Understanding the Four Momentum of a Relativistic Field

[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?

 

[pervect]

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.

 

[olgranpappy]

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?

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.

 

[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.

 

[olgranpappy]

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

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

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

 

[pmb_phy]

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.

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

 

[jostpuur]

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

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

 

[olgranpappy]

ah, I see.

 

[samalkhaiat]

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).

If

$$\partial_{a}T^{ab} = 0$$

then, it is an easy exercise to prove that

$$P^{a} = \int dx^{3} T^{0a}$$

is a 4-vector.

regards

sam

 

[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?

 

[samalkhaiat]

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.

In post #23 of

https://www.physicsforums.com/showthread.php?t=114620&page=2

I proved that

$$Q = \int dx^{3} \ J_{0}(x)$$

is Lorentz scalar, if the current J is conserved, i.e.,

$$\partial_{a}J^{a} = 0$$

Applying my method on the conserved stress tensor gives you the result you are after.
Indeed you can (by the same method) show that

$$J^{a_{1}a_{2}...a_{n-1}} = \int dx^{3} \ T^{0a_{1}a_{2}...a_{n-1}}$$

is a (n-1)-rank tensor, if the n-rank tensor T is conserved,i.e.,

$$\partial_{a}T^{aa_{1}a_{2}...a_{n-1}} = 0$$

regards

sam

 

[pervect]

If

$$\partial_{a}T^{ab} = 0$$

then, it is an easy exercise to prove that

$$P^{a} = \int dx^{3} T^{0a}$$

is a 4-vector.

regards

sam

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 transferred 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.

 

[pervect]

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.

 

[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.

 

[samalkhaiat]

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 $T^{ab}=0$.

There is no physics involved in the proof. It is just a mathematical theorem. So, if your physical system (whatever it is) possesses a conserved 2nd rank tensor;
$$\partial_{a}T^{ab}= 0$$

then

$$P^{b}= \int dx^{3} T^{0b}$$

is time-independent Lorentz scalar.

PROOF:

Let $C_{b}$ be an arbitrary constant vector. Since T is a tensor, then

$$J^{a} = T^{ab}C_{b}$$

is a vector. Since T is conserved and C is constant, then;

$$\partial_{a}J^{a} = 0$$

In
https://www.physicsforums.com/showthread.php?t=114620&page=2

I proved that

$$\partial .J = 0 \ \Rightarrow \int dx^{3}J^{0}= \mbox{scalar}$$

i.e.,

$$C_{b} \int dx^{3} T^{0b} \equiv C_{b}P^{b} = \mbox{scalar}$$

Thus, $P^{a}$ is a vector. qed.

Where is the physics? There is non, it is just math.

regards

sam

 

[olgranpappy]

There is no physics involved in the proof. It is just a mathematical theorem. So, if your physical system (whatever it is) possesses a conserved 2nd rank tensor;
$$\partial_{a}T^{ab}= 0$$

then...

But, I think that what the OP is asking for is whether or not you can start from the end of his post (#14) and show in a simple way that Q' = Q...

 

[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'

 

[pervect]

I'm sure that there's more elegant ways of talking about this, but here's my $.02.

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.

 

[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.

 

[jostpuur]

Do you agree with this?

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

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.

 

[pervect]

I think you're on the right track - it might be better to think of the charge density to be a tensor density rather than a tensor. This approach gives one a more formal justification for saying that the volume element transforms as sqrt(-g), g being the determinant of g_uv in some particular basis (with -+++ signature giving the minus sign).

I don't have any textbooks that address this particular issue in detail (they all sort of skim over it).

I'm more comfortable talking about some of the physical situations than going into the details of the math (even though the math isn't that hard, really). Consider a ring of highly conductive wire with a current flowing through it, for instance.

The ring will have a total charge, which will not depend on the frame of reference. Because charges come physically in discrete packets, we can even count them.

When we look at the charge density, however we can see that in the lab frame the charges appear evenly distributed (with a high conductivity there is no electric field in the wire, so there are only magnetic fields) while in some other frames (frames boosted relative to the lab frame) the charge distribution won't be uniform. In the later case a boost has turned the magnetic field into an electric field, - or one might also say that the relativity of simultaneity has caused the charges to "bunch together" in one part of the wire.

Even though the charge distribution changes from uniform to non-uniform after a boost, the total charge of the loop of wire is still invariant (the same before and after a boost) as long as we integrate over the whole wire. Given any frame, we can always assign a location to every charge, and we can count the charges and always get the same number regardless of frame. But we have to count the charges in the whole wire for the argument to work. The number of charges in the "left half" of the wire is not necessarily the same before and after a boost, for instance.

 

[olgranpappy]

Oh, I have another few cents to add:

One can "prove" using a simple picture (space-time diagram) that (as long as there are no "sources" or "sinks") the total charge measured at any fixed time in one frame is equal to the total charge measured at any other fixed time in any other frame related to the first by a boost.

The picture has "two" axes: a time axis running vertically, and a space axis running horizontally.

A further two axes can be drawn for the boosted time and space axes and, most importantly, the line of fixed time in the unboosted frame (a horizontal line) makes an angle of less than 45 degree with the line of fixed time in the boosted frame.

Thus any charged world lines (necessarily having mass) that run through the first line must also run through the second line.

Except for the case of annihilation of two particles of opposite charge, but in this case the sum of the charges of the annihilating particles must be zero so still there is no change in the total charge measured by either observer.

Cheers.

 

[jostpuur]

Yeah, if we choose a dust kind of four current, where the total charge is more or less simply a number of particles in the dust, then it is clear that the total charge is invariant. But a proof based on the transformation properties of the current would help approach the higher rank tensors also.

 

[samalkhaiat]

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.

Sooner or later you will be introduced to them.

We write the most general Lorentz transformations of the coordinates as

$$\bar{x}^{a} = \Lambda^{a}{}_{b}x^{b} \ \ (1)$$

where $\Lambda$ is the Lorentz matrix which contains 3 rotation angles and 3 boosts, we call these 6 numbers the Lorentz group parameters. Infinitesimal L.T. means expanding Lorentz matrix (to 1st order in the parameters) very close to the unit matrix:

$$\Lambda = I + \omega$$

or

$$\Lambda^{a}{}_{b} = \delta^{a}_{b} + \omega^{a}{}_{b}, \ \omega^{a}{}_{b} = - \omega_{b}{}^{a}$$

So, Eq(1) becomes a small deviation from the identity transformations

$$\bar{x}^{a} = x^{a} + \omega^{a}{}_{b}x^{b}$$

Similarly, the infinitesimal LT for (contravariant) vector is given by

$$\bar{J}^{a}(\bar{x}) = J^{a}(x) + \omega^{a}{}_{b}J^{b}(x)$$

This can be written as

$$\bar{J}^{a}(x + \omega x) = J^{a} + \partial_{b}(\omega^{a}{}_{c}x^{c})J^{b}(x)$$

By expanding the LHS to the 1st order in $\omega$ we find

$$\bar{J}^{a}(x) + \omega^{b}{}_{c}x^{c} \partial_{b}J^{a} = J^{a}(x) + \partial_{b}(\omega^{a}{}_{c}x^{c})J^{b}(x)$$

Using
$$\partial_{b}J^{b}= 0 \ \ \mbox{and} \ \omega^{b}_{b} = 0$$

we find

$$\bar{J}^{a}(x) - J^{a}(x) = \partial_{b}\left[\omega^{a}{}_{c}x^{c}J^{b} - \omega^{b}{}_{c}x^{c}J^{a}\right] \ \ (2)$$

We write the LHS of this as

$$\delta J^{a}(x) \equiv \bar{J}^{a}(x) - J^{a}(x)$$

which is the functional change of the vector J under infinitesimal LT ( basically it is a Lie derivative). Notice that $\bar{J}^{a}(x)$ is the value of $\bar{J}^{a}$ at the point $\bar{x} = x$; i.e.,

$$\bar{J}^{a}(x) = \bar{J}^{a}(\bar{x})|_{\bar{x} = x} \ \ (3)$$

So, $\delta J^{a}$ compares the values at two different points.

Now a = 0 in Eq(2) gives

$$\delta J^{0}(x) = \partial_{i}\left[x^{c}\omega^{0}{}_{c}J^{i} - x^{c}\omega^{i}{}_{c}J^{0} \right]$$

Integrating and using the fact that J vanishes sufficiently fast at spatial infinity leads to

$$\delta \int d^{3}x J^{0}(x) \equiv \bar{Q} - Q = 0$$

qed

************
Second method [The tube lemma and Klien-Abraham's theorem]

Let us assume that the conserved vector field is confined to a tube in spacetime;i.e., the current vanishes out side the tube. Practically, this fact (for a very wide tube) means that the current vanishes sufficiently fast at spatial infinity.
Let $\Omega_{1}$ and $\Omega_{2}$ be 2 (far apart) hypersurfaces intersected by the tube. Then, together with the wall of the tube, they constitute the boundary of a 4-dimensional domain D. From the 4-dimensional Gauss's theorem we now obtain

$$0 = \int_{D} d^{4}x \partial_{a} J^{a} = \int_{\partial D} dS_{a}J^{a} = \int_{\Omega_{2}} dS_{a}J^{a} - \int_{\Omega_{1}} dS_{a}J^{a}$$

or, if you know forms,

$$0 = \int_{D}d{}^{*}J = \int_{\partial D} {}^{*}J = \int_{\Omega_{2}} {}^{*}J - \int_{\Omega_{1}} {}^{*}J$$

This means,[the tube lemma] A conserved vector field (coclosed 1-form) that vanishes suficiently fast at spatial infinity produces the same flux
$$\int_{\Omega} dS_{a}J^{a} \ \ ( \int_{\Omega} {}^{*}J )$$
through any closed hypersurface $\Omega$ itersected by the tube.

Let us now put some physics into this lemma. Suppose S is an inertial frame of reference, and let $\Omega_{1}$ and $\Omega_{2}$ be 3-dimensional domains contained in space slices relative to S such that $\Omega_{1}$ contains the charge at time $t_{1}$ and similarly for $\Omega_{2}$ at time $t_{2}$;i.e.,

$$Q(t_{1}) = \int_{\Omega_{1}} d^{3}x J^{0}$$
$$Q(t_{2}) = \int_{\Omega_{2}} d^{3}x J^{0}$$

But, according to the Tube Lemma, these are identical;i.e., the charge measured by an observer in S is independent of time (constant).
Now, suppose S and S' are 2 different inertial frames. Then we can take $\Omega_{1}$ to be a domain contained in a space slice relative to S and $\Omega_{2}$ a domain in a space slice relative to S'. Thus, the charge measured in S respectively S', is given by
$$Q = \int_{\Omega_{1}} d^{3}x J^{0}$$
$$Q' = \int_{\Omega_{2}} d^{3}x J^{0}$$

Again by the tube lemma, these are identical. Therefore, the charge is independent of observer too (invariant). This completes the proof of Klein-Abraham's theorem:
If J is a conserved vector field that vanishes sufficiently fast at spatial infinity, then the charge
$$Q = \int d^{3}x J^{0}$$
is independent of time and the observer.


regards

sam

 

[jostpuur]

Define $f(x)=1+x^2$. Now $f(x)-1=O(x^2)$, but that does not mean that f would be constant $f(x)=1$, even though it doesn't change in an infinitesimal translation around origo. I hate to say such simple thing, because everybody knows that, but the idea of the infinitesimal proof is not clear to me, and I doubt if it is clear to anyone else either.

samalkhaiat, I think your (first) calculation is the same one that I did in the posts #19 and #20, but with different notation. Did you notice this

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.

That looks like a problem.

 

[samalkhaiat]

Define $f(x)=1+x^2$. Now $f(x)-1=O(x^2)$, but that does not mean that f would be constant $f(x)=1$, even though it doesn't change in an infinitesimal translation around origo. I hate to say such simple thing, because everybody knows that

If $f(a) = a^{2} + 1$ and $a = 10^{-999999999999}$, then yes
$f(a) = 1$.
We ignore terms that are 2nd and higher order in the infinitesimal parameters not the variables! So, if $f(x) = a^{2}x + ax^{2}$ and a is an infinitesimal parameter, then we can safely work with $f(x) = ax^{2}$.
If you want to know what happens to "your" function under an infinitesimal translation;
$$\bar{x} = x + a$$
then you should write
$$f(x) = f(\bar{x} - a) \equiv \bar{f}(\bar{x})$$
i.e.,
$$\bar{f}(\bar{x}) = (\bar{x} - a)^{2} + 1 = \bar{x}^{2} - 2a \bar{x} + 1 + o(a^{2}) \approx \bar{x}^{2} - 2a \bar{x} + 1$$
at $\bar{x} = x$, we have
$$\bar{f}(x) \approx x^{2} + 1 - 2ax = f(x) - 2ax$$

Thus, the infinitesimal change in the form of "your" function (Lie derivative) is

$$\delta f = \bar{f}(x) - f(x) \approx -2ax$$

and
$$a \partial_{x}\bar{f}(x) = a \partial_{x}f(x) + o(a^{2}) \approx a \partial_{x}f(x)$$
I hope this explains the infinitesimal "stuff".

but the idea of the infinitesimal proof is not clear to me, and I doubt if it is clear to anyone else either.

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.

samalkhaiat, I think your (first) calculation is the same one that I did in the posts #19 and #20, but with different notation.

As contrary to your understanding, it will be shown in the following that there is more to Lorentz invariance than just algebric manipulation on the transformations.

Did you notice this



That looks like a problem.

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

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

In my previous post(#24), we have seen that

$$\bar{J}^{0} - J^{0} = \partial_{i} (...)^{i}$$

Therefore
$$\bar{Q} = Q$$
qed,
or
2) transform the time as in

$$\bar{Q} = \int d^{3}\bar{x} \bar{J}^{0}(\bar{x}, \bar{t}) = \int d^{3}x \bar{J}^{0}(x, \bar{t}) = \int d^{3}x\bar{J}^{0}(x, t + \omega^{0}{}_{j}x^{j})$$

Expanding $\bar{J}^{0}$ to the 1st order in the parameter leads to

$$\bar{Q} = \int d^{3}x \bar{J}^{0}(x,t) + \int d^{3}x \omega^{0}{}_{j}x^{j} \partial_{0}J^{0}(x,t) \ \ (1)$$

From post #24 we have

$$\bar{J}^{0} = J^{0} + \omega^{0}{}_{i}J^{i} + \omega^{0}{}_{j}x^{j} \partial_{i}J^{i} + \partial_{i}(...)^{i}$$

Putting this in Eq(1), we find, after using Gauss's theorem and current conservation;

$$\bar{Q} = \int d^{3}x J^{0}(x,t) + \int d^{3}x \omega^{0}{}_{i}J^{i}(x,t)$$

or

$$\bar{Q} = Q + \int d^{3}x \omega^{0}{}_{i}J^{i}(x,t) \ \ (2)$$

So, Does this equation mean that I am stuck (like you did)? and say that there is a problem?
NO, on the contrary, I will use eq(2) to show you how to "hit two birds by one stone"!

1) Since Q is independent of time, and since this is true in all Lorentz frames, then
from eq(2), we get

$$\omega^{0}{}_{i} \int d^{3}x \partial_{0}J^{i} = 0 \ \ (3)$$

Now, if I, like you do, work with boosts; i.e., if at least one of the

$$\omega^{0}{}_{i} \neq 0$$

then

$$\partial_{0}J^{i} = 0 \ \ (4)$$

I can,now, lend you eq(4) to use in the homotopy derivative and complete your proof! This is the bird number one.

2) To finish off my current proof (bird number two), I reason as follows:
Since Q is independent of time, then full Lorentz invariance follows from the invariance under spatial rotations, i.e., we put;

$$\omega^{0}{}_{i} = 0$$

thus eq(2) becomes

$$\bar{Q} = Q$$
qed

regards

sam

 

[olgranpappy]

[edit: oops... forum's slow, I double posted instead of editting... sorry. copy of post is below]

 

[olgranpappy]

...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.
As contrary to your understanding, it will be shown in the following that there is more to Lorentz invariance than just algebric manipulation on the transformations.



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

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

In my previous post(#24), we have seen that

$$\bar{J}^{0} - J^{0} = \partial_{i} (...)^{i}$$

Therefore
$$\bar{Q} = Q$$
qed,
or
2) transform the time as in

$$\bar{Q} = \int d^{3}\bar{x} \bar{J}^{0}(\bar{x}, \bar{t}) = \int d^{3}x \bar{J}^{0}(x, \bar{t}) = \int d^{3}x\bar{J}^{0}(x, t + \omega^{0}{}_{j}x^{j})$$

Expanding $\bar{J}^{0}$ to the 1st order in the parameter leads to

$$\bar{Q} = \int d^{3}x \bar{J}^{0}(x,t) + \int d^{3}x \omega^{0}{}_{j}x^{j} \partial_{0}J^{0}(x,t) \ \ (1)$$

From post #24 we have

$$\bar{J}^{0} = J^{0} + \omega^{0}{}_{i}J^{i} + \omega^{0}{}_{j}x^{j} \partial_{i}J^{i} + \partial_{i}(...)^{i}$$

Putting this in Eq(1), we find, after using Gauss's theorem and current conservation;

$$\bar{Q} = \int d^{3}x J^{0}(x,t) + \int d^{3}x \omega^{0}{}_{i}J^{i}(x,t)$$

or

$$\bar{Q} = Q + \int d^{3}x \omega^{0}{}_{i}J^{i}(x,t) \ \ (2)$$

So, Does this equation mean that I am stuck (like you did)? and say that there is a problem?
NO, on the contrary, I will use eq(2) to show you how to "hit two birds by one stone"!

1) Since Q is independent of time, and since this is true in all Lorentz frames, then
from eq(2), we get

$$\omega^{0}{}_{i} \int d^{3}x \partial_{0}J^{i} = 0 \ \ (3)$$

Now, if I, like you do, work with boosts; i.e., if at least one of the

$$\omega^{0}{}_{i} \neq 0$$

then

$$\partial_{0}J^{i} = 0 \ \ (4)$$

Thanks for going into more detail, but (4) does not follow from (3). All one can say is that the time-derivative of the dipole moment of the system is time-independent
$$\frac{\partial}{\partial t}\int d^3x J_i=\frac{\partial}{\partial t}\int d^3x x_i -\nabla\cdot\vec j=\frac{\partial^2}{\partial t^2}\int d^3x x_i\rho$$

I can,now, lend you eq(4) to use in the homotopy derivative and complete your proof! This is the bird number one.

2) To finish off my current proof (bird number two), I reason as follows:
Since Q is independent of time, then full Lorentz invariance follows from the invariance under spatial rotations,

This reasoning doesn't seem to make sense. We are explicitly interested in the case of boost and can not only consider rotations. No one has queried the constancy of charge under rotations.

Anyways, let me restate what has already been said in maybe a more understandable form, we just want to show that the boosted charge density when integrated over the some region, all of space at time t according to some observer, gives the same value as the unboosted charge density when integrated over the same region. I.e., we want

$$\delta Q =\int d^3x \delta \rho (\vec x, t)=\int d^3x (\bar\rho(\vec x,t)-\rho(\vec x,t))=0\;.$$For an infintesimal boost

$$\bar\rho(\vec x,t)-\rho(\vec x,t)=(\delta^0_\nu-\omega^0_\nu)(j^\nu(x)+\omega^\alpha_\beta x^\beta\frac{\partial \rho}{\partial x^\alpha}) - \rho(x)$$

and, after some algebra this equals

$$\omega^\alpha_\beta\frac{\partial}{\partial x^\alpha}(\rho(x) x^\beta) - \omega^0_i j^i(x)$$

thus

$$\delta Q = \omega^0_i\int d^3 x\left( x^i\frac{\partial \rho}{\partial t} -j^i \right)$$

but $$\frac{\partial \rho}{\partial t}=-\nabla\cdot\vec j$$ and so that the first term above is equal to the 2nd term after an integration by part and they cancel giving
$$\delta Q=0$$

 

[samalkhaiat]

Thanks for going into more detail, but (4) does not follow from (3).

Ok, it doesn't, but

$$\int d^{3}x \partial_{0}J^{i} = 0 \ \ (4')$$

does. The point is this: The OP had an integral of the form

$$\int d^{3}x \ (au.x) \ \partial_{0}J^{i}(au.x,x)$$

"He", also, treated $au.x$ as the time argument in $J^{i}$. So, his integral vanishes because of (4');

$$t \int d^{3}x \partial_{0}J^{i}(t,x) = 0$$

Otherwise,

$$J^{i}(au.x,x) \equiv f^{i}(x)$$

has no time dependence and, therefore

$$\partial_{0}J^{i} = 0$$

This reasoning doesn't seem to make sense.

It makes perfect sense. Since the charge is not a component in some higher rank (spacetime) tensor (it is just a single-component quantity), and since it is time-independent in all Lorentz frames. Therefore, the invariance of the charge under the full Lorentz group, SO(1,3), follows from the invariance under the rotational subgroup, SO(3), of the Lorentz group SO(1,3).

We are explicitly interested in the case of boost and can not only consider rotations.

This thread is about the covariance/invariance of the quantities $P^{\mu}/Q$ under the Lorentz group SO(1,3). Lorentz invariance/covariance of some quantity means the behaviour of that quantity under ARBITRARY Lorentz transformations NOT just boosts. The 3 boosts transformations represent only half of the story. Lorentz group has 6 parameters, so to study the Lorentz invariance one has to consider 3 boosts plus 3 rotations.

the constancy of charge under rotations.

What does this mean?

For an infintesimal boost

$$\bar\rho(\vec x,t)-\rho(\vec x,t)=(\delta^0_\nu-\omega^0_\nu)(j^\nu(x)+\omega^\alpha_\beta x^\beta\frac{\partial \rho}{\partial x^\alpha}) - \rho(x)$$

Where did you get this from?

$$\omega^\alpha_\beta\frac{\partial}{\partial x^\alpha}(\rho(x) x^\beta) - \omega^0_i j^i(x)$$

thus

$$\delta Q = \omega^0_i\int d^3 x\left( x^i\frac{\partial \rho}{\partial t} -j^i \right)$$

but $$\frac{\partial \rho}{\partial t}=-\nabla\cdot\vec j$$ and so that the first term above is equal to the 2nd term after an integration by part and they cancel giving
$$\delta Q=0$$

Eighteen months ago, I produced (on these forums) a very simple, 3 lines long proof for the invariance of Q under arbitrary lorentz transformations [see link in post #11]. I also reproduced the same proof twice on this thread [post #24 & 26]. So, why are you bothered in doing the "same thing" ?


regards

sam

 

[olgranpappy]

It makes perfect sense.

Unfortunately, I must continue to disagree ...

Since the charge is not a component in some higher rank (spacetime) tensor (it is just a single-component quantity)

in fact, the charge is not a component of any four vector or any higher rank tensor--obviously not--since the whole point of this discussion has been to demonstrate that the charge does not transform (which the component of any tensor regardless of the rank would); the charge is the space integral of a component of a four-vector.

, and since it is time-independent in all Lorentz frames. Therefore, the invariance of the charge under the full Lorentz group, SO(1,3), follows from the invariance under the rotational subgroup, SO(3), of the Lorentz group SO(1,3).

again, I restate my disagreement, it seems like this argument would then hold for *any* time-independent quantity, which must not be the case.

Furthermore, in a previous post I stated that, in an initial frame where the current is zero, the invariance under boosts of the charge follows from the fact that
$$d^3x\to d^3x/\gamma$$
and
$$\rho\to\rho\gamma$$

To say that the above somehow can be derived entirely from rotational invariance doesn't even make sense. There is content to the above equations which can not possibly follow from considering rotations.

 

[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

$$D_{\alpha} e^{\alpha \lambda_k u^k}\Big|_{\alpha = 0} = 0$$

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

$$D_{\alpha} e^{\alpha \lambda_k u^k}\Big|_{\alpha = 0} = 0$$

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