Delta/metric question (context commutator poincare transf.)

  • Thread starter binbagsss
  • Start date
  • #1
1,232
10

Homework Statement



qq2.jpg

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:
solq2.jpg


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.
 

Attachments

  • solq2.jpg
    solq2.jpg
    38.1 KB · Views: 522
  • qq2.jpg
    qq2.jpg
    20.6 KB · Views: 553

Answers and Replies

  • #2
14,617
11,973
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.
 
  • #3
1,232
10
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 dont get how the delta has been replaced with the metric
 
  • #4
14,617
11,973
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 dont 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.
 
  • #5
1,232
10
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?
 
  • #7
1,232
10
i dont care about the metric? the answer has the metric whereas i have deltas
 
  • #8
14,617
11,973
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.
 
  • #11
nrqed
Science Advisor
Homework Helper
Gold Member
3,764
291

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).
 

Related Threads on Delta/metric question (context commutator poincare transf.)

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
507
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
868
Replies
1
Views
609
  • Last Post
Replies
2
Views
674
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
12
Views
13K
Top