Physics, asked by DNYANEAHWRI1692, 11 months ago

Current and charges conserved in Klein Gordon equation. What are the links between them?

Answers

Answered by arbabali12
0
When we deal with the Klein Gordon Lagrangian, we can find some conserved quantities.

For example, we remark that under a space time translation aμaμ we can find that the quantity: aρTμρaρTμρ is a conserved current where TμρTμρ is the energy impulsion tensor.

That means that ∂μ(aρTμρ)∂μ(aρTμρ).

My questions:

First question:

In fact we can show that ∂μTμρ=0∂μTμρ=0. Could we have guess it without explicitely calculating this quantity? I mean is it obvious that if the 4-current conserved is aρTμρaρTμρ then ∂μTμρ=0∂μTμρ=0?

[edit] By asking the question I just thought it could be because aρ∂μTμρ=0aρ∂μTμρ=0 is true for any, aa thus ∂μTμρ=0∂μTμρ=0. Am I right?

Second question:

We also can prove that there are charged conserved associated with this symmetry.

For example we have : Pμ=∫R3d3xT0μPμ=∫R3d3xT0μ that is a conserved charged (it means that dPμdt=0dPμdt=0).

It is almost the same question : could we have immediately guess from the conserved current that we would have this conserved charge ?

Is it because as the fields go to zero at infinity, ∫div(j⃗ )d3x=0∫div(j→)d3x=0 and as ∂j0∂t=div(j⃗ )∂j0∂t=div(j→), we have ∫∂j0∂td3x∫∂j0∂td3x

So because the field go to zero at infinity we immediately deduce a conserved charge when we have a conserved 4-current.

[edit]: In fact I don't think it is that obvious as we have the fields ϕϕ that goes to zero at infinity but not necessarily the spatial part of the 44-current conserved in a general case. Am I right.

Answered by RockyAk47
0
We also can prove that there are charged conserved associated with this symmetry.

For example we have : Pμ=∫R3d3xT0μPμ=∫R3d3xT0μ that is a conserved charged (it means that dPμdt=0dPμdt=0).

It is almost the same question : could we have immediately guess from the conserved current that we would have this conserved charge ?

Is it because as the fields go to zero at infinity, ∫div(j⃗ )d3x=0∫div(j→)d3x=0 and as ∂j0∂t=div(j⃗ )∂j0∂t=div(j→), we have∫∂j0∂td3x∫∂j0∂td3x

So because the field go to zero at infinity we immediately deduce a conserved charge when we have a conserved 4-current?

[edit]: In fact I don't think it is that obvious as we have the fields ϕϕ that goes to zero at infinity but not necessarily the spatial part of the 44-current conserved in a general case.

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