Delta/metric question (context commutator poincare transf.)

[binbagsss]

Homework Statement

Homework Equations

[/B]
I believe that $\frac{\partial x^u}{\partial x^p} =\delta ^u_p$ (1)

$\implies$ (if $\delta^a_b$ is a tensor, I'm not sure it is?) : $\frac{\partial x_u}{\partial x^p} = g_{au} \delta ^a_p$ (2)

The Attempt at a Solution

[/B]
sol attached:

because partial derivatives commute four of the terms cancel and I agree with the solution to get the remaining two terms the same, however, I have a delta in place where the solution has a metric so instead I would get(using (2)):

$g_{av}\delta^a_u \frac{\partial}{\partial x^p}-g_{ap}\delta^a_u\frac{\partial}{\partial x^v}$

Many thanks for your help.

 

[fresh_42]

The picture already contains the answer. What do you need a metric for in a vector space? It's Euclidean flat, you operate in a tangent space.

 

[binbagsss]

The picture already contains the answer. What do you need a metric for in a vector space? It's Euclidean flat, you operate in a tangent space.

sorry, I don't understand your comment.

I believe we have that $g^u_v=\delta^u_v$ from that $g_{uv}g^{uv}=1$, however this is with one index raised and one lowered whereas the question above has both low, so i don't get how the delta has been replaced with the metric

 

[fresh_42]

sorry, I don't understand your comment.

I believe we have that $g^u_v=\delta^u_v$ from that $g_{uv}g^{uv}=1$, however this is with one index raised and one lowered whereas the question above has both low, so i don't get how the delta has been replaced with the metric

What do you care about the metric? You have the differential operators explicitly given, so you can apply them on a suitable function and calculate the commutators. There is no need for $g\; . \;\frac{\partial x_\mu}{\partial x_\nu}= \delta_{\mu \nu}\,.$ It's an ONS.

 

[binbagsss]

What do you care about the metric? You have the differential operators explicitly given, so you can apply them on a suitable function and calculate the commutators. There is no need for $g\; . \;\frac{\partial x_\mu}{\partial x_\nu}= \delta_{\mu \nu}\,.$ It's an ONS.

one night stand? what is ONS?

 

[fresh_42]

Orthonormal system.

 

[binbagsss]

i don't care about the metric? the answer has the metric whereas i have deltas

 

[fresh_42]

Which answer do you mean? I don't know how the $\eta$ are defined and on Wikipedia it seems, that they use $M_{\mu \nu}=-J_{\mu \nu}$ but that's it. I read the $\eta$ as $\eta_{\mu \nu} = \delta_{\mu \nu}\,.$

A simple calculation will tell. Of course you can assume any basis, but why to complicate things? It's all about the product rule in the end.

 

[binbagsss]

Which answer do you mean?[ /QUOTE] 'sol attached'

 

[fresh_42]

Question answered:

The picture already contains the answer.

Just differentiate $f(x)=f(x_1,\ldots,x_4)$.

 

[nrqed]

Homework Statement

View attachment 227265

Homework Equations

[/B]
I believe that $\frac{\partial x^u}{\partial x^p} =\delta ^u_p$ (1)

$\implies$ (if $\delta^a_b$ is a tensor, I'm not sure it is?) : $\frac{\partial x_u}{\partial x^p} = g_{au} \delta ^a_p$ (2)

The Attempt at a Solution

[/B]
sol attached:View attachment 227264

because partial derivatives commute four of the terms cancel and I agree with the solution to get the remaining two terms the same, however, I have a delta in place where the solution has a metric so instead I would get(using (2)):

$g_{av}\delta^a_u \frac{\partial}{\partial x^p}-g_{ap}\delta^a_u\frac{\partial}{\partial x^v}$

Many thanks for your help.

Be careful, we get a delta if we differentiate, say, $x^\mu$ with respect to $x^\nu$, i.e the indices must both be upstairs (and we get a delta if they are both downstairs). However, here is what to do when they are not both up or down. For example consider$\frac{\partial}{\partial x^\mu} x_\nu = \frac{\partial}{\partial x^\mu} \eta_{\nu \alpha} x^\alpha = \eta_{\nu \alpha} \delta_\mu^\alpha = \eta_{\nu \mu}$

In the end the reason is that the zeroth component $x_0$ has opposite sign from $x^0$ (well, assuming the mostly plus metric, in the other case it is the three others that differ in sign).