# Covariant and noncovariant Lorentz invariants

1. Mar 1, 2012

### Heirot

It recently came to my attention that there exists two "kinds" of Lorentz invariants: the covariant and the noncovariant ones.

The covariant ones would be Lorentz scalars e.g. fully contracted Lorentz tensors. If one applies the Lorentz transformation to a covariant Lorentz scalar, one would find it is invariant in form and in numerical value.

On the other hand, quantities like EV (volume times energy) are certainly not Lorentz scalars and they are most certainly not form invariant under Lorentz transformations (energy goes to energy and momentum, volume goes to volume and time), yet they appear to be numerically identical in all Lorentz frames. One can show this by noticing that the volume gets contracted: V = V_0/gamma, while the energy is related to the mass by E = gamma * m. (c=1) Therefore EV=mV_0 which is obviously the same number in every Lorentz frame.

I'm still not quite sure what to make of these quantities. None of the standard textbooks explicitly deals with it. Any thoughts?

2. Mar 1, 2012

### bcrowell

Staff Emeritus
EV is not identical in all Lorentz frames. For example, let E be the energy of a certain light wave, and V the volume of George W. Bush's cranium. Then when we transform out of the rest frame of GWB's head, V picks up a factor of 1/gamma, but E changes by the relativistic Doppler shift factor, which is not gamma.

3. Mar 2, 2012

### tom.stoer

Lorentz invariants are not only covariant i.e. change w.r.t. some rep. of the Lorentz group, but they are invariant i.e. they change w.r.t. the trivial rep. of the Lorentz group, i.e. they don't change at all.

So there is only one class of Lorentz invariants, namely invariants.

4. Mar 2, 2012

### Heirot

OK, let's try with another example. Consider Maxwell equation div B = 0. It is true in every Lorentz frame. So, one would conclude that the number div B = div' B' = 0 is Lorentz invariant. But if one Lorentz transforms div B = 0 to another frame, they obtain something like a*(div' B') + b*(d/dt' B' + curl' E') = 0, where a and b are some velocity dependent constants. Therefore, div B is not Lorentz invariant but it has the same numerical value in every frame (zero). Would you agree?

5. Mar 2, 2012

### Bill_K

I believe what Heirot is talking about is the difference between a Lorentz invariant which is a local quantity, i.e. a scalar field, and one which is a global quantity, i.e. the integral of a density.

An example of the latter is the total charge Q, which is obtained by integrating the 4th component of jμ over a hypersurface. Q is invariant in the sense that it doesn't depend on the choice of hypersurface.

6. Mar 2, 2012

### tom.stoer

I think Heirot is talking about manifest and not manifest Lorentz covariance.

$$\nabla B = 0$$

is not manifest Lorentz covariant but

$$dF = 0$$

is.

7. Mar 2, 2012

### Heirot

@Bill K: I'm not sure that's what I'm talking about. Maxwell equation div B should be a local invarinat quantity, although not a scalar field because it doesn't have the transformation properties of a scalar field.

8. Mar 2, 2012

### Heirot

OK, so you agree there exists two types (manifest and non manifest) of invariant quantities. While it's trivial to construct manifest invariants, the question remains whether it is possible to reduce the non manifest invariant (obtained ad hoc) to some manifest invariant expression?

9. Mar 2, 2012

### tom.stoer

No. What I am saying is that you can have manifest covariant equations (with covariant quantities like F or even invariant i.e. scalar quantities) which you can re-write in a not manifest covariant form. But that does not introduce any new quantity! B is already contained in F, so the equation div B = 0 does not introduce a new quantity B.

10. Mar 2, 2012

### Heirot

OK, I agree with that. My final question is: can we write div B = 0 in a manifest covariant way which does not include Faraday's law?

11. Mar 2, 2012

### tom.stoer

no; using forms the covariant equation reads

$$dF = 0$$

$$\partial_\mu F^{\mu\nu} = 0$$

and these equations always contain both Gauss's law for magnetism and the Maxwell–Faraday equation (= Faraday's law of induction)

12. Mar 2, 2012

### Heirot

That's what I thought. Thanks!

13. Mar 2, 2012

### robphy

I think you mean
$dF=0$ (ie, Bianchi) becomes $d_{[a}F_{bc]}=0$ tensorially.

14. Mar 3, 2012

### tom.stoer

oh, of course, sorry, thanks for the correction, it was nonsense! it's the other Maxwell equation, and dF= 0 must read

$$\epsilon^{\mu\nu\lambda\rho}\,\partial_\lambda \, F_{\mu\nu} = 0$$