Mixing
Orthogonal Projection Tensor
$begingroup$
Hi my question is about this orthogonal projection tensor $P^sigma_nuequivdelta^sigma_nu+U^sigma U_nu$.
It should have the following properties when $V,W$ are vectors parallel and perpendicular to U, and $U$ is a 4-velocity in spacetime. These equations come from (1.121) and (1.122) in Sean Carroll's gravity book.
$
P^sigma_nu V^nu_parallel ~~= ~~0
$
$
P^sigma_nu W^nu_perp ~~=~~ W^sigma_perp
$
However, when I test for the first property I do not get the correct answer. I suppose I am messing up how I am using the Kronecker delta but I'm not sure how. Lets operate on and a vector $V$ that is parallel to $U$ to get the "equal zero" condition above.
$
P^sigma_nu V^nu_parallel ~~=~~delta^sigma_nu V^nu_parallel+U^sigma U_nu V^nu_parallel
$
If $sigma=nu$ then $delta^sigma_nu=1$.
$
P^sigma_nu V^nu_parallel~~=~~ V^sigma_parallel +U^sigma U_sigma V^sigma_parallel qquad qquad [U^sigma U_sigma=-1]\
quadqquad=~~ V^sigma_parallel - V^sigma_parallel \
quadqquad=~~0 quadcheckmark
$
That looks good, now let's check when $sigmaneqnu$ and $delta^sigma_nu=0$.
$
P^sigma_nu V^nu_parallel~~= ~~0 +U^sigma U_nu V^nu_parallel qquad qquad~~~ [Vparallel Uimplies U_nu V^nuneq0]\
quadqquad~neq~~ 0 qquad qquad qquad qquad~~~~~~ [mathrm{contradicts~Eq.}(2)]
$
It looks like I should totally disregard the case where $sigmaneqnu$ but I don't see why.
tensors
asked Aug 30 '16 at 20:14
Jonathan TookerJonathan Tooker
62
$endgroup$
add a comment |
$begingroup$
Hi my question is about this orthogonal projection tensor $P^sigma_nuequivdelta^sigma_nu+U^sigma U_nu$.
It should have the following properties when $V,W$ are vectors parallel and perpendicular to U, and $U$ is a 4-velocity in spacetime. These equations come from (1.121) and (1.122) in Sean Carroll's gravity book.
$
P^sigma_nu V^nu_parallel ~~= ~~0
$
$
P^sigma_nu W^nu_perp ~~=~~ W^sigma_perp
$
However, when I test for the first property I do not get the correct answer. I suppose I am messing up how I am using the Kronecker delta but I'm not sure how. Lets operate on and a vector $V$ that is parallel to $U$ to get the "equal zero" condition above.
$
P^sigma_nu V^nu_parallel ~~=~~delta^sigma_nu V^nu_parallel+U^sigma U_nu V^nu_parallel
$
If $sigma=nu$ then $delta^sigma_nu=1$.
$
P^sigma_nu V^nu_parallel~~=~~ V^sigma_parallel +U^sigma U_sigma V^sigma_parallel qquad qquad [U^sigma U_sigma=-1]\
quadqquad=~~ V^sigma_parallel - V^sigma_parallel \
quadqquad=~~0 quadcheckmark
$
That looks good, now let's check when $sigmaneqnu$ and $delta^sigma_nu=0$.
$
P^sigma_nu V^nu_parallel~~= ~~0 +U^sigma U_nu V^nu_parallel qquad qquad~~~ [Vparallel Uimplies U_nu V^nuneq0]\
quadqquad~neq~~ 0 qquad qquad qquad qquad~~~~~~ [mathrm{contradicts~Eq.}(2)]
$
It looks like I should totally disregard the case where $sigmaneqnu$ but I don't see why.
tensors
asked Aug 30 '16 at 20:14
Jonathan TookerJonathan Tooker
62
$endgroup$
add a comment |
$begingroup$
Hi my question is about this orthogonal projection tensor $P^sigma_nuequivdelta^sigma_nu+U^sigma U_nu$.
It should have the following properties when $V,W$ are vectors parallel and perpendicular to U, and $U$ is a 4-velocity in spacetime. These equations come from (1.121) and (1.122) in Sean Carroll's gravity book.
$
P^sigma_nu V^nu_parallel ~~= ~~0
$
$
P^sigma_nu W^nu_perp ~~=~~ W^sigma_perp
$
However, when I test for the first property I do not get the correct answer. I suppose I am messing up how I am using the Kronecker delta but I'm not sure how. Lets operate on and a vector $V$ that is parallel to $U$ to get the "equal zero" condition above.
$
P^sigma_nu V^nu_parallel ~~=~~delta^sigma_nu V^nu_parallel+U^sigma U_nu V^nu_parallel
$
If $sigma=nu$ then $delta^sigma_nu=1$.
$
P^sigma_nu V^nu_parallel~~=~~ V^sigma_parallel +U^sigma U_sigma V^sigma_parallel qquad qquad [U^sigma U_sigma=-1]\
quadqquad=~~ V^sigma_parallel - V^sigma_parallel \
quadqquad=~~0 quadcheckmark
$
That looks good, now let's check when $sigmaneqnu$ and $delta^sigma_nu=0$.
$
P^sigma_nu V^nu_parallel~~= ~~0 +U^sigma U_nu V^nu_parallel qquad qquad~~~ [Vparallel Uimplies U_nu V^nuneq0]\
quadqquad~neq~~ 0 qquad qquad qquad qquad~~~~~~ [mathrm{contradicts~Eq.}(2)]
$
It looks like I should totally disregard the case where $sigmaneqnu$ but I don't see why.
tensors
asked Aug 30 '16 at 20:14
Jonathan TookerJonathan Tooker
62
$endgroup$
Hi my question is about this orthogonal projection tensor $P^sigma_nuequivdelta^sigma_nu+U^sigma U_nu$.
It should have the following properties when $V,W$ are vectors parallel and perpendicular to U, and $U$ is a 4-velocity in spacetime. These equations come from (1.121) and (1.122) in Sean Carroll's gravity book.
$
P^sigma_nu V^nu_parallel ~~= ~~0
$
$
P^sigma_nu W^nu_perp ~~=~~ W^sigma_perp
$
However, when I test for the first property I do not get the correct answer. I suppose I am messing up how I am using the Kronecker delta but I'm not sure how. Lets operate on and a vector $V$ that is parallel to $U$ to get the "equal zero" condition above.
$
P^sigma_nu V^nu_parallel ~~=~~delta^sigma_nu V^nu_parallel+U^sigma U_nu V^nu_parallel
$
If $sigma=nu$ then $delta^sigma_nu=1$.
$
P^sigma_nu V^nu_parallel~~=~~ V^sigma_parallel +U^sigma U_sigma V^sigma_parallel qquad qquad [U^sigma U_sigma=-1]\
quadqquad=~~ V^sigma_parallel - V^sigma_parallel \
quadqquad=~~0 quadcheckmark
$
That looks good, now let's check when $sigmaneqnu$ and $delta^sigma_nu=0$.
$
P^sigma_nu V^nu_parallel~~= ~~0 +U^sigma U_nu V^nu_parallel qquad qquad~~~ [Vparallel Uimplies U_nu V^nuneq0]\
quadqquad~neq~~ 0 qquad qquad qquad qquad~~~~~~ [mathrm{contradicts~Eq.}(2)]
$
It looks like I should totally disregard the case where $sigmaneqnu$ but I don't see why.
tensors
tensors
asked Aug 30 '16 at 20:14
Jonathan TookerJonathan Tooker
62
asked Aug 30 '16 at 20:14
Jonathan TookerJonathan Tooker
62
asked Aug 30 '16 at 20:14
Jonathan TookerJonathan Tooker
62
asked Aug 30 '16 at 20:14
Jonathan TookerJonathan Tooker
62
asked Aug 30 '16 at 20:14
Jonathan TookerJonathan Tooker
62
62
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Talking about $sigma=nu$ and $neq nu$ are ill defined. $nu$ is a dummy index in the Einstein summation convention so you are, at best, describing terms in the sum. The real cases are about the inner product $V^nu U_nu$, because $delta^sigma_nu V^nu = V^sigma$ always.
answered Aug 30 '16 at 20:19
Sean LakeSean Lake
1,268517
$endgroup$
$begingroup$
Thank you. I am just confused how the delta in the polynomial can also control the indices in the term that it is not multiplied with. Could you say a little more? That seem to violate the interpretation of the delta as a tensor. I am reviewing and I don't remember too well. It's just a rule that if a delta appears anywhere, it contracts the index in $delta^sigma_nu V^nu=V^sigma$ and ALSO all it changes all the other instances of $V^nu$? That is hard for me to wrap my head around.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 20:27
$begingroup$
Nope. The other $nu$ is contracted with the $U$. So, it works like this:$$begin{align}P_nu^sigma &= delta_nu^sigma + U^sigma U_nu\ P_nu^sigma V^nu &= (delta_nu^sigma + U^sigma U_nu)V^nu\ & = delta_nu^sigma V^nu + U^sigma U_nu V^nu \ & = V^sigma + U^sigma U_nu V^nu. end{align}$$ So, the only question is how $V$ is related to $U$.
$endgroup$
– Sean Lake
Aug 30 '16 at 21:18
$begingroup$
Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^sigma_nu V^nu=V^sigma$?
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 21:41
$begingroup$
The thing is that $V$ is not always parallel to $U$. For a general vector, $V$, you can decompose it into a part that's parallel and a part that's perpendicular to $U$: $V = V_{||} + V_{perp},$ where the defining equation of $V_{||}$ is $V_{||}^mu = a U^mu$ for some constant $a$. So: how do we find $a$? Once we have $a$, how do we construct $V_{perp}$? What happens to the equation when we plug this broken down version of $V$ into it?
$endgroup$
– Sean Lake
Aug 30 '16 at 21:51
$begingroup$
LOL I'm all confused. I think I got it now despite the error in my comment above. My previous comment is so wrong! I completely forgot what I was even talking about. Thanks again. $PV=0$ is what I was looking for, not $PV=V$.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 22:00
add a comment |
$begingroup$
I think what you are missing is the following: by applying the operator $P^{sigma}_{nu}$ to the vector V this is what you are actually doing:
begin{equation*}
sum_{nu=0}^{3}P^{sigma}_{nu}V^{nu}=sum_{nu=0}^{3}(delta^{sigma}_{nu}+U^{sigma}U_{nu})V^{nu}
end{equation*}
From which follows:
begin{equation*}
P^{sigma}_{nu}V^{nu}=V^{sigma}-V^{sigma}
=0end{equation*}
As you correctly stated in the beginning of your question. The point is that $nu$ becomes a dummy index only $textbf{after}$ the operator is contracted on the vector.
answered Jul 21 '18 at 9:56
user9862720user9862720
1
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Talking about $sigma=nu$ and $neq nu$ are ill defined. $nu$ is a dummy index in the Einstein summation convention so you are, at best, describing terms in the sum. The real cases are about the inner product $V^nu U_nu$, because $delta^sigma_nu V^nu = V^sigma$ always.
answered Aug 30 '16 at 20:19
Sean LakeSean Lake
1,268517
$endgroup$
$begingroup$
Thank you. I am just confused how the delta in the polynomial can also control the indices in the term that it is not multiplied with. Could you say a little more? That seem to violate the interpretation of the delta as a tensor. I am reviewing and I don't remember too well. It's just a rule that if a delta appears anywhere, it contracts the index in $delta^sigma_nu V^nu=V^sigma$ and ALSO all it changes all the other instances of $V^nu$? That is hard for me to wrap my head around.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 20:27
$begingroup$
Nope. The other $nu$ is contracted with the $U$. So, it works like this:$$begin{align}P_nu^sigma &= delta_nu^sigma + U^sigma U_nu\ P_nu^sigma V^nu &= (delta_nu^sigma + U^sigma U_nu)V^nu\ & = delta_nu^sigma V^nu + U^sigma U_nu V^nu \ & = V^sigma + U^sigma U_nu V^nu. end{align}$$ So, the only question is how $V$ is related to $U$.
$endgroup$
– Sean Lake
Aug 30 '16 at 21:18
$begingroup$
Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^sigma_nu V^nu=V^sigma$?
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 21:41
$begingroup$
The thing is that $V$ is not always parallel to $U$. For a general vector, $V$, you can decompose it into a part that's parallel and a part that's perpendicular to $U$: $V = V_{||} + V_{perp},$ where the defining equation of $V_{||}$ is $V_{||}^mu = a U^mu$ for some constant $a$. So: how do we find $a$? Once we have $a$, how do we construct $V_{perp}$? What happens to the equation when we plug this broken down version of $V$ into it?
$endgroup$
– Sean Lake
Aug 30 '16 at 21:51
$begingroup$
LOL I'm all confused. I think I got it now despite the error in my comment above. My previous comment is so wrong! I completely forgot what I was even talking about. Thanks again. $PV=0$ is what I was looking for, not $PV=V$.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 22:00
add a comment |
$begingroup$
Talking about $sigma=nu$ and $neq nu$ are ill defined. $nu$ is a dummy index in the Einstein summation convention so you are, at best, describing terms in the sum. The real cases are about the inner product $V^nu U_nu$, because $delta^sigma_nu V^nu = V^sigma$ always.
answered Aug 30 '16 at 20:19
Sean LakeSean Lake
1,268517
$endgroup$
$begingroup$
Thank you. I am just confused how the delta in the polynomial can also control the indices in the term that it is not multiplied with. Could you say a little more? That seem to violate the interpretation of the delta as a tensor. I am reviewing and I don't remember too well. It's just a rule that if a delta appears anywhere, it contracts the index in $delta^sigma_nu V^nu=V^sigma$ and ALSO all it changes all the other instances of $V^nu$? That is hard for me to wrap my head around.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 20:27
$begingroup$
Nope. The other $nu$ is contracted with the $U$. So, it works like this:$$begin{align}P_nu^sigma &= delta_nu^sigma + U^sigma U_nu\ P_nu^sigma V^nu &= (delta_nu^sigma + U^sigma U_nu)V^nu\ & = delta_nu^sigma V^nu + U^sigma U_nu V^nu \ & = V^sigma + U^sigma U_nu V^nu. end{align}$$ So, the only question is how $V$ is related to $U$.
$endgroup$
– Sean Lake
Aug 30 '16 at 21:18
$begingroup$
Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^sigma_nu V^nu=V^sigma$?
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 21:41
$begingroup$
The thing is that $V$ is not always parallel to $U$. For a general vector, $V$, you can decompose it into a part that's parallel and a part that's perpendicular to $U$: $V = V_{||} + V_{perp},$ where the defining equation of $V_{||}$ is $V_{||}^mu = a U^mu$ for some constant $a$. So: how do we find $a$? Once we have $a$, how do we construct $V_{perp}$? What happens to the equation when we plug this broken down version of $V$ into it?
$endgroup$
– Sean Lake
Aug 30 '16 at 21:51
$begingroup$
LOL I'm all confused. I think I got it now despite the error in my comment above. My previous comment is so wrong! I completely forgot what I was even talking about. Thanks again. $PV=0$ is what I was looking for, not $PV=V$.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 22:00
add a comment |
$begingroup$
Talking about $sigma=nu$ and $neq nu$ are ill defined. $nu$ is a dummy index in the Einstein summation convention so you are, at best, describing terms in the sum. The real cases are about the inner product $V^nu U_nu$, because $delta^sigma_nu V^nu = V^sigma$ always.
answered Aug 30 '16 at 20:19
Sean LakeSean Lake
1,268517
$endgroup$
Talking about $sigma=nu$ and $neq nu$ are ill defined. $nu$ is a dummy index in the Einstein summation convention so you are, at best, describing terms in the sum. The real cases are about the inner product $V^nu U_nu$, because $delta^sigma_nu V^nu = V^sigma$ always.
answered Aug 30 '16 at 20:19
Sean LakeSean Lake
1,268517
answered Aug 30 '16 at 20:19
Sean LakeSean Lake
1,268517
answered Aug 30 '16 at 20:19
Sean LakeSean Lake
1,268517
answered Aug 30 '16 at 20:19
Sean LakeSean Lake
1,268517
1,268517
$begingroup$
Thank you. I am just confused how the delta in the polynomial can also control the indices in the term that it is not multiplied with. Could you say a little more? That seem to violate the interpretation of the delta as a tensor. I am reviewing and I don't remember too well. It's just a rule that if a delta appears anywhere, it contracts the index in $delta^sigma_nu V^nu=V^sigma$ and ALSO all it changes all the other instances of $V^nu$? That is hard for me to wrap my head around.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 20:27
$begingroup$
Nope. The other $nu$ is contracted with the $U$. So, it works like this:$$begin{align}P_nu^sigma &= delta_nu^sigma + U^sigma U_nu\ P_nu^sigma V^nu &= (delta_nu^sigma + U^sigma U_nu)V^nu\ & = delta_nu^sigma V^nu + U^sigma U_nu V^nu \ & = V^sigma + U^sigma U_nu V^nu. end{align}$$ So, the only question is how $V$ is related to $U$.
$endgroup$
– Sean Lake
Aug 30 '16 at 21:18
$begingroup$
Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^sigma_nu V^nu=V^sigma$?
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 21:41
$begingroup$
The thing is that $V$ is not always parallel to $U$. For a general vector, $V$, you can decompose it into a part that's parallel and a part that's perpendicular to $U$: $V = V_{||} + V_{perp},$ where the defining equation of $V_{||}$ is $V_{||}^mu = a U^mu$ for some constant $a$. So: how do we find $a$? Once we have $a$, how do we construct $V_{perp}$? What happens to the equation when we plug this broken down version of $V$ into it?
$endgroup$
– Sean Lake
Aug 30 '16 at 21:51
$begingroup$
LOL I'm all confused. I think I got it now despite the error in my comment above. My previous comment is so wrong! I completely forgot what I was even talking about. Thanks again. $PV=0$ is what I was looking for, not $PV=V$.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 22:00
add a comment |
$begingroup$
Thank you. I am just confused how the delta in the polynomial can also control the indices in the term that it is not multiplied with. Could you say a little more? That seem to violate the interpretation of the delta as a tensor. I am reviewing and I don't remember too well. It's just a rule that if a delta appears anywhere, it contracts the index in $delta^sigma_nu V^nu=V^sigma$ and ALSO all it changes all the other instances of $V^nu$? That is hard for me to wrap my head around.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 20:27
$begingroup$
Nope. The other $nu$ is contracted with the $U$. So, it works like this:$$begin{align}P_nu^sigma &= delta_nu^sigma + U^sigma U_nu\ P_nu^sigma V^nu &= (delta_nu^sigma + U^sigma U_nu)V^nu\ & = delta_nu^sigma V^nu + U^sigma U_nu V^nu \ & = V^sigma + U^sigma U_nu V^nu. end{align}$$ So, the only question is how $V$ is related to $U$.
$endgroup$
– Sean Lake
Aug 30 '16 at 21:18
$begingroup$
Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^sigma_nu V^nu=V^sigma$?
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 21:41
$begingroup$
The thing is that $V$ is not always parallel to $U$. For a general vector, $V$, you can decompose it into a part that's parallel and a part that's perpendicular to $U$: $V = V_{||} + V_{perp},$ where the defining equation of $V_{||}$ is $V_{||}^mu = a U^mu$ for some constant $a$. So: how do we find $a$? Once we have $a$, how do we construct $V_{perp}$? What happens to the equation when we plug this broken down version of $V$ into it?
$endgroup$
– Sean Lake
Aug 30 '16 at 21:51
$begingroup$
LOL I'm all confused. I think I got it now despite the error in my comment above. My previous comment is so wrong! I completely forgot what I was even talking about. Thanks again. $PV=0$ is what I was looking for, not $PV=V$.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 22:00
$begingroup$
Thank you. I am just confused how the delta in the polynomial can also control the indices in the term that it is not multiplied with. Could you say a little more? That seem to violate the interpretation of the delta as a tensor. I am reviewing and I don't remember too well. It's just a rule that if a delta appears anywhere, it contracts the index in $delta^sigma_nu V^nu=V^sigma$ and ALSO all it changes all the other instances of $V^nu$? That is hard for me to wrap my head around.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 20:27
$begingroup$
Thank you. I am just confused how the delta in the polynomial can also control the indices in the term that it is not multiplied with. Could you say a little more? That seem to violate the interpretation of the delta as a tensor. I am reviewing and I don't remember too well. It's just a rule that if a delta appears anywhere, it contracts the index in $delta^sigma_nu V^nu=V^sigma$ and ALSO all it changes all the other instances of $V^nu$? That is hard for me to wrap my head around.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 20:27
$begingroup$
Nope. The other $nu$ is contracted with the $U$. So, it works like this:$$begin{align}P_nu^sigma &= delta_nu^sigma + U^sigma U_nu\ P_nu^sigma V^nu &= (delta_nu^sigma + U^sigma U_nu)V^nu\ & = delta_nu^sigma V^nu + U^sigma U_nu V^nu \ & = V^sigma + U^sigma U_nu V^nu. end{align}$$ So, the only question is how $V$ is related to $U$.
$endgroup$
– Sean Lake
Aug 30 '16 at 21:18
$begingroup$
Nope. The other $nu$ is contracted with the $U$. So, it works like this:$$begin{align}P_nu^sigma &= delta_nu^sigma + U^sigma U_nu\ P_nu^sigma V^nu &= (delta_nu^sigma + U^sigma U_nu)V^nu\ & = delta_nu^sigma V^nu + U^sigma U_nu V^nu \ & = V^sigma + U^sigma U_nu V^nu. end{align}$$ So, the only question is how $V$ is related to $U$.
$endgroup$
– Sean Lake
Aug 30 '16 at 21:18
$begingroup$
Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^sigma_nu V^nu=V^sigma$?
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 21:41
$begingroup$
Thank you for your patience. $V$ is parallel to $U$. How does that ensure that the second term is zero giving the expected $P^sigma_nu V^nu=V^sigma$?
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 21:41
$begingroup$
The thing is that $V$ is not always parallel to $U$. For a general vector, $V$, you can decompose it into a part that's parallel and a part that's perpendicular to $U$: $V = V_{||} + V_{perp},$ where the defining equation of $V_{||}$ is $V_{||}^mu = a U^mu$ for some constant $a$. So: how do we find $a$? Once we have $a$, how do we construct $V_{perp}$? What happens to the equation when we plug this broken down version of $V$ into it?
$endgroup$
– Sean Lake
Aug 30 '16 at 21:51
$begingroup$
The thing is that $V$ is not always parallel to $U$. For a general vector, $V$, you can decompose it into a part that's parallel and a part that's perpendicular to $U$: $V = V_{||} + V_{perp},$ where the defining equation of $V_{||}$ is $V_{||}^mu = a U^mu$ for some constant $a$. So: how do we find $a$? Once we have $a$, how do we construct $V_{perp}$? What happens to the equation when we plug this broken down version of $V$ into it?
$endgroup$
– Sean Lake
Aug 30 '16 at 21:51
$begingroup$
LOL I'm all confused. I think I got it now despite the error in my comment above. My previous comment is so wrong! I completely forgot what I was even talking about. Thanks again. $PV=0$ is what I was looking for, not $PV=V$.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 22:00
$begingroup$
LOL I'm all confused. I think I got it now despite the error in my comment above. My previous comment is so wrong! I completely forgot what I was even talking about. Thanks again. $PV=0$ is what I was looking for, not $PV=V$.
$endgroup$
– Jonathan Tooker
Aug 30 '16 at 22:00
add a comment |
$begingroup$
I think what you are missing is the following: by applying the operator $P^{sigma}_{nu}$ to the vector V this is what you are actually doing:
begin{equation*}
sum_{nu=0}^{3}P^{sigma}_{nu}V^{nu}=sum_{nu=0}^{3}(delta^{sigma}_{nu}+U^{sigma}U_{nu})V^{nu}
end{equation*}
From which follows:
begin{equation*}
P^{sigma}_{nu}V^{nu}=V^{sigma}-V^{sigma}
=0end{equation*}
As you correctly stated in the beginning of your question. The point is that $nu$ becomes a dummy index only $textbf{after}$ the operator is contracted on the vector.
answered Jul 21 '18 at 9:56
user9862720user9862720
1
$endgroup$
add a comment |
$begingroup$
I think what you are missing is the following: by applying the operator $P^{sigma}_{nu}$ to the vector V this is what you are actually doing:
begin{equation*}
sum_{nu=0}^{3}P^{sigma}_{nu}V^{nu}=sum_{nu=0}^{3}(delta^{sigma}_{nu}+U^{sigma}U_{nu})V^{nu}
end{equation*}
From which follows:
begin{equation*}
P^{sigma}_{nu}V^{nu}=V^{sigma}-V^{sigma}
=0end{equation*}
As you correctly stated in the beginning of your question. The point is that $nu$ becomes a dummy index only $textbf{after}$ the operator is contracted on the vector.
answered Jul 21 '18 at 9:56
user9862720user9862720
1
$endgroup$
add a comment |
$begingroup$
I think what you are missing is the following: by applying the operator $P^{sigma}_{nu}$ to the vector V this is what you are actually doing:
begin{equation*}
sum_{nu=0}^{3}P^{sigma}_{nu}V^{nu}=sum_{nu=0}^{3}(delta^{sigma}_{nu}+U^{sigma}U_{nu})V^{nu}
end{equation*}
From which follows:
begin{equation*}
P^{sigma}_{nu}V^{nu}=V^{sigma}-V^{sigma}
=0end{equation*}
As you correctly stated in the beginning of your question. The point is that $nu$ becomes a dummy index only $textbf{after}$ the operator is contracted on the vector.
answered Jul 21 '18 at 9:56
user9862720user9862720
1
$endgroup$
I think what you are missing is the following: by applying the operator $P^{sigma}_{nu}$ to the vector V this is what you are actually doing:
begin{equation*}
sum_{nu=0}^{3}P^{sigma}_{nu}V^{nu}=sum_{nu=0}^{3}(delta^{sigma}_{nu}+U^{sigma}U_{nu})V^{nu}
end{equation*}
From which follows:
begin{equation*}
P^{sigma}_{nu}V^{nu}=V^{sigma}-V^{sigma}
=0end{equation*}
As you correctly stated in the beginning of your question. The point is that $nu$ becomes a dummy index only $textbf{after}$ the operator is contracted on the vector.
answered Jul 21 '18 at 9:56
user9862720user9862720
1
answered Jul 21 '18 at 9:56
user9862720user9862720
1
answered Jul 21 '18 at 9:56
user9862720user9862720
1
answered Jul 21 '18 at 9:56
user9862720user9862720
1
1
add a comment |
add a comment |
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