General Relativity and the curvature of space: more space or less than flat?

  • #1
Martian2020
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I could not find answer on simple question: is there more or less space near massive objects in accordance with general relativity?
General relativity. Curvature of spacetime: ok. time dilation: ok. What about space? Curvature is intrinsic and given by complex equations. But could we definitely say is there more space between 2 points along curved space through the star than would be through flat space (no star there) or less? Space is usually visualized ( in "pop-science" places) as stretched sheet, but maybe it is actually so curved intrinsically that there is less space, not more... Could not easily find an answer with web search. thank you!

I'm looking for theoretical answer from general relativity. As humans now officially define length using light speed, time dilation will result in more space in the experiment than with flat space, as I understand it.

BTW, could somebody confirm (or disprove) that speed of light is same as measured by local clocks even in case of acceleration?
 
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  • #2
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I think you need to carefully define your terms. You mean something, for sure, but it's not what you said. There is the same space in a cubic meter no matter where it is.
 
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Summary:: I could not find answer on simple question: is there more or less space near massive objects in accordance with general relativity?

General relativity. Curvature of spacetime: ok. time dilation: ok. What about space? Curvature is intrinsic and given by complex equations. But could we definitely say is there more space between 2 points along curved space through the star than would be through flat space (no star there) or less? Space is usually visualized ( in "pop-science" places) as stretched sheet, but maybe it is actually so curved intrinsically that there is less space, not more... Could not easily find an answer with web search. thank you!
Is New York further from London than it would be if the Earth were flat? It's a meaningless question.

If you project the surface of the Earth onto a 2D map, then the distance between New York and London on the flat map depends on the projection. You could have them exactly the "right" distance apart or closer together or further apart.

The curved spacetime near a star is what it is and cannot be directly compared to a region of flat spacetime in an unambiguously meaningful way.
 
  • #4
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Summary:: I could not find answer on simple question: is there more or less space near massive objects in accordance with general relativity?
Well, space is a frame-dependent concept. So you would need to define things a little better.

However, the Shapiro delay is an invariant concept that may answer your underlying question:

https://www.mathpages.com/home/kmath750/kmath750.htm
 
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  • #5
A.T.
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I could not find answer on simple question: is there more or less space near massive objects in accordance with general relativity?
The diameter through the massive object is more than the circumference / pi. So there is more space within the same perimeter, if it contains mass.

See also chapter 11 (PDF page 177):
https://archive.org/download/L.EpsteinRelativityVisualizedelemTxt1994Insight/L. Epstein - Relativity Visualized [elem txt] (1994, Insight)_text.pdf

Space is usually visualized ( in "pop-science" places) as stretched sheet
For the spatial geometry, that analogy is OK-

BTW, could somebody confirm (or disprove) that speed of light is same as measured by local clocks even in case of acceleration?
In non-inertial frames light doesn't have to travel at c.
 
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  • #6
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In non-inertial frames light doesn't have to travel at c.

More precisely, in non-inertial frames the coordinate speed of light does not have to be ##c##. But actual measurements of the local speed of light will still give ##c##. The coordinate speed is not something that gets directly measured; it is calculated.
 
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  • #7
Martian2020
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Well, space is a frame-dependent concept. So you would need to define things a little better.

However, the Shapiro delay is an invariant concept that may answer your underlying question:

https://www.mathpages.com/home/kmath750/kmath750.htm
Thank you for the link!

"The first two of these are fairly straight-forward differences that are easily reconciled, but the third – involving the use of a “straight” path versus the geodesic path – is actually somewhat subtle, and there are many erroneous claims in the literature that the “straight” path approximation makes no difference in the result to the first order in m/r. Surprisingly, it actually does make a first-order difference (as explained below), and the failure to correctly account for this is responsible for much confusion on this topic."

The article is a highly mathematical read (for me). I could not easily find if geodesic vs "straight" path gives more or less time? That is what I was asking for, I think.
 
  • #8
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I could not easily find if geodesic vs "straight" path gives more or less time?

The "straight" path as that term is used in the article is a mathematical fiction; it doesn't exist.

The "time delay" is based on a comparison of two situations:

(1) The situation pictured in the article linked to: Mercury and Earth with the Sun in between.

(2) A situation with Mercury and Earth in the same relative positions, but the Sun not in between.

Strictly speaking, we can't have the Sun not in between at all; but we can look at situations where the location of the Sun relative to Mercury and the Earth varies, and look at the dependence of the round-trip travel time of the light signal on that variation and compare it with what GR predicts.
 
  • #9
Martian2020
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Is New York further from London than it would be if the Earth were flat? It's a meaningless question.

If you project the surface of the Earth onto a 2D map, then the distance between New York and London on the flat map depends on the projection. You could have them exactly the "right" distance apart or closer together or further apart.

The curved spacetime near a star is what it is and cannot be directly compared to a region of flat spacetime in an unambiguously meaningful way.
How could you project so that distance on 2D would be more than geodesic on a sphere? (I really could not see a way). Surface of a disk projection is smaller than surface of semi-sphere.
 
  • #10
Martian2020
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Strictly speaking, we can't have the Sun not in between at all; but we can look at situations where the location of the Sun relative to Mercury and the Earth varies, and look at the dependence of the round-trip travel time of the light signal on that variation and compare it with what GR predicts.
"As a practical matter, only the variation in the transit time as the perihelion distance changes (e.g., as the planet approaches superior conjunction) can be measured, but in theory we can evaluate the absolute transit time given the spatial distances in terms of suitable coordinates."
When I've read that I understood GR can predict in general way. Doesn't it predict exact curvature?
In my question I specified I am looking for GR prediction. Experiment is another way, useful, but if cannot be done some way, that at least we have theory. Does my question have definitive theoretical answer? ;-)
 
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  • #11
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I could not easily find if geodesic vs "straight" path gives more or less time? That is what I was asking for, I think.
In, for example, the case of Mercury as described by @PeterDonis, the light transit time takes longer the closer the path passes to the sun. This is why it is called Shapiro “delay”.

In my question I specified I am looking for GR prediction, I thought I made that clear. Experiment is another way, usuful, but if cannot be done some way, that at least we have theory. Does my question have definitive theoretical answer? ;-)
The Shapiro delay has both a definitive theoretical derivation and experimental confirmation. So insofar as your question reduces to the Shapiro delay, yes.
 
  • #12
Martian2020
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The Shapiro delay has both a definitive theoretical derivation and experimental confirmation. So insofar as your question reduces to the Shapiro delay, yes.
Surely not, as I noted in the question, there is a delay expected just from time dilation. I wanted to know space part of curvature. However paper linked by @PeterDonis actually gives an answer, though I do not understand it that well so see it is correct:
"For a Sun-grazing pulse between Earth and Mercury (round trip) this yields a coordinate delay time of 239 μsec, compared with only 199 μsec given by equation (1). This is a difference of 8m/c, even though both are expressed in terms of Schwarzschild coordinates. The only difference is that (1) was based on the “straight path” approximation” and (2) was based on the actual geodesic path."
Quite a large difference... 20% longer just for small part of path was stretched! It means diameter of the Sun is stretched many times circumference/pi (assuming stretch is "uniform"). Could somebody quickly calculate how many?
P.S. measurements given is the article have large margin of error, so they do not show which way is more correct.
 
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  • #13
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as I noted in the question, there is a delay expected just from time dilation.

"Time dilation" is an ambiguous term. As the article linked to is using the term, "time dilation" does not contribute to the Shapiro delay. Time dilation is just what you need to account for to convert the calculated result in coordinate time into the predicted observation in Earth clock time.

I wanted to know space part of curvature

The Shapiro time delay is not due to "space" curvature. It's due to spacetime curvature.

20% longer just for small part of path was stretched!

You are misunderstanding what the "straight path" approximation means. It doesn't mean "we include time dilation but ignore space curvature". It means we ignore the fact that the Sun's gravity bends the path that the light follows in space.
 
  • #14
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I wanted to know space part of curvature.
There is no invariant definition of space so there is no invariant answer to that aspect of your question.
 
  • #15
Martian2020
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The diameter through the massive object is more than the circumference / pi. So there is more space within the same perimeter, if it contains mass.

See also chapter 11 (PDF page 177):
https://archive.org/download/L.EpsteinRelativityVisualizedelemTxt1994Insight/L. Epstein - Relativity Visualized [elem txt] (1994, Insight)_text.pdf
Thank you!
I had difficulties finding in the web size of the diameter of the Sun and the Earth (in circumference / pi). Maybe you were interested too and know?
 
  • #16
Martian2020
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There is no invariant definition of space so there is no invariant answer to that aspect of your question.
Interesting... Spacetime is invariant though?
 
  • #17
Martian2020
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You are misunderstanding what the "straight path" approximation means. It doesn't mean "we include time dilation but ignore space curvature". It means we ignore the fact that the Sun's gravity bends the path that the light follows in space.
maybe, have to think about it.
Do you understand math in the article to see for yourself if conclusions are correct? I found it contradicts space curvature as commonly seen (in fact it states its' view is more correct):
"Integrating the spatial lengths of the straight and geodesic paths shows the same first order difference, so the claims that the path lengths differ only in the second order are incorrect… but how can this be?

The answer is interesting. People have been misled by the drawings that show a smooth bow shape for the geodesic path, as if the path is gradually curved all along its length. If that were accurate, the lengths would indeed differ only in the second or higher order of the deflection angle. But in fact the path consists of two virtually straight legs, with almost all of the angular deflection occurring very near the perihelion. This is shown (in exaggerated form) in the figure below."

https://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/general_relativity_massive/index.html
"According to general relativity, for each mile that we come closer to the sun, the circle does not lose 2π miles in circumference; it loses only (0.99999999)x2π miles."
The "infamous" rubber sheet analogy: "Just about the only thing right in the rubber sheet model is that the surface of the membrane is similar to the surface of the embedding diagram."

Do you agree to more "traditional" view of gradual curvature or as described in the article negligable curvature far from the Sun and lots of it near the Sun?
 
  • #18
Martian2020
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Thank you!
I had difficulties finding in the web size of the diameter of the Sun and the Earth (in circumference / pi). Maybe you were interested too and know?
I've found http://www.johnstonsarchive.net/relativity/stcurve.pdf
"This is the outcome of a small exercise I engaged in with my tiny under-
standing of relativity; serious relativists please forgive me (and set me straight
as necessary!"
If correct, differences are small:
"The true diameters of the Sun and Earth are 4.1 km and 4.4 mm
greater, respectively, than one would expect from applying Euclidean geometry
(C=πd) to the observed surface of these bodies."
@A.T., seems question is answered, thank you!
 
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  • #19
Martian2020
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As of now picture is clear on the original question.

And I was incorrect at assuming times difference of diameter due to Shapiro delays differences. The paper I mentioned above stated only 4.`1 km more diameter for the Sun (of 695,990 km, 0.001%), whereas paper on Shapiro delay states 239 vs 199 mc difference (20%) between geodesic vs "straight" path for light. However this 20% is not differences in total time for light travel, only effect of the Sun.
So time dilation gives 199 mc additionally and space curvature adds 40mc to many minutes of travel.
 
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  • #20
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I found it contradicts space curvature as commonly seen

You are still misunderstanding what the "straight path" in the article means. It has nothing to do with ignoring vs. including "space curvature". Go back and read the last part of my post #13.
 
  • #21
Martian2020
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You are misunderstanding what the "straight path" approximation means. It doesn't mean "we include time dilation but ignore space curvature". It means we ignore the fact that the Sun's gravity bends the path that the light follows in space.
Per GR there is no gravity, it is a manifestation of curvature of spacetime. Therefor your statement is saying the same just not in GR terms. Or please explain in terms of GR.
 
  • #22
Martian2020
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As of now picture is clear on the original question.

And I was incorrect at assuming times difference of diameter due to Shapiro delays differences. The paper I mentioned above stated only 4.`1 km more diameter for the Sun (of 695,990 km, 0.001%), whereas paper on Shapiro delay states 239 vs 199 mc difference (20%) between geodesic vs "straight" path for light. However this 20% is not differences in total time for light travel, only effect of the Sun.
So time dilation gives 199 mc additionally and space curvature adds 40mc to many minutes of travel.
http://thescienceexplorer.com/universe/earth-s-core-younger-its-surface-due-curving-space-time
Time dilation of the core of Earth compared to center (0.0000000003) is 4.77 times larger then difference of "real" Earth diameter from Euclidean per paper I linked. So the difference is very close to differences from Shapiro delay calculations (4.97 times). More things match up...
 
  • #23
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please explain in terms of GR.

I did. "The Sun's gravity bends the path that the light follows in space" is a GR explanation. If you are hung up on the words "gravity" and "space" since you think you know that "gravity isn't a force" in GR and that "space" is frame-dependent (but if you know the latter, why do you keep talking about "space curvature"?), fine, I'll rephrase the GR expanation: "The spacetime curvature due to the presence of the Sun means that the path that the light follows, when projected into a spacelike hypersurface of constant Schwarzschild coordinate time, is not exactly a straight line in terms of the spatial coordinates on that hypersurface. But it's very close to a straight line in those spatial coordinates; so the straight line approximation uses that straight line that it's very close to, as an approximation to the actual path."

None of that changes what I said before, that the straight line approximation has nothing to do with ignoring "space curvature". It doesn't. "The spacetime curvature due to the Sun" making the path that the light follows, projected in space, not exactly a straight line, is not a manifestation of "space curvature". It's a manifestation of spacetime curvature, the same kind that Eddington did his eclipse expeditions to measure.
 
  • #24
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How could you project so that distance on 2D would be more than geodesic on a sphere?
Consider a Mercator projection map of the earth. On that map Greenland appears larger than the entire continent of Africa.
 
  • #25
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The paper I mentioned above

Is not a valid source. It's someone's personal calculation, and that someone admits they know very little relativity.

You need to look at some GR textbooks.
 
  • #26
Martian2020
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"space" is frame-dependent (but if you know the latter, why do you keep talking about "space curvature"?)
I'm thinking about your last post. Please answer to help me understand:

@Dale has not answered.
Is Spacetime invariant? (I know space is not, and time is not, but the "total" I am not so sure).

And
Can we select 2 frames in one of which space would be intrinsically "stretched", but "shrunk" in the other? (I do not see inertial ones, but not sure about ones with acceleration...)
 
  • #27
PeroK
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I'm thinking about your last post. Please answer to help me understand:

@Dale has not answered.
Is Spacetime invariant? (I know space is not, and time is not, but the "total" I am not so sure).

And
Can we select 2 frames in one of which space would be intrinsically "stretched", but "shrunk" in the other? (I do not see inertial ones, but not sure about ones with acceleration...)

A spacetime interval between two events along a particular path has invariant length. Where invariant means the same in all coordinate systems.

This extends to concepts like a spacelike hypersurface, which is a 3D surface that in some sense represents a slice of spacetime at a particular moment in time. And is spacelike in all coordinates systems. In other words, being spacelike is an invariant (or coordinate-independent) property.

The most important defining property of a region of spacetime is its curvature, as described by the Riemann curvature tensor. In particular, you can generate the Ricci curvature scalar, which is an invariant quantity.

The fact that you put words like "stretched" and "shrunk" in quotation marks indicates that these are too imprecise to have any meaning in the context of GR.
 
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  • #28
Martian2020
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The spacetime curvature due to the presence of the Sun means that the path that the light follows, when projected into a spacelike hypersurface of constant Schwarzschild coordinate time, is not exactly a straight line in terms of the spatial coordinates on that hypersurface. But it's very close to a straight line in those spatial coordinates; so the straight line approximation uses that straight line that it's very close to, as an approximation to the actual path.
Thank you. Now I sense there are different things. However I do not understand in what "situaltions" these can amount for same results and in which to different ones. As I see it, I need to either:
1) learn more GR calculus
2) learn to visualize internal curvatures w/out embedding them, starting with 2d
3) see clear example where effects amount to different results.
As of now I continue to think about the issue.
 
  • #29
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@Dale has not answered.
Is Spacetime invariant?

Oh, is that your question? The answer is yes.

Now I sense there are different things.

No, there is only one thing: spacetime curvature.

1) learn more GR calculus

Definitely.

2) learn to visualize internal curvatures w/out embedding them, starting with 2d

I don't see how this will help you.

3) see clear example where effects amount to different results.

I don't know what you mean by "effects amount to different results". As above, there is only one "effect" involved: spacetime curvature.
 
  • #30
Martian2020
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A spacetime interval between two events along a particular path has invariant length. Where invariant means the same in all coordinate systems.
...
The fact that you put words like "stretched" and "shrunk" in quotation marks indicates that these are too imprecise to have any meaning in the context of GR.
Thank you.
I see your view (or GR as per your understanding view) on space "stretched". Does "Gravitational time dilation" bears same impreciseness? Otherwise, how can we synchronize clocks between accelerated frames of reference (even if space distance remain the same - after above discussion I am not sure about it for center and surface of Earth, I just recall reading argument that two points can be rigid to each other even having different accelerations in space)?
 
  • #31
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Thank you.
I see your view (or GR as per your understanding view) on space "stretched". Does "Gravitational time dilation" bears same impreciseness? Otherwise, how can we synchronize clocks between accelerated frames of reference (even if space distance remain the same - after above discussion I am not sure about it for center and surface of Earth, I just recall reading argument that two points can be rigid to each other even having different accelerations in space)?
Gravitational time dilation has a context in which it is precise. I can't follow the rest of your post.
 
  • #32
Martian2020
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Gravitational time dilation has a context in which it is precise. I can't follow the rest of your post.
Can Gravitational time dilation be explained as a result of spacetime curvature?
Could you state that context?
 
  • #33
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Surely not, as I noted in the question, there is a delay expected just from time dilation. I wanted to know space part of curvature.

The curvature (Riemann curvature) is a 4d entity. If one wants to talk about the space part of the curvature, one need a mechanism to identify what is time and what is space.

There are a number of conventions to do this, but they are, in the end, conventions. Probably you don't think of space-time as a unified 4-dimensional object. While it is possible to split spacetime into 3+1 dimensions, the split isn't unique.

For example, take special relativity and Lorentz contraction. Different observers have different notions of distances / lengths.

My favorite technique for splitting the Riemann tensor into a space part and a time part is the Bel decomposition. Mathematically that technique requires one to specify a unit vector field at every point in spacetime. This unit vector field can be regarded as the "direction of time", the integral curves of this vector field are the worldlines of "stationary observers". Another way of putting this is that you imagine you have an infinite congruence of worldlines so that one and only one worldline passes through every event in space-time. These are the worldlines of the "stationary observers" I referred to. Then this set of worldlines defines what one means by "space", by picking out one particular worldline one specifies the "spatial location" of an event. The distance along the worldline represents time. There are some subtle parts to the later that I will skip over.

There are other techniques for splitting space-time into space+time, such as choosing a particular set of coordinates. But any attempt to talk rigorously about "spatial curvature" needs to identify some such mechanism to split apart the 4d spactime into a 3d space and a 1d time.
 
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  • #34
Martian2020
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The curvature (Riemann curvature) is a 4d entity. If one wants to talk about the space part of the curvature, one need a mechanism to identify what is time and what is space.

There are a number of conventions to do this, but they are, in the end, conventions. Probably you don't think of space-time as a unified 4-dimensional object. While it is possible to split spacetime into 3+1 dimensions, the split isn't unique.

For example, take special relativity and Lorentz contraction. Different observers have different notions of distances / lengths.
Thank you for interesting info.
Just so to be sure (myself), spacetime (4d) is invariant for all observers? Curvature is same? Why then is it called GR? Because different observers see/experience differently space (3d) and time (1d)?

Related question that bothers me: do we draw geodesics in spacetime because of law of conservation of energy-momentum? In wiki (I know some say it is not authoritative source): https://en.wikipedia.org/wiki/Physi...al_relativity#Conservation_of_energy–momentum
"Unlike classical mechanics and special relativity, it is not usually possible to unambiguously define the total energy and momentum in general relativity, so the tensorial conservation laws are local statements only (see ADM energy, though)."
Does above mean separately energy and momentum could be not conserved, but as a whole tensor remains same? So that energy-momentum tensor is conserved (when there is no acceleration) and that is the reason for geodesics in spacetime?
 
  • #35
Martian2020
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The curvature (Riemann curvature) is a 4d entity. If one wants to talk about the space part of the curvature, one need a mechanism to identify what is time and what is space.

There are a number of conventions to do this, but they are, in the end, conventions.
...
My favorite technique for splitting the Riemann tensor into a space part and a time part is the Bel decomposition.
The way technology of humanity is now doing it (as I understand): atomic clocks are used to measure time and speed of light as traveled per unit of time of above clocks measure distance in space. What convention does this method use?
 

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