Tensor summation help - einstein notation
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0
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If $(x)^2 = x_u x^u$ then what is $x^4$ ?
I have thought about
$(x)^4 = x_{vu} x^{vu}$
and
$(x)^4 = x_{uu} x^{uu}$
But I am honestly not sure...
differential-geometry
asked Sep 15 '16 at 16:14
Alex Santeri
11
|
show 1 more comment
up vote
0
down vote
favorite
If $(x)^2 = x_u x^u$ then what is $x^4$ ?
I have thought about
$(x)^4 = x_{vu} x^{vu}$
and
$(x)^4 = x_{uu} x^{uu}$
But I am honestly not sure...
differential-geometry
asked Sep 15 '16 at 16:14
Alex Santeri
11
Isn't it simply $(x^2)^2$ since $x^2$ is scalar ? (I'm reaaaaaally not sure about this)
– mwoua
Sep 15 '16 at 16:37
I have thought about it but too, I dont know how I would write it out properly
– Alex Santeri
Sep 15 '16 at 16:44
I would have written it $(x_ux^u)^2$. If $x$ is a vector, I don't see any other way to write it.
– mwoua
Sep 15 '16 at 16:55
1
Horrible notation from the outset!
– Ted Shifrin
Sep 16 '16 at 3:53
Is that what retired university lecturers do on their free time? Come complain about notation without any hints of help? nice life
– Alex Santeri
Sep 16 '16 at 14:40
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $(x)^2 = x_u x^u$ then what is $x^4$ ?
I have thought about
$(x)^4 = x_{vu} x^{vu}$
and
$(x)^4 = x_{uu} x^{uu}$
But I am honestly not sure...
differential-geometry
asked Sep 15 '16 at 16:14
Alex Santeri
11
If $(x)^2 = x_u x^u$ then what is $x^4$ ?
I have thought about
$(x)^4 = x_{vu} x^{vu}$
and
$(x)^4 = x_{uu} x^{uu}$
But I am honestly not sure...
differential-geometry
differential-geometry
asked Sep 15 '16 at 16:14
Alex Santeri
11
asked Sep 15 '16 at 16:14
Alex Santeri
11
edited Nov 20 at 17:59
asked Sep 15 '16 at 16:14
Alex Santeri
11
asked Sep 15 '16 at 16:14
Alex Santeri
11
asked Sep 15 '16 at 16:14
Alex Santeri
11
11
Isn't it simply $(x^2)^2$ since $x^2$ is scalar ? (I'm reaaaaaally not sure about this)
– mwoua
Sep 15 '16 at 16:37
I have thought about it but too, I dont know how I would write it out properly
– Alex Santeri
Sep 15 '16 at 16:44
I would have written it $(x_ux^u)^2$. If $x$ is a vector, I don't see any other way to write it.
– mwoua
Sep 15 '16 at 16:55
1
Horrible notation from the outset!
– Ted Shifrin
Sep 16 '16 at 3:53
Is that what retired university lecturers do on their free time? Come complain about notation without any hints of help? nice life
– Alex Santeri
Sep 16 '16 at 14:40
|
show 1 more comment
Isn't it simply $(x^2)^2$ since $x^2$ is scalar ? (I'm reaaaaaally not sure about this)
– mwoua
Sep 15 '16 at 16:37
I have thought about it but too, I dont know how I would write it out properly
– Alex Santeri
Sep 15 '16 at 16:44
I would have written it $(x_ux^u)^2$. If $x$ is a vector, I don't see any other way to write it.
– mwoua
Sep 15 '16 at 16:55
1
Horrible notation from the outset!
– Ted Shifrin
Sep 16 '16 at 3:53
Is that what retired university lecturers do on their free time? Come complain about notation without any hints of help? nice life
– Alex Santeri
Sep 16 '16 at 14:40
Isn't it simply $(x^2)^2$ since $x^2$ is scalar ? (I'm reaaaaaally not sure about this)
– mwoua
Sep 15 '16 at 16:37
Isn't it simply $(x^2)^2$ since $x^2$ is scalar ? (I'm reaaaaaally not sure about this)
– mwoua
Sep 15 '16 at 16:37
I have thought about it but too, I dont know how I would write it out properly
– Alex Santeri
Sep 15 '16 at 16:44
I have thought about it but too, I dont know how I would write it out properly
– Alex Santeri
Sep 15 '16 at 16:44
I would have written it $(x_ux^u)^2$. If $x$ is a vector, I don't see any other way to write it.
– mwoua
Sep 15 '16 at 16:55
I would have written it $(x_ux^u)^2$. If $x$ is a vector, I don't see any other way to write it.
– mwoua
Sep 15 '16 at 16:55
1
1
Horrible notation from the outset!
– Ted Shifrin
Sep 16 '16 at 3:53
Horrible notation from the outset!
– Ted Shifrin
Sep 16 '16 at 3:53
Is that what retired university lecturers do on their free time? Come complain about notation without any hints of help? nice life
– Alex Santeri
Sep 16 '16 at 14:40
Is that what retired university lecturers do on their free time? Come complain about notation without any hints of help? nice life
– Alex Santeri
Sep 16 '16 at 14:40
|
show 1 more comment
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Isn't it simply $(x^2)^2$ since $x^2$ is scalar ? (I'm reaaaaaally not sure about this)
– mwoua
Sep 15 '16 at 16:37
I have thought about it but too, I dont know how I would write it out properly
– Alex Santeri
Sep 15 '16 at 16:44
I would have written it $(x_ux^u)^2$. If $x$ is a vector, I don't see any other way to write it.
– mwoua
Sep 15 '16 at 16:55
1
Horrible notation from the outset!
– Ted Shifrin
Sep 16 '16 at 3:53
Is that what retired university lecturers do on their free time? Come complain about notation without any hints of help? nice life
– Alex Santeri
Sep 16 '16 at 14:40