Understanding the Four Momentum of a Relativistic Field

[jostpuur]

Look, don't blame the "infinitesimal proof"! The local structure of any Lie group is determined by studying the group for values of the parameters in the neighbourhood of the identity; i.e., infinitesimal transformations. This is the essential feature of all Lie group including Lorentz group.

I understand that when a members of a group are written in a form e^{\lambda_k u^k} where \lambda_k are some constant objects, and u^k the parameters that are being mapped into the group by this expression, then for example

<br /> D_{\alpha} e^{\alpha \lambda_k u^k}\Big|_{\alpha = 0} = 0<br />

will imply e^{\lambda_k u^k} = 1 for all u^k, and in this sense the infinitesimal behaviour controls the whole group.

Now I want to prove that D_{\alpha} Q(\alpha)=0 for all \alpha\in [0,1], and I don't see how proving this for \alpha=0 only could be sufficient. I don't see how this Q could be even transformed by some explicit representation of some Lie group.

You are missing the essential point and that is : In all Lorentz frames, the charge is independent of time.
So, you could either
1) fix the time from the start and write

\bar{Q} = \int d^{3}\bar{x} \bar{J}^{0}(\bar{x}, \bar{t}) = \int d^{3}x \bar{J}^{0}(x, \bar{t}) = \int d^{3}x \bar{J}^{0}(x;T)

and

Q = \int d^{3}x J^{0}(x;T)

thus

\bar{Q} - Q = \int d^{3}x \left( \bar{J}^{0}(x;T) - J^{0}(x;T) \right)

You are integrating over some three dimensional subsets of the four dimensional spacetime. What are these integration domains?

 

[jostpuur]

J. D. Jackson addresses this very issue at the bottom of page 607 in his book "Classical Electrodynamics (Third Edition)."

The gist being that, for source free fields, E and \vec P do transform properly. Cheers.

Jackson does not prove this, but only mentions it. Does anyone know if the proof relies only on the transformation properties of the fields, or does it use the equations of motion too? That is relevant for the problem.

 

[samalkhaiat]

I understand that when a members of a group are written in a form e^{\lambda_k u^k} where \lambda_k are some constant objects, and u^k the parameters that are being mapped into the group by this expression, then for example

<br /> D_{\alpha} e^{\alpha \lambda_k u^k}\Big|_{\alpha = 0} = 0<br />

will imply e^{\lambda_k u^k} = 1 for all u^k, and in this sense the infinitesimal behaviour controls the whole group.

I think, you need to read about the properties of Lie groups and Algebras; start by looking at the importance of infinitesimal transformations for deriving Lie Agebras; see post#2 and post#3 in

https://www.physicsforums.com/showthread.php?t=172461

You are integrating over some three dimensional subsets of the four dimensional spacetime. What are these integration domains?

Ok, I thought you would understand the reason for ignoring the Jacobian! Let us start again; the variation of any integral is made of the sum of the variation of the integrand and of the variation in the region of integration;

\delta \int d^{3}x \ J^{0}(x,T) = \int d^{3}x \ \delta J^{0} + \int \delta (d^{3}x) \ J^{0}(x,T)

The first integral vanishes because of Gauss' theorem (we saw that in many posts), so we are left with

\delta Q = \int \delta (d^{3}x) \ J^{0}(x,T)

At fixed time (t = T), we have

\bar{x}^{i} = x^{i} + \omega^{i}{}_{0}T + \omega^{i}{}_{j}x^{j}

And

\frac{\partial \bar{x}^{i}}{\partial x^{j}} = \delta^{i}{}_{j} + \omega^{i}{}_{j}

Thus the Jacobian becomes identical to that of the O(3) transformations;

\mathcal{J}(\frac{\bar{x}}{x}) = | \delta^{i}_{j} + \omega^{i}{}_{j}| \approx 1 + Tr(\omega) = 1

And

d^{3}\bar{x} = \mathcal{J}(\frac{\bar{x}}{x})d^{3}x \approx d^{3}x

Therefore the variation in the region of integration; \delta (d^{3}x) = 0 and we arrive at \delta Q = 0.

sorry for the late reply.


regargs

sam