0 rank tensor vs 1D vector The Next CEO of Stack OverflowHistory of Electromagnetic Field TensorIn field theory, why are some symmetry transformations applied to the field values while other act on the space that the fields are defined on?Possible confusion, the inertia of something yields a tensor? (trying to understand an example)Confusion about the mathematical nature of Elecromagnetic tensor end the E, B fieldsWhat exactly is the Parity transformation? Parity in spherical coordinatesHow to represent tensors in a base? And some questions about indicesA fundamental question about tensors and vectors4-Vector DefinitionDoubts on covariant and contravariant vectors and on double tensorsZero order Tensor
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0 rank tensor vs 1D vector
The Next CEO of Stack OverflowHistory of Electromagnetic Field TensorIn field theory, why are some symmetry transformations applied to the field values while other act on the space that the fields are defined on?Possible confusion, the inertia of something yields a tensor? (trying to understand an example)Confusion about the mathematical nature of Elecromagnetic tensor end the E, B fieldsWhat exactly is the Parity transformation? Parity in spherical coordinatesHow to represent tensors in a base? And some questions about indicesA fundamental question about tensors and vectors4-Vector DefinitionDoubts on covariant and contravariant vectors and on double tensorsZero order Tensor
$begingroup$
What is the difference between zero-rank tensor $x$ (scalar) and 1D vector $[x]$?
As far as I understand tensor is anything which can be measured and different measures can be transformed into eachother. That is, there are different basises for looking at one object.
Is lengh a scalar (zero rank tensor)?
I think it is not.
ex.:
- physical parameter: writing pen's length
- tensor: $l$
- length in inches: $[5.511811023622]$
- length in centimeters: $[14]$
- transformation law: 1cm = 2.54inch
so $l$ is a scalar, but on the other hand it's a tensor of rank 1 since "physical parameter of length is invariant, only it's measures (in different units) are".
The same example can be made with classical example of temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.
tensor-calculus
edited 2 hours ago
Szabolcs Berecz
1031
asked 2 hours ago
coobitcoobit
340110
$endgroup$
add a comment |
$begingroup$
What is the difference between zero-rank tensor $x$ (scalar) and 1D vector $[x]$?
As far as I understand tensor is anything which can be measured and different measures can be transformed into eachother. That is, there are different basises for looking at one object.
Is lengh a scalar (zero rank tensor)?
I think it is not.
ex.:
- physical parameter: writing pen's length
- tensor: $l$
- length in inches: $[5.511811023622]$
- length in centimeters: $[14]$
- transformation law: 1cm = 2.54inch
so $l$ is a scalar, but on the other hand it's a tensor of rank 1 since "physical parameter of length is invariant, only it's measures (in different units) are".
The same example can be made with classical example of temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.
tensor-calculus
edited 2 hours ago
Szabolcs Berecz
1031
asked 2 hours ago
coobitcoobit
340110
$endgroup$
add a comment |
$begingroup$
What is the difference between zero-rank tensor $x$ (scalar) and 1D vector $[x]$?
As far as I understand tensor is anything which can be measured and different measures can be transformed into eachother. That is, there are different basises for looking at one object.
Is lengh a scalar (zero rank tensor)?
I think it is not.
ex.:
- physical parameter: writing pen's length
- tensor: $l$
- length in inches: $[5.511811023622]$
- length in centimeters: $[14]$
- transformation law: 1cm = 2.54inch
so $l$ is a scalar, but on the other hand it's a tensor of rank 1 since "physical parameter of length is invariant, only it's measures (in different units) are".
The same example can be made with classical example of temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.
tensor-calculus
edited 2 hours ago
Szabolcs Berecz
1031
asked 2 hours ago
coobitcoobit
340110
$endgroup$
What is the difference between zero-rank tensor $x$ (scalar) and 1D vector $[x]$?
As far as I understand tensor is anything which can be measured and different measures can be transformed into eachother. That is, there are different basises for looking at one object.
Is lengh a scalar (zero rank tensor)?
I think it is not.
ex.:
- physical parameter: writing pen's length
- tensor: $l$
- length in inches: $[5.511811023622]$
- length in centimeters: $[14]$
- transformation law: 1cm = 2.54inch
so $l$ is a scalar, but on the other hand it's a tensor of rank 1 since "physical parameter of length is invariant, only it's measures (in different units) are".
The same example can be made with classical example of temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.
tensor-calculus
tensor-calculus
edited 2 hours ago
Szabolcs Berecz
1031
asked 2 hours ago
coobitcoobit
340110
edited 2 hours ago
Szabolcs Berecz
1031
asked 2 hours ago
coobitcoobit
340110
edited 2 hours ago
Szabolcs Berecz
1031
edited 2 hours ago
Szabolcs Berecz
1031
edited 2 hours ago
Szabolcs Berecz
1031
1031
asked 2 hours ago
coobitcoobit
340110
asked 2 hours ago
coobitcoobit
340110
asked 2 hours ago
coobitcoobit
340110
340110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.
The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.
But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.
Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.
Under any other transformation group, the distinction between scalars and vectors is similar.
answered 1 hour ago
G. SmithG. Smith
10.2k11428
$endgroup$
$begingroup$
I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
$endgroup$
– coobit
49 mins ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
37 mins ago
$begingroup$
Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
$endgroup$
– G. Smith
13 mins ago
add a comment |
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1 Answer
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1 Answer
1
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oldest
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active
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votes
$begingroup$
“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.
The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.
But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.
Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.
Under any other transformation group, the distinction between scalars and vectors is similar.
answered 1 hour ago
G. SmithG. Smith
10.2k11428
$endgroup$
$begingroup$
I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
$endgroup$
– coobit
49 mins ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
37 mins ago
$begingroup$
Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
$endgroup$
– G. Smith
13 mins ago
add a comment |
$begingroup$
“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.
The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.
But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.
Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.
Under any other transformation group, the distinction between scalars and vectors is similar.
answered 1 hour ago
G. SmithG. Smith
10.2k11428
$endgroup$
$begingroup$
I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
$endgroup$
– coobit
49 mins ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
37 mins ago
$begingroup$
Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
$endgroup$
– G. Smith
13 mins ago
add a comment |
$begingroup$
“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.
The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.
But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.
Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.
Under any other transformation group, the distinction between scalars and vectors is similar.
answered 1 hour ago
G. SmithG. Smith
10.2k11428
$endgroup$
“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.
The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.
But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.
Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.
Under any other transformation group, the distinction between scalars and vectors is similar.
answered 1 hour ago
G. SmithG. Smith
10.2k11428
edited 1 hour ago
answered 1 hour ago
G. SmithG. Smith
10.2k11428
answered 1 hour ago
G. SmithG. Smith
10.2k11428
answered 1 hour ago
G. SmithG. Smith
10.2k11428
10.2k11428
$begingroup$
I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
$endgroup$
– coobit
49 mins ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
37 mins ago
$begingroup$
Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
$endgroup$
– G. Smith
13 mins ago
add a comment |
$begingroup$
I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
$endgroup$
– coobit
49 mins ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
37 mins ago
$begingroup$
Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
$endgroup$
– G. Smith
13 mins ago
$begingroup$
I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
$endgroup$
– coobit
49 mins ago
$begingroup$
I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar?
$endgroup$
– coobit
49 mins ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
37 mins ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
37 mins ago
$begingroup$
Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
$endgroup$
– G. Smith
13 mins ago
$begingroup$
Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number.
$endgroup$
– G. Smith
13 mins ago
add a comment |
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