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).