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What unit of measurement is used to indicate the warping of space
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It is my understanding that planetary bodies bend/warp space and that warping results in gravitational lensing (among other things).
My question is what unit of measurement is used to describe the degree to which space is warped? Is there a maximum limit that space can be bent?
general-relativity gravity
edited Aug 17 at 3:40
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3,04741431
asked Aug 17 at 1:58
MLE
884
add a comment |Â
up vote
17
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It is my understanding that planetary bodies bend/warp space and that warping results in gravitational lensing (among other things).
My question is what unit of measurement is used to describe the degree to which space is warped? Is there a maximum limit that space can be bent?
general-relativity gravity
edited Aug 17 at 3:40
Chair
3,04741431
asked Aug 17 at 1:58
MLE
884
To add to Mike's answer, the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). Time dilation causes gravity and redshift. Length dilation makes distances larger. If you hover near a black hole, you see it farther than it actually is. As you get closer to the event horizon (while hovering, but not falling), you see it moving farther from you. If you get really close, you would see yourself light years away from the black hole, except its gravity is only getting stronger.
– safesphere
Aug 17 at 3:29
Without contradicting Mike's answer, the above mentioned physical effects of the spacetime curvature can be viewed as dimensionless. For example, time may be dilated twice (with the redshift of $z+1=2$).
– safesphere
Aug 17 at 3:41
1
@safesphere: the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). No, this is wrong. You can't interpret GR in this way. * For example, time may be dilated twice (with the redshift of z+1=2).* No, Doppler shifts have nothing to do with curvature.
– Ben Crowell
Aug 22 at 0:32
@BenCrowell You saying that something is "wrong" doesn't make it wrong only because you said it. You need to do a bit better than that. If you don't like length "dilation", you can call it "spagnetification" instead or any term you prefer as the opposite of "contraction". On your second point, the Schwarzschild static time dilation matched the redshift last time I checked.
– safesphere
Aug 22 at 3:43
add a comment |Â
up vote
17
down vote
favorite
up vote
17
down vote
favorite
It is my understanding that planetary bodies bend/warp space and that warping results in gravitational lensing (among other things).
My question is what unit of measurement is used to describe the degree to which space is warped? Is there a maximum limit that space can be bent?
general-relativity gravity
edited Aug 17 at 3:40
Chair
3,04741431
asked Aug 17 at 1:58
MLE
884
It is my understanding that planetary bodies bend/warp space and that warping results in gravitational lensing (among other things).
My question is what unit of measurement is used to describe the degree to which space is warped? Is there a maximum limit that space can be bent?
general-relativity gravity
edited Aug 17 at 3:40
Chair
3,04741431
asked Aug 17 at 1:58
MLE
884
edited Aug 17 at 3:40
Chair
3,04741431
edited Aug 17 at 3:40
Chair
3,04741431
edited Aug 17 at 3:40
Chair
3,04741431
3,04741431
asked Aug 17 at 1:58
MLE
884
asked Aug 17 at 1:58
MLE
884
asked Aug 17 at 1:58
MLE
884
884
To add to Mike's answer, the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). Time dilation causes gravity and redshift. Length dilation makes distances larger. If you hover near a black hole, you see it farther than it actually is. As you get closer to the event horizon (while hovering, but not falling), you see it moving farther from you. If you get really close, you would see yourself light years away from the black hole, except its gravity is only getting stronger.
– safesphere
Aug 17 at 3:29
Without contradicting Mike's answer, the above mentioned physical effects of the spacetime curvature can be viewed as dimensionless. For example, time may be dilated twice (with the redshift of $z+1=2$).
– safesphere
Aug 17 at 3:41
1
@safesphere: the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). No, this is wrong. You can't interpret GR in this way. * For example, time may be dilated twice (with the redshift of z+1=2).* No, Doppler shifts have nothing to do with curvature.
– Ben Crowell
Aug 22 at 0:32
@BenCrowell You saying that something is "wrong" doesn't make it wrong only because you said it. You need to do a bit better than that. If you don't like length "dilation", you can call it "spagnetification" instead or any term you prefer as the opposite of "contraction". On your second point, the Schwarzschild static time dilation matched the redshift last time I checked.
– safesphere
Aug 22 at 3:43
add a comment |Â
To add to Mike's answer, the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). Time dilation causes gravity and redshift. Length dilation makes distances larger. If you hover near a black hole, you see it farther than it actually is. As you get closer to the event horizon (while hovering, but not falling), you see it moving farther from you. If you get really close, you would see yourself light years away from the black hole, except its gravity is only getting stronger.
– safesphere
Aug 17 at 3:29
Without contradicting Mike's answer, the above mentioned physical effects of the spacetime curvature can be viewed as dimensionless. For example, time may be dilated twice (with the redshift of $z+1=2$).
– safesphere
Aug 17 at 3:41
1
@safesphere: the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). No, this is wrong. You can't interpret GR in this way. * For example, time may be dilated twice (with the redshift of z+1=2).* No, Doppler shifts have nothing to do with curvature.
– Ben Crowell
Aug 22 at 0:32
@BenCrowell You saying that something is "wrong" doesn't make it wrong only because you said it. You need to do a bit better than that. If you don't like length "dilation", you can call it "spagnetification" instead or any term you prefer as the opposite of "contraction". On your second point, the Schwarzschild static time dilation matched the redshift last time I checked.
– safesphere
Aug 22 at 3:43
To add to Mike's answer, the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). Time dilation causes gravity and redshift. Length dilation makes distances larger. If you hover near a black hole, you see it farther than it actually is. As you get closer to the event horizon (while hovering, but not falling), you see it moving farther from you. If you get really close, you would see yourself light years away from the black hole, except its gravity is only getting stronger.
– safesphere
Aug 17 at 3:29
To add to Mike's answer, the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). Time dilation causes gravity and redshift. Length dilation makes distances larger. If you hover near a black hole, you see it farther than it actually is. As you get closer to the event horizon (while hovering, but not falling), you see it moving farther from you. If you get really close, you would see yourself light years away from the black hole, except its gravity is only getting stronger.
– safesphere
Aug 17 at 3:29
Without contradicting Mike's answer, the above mentioned physical effects of the spacetime curvature can be viewed as dimensionless. For example, time may be dilated twice (with the redshift of $z+1=2$).
– safesphere
Aug 17 at 3:41
Without contradicting Mike's answer, the above mentioned physical effects of the spacetime curvature can be viewed as dimensionless. For example, time may be dilated twice (with the redshift of $z+1=2$).
– safesphere
Aug 17 at 3:41
1
1
@safesphere: the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). No, this is wrong. You can't interpret GR in this way. * For example, time may be dilated twice (with the redshift of z+1=2).* No, Doppler shifts have nothing to do with curvature.
– Ben Crowell
Aug 22 at 0:32
@safesphere: the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). No, this is wrong. You can't interpret GR in this way. * For example, time may be dilated twice (with the redshift of z+1=2).* No, Doppler shifts have nothing to do with curvature.
– Ben Crowell
Aug 22 at 0:32
@BenCrowell You saying that something is "wrong" doesn't make it wrong only because you said it. You need to do a bit better than that. If you don't like length "dilation", you can call it "spagnetification" instead or any term you prefer as the opposite of "contraction". On your second point, the Schwarzschild static time dilation matched the redshift last time I checked.
– safesphere
Aug 22 at 3:43
@BenCrowell You saying that something is "wrong" doesn't make it wrong only because you said it. You need to do a bit better than that. If you don't like length "dilation", you can call it "spagnetification" instead or any term you prefer as the opposite of "contraction". On your second point, the Schwarzschild static time dilation matched the redshift last time I checked.
– safesphere
Aug 22 at 3:43
add a comment |Â
1 Answer
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up vote
27
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While "warping of space" is a perfectly good way to think about it, to be precise we talk about the "curvature of spacetime". General relativity is the standard modern theory of gravity, and it describes this curvature with something called the Riemann tensor (although there are numerous other good ways to talk about curvature). A tensor is — to put it very roughly — a collection of numbers, and you can't boil it down to just one number. But each of the numbers in this Riemann tensor has units of $1/mathrmdistance^2$. Basically, the Riemann tensor measures the rate of change of the rate of change (yes, both) of space's shape as you move along different directions — so each of those rates of change brings in one factor of $1/mathrmdistance$, which is how you get a total of $1/mathrmdistance^2$.
There's no known maximum limit to how big the curvature can be. In our current theory, at least, there are places where it can become infinite. These are called gravitational singularities. For example, we believe that singularities are found inside of black holes. Of course, we also believe that the theory of general relativity is incomplete, and needs to be combined with quantum theory somehow. When this is done, most physicists expect that singularities will disappear and be replaced by some crazy quantum phenomenon. As just a vague ballpark guess, most physicists would expect that craziness to kick in once the curvature gets anywhere close to $1/ell_mathrmP^2$, where $ell_mathrmP$ is the Planck length — which might be something like a limit to the curvature.
In certain special cases, you can measure warping in different ways. Relatively recently, one important measurement of warping has been the detection of gravitational waves. Because they are very weak, we don't have to deal with a lot of the headaches that you normally run into in general relativity, which means that we can get away with measuring the warpage with something called the strain. The strain is the ratio between how much the gravitational wave changes the length of an object it passes and the normal length of that object. The units of strain, therefore, are just $mathrmdistance/mathrmdistance$ — which is to say that it's dimensionless.
answered Aug 17 at 2:24
Mike
11.6k13753
This is probably a reasonable answer at the level of the OP's question, but really this all assumes that the metric is dimensionless, which is not necessarily true. Different components of a tensor do not even have to have the same units. I have a detailed discussion of this sort of thing in section 5.11 of my GR book, lightandmatter.com/genrel .
– Ben Crowell
Aug 22 at 0:35
add a comment |Â
protected by ACuriousMind♦ Aug 17 at 15:04
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Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
27
down vote
accepted
While "warping of space" is a perfectly good way to think about it, to be precise we talk about the "curvature of spacetime". General relativity is the standard modern theory of gravity, and it describes this curvature with something called the Riemann tensor (although there are numerous other good ways to talk about curvature). A tensor is — to put it very roughly — a collection of numbers, and you can't boil it down to just one number. But each of the numbers in this Riemann tensor has units of $1/mathrmdistance^2$. Basically, the Riemann tensor measures the rate of change of the rate of change (yes, both) of space's shape as you move along different directions — so each of those rates of change brings in one factor of $1/mathrmdistance$, which is how you get a total of $1/mathrmdistance^2$.
There's no known maximum limit to how big the curvature can be. In our current theory, at least, there are places where it can become infinite. These are called gravitational singularities. For example, we believe that singularities are found inside of black holes. Of course, we also believe that the theory of general relativity is incomplete, and needs to be combined with quantum theory somehow. When this is done, most physicists expect that singularities will disappear and be replaced by some crazy quantum phenomenon. As just a vague ballpark guess, most physicists would expect that craziness to kick in once the curvature gets anywhere close to $1/ell_mathrmP^2$, where $ell_mathrmP$ is the Planck length — which might be something like a limit to the curvature.
In certain special cases, you can measure warping in different ways. Relatively recently, one important measurement of warping has been the detection of gravitational waves. Because they are very weak, we don't have to deal with a lot of the headaches that you normally run into in general relativity, which means that we can get away with measuring the warpage with something called the strain. The strain is the ratio between how much the gravitational wave changes the length of an object it passes and the normal length of that object. The units of strain, therefore, are just $mathrmdistance/mathrmdistance$ — which is to say that it's dimensionless.
answered Aug 17 at 2:24
Mike
11.6k13753
This is probably a reasonable answer at the level of the OP's question, but really this all assumes that the metric is dimensionless, which is not necessarily true. Different components of a tensor do not even have to have the same units. I have a detailed discussion of this sort of thing in section 5.11 of my GR book, lightandmatter.com/genrel .
– Ben Crowell
Aug 22 at 0:35
add a comment |Â
up vote
27
down vote
accepted
While "warping of space" is a perfectly good way to think about it, to be precise we talk about the "curvature of spacetime". General relativity is the standard modern theory of gravity, and it describes this curvature with something called the Riemann tensor (although there are numerous other good ways to talk about curvature). A tensor is — to put it very roughly — a collection of numbers, and you can't boil it down to just one number. But each of the numbers in this Riemann tensor has units of $1/mathrmdistance^2$. Basically, the Riemann tensor measures the rate of change of the rate of change (yes, both) of space's shape as you move along different directions — so each of those rates of change brings in one factor of $1/mathrmdistance$, which is how you get a total of $1/mathrmdistance^2$.
There's no known maximum limit to how big the curvature can be. In our current theory, at least, there are places where it can become infinite. These are called gravitational singularities. For example, we believe that singularities are found inside of black holes. Of course, we also believe that the theory of general relativity is incomplete, and needs to be combined with quantum theory somehow. When this is done, most physicists expect that singularities will disappear and be replaced by some crazy quantum phenomenon. As just a vague ballpark guess, most physicists would expect that craziness to kick in once the curvature gets anywhere close to $1/ell_mathrmP^2$, where $ell_mathrmP$ is the Planck length — which might be something like a limit to the curvature.
In certain special cases, you can measure warping in different ways. Relatively recently, one important measurement of warping has been the detection of gravitational waves. Because they are very weak, we don't have to deal with a lot of the headaches that you normally run into in general relativity, which means that we can get away with measuring the warpage with something called the strain. The strain is the ratio between how much the gravitational wave changes the length of an object it passes and the normal length of that object. The units of strain, therefore, are just $mathrmdistance/mathrmdistance$ — which is to say that it's dimensionless.
answered Aug 17 at 2:24
Mike
11.6k13753
This is probably a reasonable answer at the level of the OP's question, but really this all assumes that the metric is dimensionless, which is not necessarily true. Different components of a tensor do not even have to have the same units. I have a detailed discussion of this sort of thing in section 5.11 of my GR book, lightandmatter.com/genrel .
– Ben Crowell
Aug 22 at 0:35
add a comment |Â
up vote
27
down vote
accepted
up vote
27
down vote
accepted
While "warping of space" is a perfectly good way to think about it, to be precise we talk about the "curvature of spacetime". General relativity is the standard modern theory of gravity, and it describes this curvature with something called the Riemann tensor (although there are numerous other good ways to talk about curvature). A tensor is — to put it very roughly — a collection of numbers, and you can't boil it down to just one number. But each of the numbers in this Riemann tensor has units of $1/mathrmdistance^2$. Basically, the Riemann tensor measures the rate of change of the rate of change (yes, both) of space's shape as you move along different directions — so each of those rates of change brings in one factor of $1/mathrmdistance$, which is how you get a total of $1/mathrmdistance^2$.
There's no known maximum limit to how big the curvature can be. In our current theory, at least, there are places where it can become infinite. These are called gravitational singularities. For example, we believe that singularities are found inside of black holes. Of course, we also believe that the theory of general relativity is incomplete, and needs to be combined with quantum theory somehow. When this is done, most physicists expect that singularities will disappear and be replaced by some crazy quantum phenomenon. As just a vague ballpark guess, most physicists would expect that craziness to kick in once the curvature gets anywhere close to $1/ell_mathrmP^2$, where $ell_mathrmP$ is the Planck length — which might be something like a limit to the curvature.
In certain special cases, you can measure warping in different ways. Relatively recently, one important measurement of warping has been the detection of gravitational waves. Because they are very weak, we don't have to deal with a lot of the headaches that you normally run into in general relativity, which means that we can get away with measuring the warpage with something called the strain. The strain is the ratio between how much the gravitational wave changes the length of an object it passes and the normal length of that object. The units of strain, therefore, are just $mathrmdistance/mathrmdistance$ — which is to say that it's dimensionless.
answered Aug 17 at 2:24
Mike
11.6k13753
While "warping of space" is a perfectly good way to think about it, to be precise we talk about the "curvature of spacetime". General relativity is the standard modern theory of gravity, and it describes this curvature with something called the Riemann tensor (although there are numerous other good ways to talk about curvature). A tensor is — to put it very roughly — a collection of numbers, and you can't boil it down to just one number. But each of the numbers in this Riemann tensor has units of $1/mathrmdistance^2$. Basically, the Riemann tensor measures the rate of change of the rate of change (yes, both) of space's shape as you move along different directions — so each of those rates of change brings in one factor of $1/mathrmdistance$, which is how you get a total of $1/mathrmdistance^2$.
There's no known maximum limit to how big the curvature can be. In our current theory, at least, there are places where it can become infinite. These are called gravitational singularities. For example, we believe that singularities are found inside of black holes. Of course, we also believe that the theory of general relativity is incomplete, and needs to be combined with quantum theory somehow. When this is done, most physicists expect that singularities will disappear and be replaced by some crazy quantum phenomenon. As just a vague ballpark guess, most physicists would expect that craziness to kick in once the curvature gets anywhere close to $1/ell_mathrmP^2$, where $ell_mathrmP$ is the Planck length — which might be something like a limit to the curvature.
In certain special cases, you can measure warping in different ways. Relatively recently, one important measurement of warping has been the detection of gravitational waves. Because they are very weak, we don't have to deal with a lot of the headaches that you normally run into in general relativity, which means that we can get away with measuring the warpage with something called the strain. The strain is the ratio between how much the gravitational wave changes the length of an object it passes and the normal length of that object. The units of strain, therefore, are just $mathrmdistance/mathrmdistance$ — which is to say that it's dimensionless.
answered Aug 17 at 2:24
Mike
11.6k13753
edited Aug 21 at 19:53
answered Aug 17 at 2:24
Mike
11.6k13753
answered Aug 17 at 2:24
Mike
11.6k13753
answered Aug 17 at 2:24
Mike
11.6k13753
11.6k13753
This is probably a reasonable answer at the level of the OP's question, but really this all assumes that the metric is dimensionless, which is not necessarily true. Different components of a tensor do not even have to have the same units. I have a detailed discussion of this sort of thing in section 5.11 of my GR book, lightandmatter.com/genrel .
– Ben Crowell
Aug 22 at 0:35
add a comment |Â
This is probably a reasonable answer at the level of the OP's question, but really this all assumes that the metric is dimensionless, which is not necessarily true. Different components of a tensor do not even have to have the same units. I have a detailed discussion of this sort of thing in section 5.11 of my GR book, lightandmatter.com/genrel .
– Ben Crowell
Aug 22 at 0:35
This is probably a reasonable answer at the level of the OP's question, but really this all assumes that the metric is dimensionless, which is not necessarily true. Different components of a tensor do not even have to have the same units. I have a detailed discussion of this sort of thing in section 5.11 of my GR book, lightandmatter.com/genrel .
– Ben Crowell
Aug 22 at 0:35
This is probably a reasonable answer at the level of the OP's question, but really this all assumes that the metric is dimensionless, which is not necessarily true. Different components of a tensor do not even have to have the same units. I have a detailed discussion of this sort of thing in section 5.11 of my GR book, lightandmatter.com/genrel .
– Ben Crowell
Aug 22 at 0:35
add a comment |Â
protected by ACuriousMind♦ Aug 17 at 15:04
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
To add to Mike's answer, the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). Time dilation causes gravity and redshift. Length dilation makes distances larger. If you hover near a black hole, you see it farther than it actually is. As you get closer to the event horizon (while hovering, but not falling), you see it moving farther from you. If you get really close, you would see yourself light years away from the black hole, except its gravity is only getting stronger.
– safesphere
Aug 17 at 3:29
Without contradicting Mike's answer, the above mentioned physical effects of the spacetime curvature can be viewed as dimensionless. For example, time may be dilated twice (with the redshift of $z+1=2$).
– safesphere
Aug 17 at 3:41
1
@safesphere: the physical effects of the spacetime curvature are gravitational time dilation and length dilation (as opposed to length contraction in Special Relativity). No, this is wrong. You can't interpret GR in this way. * For example, time may be dilated twice (with the redshift of z+1=2).* No, Doppler shifts have nothing to do with curvature.
– Ben Crowell
Aug 22 at 0:32
@BenCrowell You saying that something is "wrong" doesn't make it wrong only because you said it. You need to do a bit better than that. If you don't like length "dilation", you can call it "spagnetification" instead or any term you prefer as the opposite of "contraction". On your second point, the Schwarzschild static time dilation matched the redshift last time I checked.
– safesphere
Aug 22 at 3:43