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
I want to delete every two lines after 3rd lines in file contain very large number of lines :
Is there always a complete, orthogonal set of unitary matrices?
Is it okay to majorly distort historical facts while writing a fiction story?
0 rank tensor vs 1D vector
Is French Guiana a (hard) EU border?
Domestic-to-international connection at Orlando (MCO)
What is the value of α and β in a triangle?
How many extra stops do monopods offer for tele photographs?
What connection does MS Office have to Netscape Navigator?
Can a Bladesinger Wizard use Bladesong with a Hand Crossbow?
Chain wire methods together in Lightning Web Components
Writing differences on a blackboard
Why the difference in type-inference over the as-pattern in two similar function definitions?
Method for adding error messages to a dictionary given a key
Prepend last line of stdin to entire stdin
How to invert MapIndexed on a ragged structure? How to construct a tree from rules?
Plot of histogram similar to output from @risk
Why isn't acceleration always zero whenever velocity is zero, such as the moment a ball bounces off a wall?
Is this "being" usage is essential?
WOW air has ceased operation, can I get my tickets refunded?
Why do remote US companies require working in the US?
Can MTA send mail via a relay without being told so?
Unclear about dynamic binding
Solving system of ODEs with extra parameter
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
350110
$endgroup$
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
350110
$endgroup$
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
350110
$endgroup$
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
350110
edited 2 hours ago
Szabolcs Berecz
1031
asked 2 hours ago
coobitcoobit
350110
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
350110
asked 2 hours ago
coobitcoobit
350110
asked 2 hours ago
coobitcoobit
350110
350110
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 2 hours 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
1 hour ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
50 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
26 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f469598%2f0-rank-tensor-vs-1d-vector%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 2 hours 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
1 hour ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
50 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
26 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 2 hours 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
1 hour ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
50 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
26 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 2 hours 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 2 hours ago
G. SmithG. Smith
10.2k11428
edited 1 hour ago
answered 2 hours ago
G. SmithG. Smith
10.2k11428
answered 2 hours ago
G. SmithG. Smith
10.2k11428
answered 2 hours 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
1 hour ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
50 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
26 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
1 hour ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
50 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
26 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
1 hour 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
1 hour ago
$begingroup$
@coobit Consider the group of reflections along that one dimension. A vector changes sign, but a scalar doesn't.
$endgroup$
– Chiral Anomaly
50 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
50 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
26 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
26 mins ago
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f469598%2f0-rank-tensor-vs-1d-vector%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
