r. The "difference" minus zero then becomes the "distance". Does this not qualify the use of "distance" under the circumstances?

I am not sure we are on the same page here. Usually when I refer to the coordinate distance (or difference) I am referring to measurements made by the stationary Schwarzschild observer at infinity. Local measurements refer to measurements made by an observer that is stationary with respect to the Schwarzschild metric or stationary relative to a shell. Having said that, Passionflower has demonstrated in another thread, that a freefalling observer will make the same length measurements when passing r, as the stationary observer that remains at infinity, (if the free falling observer was initially stationary at infinity).

To make this clearer here is a practical example. A vertical test meter rod is calibrated at infinity and then transported down to r=4m. A local stationary shell observer (Bob) considers the test rod to still be 1 metre long using local radar measurements when the rod is at rest on the surface of the shell. The observer at infinity (Clare) considers the coordinate difference length of the rod to be 1*sqrt(1-2m/r) = 0.7071 metres. This is what I mean by gravitational length contraction. An observer that freefalls from initially being at rest at infinity (Alice) will also measure the length of the test metre rod at rest at r to be 0.7071 metres long as she falls past r, in agreement with what Clare considers the length of the test rod to be. This measurement made by Alice is local, but it is not the proper length of the test rod, because Alice is not at rest with the test rod. Now if Alice had a vertical test metre rod in her hand when she jumped from infinity, then Bob would consider the length of the rod in Alice's hand to be 0.7071 metres long as she passed Bob. Clare at infinity would say the coordinate difference length of the rod in Alice's hand as she passes Bob is 0.7071*0.7071 = 0.5 metres long.

Question for you yuiop, when you are talking about lengths changing under gravity do you mean:

- a change in ruler distance caused by tidal effects.
- a change in radar distance.
- proper lengths (e.g. as measured by a local co-moving observer) v.s. Lorentz adjusted lengths.​

Or perhaps something else?

All my discussions in this thread have ignored tidal effects and either treats them as insignificant over the short distances under consideration, or the rods are considered sufficiently rigid to resist tidal compression/stretching to an acceptable accuracy. When I say "gravitational length contraction" I mean changes in length purely due to the local gravitational potential, in a manner that analogous to time dilation due to gravitational potential. This gravitational length contraction is observed in the measurements of the distant Schwarzschild observer "at infinity" or by the locally passing freefalling observer. As in SR this length contraction of the rod is not measured by the local observer that is at rest with the rod and for such an observer the proper length of the rod is the same at any gravitational potential.

I do not necessarily mean "a change in radar distance", but that is the simplest way to define short lengths. When Bob on the shell surface says the rod in his hand is one metre long, how does he know that it is in fact one metre long? He can check that by measuring its radar length locally, which is essentially how the metre is defined in SI units. He could transport a metre rule from flat space and compare it with his local rod. the results are the same either way and ultimately the "master ruler" that defines the metre in flat space has to be calibrated using radar distance. Perhaps you are considering the differences between radar distances over extended distances compared to proper lengths using short rulers? As you already know this causes discrepancies over extended distances and the radar distances will be different depending on which end the radar distance is measured from, but in this thread I am mainly considering short distances where proper measurements made by a rod should substantially agree with local radar measurements. Hopefully the comments I have made in response to Pierre's questions, will also clarify to you what I am getting at.

 

[Pierre007080]

Passionflower has demonstrated in another thread, that a freefalling observer will make the same length measurements when passing r, as the stationary observer that remains at infinity, (if the free falling observer was initially stationary at infinity).

Clare at infinity would say the coordinate difference length of the rod in Alice's hand as she passes Bob is 0.7071*0.7071 = 0.5 metres long.

Hi Yuiop,
Well explained. I follow and agree with everything except the last reference to the double transformation that Claire observes. Let's consider the two quotes included. My understanding is that the free falling observer is the "inertial reference frame" and that the Swarzschild observer at infinity (Claire) would see the same a Alice (free falling frame). This confirms what Passionflower demonstrated, but goes against your double transformation (0.7071*0.7071 = 0.5 metres long). Do you mean that Special Relativity transformations apply between Alice and Claire on top of the GR transformaton? Surely this would only apply if Alice were accelerated faster than the free falling test particle?

 

[Passionflower]

This is getting complicated, it seems I have to disagree with some things you write, but perhaps I simply misunderstand them. I will first comment on what you write and then write my views then we can see if there is still anything we disagree on.

Having said that, Passionflower has demonstrated in another thread, that a freefalling observer will make the same length measurements when passing r, as the stationary observer that remains at infinity, (if the free falling observer was initially stationary at infinity).

That is not exactly what I said, I hope you do not interpret this as nitpicking let me try to be more exact.

Basically a free falling observer's measure of distance of objects stationary in the field (e.g. moving wrt this observer) happens to match the Schwarzschild's r coordinate, it does not mean that this coordinate represents anything physical it just happens to numerically match due to the fact that the free falling observer is not stationary, e.g. in motion wrt the field and that the Lorentz contraction of the escape velocity cancels out the integrand that is used to calculate the physical distance between two r coordinate values. But, neutralizing for tidal effects, a co-moving observer on such a free falling rod will obviously measure the rod still to be one meter.

I try to avoid the 'observer' at infinity because realistically such an observer cannot measure anything and an observer that stays at infinity is obviously not free falling but stationary.

By the way the reason I write 'free falling from infinity' has nothing to do with infinity per se, it is a way to express that the energy condition is 1, e.g. that the observer has no other speed wrt to a stationary observer than the escape velocity. Perhaps using the phrase 'traveling at escape velocity' is more to the point and less confusing.

Yes, the Schwarzschild metric has spherical symmetry so a great circle on the surface of a shell with uniform radius r, will have the same circumference as any other great circle chosen at random that lies of the same shell.

I would rather state it differently: "a great circle on the surface of a symmetric shell with a given area". I know you understand this but let's attempt to make clear for everybody that the r coordinate in Schwarzschild coordinates is a function.

In fact it is defined as:

$$r = 1/2\,{\frac {\sqrt {A}}{\sqrt {\pi }}}$$

So, in other words the r coordinate is simply a number it represents nothing physical. In other words this coordinate by itself cannot be used in any way to 'prove' that rods shrink or expand in a gravitational field.

The circumference of the great circle on the shell, using a series of such short horizontal rulers laid end to end, would be 2*pi*r according to the local observer at rest on the shell in agreement with what the observer at infinity considers the circumference of the great circle on the shell to be.

That is true because it is defined as such, again using:

$$r = 1/2\,{\frac {\sqrt {A}}{\sqrt {\pi }}}$$

I understand you know this I just want to be explicit here.
So basically what you wrote is correct but it is also a tautology.

To make this clearer here is a practical example. A vertical test meter rod is calibrated at infinity and then transported down to r=4m. A local stationary shell observer (Bob) considers the test rod to still be 1 metre long using local radar measurements when the rod is at rest on the surface of the shell. The observer at infinity (Clare) considers the coordinate difference length of the rod to be 1*sqrt(1-2m/r) = 0.7071 metres. This is what I mean by gravitational length contraction.

And

When I say "gravitational length contraction" I mean changes in length purely due to the local gravitational potential, in a manner that analogous to time dilation due to gravitational potential.

I stop here because much what you write below this hinges on these statements.

My question is: how do you conclude that rods shrink?

So now from my perspective:
Let's first take a look at space in a Schwarzschild solution.

Suppose you have magic bag which has a given area A. The magical thing about this bag is that you can stick more volume in there than:

$$V >1/6\,{\frac {{A}^{3/2}}{\sqrt {\pi }}}$$

That is exactly the case for a Schwarzschild solution (r> r_s), the volume between two static shells of area A1 and A2 (observe that the situation is described without using r!) is larger than the volume using the above mentioned formula for each shell and after proper subtraction.

Now one could claim something else and say, no the bag is not magic because it 'creates' extra volume, it is magic because everything you stick in there magically shrinks. Now it would be helpful to see if we are on the same page wrt this issue. I do not think the bag is magical because things shrink but instead I think it is magical because there is simply more volume within this area which is basically an effect of curvature. Do you agree with this analogy for the Schwarzschild solution?

Second point is that I see a clear and complete distinction between ruler and radar distance. I am not saying you don't but I just want to be explicit here. I think we all agree that the coordinate speed of light slows down in a Schwarzschild solution, that effectively means that for all observers, for all points and intervals the speed of light will be measured at <c except measured at the point where the observer is actually located. Then it follows that, because distances are spacelike and not timelike, radar distances will always be greater than those done in flat space.

With regards to ruler distance I think a rod with a ruler distance of 1 meter which is stationary in the field has a proper length of 1 meter everywhere (r> r_s). Observers in relative motion wrt stationary observers will observe this meter to be less than 1 meter due to Lorentz contraction, just like in flat space. How much less seems to be a matter of discussion, as it depends how we define it (e.g. for a free falling at escape velocity it is r2-r1 or, what we called the 'Fermi distance' which is slightly more than this).

 

[Mentz114]

PassionFlower, that is a very interesting post and throws some light on something that I should know, but was bothering me. Am I right that the expressions you've written are derived from the volume element of the spatial slices ? In flat space we have

$$dV_f=r^2sin(\theta)dr d\theta d\phi \Rightarrow \frac{dV_f}{dr}=r^2sin(\theta) d\theta d\phi$$

which gives on integration ( I assume) $V=4 \pi r^3/3$. For a spatial section in Schwarzschild spacetime the volume element is ( I conjecture)

$$dV_s=r^2\sqrt{\frac{r}{r-2m}}sin(\theta)dr d\theta d\phi \Rightarrow \frac{dV_s}{dr}=r^2\sqrt{\frac{r}{r-2m}}sin(\theta) d\theta d\phi$$

This looks about right to give the relations you've written. But when I integrate $dV_s/dr$ I get a horrendous expression.

 

[yuiop]

PassionFlower, that is a very interesting post and throws some light on something that I should know, but was bothering me. Am I right that the expressions you've written are derived from the volume element of the spatial slices ? In flat space we have

$$dV_f=r^2sin(\theta)dr d\theta d\phi \Rightarrow \frac{dV_f}{dr}=r^2sin(\theta) d\theta d\phi$$

which gives on integration ( I assume) $V=4 \pi r^3/3$. For a spatial section in Schwarzschild spacetime the volume element is ( I conjecture)

$$dV_s=r^2\sqrt{\frac{r}{r-2m}}sin(\theta)dr d\theta d\phi \Rightarrow \frac{dV_s}{dr}=r^2\sqrt{\frac{r}{r-2m}}sin(\theta) d\theta d\phi$$

This looks about right to give the relations you've written. But when I integrate $dV_s/dr$ I get a horrendous expression.

The method to get the area and volume of a sphere by integration is described here: http://en.wikipedia.org/wiki/Sphere

As far as passionflower's equations are concerned, they can be obtained by a much simpler method.

The area of a sphere in flat space is $$A = 4\pi r^2$$ which can be solved for r in terms of surface area, to obtain $$r = (1/2)(\sqrt{A/\pi})$$.

The volume of a sphere in flat space is $$(4/3)\pi r^3$$.

Substituting the area formula for r into the volume equation gives the volume in terms of area as:

$$V = (A/6)\sqrt{A/\pi}$$

which is the equation given by Passionflower.

 

[yuiop]

My understanding is that the free falling observer is the "inertial reference frame" and that the Swarzschild observer at infinity (Claire) would see the same a Alice (free falling frame). This confirms what Passionflower demonstrated, but goes against your double transformation (0.7071*0.7071 = 0.5 metres long). Do you mean that Special Relativity transformations apply between Alice and Claire on top of the GR transformaton? Surely this would only apply if Alice were accelerated faster than the free falling test particle?

I stand by this calculation and hopefully I will have time in the near future to post a proof or at least a demonstration of the validity of that conclusion. For now it is worth considering the fact that although both Alice and Claire see Bob in the same way, they do not necessarily see each other in the same way. Although me and Passionflower obviously differ on the semantics, maybe Passionflower can comment on whether he agrees or disagrees with this numerical result, where philosophical semantic differences shouldn't be an issue?

I would also like to clarify that length contraction of a falling object is the product of gravitational length contraction and Lorentz velocity length contraction factors, just as the time dilation of a falling object is the product of the gravitational time dilation and Lorentz velocity time dilation factors, which was something I demonstrated in this forum a long time ago.

 

[yuiop]

We obviously differ on a lot of semantic and philosophical issues regarding coordinate radius, but for what it is worth, I understand all you say and basically we are seeing the same situation from different points of view which is what relativity is all about. However, I will not sweep your comments under the carpet and will come back to them later. For now, I will concentrate on a quantifiable predication you make, which is where we should agree, but unfortunately we do not.

Second point is that I see a clear and complete distinction between ruler and radar distance. I am not saying you don't but I just want to be explicit here. I think we all agree that the coordinate speed of light slows down in a Schwarzschild solution, that effectively means that for all observers, for all points and intervals the speed of light will be measured at <c except measured at the point where the observer is actually located. Then it follows that, because distances are spacelike and not timelike, radar distances will always be greater than those done in flat space.

Using units of GM/c^2=c=1, consider the measurements made by a stationary shell observer at r=4.

The radar distance to r=3 is -1.68736492.
The average speed of light in terms of coordinate distance, between r=4 and r=3 according to this observer is -1/-1.68736492 = 0.592640032c.

The ruler distance to r=3 is -1.54216559 according to the observer at r=4.
The average speed of light in terms of ruler distance, between r=4 and r=3 according to this observer is -1.54216559/-1.68736492 = 0.913949064c.

So in terms of ruler or proper distance the average speed of light below the observer is <c using either method as you claim, but...

The local radar distance to r=5 is 1.28052104 according to the observer at r=4.
The average speed of light in terms of coordinate distance, between r=4 and r=5 according to this observer is 1/1.28052104 = 0.780932112c.

The ruler distance to r=5 is 1.34524612.
The average speed of light in terms of ruler distance, between r=4 and r=5 according to this observer is 1.34524612/1.28052104 = 1.05054589c.

So the average speed of light is >c above the observer when using ruler distances, so your claim that all observers will measure the speed of light to be less than c is only true when using coordinate distances.

[EDIT] I have to edit this post to correct an error in the original equations and now the radar distance is greater than the coordinate distance above and below the observer in this particular case, due to a poor choice of initial conditions, but I hope to show in a following post that with a different choice of parameters the radar distance above the observer is shorter than both the coordinate and ruler distance and the measured speed of light is greater than c in both cases.

 

[Passionflower]

We obviously differ on a lot of semantic and philosophical issues regarding coordinate radius, but for what it is worth, I understand all you say and basically we are seeing the same situation from different points of view which is what relativity is all about. However, I will not sweep your comments under the carpet and will come back to them later.

Please do as my views are not carved in stone and I would not be proven wrong the first time.

For now, I will concentrate on a quantifiable predication you make, which is where we should agree, but unfortunately we do not.

A good approach.

When I spoke about light being slower in a gravitational field, and it gets slower for lower r values, I am talking about the coordinate speed of light, your calculations seem to relate to the speed for a particular observer. However I did mention the word 'measure' and that is definitely a mistake, because different observer will obviously measure this speed to be different. So you are completely right about that.

I also will get back on your calculations in a short while, by the way in the mean time it would be helpful to provide the formulas as well on how you obtained these numbers and when you use ruler distance what the size of that ruler would be if it would be transported our all the way to infinity away from the field.

So with regards to the coordinate speed of light both wrt r values and to physical distances if we make some adjustments (as measured by a stationary observer (but not timed by his clock) we have:

$$c \left( 1-2\,{\frac {M}{r}} \right)$$

Clearly this formula indicates that the speed is always < c except for r is infinity.

Plotting the light travel time (not for an observer) for the pairs r and r+1 and r and r+a physical distance 1 using:

$$r_{{2}}-r_{{1}}+2\,M\ln \left( {\frac {r_{{2}}-2\,M}{r_{{1}}-2\,M}} \right)$$

We get:
[PLAIN]http://img812.imageshack.us/img812/8347/lightspeedcoordinate.gif

 

[pervect]

I

I think what Pierre meant was "horizontal distance" in the infinitesimal limit for a short vector that is at right angles to where the radial geodesic intersects the shell sphere surface and tangential to the surface of the sphere.

I really can't tell what Pierre meant by his original quesition, hopefully he'll clarify it - perhaps he already did, there are a number of posts that have gone by in the meantime. The original post did seem vague and to contain a number of assumptions, making it hard to answer as written.

Pierre seems to be enthusiastic about your response, so perhaps you are addressing his question.

Perhaps you will agree that the following is true?

The circumference of ring with Schwarzschild radius r as measured by a local observer r is 2*Pi*r whether the local observer measures the circumference using a tape measure or using radar measurements. This circumference will agree with circumference according to the Schwarzschild observer at infinity.

Yes, I'll agree with that, assuming our local obsever is a static observer, i.e. they are holding station at some constant schwarzschild r.

Well, I think this is interesting subject in itself and I will start a new thread on the subject in the near future. I hope you will agree that "gravity" has an effect on lengths and distances that makes distances measured in curved spacetime different to lengths and distances measured in flat space, in a way that has more physical significance than just a choice of coordinates. In other words, intrinsic curvature of curved spacetime is not just something that can be transformed away.

I'll agree that the space-time near a massive body is curved, of course. Less clear is whether you need to have a massive body to have "gravity" - i.e. if you're in an accelerating rocketship, is what you feel "gravity"?

 

[yuiop]

A good approach.

When I spoke about light being slower in a gravitational field, and it gets slower for lower r values, I am talking about the coordinate speed of light, your calculations seem to relate to the speed for a particular observer. However I did mention the word 'measure' and that is definitely a mistake, because different observer will obviously measure this speed to be different. So you are completely right about that.

I also will get back on your calculations in a short while, by the way in the mean time it would be helpful to provide the formulas as well on how you obtained these numbers and when you use ruler distance what the size of that ruler would be if it would be transported our all the way to infinity away from the field.

Yes, I am talking about the speed of light for a particular observer, rather than just the coordinate observer at infinity, but you did say for "all observers".

The ruler distance is the distance measured using very short rods of say 1 meter in length. This rod size is the proper length and is what the length of the rod is when it stationary at r according to a local observer that is also stationary at r and if this ruler is transported to infinity, the observer at infinity will agree that the length of the ruler is 1 meter.

I have had to edit some of the numbers in my last post because I noticed some errors in my calculations when going over the equations again to post them here as requested. Good thing you asked!

The coordinate speed of light according to the observer at infinity is:

$$\frac{dr}{dt} = (1-2/r)$$

using units of G=m=c=1 so the Schwarzschild radius is at r=2.

The radar distance from r0 to r1 is obtained by integrating:

$$t = \int_{r0}^{r1} \frac{1}{(1-2/r)} dr = (r1-r0) + ln\left(\frac{r1-2}{r0-2}\right)$$

This is the coordinate time it takes light to travel that distance. If the local observer is located at r0, then the radar distance according to that observer (allowing for gravitational time dilation) is:

$$t = \sqrt{1-2/r0} \left((r1-r0) + ln\left(\frac{r1-2}{r0-2}\right)\right)$$

This is entirely valid and not an approximation, because the observer at r0 is using a single clock to measure the up and back down time of the radar signal.

The gravitational length contraction factor for a short stationary rod is:

$$ds = \frac{dr}{\sqrt{1-2/r}}$$

where ds is the length of the rod according to the local observer and dr is the coordinate difference length of the rod according to the Schwarzschild observer at infinity. By integration, the ruler distance between r0 and r1 is:

$$s = \int_{r0}^{r1} \frac{dr}{\sqrt{1-2/r}} = \sqrt{r1(r1-2)} - \sqrt{r0(r0-2)} + ln\left(\frac{\sqrt{r0} + \sqrt{r0-2}}{\sqrt{r1} + \sqrt{r1-2}} \right)$$

O.K. assuming the above is correct, here is another numerical example that demonstrates that the speed of light above an observer can be measured to be greater than c.

The local stationary observer is located at r=4 (again).
The target mirror is located at r=50.

The ruler distance from r=4 to r=50 is 48.9834839.
The radar distance from r=4 to r=50, according to the observer at r=4 is 37.0213588.
The average speed of light in terms of coordinate distance is 46/37.0213588 = 1.24252598c.
The average speed of light in terms of ruler distance is 48.9834839/37.0213588 = 1.32311416c.

In the above example the radar distance is less than both the coordinate and ruler distances and the average speed of light according to the local observer is >c whether measured in terms of coordinate distance or proper ruler distance.

 

[yuiop]

Hi Yuiop,
Well explained. I follow and agree with everything except the last reference to the double transformation that Claire observes. Let's consider the two quotes included. My understanding is that the free falling observer is the "inertial reference frame" and that the Swarzschild observer at infinity (Claire) would see the same a Alice (free falling frame).

The r, dr, dt and angular derivatives of the Schwarzschild metric are defined as measurements made by a stationary observer "at infinity" and only the proper time tau is measured "locally". We have shown that in certain circumstances that the measurements made by a free falling local observer are numerically equivalent to the measurements made by the observer at infinty, but it would be a mistake to assume that their points of view are identical. For example, if a test particle has Schwarzschild parameters r=3 and dr=0, this means that is stationary with respect to the observer at infinity. To the free falling observer falling past r=3, the test particle is not stationary and $dr \ne 0$.

Infinity is a very difficult concept and in some respects it is better to think of the observer at infinity as just "very far away" where the curvature is almost flat rather than actually being at "infinity" which introduces all sorts of headaches. For example if an observer freefalls from infinity for a finite amount of coordinate or proper time, then by the definition of infinity, they are still at infinity.

 

[Passionflower]

Yes, I am talking about the speed of light for a particular observer, rather than just the coordinate observer at infinity, but you did say for "any observer".

Indeed, so that was clearly a mistake on my part.

The coordinate speed of light according to the observer at infinity is:

$$\frac{dr}{dt} = (1-2/r)$$

using units of G=m=c=1 so the Schwarzschild radius is at r=2.

Yes I agree with that. And in this context I hope you would agree that everywhere in the field this speed is < c. However most observers will measure this speed differently due to their changes in relative clock rates. Could we identify an observer for which this is unity? Perhaps a free falling observer going at escape velocity?

The radar distance from r0 to r1 is obtained by integrating:

$$t = \int_{r0}^{r1} \frac{1}{(1-2/r)} dr = (r1-r0) + ln\left(\frac{r1-2}{r0-2}\right)$$

Yes full agreement on this.

This is the coordinate time it takes light to travel that distance. If the local observer is located at r0, then the radar distance according to that observer (allowing for gravitational time dilation) is:

$$t = \sqrt{1-2/r0} \left((r1-r0) + ln\left(\frac{r1-2}{r0-2}\right)\right)$$

This is entirely valid and not an approximation, because the observer at r0 is using a single clock to measure the up and back down time of the radar signal.

Yes, again full agreement.

The gravitational length contraction factor for a short stationary rod is:

$$ds = \frac{dr}{\sqrt{1-2/r}}$$

where ds is the length of the rod according to the local observer and dr is the coordinate difference length of the rod according to the Schwarzschild observer at infinity. By integration, the ruler distance between r0 and r1 is:

$$s = \int_{r0}^{r1} \frac{dr}{\sqrt{1-2/r}} = \sqrt{r1(r1-2)} - \sqrt{r0(r0-2)} + ln\left(\frac{\sqrt{r0} + \sqrt{r-2}}{\sqrt{r1} + \sqrt{r-2}} \right)$$

Here it looks we might have some difference of opinion. It appears this hinges on the interpretation of r and if the Schwarzschild solution is 'volume expanding' or 'object reducing'. Perhaps it would help if you could give me an analog of my description of the meaning of r from your perspective and then we can see what the differences really are (if any).

O.K. assuming the above is correct

I did not check the numbers but I fully agree with the formulas, except for what you call 'gravitational length contraction'.

So, it seems, if I am not mistaken, that the gravitational length contraction is the main thing to discuss.

To calculate the physical volume(as opposed to the coordinate volume) between r1 and r2 is:

$$\begin{align*} 4/3\,\pi \,\sqrt {{r_{{2}}}^{5} \left( r_{{2}}-2\,M \right) }+10/3\, \pi \,M\sqrt {{r_{{2}}}^{3} \left( r_{{2}}-2\,M \right) }+10\,\pi \,{M }^{2}\sqrt {r_{{2}} \left( r_{{2}}-2\,M \right) }+ \\ 20\,\pi \,{M}^{3} \ln \left( 1/2\,{\frac {r_{{2}}}{M}}+1/2\,\sqrt {2\,{\frac {r_{{2}}}{ M}}-4} \right) -4/3\,\pi \,\sqrt {{r_{{1}}}^{5} \left( r_{{1}}-2\,M \right) }- \\ 10/3\,\pi \,M\sqrt {{r_{{1}}}^{3} \left( r_{{1}}-2\,M \right) }-10\,\pi \,{M}^{2}\sqrt {r_{{1}} \left( r_{{1}}-2\,M \right) }- \\ 20\,\pi \,{M}^{3}\ln \left( 1/2\,{\frac {r_{{1}}}{M}}+1/2 \,\sqrt {2\,{\frac {r_{{1}}}{M}}-4} \right) \end{align}$$

(Sorry for the alignment, perhaps someone can suggest how to left align. The formula is too long and I had to break it up)

To work out the 'magic bag' metaphor a little bit more I plotted the volume between the event horizon and increasing areas (area * pi) both for an Euclidean space and for a Schwarzschild solution with m=1/2. Note that the plot does not use the r coordinate in any way, the volumes are based on the areas.

[PLAIN]http://img830.imageshack.us/img830/4332/volumeeuclideanvsproper.gif

It is clear the there is more volume between these areas than expected based on Euclidean calculations.

Note that for those who have an objection taking the area of the event horizon, an objection which I do not share by the way, taking the plots between any two larger areas will give the same results, the volume is larger than in the Euclidean calculation.

 

[pervect]

MTW gives a formula for a closely related case on pg 604 which agrees with Passionflower's formula as far as the setup goes (though they don't work out the integral, and I'm not going to either).

It's the volume element for the interior volume for a relativistic star using the interior Schwarzschild solution. Basically it's the same except m(r) isn't a constant in that case, m(r) is a function of r, which can be thought of as the mass within the radius r. (MTW gives more details, but their explanation of how m(r) is computed is a bit on the long side, perhaps a bit hand-wavy, and I don't see a need to go into it. Which may or may not mean it'll turn into a hot topic :-) ).

At any rate the formula is in agreement, i.e.:

dV = 4 pi r^2 dr / sqrt(1 - 2m(r)/r)

Also, they have a sketch of an embedding diagram of the equatorial spatial slice (a lice through the equator of constant Schwarzschild time) on pg 614 - a lot of authors have similar embedding diagrams, it's one of the more popular ones. From the sketch it's pretty clear that the enclosed volume is greater than it would be in Euclidean space.

Of course the volume element depends on the space-time slice, but the obvious spatial slice (orthogonal to Schwarzschild time) is used.

Finding the geometry of a space-slice for a massive object which is not a black hole is only slightly more complicated mathematically, is of great physical interest, and avoids dealing with any need to consider complications that could be introduced by the event horizon.

 

[Mentz114]

$$\begin{align*} & 4/3\,\pi \,\sqrt {{r_{{2}}}^{5} \left( r_{{2}}-2\,M \right) }+10/3\, \pi \,M\sqrt {{r_{{2}}}^{3} \left( r_{{2}}-2\,M \right) }+10\,\pi \,{M }^{2}\sqrt {r_{{2}} \left( r_{{2}}-2\,M \right) } \\ & + 20\,\pi \,{M}^{3} \ln \left( 1/2\,{\frac {r_{{2}}}{M}}+1/2\,\sqrt {2\,{\frac {r_{{2}}}{ M}}-4} \right) -4/3\,\pi \,\sqrt {{r_{{1}}}^{5} \left( r_{{1}}-2\,M \right) } \\ & - 10/3\,\pi \,M\sqrt {{r_{{1}}}^{3} \left( r_{{1}}-2\,M \right) }-10\,\pi \,{M}^{2}\sqrt {r_{{1}} \left( r_{{1}}-2\,M \right) } \\ & - 20\,\pi \,{M}^{3}\ln \left( 1/2\,{\frac {r_{{1}}}{M}}+1/2 \,\sqrt {2\,{\frac {r_{{1}}}{M}}-4} \right) \end{align}$$

This is the very integral I mentioned in post#25. I'm glad to see you're paying attention to the crucial facts - i.e. that the volume enclosed by a 2D surface of equal curvature is not dependent on r the way it is in flat space.

[PassionFlower, the alignment in Latex is done with the '&' character.]