Excluding basis in tensor notationConversion of mixed tensors into mixed tensors and into covariant (or contravariant) onessum representation by/ determinant of elementary tensorsWhat is contracting a tensor actually doing?How to identifiy $V wedge V$ with the space of all alternating bilinear formsCoordinate-free notation for tensor contraction?How to represent matrix multiplication in tensor algebra?Question about abstract notation of Maxwell Stress TensorHow to interpret stress tensor as a contravariant second order tensorEquivalence between dual vectors and dual covectorsAre all bivectors in three dimensions simple?In what sense is the cross-product not a tensor?
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Excluding basis in tensor notation
Conversion of mixed tensors into mixed tensors and into covariant (or contravariant) onessum representation by/ determinant of elementary tensorsWhat is contracting a tensor actually doing?How to identifiy $V wedge V$ with the space of all alternating bilinear formsCoordinate-free notation for tensor contraction?How to represent matrix multiplication in tensor algebra?Question about abstract notation of Maxwell Stress TensorHow to interpret stress tensor as a contravariant second order tensorEquivalence between dual vectors and dual covectorsAre all bivectors in three dimensions simple?In what sense is the cross-product not a tensor?
$begingroup$
My question is: is there anywhere in tensors that we lose something by dropping the basis, or where it makes something more difficult? Like by saying the a tensor $T^ije_iotimes e_j$ is represented entirely by the notation $T^ij$?
Another thing, because there is only one i and j in the previous expression, does the summation convention kick in?
I heard this on the YouTube channel XylyXylyX (that's actually the name). Correct me if this is wrong because I've just started learning this myself.
tensors
asked Mar 12 at 20:00
Benjamin ThoburnBenjamin Thoburn
356313
$endgroup$
add a comment |
$begingroup$
My question is: is there anywhere in tensors that we lose something by dropping the basis, or where it makes something more difficult? Like by saying the a tensor $T^ije_iotimes e_j$ is represented entirely by the notation $T^ij$?
Another thing, because there is only one i and j in the previous expression, does the summation convention kick in?
I heard this on the YouTube channel XylyXylyX (that's actually the name). Correct me if this is wrong because I've just started learning this myself.
tensors
asked Mar 12 at 20:00
Benjamin ThoburnBenjamin Thoburn
356313
$endgroup$
$begingroup$
You might find part of this answer helpful.
$endgroup$
– Michael Albanese
Mar 12 at 21:13
$begingroup$
@Michael Albanese Thanks, which part exactly?
$endgroup$
– Benjamin Thoburn
Mar 14 at 8:50
$begingroup$
@Michael Albanese also it's confusing me that in one case $T^ij$ is a tensor valued quantity and in the other it's a scalar quantity being summed over.
$endgroup$
– Benjamin Thoburn
Mar 14 at 9:24
$begingroup$
Basically all of it. $T^ij$ is not a tensor-valued quantity. As is mentioned in the answer below, $T^ij$ are the coefficients of the tensor with respect to a fixed basis of the vector space. If the basis changes, the coefficients change in a certain way. Often in physics, people regard the collection of coefficients $T^ij$, together with the knowledge of how it transforms under a change of basis, as the tensor itself.
$endgroup$
– Michael Albanese
Mar 14 at 14:03
add a comment |
$begingroup$
My question is: is there anywhere in tensors that we lose something by dropping the basis, or where it makes something more difficult? Like by saying the a tensor $T^ije_iotimes e_j$ is represented entirely by the notation $T^ij$?
Another thing, because there is only one i and j in the previous expression, does the summation convention kick in?
I heard this on the YouTube channel XylyXylyX (that's actually the name). Correct me if this is wrong because I've just started learning this myself.
tensors
asked Mar 12 at 20:00
Benjamin ThoburnBenjamin Thoburn
356313
$endgroup$
My question is: is there anywhere in tensors that we lose something by dropping the basis, or where it makes something more difficult? Like by saying the a tensor $T^ije_iotimes e_j$ is represented entirely by the notation $T^ij$?
Another thing, because there is only one i and j in the previous expression, does the summation convention kick in?
I heard this on the YouTube channel XylyXylyX (that's actually the name). Correct me if this is wrong because I've just started learning this myself.
tensors
tensors
asked Mar 12 at 20:00
Benjamin ThoburnBenjamin Thoburn
356313
asked Mar 12 at 20:00
Benjamin ThoburnBenjamin Thoburn
356313
edited Mar 14 at 8:53
Benjamin Thoburn
asked Mar 12 at 20:00
Benjamin ThoburnBenjamin Thoburn
356313
asked Mar 12 at 20:00
Benjamin ThoburnBenjamin Thoburn
356313
asked Mar 12 at 20:00
Benjamin ThoburnBenjamin Thoburn
356313
356313
$begingroup$
You might find part of this answer helpful.
$endgroup$
– Michael Albanese
Mar 12 at 21:13
$begingroup$
@Michael Albanese Thanks, which part exactly?
$endgroup$
– Benjamin Thoburn
Mar 14 at 8:50
$begingroup$
@Michael Albanese also it's confusing me that in one case $T^ij$ is a tensor valued quantity and in the other it's a scalar quantity being summed over.
$endgroup$
– Benjamin Thoburn
Mar 14 at 9:24
$begingroup$
Basically all of it. $T^ij$ is not a tensor-valued quantity. As is mentioned in the answer below, $T^ij$ are the coefficients of the tensor with respect to a fixed basis of the vector space. If the basis changes, the coefficients change in a certain way. Often in physics, people regard the collection of coefficients $T^ij$, together with the knowledge of how it transforms under a change of basis, as the tensor itself.
$endgroup$
– Michael Albanese
Mar 14 at 14:03
add a comment |
$begingroup$
You might find part of this answer helpful.
$endgroup$
– Michael Albanese
Mar 12 at 21:13
$begingroup$
@Michael Albanese Thanks, which part exactly?
$endgroup$
– Benjamin Thoburn
Mar 14 at 8:50
$begingroup$
@Michael Albanese also it's confusing me that in one case $T^ij$ is a tensor valued quantity and in the other it's a scalar quantity being summed over.
$endgroup$
– Benjamin Thoburn
Mar 14 at 9:24
$begingroup$
Basically all of it. $T^ij$ is not a tensor-valued quantity. As is mentioned in the answer below, $T^ij$ are the coefficients of the tensor with respect to a fixed basis of the vector space. If the basis changes, the coefficients change in a certain way. Often in physics, people regard the collection of coefficients $T^ij$, together with the knowledge of how it transforms under a change of basis, as the tensor itself.
$endgroup$
– Michael Albanese
Mar 14 at 14:03
$begingroup$
You might find part of this answer helpful.
$endgroup$
– Michael Albanese
Mar 12 at 21:13
$begingroup$
You might find part of this answer helpful.
$endgroup$
– Michael Albanese
Mar 12 at 21:13
$begingroup$
@Michael Albanese Thanks, which part exactly?
$endgroup$
– Benjamin Thoburn
Mar 14 at 8:50
$begingroup$
@Michael Albanese Thanks, which part exactly?
$endgroup$
– Benjamin Thoburn
Mar 14 at 8:50
$begingroup$
@Michael Albanese also it's confusing me that in one case $T^ij$ is a tensor valued quantity and in the other it's a scalar quantity being summed over.
$endgroup$
– Benjamin Thoburn
Mar 14 at 9:24
$begingroup$
@Michael Albanese also it's confusing me that in one case $T^ij$ is a tensor valued quantity and in the other it's a scalar quantity being summed over.
$endgroup$
– Benjamin Thoburn
Mar 14 at 9:24
$begingroup$
Basically all of it. $T^ij$ is not a tensor-valued quantity. As is mentioned in the answer below, $T^ij$ are the coefficients of the tensor with respect to a fixed basis of the vector space. If the basis changes, the coefficients change in a certain way. Often in physics, people regard the collection of coefficients $T^ij$, together with the knowledge of how it transforms under a change of basis, as the tensor itself.
$endgroup$
– Michael Albanese
Mar 14 at 14:03
$begingroup$
Basically all of it. $T^ij$ is not a tensor-valued quantity. As is mentioned in the answer below, $T^ij$ are the coefficients of the tensor with respect to a fixed basis of the vector space. If the basis changes, the coefficients change in a certain way. Often in physics, people regard the collection of coefficients $T^ij$, together with the knowledge of how it transforms under a change of basis, as the tensor itself.
$endgroup$
– Michael Albanese
Mar 14 at 14:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by
$mathbf T^ij mathbf e_i otimes mathbf e_j$
which is short for
$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$
But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.
answered Mar 14 at 10:06
gandalf61gandalf61
9,109825
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by
$mathbf T^ij mathbf e_i otimes mathbf e_j$
which is short for
$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$
But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.
answered Mar 14 at 10:06
gandalf61gandalf61
9,109825
$endgroup$
add a comment |
$begingroup$
Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by
$mathbf T^ij mathbf e_i otimes mathbf e_j$
which is short for
$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$
But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.
answered Mar 14 at 10:06
gandalf61gandalf61
9,109825
$endgroup$
add a comment |
$begingroup$
Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by
$mathbf T^ij mathbf e_i otimes mathbf e_j$
which is short for
$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$
But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.
answered Mar 14 at 10:06
gandalf61gandalf61
9,109825
$endgroup$
Strictly speaking $mathbf T^ij$ represents the $(i,j)$th component of the $(2,0)$ tensor $mathbf T$ relative to a particular basis of $V times V$, and the whole tensor would be represented by
$mathbf T^ij mathbf e_i otimes mathbf e_j$
which is short for
$sum_i=1^nsum_j=1^nmathbf T^ij mathbf e_i otimes mathbf e_j$
But $mathbf T^ij$ is often used informally to represent all of the $n^2$ components of $mathbf T$.
answered Mar 14 at 10:06
gandalf61gandalf61
9,109825
answered Mar 14 at 10:06
gandalf61gandalf61
9,109825
answered Mar 14 at 10:06
gandalf61gandalf61
9,109825
answered Mar 14 at 10:06
gandalf61gandalf61
9,109825
9,109825
add a comment |
add a comment |
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$begingroup$
You might find part of this answer helpful.
$endgroup$
– Michael Albanese
Mar 12 at 21:13
$begingroup$
@Michael Albanese Thanks, which part exactly?
$endgroup$
– Benjamin Thoburn
Mar 14 at 8:50
$begingroup$
@Michael Albanese also it's confusing me that in one case $T^ij$ is a tensor valued quantity and in the other it's a scalar quantity being summed over.
$endgroup$
– Benjamin Thoburn
Mar 14 at 9:24
$begingroup$
Basically all of it. $T^ij$ is not a tensor-valued quantity. As is mentioned in the answer below, $T^ij$ are the coefficients of the tensor with respect to a fixed basis of the vector space. If the basis changes, the coefficients change in a certain way. Often in physics, people regard the collection of coefficients $T^ij$, together with the knowledge of how it transforms under a change of basis, as the tensor itself.
$endgroup$
– Michael Albanese
Mar 14 at 14:03