Four momentum of a field

1. Aug 17, 2007

jostpuur

Suppose that some relativistic field has energy density and momentum density. Integrating these, we get the total energy E and momentum p of the field. But does it make sense to call (E,p) a four momentum of the field? At quick glance it looks like it would be impossible to derive any transformation properties for this object, because of the simultaneity stuff in relativity. Is that a four vector?

2. Aug 17, 2007

pervect

Staff Emeritus
The energy density and momentum density of a field can be described by the "stress energy tensor" of the field. This stress-energy tensor field is a geometric object. It's not a 4-vector, but 4 dimensional rank 2 tensor, which you can think of as being like a 4x4 matrix if you are not familiar with tensors. (It has 16 components, but because it's symmetric, only 10 of them are unique).

The stress-energy tensor has amongst its components the energy density per unit volume and the momentum density per unit volume, but it also contains additional terms, the pressure terms.

The energy and momentum of a point particle, i.e. a system with zero volume will transform as a 4-vector, but in general the energy and momentum of a system with a nonzero volume will not. Fields have a non-zero volume almost by definition (though possibly there might be some very weird case that I haven't thought of).

The total energy and total momentum of an isolated system will, however, transform as a 4-vector. I believe that you'd have to include the energy momentum of the field and the physical object generating the field to have a closed system, I'd have to work some examples to convince myself fully on this point.

The total energy and total momentum of a non-isolated system that has a nonzero volume is in general not a 4-vector. So if you mark out some arbitrary volume of the field, the energy and momentum contained with that volume will probably not transform as a 4-vector, because it's interacting with the rest of the field and you don't have a closed system.

Last edited: Aug 17, 2007
3. Aug 17, 2007

olgranpappy

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.

4. Aug 18, 2007

jostpuur

I was aware of the fact that $T^{\mu 0}$ does not transform as a four vector, but was wondering about

$$\big(\int d^3x\; T^{00}, \int d^3x\; T^{10}, \int d^3x\; T^{20}, \int d^3x\; T^{30}\big).$$

Well, very nice, if this always is a four vector (for isolated systems). I have difficulty imagining how this could be proven though.

5. Aug 18, 2007

olgranpappy

...and, I just told you the answer in my previous post. What question did you think I was answering?

6. Aug 18, 2007

pmb_phy

Don't worry about it. You're correct. I recall going through this derivation before. I can be found in Ohanian's text on GR.

Pete

7. Aug 18, 2007

jostpuur

It was a response to pervect's post, where he explained about energy momentum tensor. Quotation could have made it clearer of course.

8. Aug 18, 2007

olgranpappy

ah, I see.

9. Aug 19, 2007

samalkhaiat

10. Aug 19, 2007

jostpuur

After a boost, when a total energy of a field is calculated, the integration is carried out along the plane

$$t' = \gamma (t-xu/c^2) = 0.$$

But if I first translate the field before the boost, or choose the origo differently, then the integration domain becomes something different. How could the energy in boosted frame be independent of the original choice of origo?

11. Aug 19, 2007

samalkhaiat

Last edited: Aug 19, 2007
12. Aug 19, 2007

pervect

Staff Emeritus
Of course, $$\partial_{a}T^{ab} = 0$$ for the combined system of particles and fields (see Jackson, page 611 for instance). However, if one considers the field only, and ignores energy and momentum of particles or media with which the field is ineteracting, there can be a non-zero divergence of the stress energy tensor of the field alone. This gives rise to what Jackson calls the "Lorentz force density" of the field. Basically, energy and momentum can be transfered from the fields to the particles (media), and vica versa, so to get conservation one must include both in the stress-energy tensor.

Thus the remarks about how the divergence of the field is zero in a source-free field, but not necessarily zero in a field with sources.

By the way, do you have a reference where the "easy proof" is worked out in detail? I think you are going to have to include some statements that the volume in question is surrounded by a vacuum region to make it a valid proof. This is another place where the idea of an "isolated system" comes in, we can in fact think of an isolated system as one that is surrounded by a vacuum where $T^{ab}=0$.

[add]I see you did post a link while I was working on this post, I'll go look it up.

Last edited: Aug 19, 2007
13. Aug 19, 2007

pervect

Staff Emeritus
It might be helpful to mention the example of a unit cube of material in an environment where it has an isotropic pressure of P.

The total energy-momentum in this cube of material will not transform as a 4-vector.

Let the density of the cube be $\rho$ and the isotropic pressure be P, then in the rest frame of the cube, the volume is unity and

E =$\rho$, p=0

However, upon performing a Lorentz boost by a factor $\beta$ in the x direction with $\gamma = 1/\sqrt{1-\beta^2}$ one finds

$$T^{00} = \gamma^2 ( \rho +\beta^2 P)$$
$$T^{01} = \beta \gamma^2 (\rho + P)$$

see for instance Rindler, Intro to SR 2nd ed, pg 132. (Actually, this was an inverse Lorentz boost, i.e. x' = x + beta t, but that doesn't affect the argument).

Since the volume of the cube is multipled by 1/gamma one has the end result

$$E = \gamma (\rho + \beta^2 P)$$
$$p = \gamma \beta (\rho + P)$$

and E^2-p^2 = $\rho^2 - \beta^2 P^2$

thus E^2-p^2 is not invariant under a boost, and (E,p) is not a 4-vector.

Note that I've used geometric units throughout. I'll add that it's only the component of pressure in the direction of boost that's significant to the result, but it's easier to do the calculation with the pressure being isotropic.

Last edited: Aug 19, 2007
14. Aug 19, 2007

jostpuur

It seems it could be smart for me to deal with the conserving four current first. I don't understand that infinitesimal stuff samalkhaiat does.

So there's a total charge

$$Q=\int d^3x\; j^0(0,x)$$

Suppose we then take a new frame that is boosted with velocity u. If I succeeded in doing all the Lorentz transformation stuff correctly, then the total charge Q' in the boosted frame can be written using the original frame quantities like this.

$$Q' = \int d^3x'\; (j')^0(0,x') = \int d^3x\; \Big( j^0 (x\cdot u/c^2, x) - \frac{u}{c} \cdot j (x\cdot u/c^2, x)\Big)$$

But I cannot see why this should be equal to original Q simply by the continuity equation.

15. Aug 20, 2007

samalkhaiat

Last edited: Aug 20, 2007
16. Aug 20, 2007

olgranpappy

17. Aug 20, 2007

olgranpappy

I think the problem is that we would like to do both the integrals (over d^3x and d^3x') at "a fixed time." But this is not a easy to handle concept because of the boost which mixes up the space and time part as can be seen from the far RHS of post #14.

Let me just introduce a very simple case--not at all the most general--but, perhaps useful. In truth, I can not think of an easy way to explain the general case.

Consider the case of a collection of charges at rest (in some frame). The amount of charge in some small volume $$d^3 x$$ is
$$Q=\rho d^3 x\;.$$

Now, we consider a boost in the z-direction. The boosted coordinate system has one side of the volume (the z-side) Lorentz contracted such that
$$d^3x' = \frac{d^3 x}{\gamma}$$

But, also, since the four-current is a four vector, and since there is no (three) current in the original frame, we have
$$\rho' = \gamma\rho$$

Thus
$$\rho' d^3 x' = \gamma \rho \frac{d^3 x}{\gamma}=\rho d^3 x\;.$$

I.e., Q=Q'

18. Aug 20, 2007

pervect

Staff Emeritus

Direct computation of the E-field of a charge, and Gauss's law, which gives one a way of computing a charge from a surface integral of the field, is one approach to showing that the charge of a moving charged particle is the same as the charge of a stationary charged particle, or that the charge of a system of charged particles is independent of the motion of the individual charges which comprise it.

See for instance http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf

for the E-field of a moving charge, you'll have to do the integral yourself (and it's a messy one, IIRC).

However, I think that this approach is a bit backwards, as knowledge of the invariance of charge with velocity is more fundamental than Maxwell's equations. I'm not sure how to best demonstrate this. While deriving charge conservation from Maxwell's equations is a bit like putting the horse after the cart, it does demosntrate that charge is conserved.

While the charge of a system of particles is independent of the motion of the individual particles, the mass of a system of particles DOES depend on the motion of its constituent particles. For example, a pressure vessel containing a hot gas has a higher mass (by which I mean invariant mass) than the same vessel containing a cold gas, and the only difference between them is that in one case the particles composing the gas are moving, and in the other case they are not.

19. Aug 22, 2007

jostpuur

I believe I know what should be done here, but I didn't manage in performing the calculation all the way.

If we have a function $f:X\to Y$, two manifolds $M_1\subset X$ and $M_2\subset X$, and we want to prove that two integrals

$$\int_{M_1} dx\;f(x) = \int_{M_2} dx\; f(x)$$

are equal, the standard trick is to choose some homotopy between the manifolds, a continuously parametrized manifold $M(\alpha)$ so that $M(0)=M_1$ and $M(1)=M_2$, and then prove

$$D_{\alpha} \int_{M(\alpha)} dx\;f(x) = 0$$ For all $\alpha\in[0,1]$.

Now we want to prove

$$Q=\int d^3x\; j^0(0,x) = \int d^3x\;\Big( j^0(x\cdot u/c, x) - \frac{u}{c}\cdot j(x\cdot u/c, x) \Big) = Q'$$

So we should define

$$Q(\alpha) = \int d^3x\;\Big( j^0(\alpha x\cdot u/c, x) - \frac{\alpha u}{c}\cdot j(\alpha x\cdot u/c, x)\Big)$$

and then prove $D_{\alpha} Q(\alpha) = 0$. The derivative of the integrand is

$$\frac{x\cdot u}{c}\partial_0 j^0(\alpha x\cdot u/c, x) - \frac{u}{c}\cdot j(\alpha x\cdot u/c, x) - \frac{\alpha (x\cdot u)u}{c^2}\cdot\partial_0 j(\alpha x\cdot u/c, x)$$

After a substitution $\partial_0 j^0 = -\nabla\cdot j$ the two first terms can be combined to be

$$-\nabla\cdot\Big(\frac{(x\cdot u)}{c} j(\alpha x\cdot u/c, x)\Big)$$

and the integral of this vanishes. But we are still left with

$$D_{\alpha} Q(\alpha) = -\frac{\alpha}{c^2} \int d^3x\; (x\cdot u) u\cdot \partial_0 j (\alpha x\cdot u/c, x),$$

and I cannot see how this vanishes. Notice, btw, that this term vanishes for $\alpha = 0$, so it doesn't show in the infinitesimal boost.

Last edited: Aug 22, 2007
20. Aug 22, 2007

jostpuur

Do you agree with this?

In boost in the 1-direction, the plane $t'=0$ is $t=ux^1/c^2$. A distance (in 1-direction) between two points in this plane is

$$\Delta (x')^1 = \frac{\Delta x^1 - u\Delta t}{\sqrt{1-u^2/c^2}} = \Delta x^1\sqrt{1- u^2/c^2}.$$

So when the integral in boosted frame is reparametrized as an integral over x (in the non-boosted frame), the integral should be weighted

$$\int d^3x' = \int d^3x\;\sqrt{1-u^2/c^2}.$$

On the other hand at any given point the current density transforms

$$(j')^0 = \frac{j^0 - uj^1/c}{\sqrt{1-u^2/c^2}}.$$

So the gammas cancel, and there it is.

Last edited: Aug 22, 2007