What does Einstein mean by saying that equations of transformation are homogeneous?
My question arises from reading Einstein's The Meaning of Relativity. In particular, on page 12:
If the three components of a vector vanish for one system of Cartesian co-ordinates, they vanish for all systems, because the equations of transformation are homogeneous.
And again on page 13:
In my notation the transformation under discussion is of the form
$$a_{overline{i}overline{j}}
=e_{:overline{i}}^{i}e_{:overline{j}}^{j}a_{ij}.$$
This transformation is homogeneous and of the first degree in the $a_{munu}.$
So, with that added background, here is the original formulation of my question: In Euclidean 3-space, we are given various rectangular Cartesian coordinate system all of the same scale, and related by transformation equations of the form
$$x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i},$$
with the determinant $left|e_{:i}^{overline{i}}right|=1.$
What does it mean to say that the equations of transformation are homogeneous? In particular what does it mean to say if the components of a vector vanish in any of these coordinate systems, they vanish in all of these coordinate systems, because the equations of transformation are homogeneous?
To me, homogeneous applies to systems of linear equations or differential equations. And the short definition is that the right-hand side is zero.
linear-algebra differential-geometry linear-transformations mathematical-physics affine-geometry
asked Nov 28 at 4:17
Steven Hatton
721315
add a comment |
My question arises from reading Einstein's The Meaning of Relativity. In particular, on page 12:
If the three components of a vector vanish for one system of Cartesian co-ordinates, they vanish for all systems, because the equations of transformation are homogeneous.
And again on page 13:
In my notation the transformation under discussion is of the form
$$a_{overline{i}overline{j}}
=e_{:overline{i}}^{i}e_{:overline{j}}^{j}a_{ij}.$$
This transformation is homogeneous and of the first degree in the $a_{munu}.$
So, with that added background, here is the original formulation of my question: In Euclidean 3-space, we are given various rectangular Cartesian coordinate system all of the same scale, and related by transformation equations of the form
$$x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i},$$
with the determinant $left|e_{:i}^{overline{i}}right|=1.$
What does it mean to say that the equations of transformation are homogeneous? In particular what does it mean to say if the components of a vector vanish in any of these coordinate systems, they vanish in all of these coordinate systems, because the equations of transformation are homogeneous?
To me, homogeneous applies to systems of linear equations or differential equations. And the short definition is that the right-hand side is zero.
linear-algebra differential-geometry linear-transformations mathematical-physics affine-geometry
asked Nov 28 at 4:17
Steven Hatton
721315
Generally speaking homogeneous means all of the same degree (when applied to graded rings generally and polynomials in particular). Here what he means is linear as opposed to affine.
– jgon
Nov 29 at 8:03
en.wikipedia.org/wiki/Homogeneous_function
– Anthony Carapetis
Nov 29 at 8:27
Ah! I should have recognized that. I don't have to follow the link. This is the meaning used in Euler's homogeneous function theorem. If you post that as an answer, I will accept it.
– Steven Hatton
Nov 29 at 8:33
BTW: Einstein apparently uses the idiomatic slang of calling all affine transformation "linear". That is, to Einstein: $x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i}$ is linear.
– Steven Hatton
Nov 29 at 8:38
add a comment |
My question arises from reading Einstein's The Meaning of Relativity. In particular, on page 12:
If the three components of a vector vanish for one system of Cartesian co-ordinates, they vanish for all systems, because the equations of transformation are homogeneous.
And again on page 13:
In my notation the transformation under discussion is of the form
$$a_{overline{i}overline{j}}
=e_{:overline{i}}^{i}e_{:overline{j}}^{j}a_{ij}.$$
This transformation is homogeneous and of the first degree in the $a_{munu}.$
So, with that added background, here is the original formulation of my question: In Euclidean 3-space, we are given various rectangular Cartesian coordinate system all of the same scale, and related by transformation equations of the form
$$x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i},$$
with the determinant $left|e_{:i}^{overline{i}}right|=1.$
What does it mean to say that the equations of transformation are homogeneous? In particular what does it mean to say if the components of a vector vanish in any of these coordinate systems, they vanish in all of these coordinate systems, because the equations of transformation are homogeneous?
To me, homogeneous applies to systems of linear equations or differential equations. And the short definition is that the right-hand side is zero.
linear-algebra differential-geometry linear-transformations mathematical-physics affine-geometry
asked Nov 28 at 4:17
Steven Hatton
721315
My question arises from reading Einstein's The Meaning of Relativity. In particular, on page 12:
If the three components of a vector vanish for one system of Cartesian co-ordinates, they vanish for all systems, because the equations of transformation are homogeneous.
And again on page 13:
In my notation the transformation under discussion is of the form
$$a_{overline{i}overline{j}}
=e_{:overline{i}}^{i}e_{:overline{j}}^{j}a_{ij}.$$
This transformation is homogeneous and of the first degree in the $a_{munu}.$
So, with that added background, here is the original formulation of my question: In Euclidean 3-space, we are given various rectangular Cartesian coordinate system all of the same scale, and related by transformation equations of the form
$$x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i},$$
with the determinant $left|e_{:i}^{overline{i}}right|=1.$
What does it mean to say that the equations of transformation are homogeneous? In particular what does it mean to say if the components of a vector vanish in any of these coordinate systems, they vanish in all of these coordinate systems, because the equations of transformation are homogeneous?
To me, homogeneous applies to systems of linear equations or differential equations. And the short definition is that the right-hand side is zero.
linear-algebra differential-geometry linear-transformations mathematical-physics affine-geometry
linear-algebra differential-geometry linear-transformations mathematical-physics affine-geometry
asked Nov 28 at 4:17
Steven Hatton
721315
asked Nov 28 at 4:17
Steven Hatton
721315
edited Nov 29 at 7:50
asked Nov 28 at 4:17
Steven Hatton
721315
asked Nov 28 at 4:17
Steven Hatton
721315
asked Nov 28 at 4:17
Steven Hatton
721315
721315
Generally speaking homogeneous means all of the same degree (when applied to graded rings generally and polynomials in particular). Here what he means is linear as opposed to affine.
– jgon
Nov 29 at 8:03
en.wikipedia.org/wiki/Homogeneous_function
– Anthony Carapetis
Nov 29 at 8:27
Ah! I should have recognized that. I don't have to follow the link. This is the meaning used in Euler's homogeneous function theorem. If you post that as an answer, I will accept it.
– Steven Hatton
Nov 29 at 8:33
BTW: Einstein apparently uses the idiomatic slang of calling all affine transformation "linear". That is, to Einstein: $x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i}$ is linear.
– Steven Hatton
Nov 29 at 8:38
add a comment |
Generally speaking homogeneous means all of the same degree (when applied to graded rings generally and polynomials in particular). Here what he means is linear as opposed to affine.
– jgon
Nov 29 at 8:03
en.wikipedia.org/wiki/Homogeneous_function
– Anthony Carapetis
Nov 29 at 8:27
Ah! I should have recognized that. I don't have to follow the link. This is the meaning used in Euler's homogeneous function theorem. If you post that as an answer, I will accept it.
– Steven Hatton
Nov 29 at 8:33
BTW: Einstein apparently uses the idiomatic slang of calling all affine transformation "linear". That is, to Einstein: $x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i}$ is linear.
– Steven Hatton
Nov 29 at 8:38
Generally speaking homogeneous means all of the same degree (when applied to graded rings generally and polynomials in particular). Here what he means is linear as opposed to affine.
– jgon
Nov 29 at 8:03
Generally speaking homogeneous means all of the same degree (when applied to graded rings generally and polynomials in particular). Here what he means is linear as opposed to affine.
– jgon
Nov 29 at 8:03
en.wikipedia.org/wiki/Homogeneous_function
– Anthony Carapetis
Nov 29 at 8:27
en.wikipedia.org/wiki/Homogeneous_function
– Anthony Carapetis
Nov 29 at 8:27
Ah! I should have recognized that. I don't have to follow the link. This is the meaning used in Euler's homogeneous function theorem. If you post that as an answer, I will accept it.
– Steven Hatton
Nov 29 at 8:33
Ah! I should have recognized that. I don't have to follow the link. This is the meaning used in Euler's homogeneous function theorem. If you post that as an answer, I will accept it.
– Steven Hatton
Nov 29 at 8:33
BTW: Einstein apparently uses the idiomatic slang of calling all affine transformation "linear". That is, to Einstein: $x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i}$ is linear.
– Steven Hatton
Nov 29 at 8:38
BTW: Einstein apparently uses the idiomatic slang of calling all affine transformation "linear". That is, to Einstein: $x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i}$ is linear.
– Steven Hatton
Nov 29 at 8:38
add a comment |
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Generally speaking homogeneous means all of the same degree (when applied to graded rings generally and polynomials in particular). Here what he means is linear as opposed to affine.
– jgon
Nov 29 at 8:03
en.wikipedia.org/wiki/Homogeneous_function
– Anthony Carapetis
Nov 29 at 8:27
Ah! I should have recognized that. I don't have to follow the link. This is the meaning used in Euler's homogeneous function theorem. If you post that as an answer, I will accept it.
– Steven Hatton
Nov 29 at 8:33
BTW: Einstein apparently uses the idiomatic slang of calling all affine transformation "linear". That is, to Einstein: $x^{overline{i}}=a^{overline{i}}+e_{:i}^{overline{i}}x^{i}$ is linear.
– Steven Hatton
Nov 29 at 8:38