# Boosting the potential

1. Nov 13, 2007

### Mentz114

The contraction of the EM field tensor is Lorentz invariant. Using the standard formulae,
the fields $$\vec{E} = ( E_x, 0, 0 )$$ and $$\vec{B} = (0, B_y, 0)$$ when bosted in the x direction go to

$$\vec{E'} = ( E_x, 0, \gamma\beta B_y )$$
$$\vec{B'} = (0, \gamma B_y, 0)$$

and it is clear that $$E_x^2-B_y^2 = E_z'^2 + E_x'^2-B_y'^2$$.

This potential $$A^{\mu} = (\phi(x), A_x(z), 0, 0 )$$ gives

$$\vec{E} = ( -\partial_x\phi(x), 0, 0 )$$
$$\vec{B} = (0, -\partial_zA_x(z), 0)$$.

I thought that if I boosted the potential as a 4-vector, then calculated the fields again, I
would get the same result. The boosted potential is

$$A'^{\mu} = (\gamma\phi(x)-\gamma\beta A_x(z) , \gamma A_x(z)-\gamma\beta\phi(x), 0, 0 )$$

On recalculating the fields, Ex is multiplied by $$\gamma$$, while the other fields are
correct. So boosting the potential seems to be not Lorentz invariant, or just wrong maybe?

Any references where I might find out more ?

2. Nov 13, 2007

### Chris Hillman

I think this is really a question about comparing the symmetry groups of the Maxwell equations in their standard formulation in terms of EM fields and reformulated in terms of scalar and vector potential. In general, the point symmetry group of a PDE (or system of PDEs) will differ from the point symmetry group of a "potential form" of that PDE. See for example the textbook by Olver, Applications of Lie Groups to Differential Equations.

3. Nov 13, 2007

### Mentz114

Thanks, Chris, but that has gone a mile over my head. I can see though, that the two ways of boosting may represent different physical situations.

4. Nov 13, 2007

### Chris Hillman

I think you missed the point. Introducing the potential means that we now have many ways of describing one and the same physical scenario; there is a many-to-one correspondence between potentials and fields.

Sorry, gotta run, if I get a chance I'll return later today and say more.

Last edited: Nov 13, 2007
5. Nov 13, 2007

### Mentz114

Of course I missed the point, and the rest ! Not to worry, I've probably just made a mistake.
[retrospective edit]

Last edited: Nov 13, 2007
6. Nov 13, 2007

### jostpuur

Looks like a calculation mistake to me.

$$F^{\mu\nu}=\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}$$

If you transform the components of the tensor, which are the components of E and B, to be

$$F'^{\alpha\beta} = \Lambda^{\alpha}{}_{\mu} \Lambda^{\beta}{}_{\nu} F^{\mu\nu}$$

you should get the same result if you first transform the potential to be

$$A'^{\alpha} = \Lambda^{\alpha}{}_{\mu} A^{\mu}$$

and then compute the derivatives using derivative operators

$$\partial'^{\alpha} = \Lambda^{\alpha}{}_{\mu} \partial^{\mu}$$

7. Nov 13, 2007

### Mentz114

Jostpuur:
" ... and then compute the derivatives using derivative operators
$$\partial '^{\alpha} = \Lambda^{\alpha}{}_{\mu} \partial^{\mu}$$ "

Aha. One must boost the differential operators too. Thank you, I'll try it with some simple stuff, although I can see immediately it will result in the correct boost to the rank 2 tensor.

Last edited: Nov 13, 2007
8. Nov 13, 2007

### jostpuur

I almost guessed that it had something to do with this, but I didn't have the time to start calculating it myself. The transformation of the derivative operators comes down to the chain rule, if you try to understand why they transform. Try checking more carefully what happens to the time and position parameters of the potentials when you boost them.

9. Nov 13, 2007

### Mentz114

I suppose if we are mixing co-ordinates in the potential, we must mix them in the operators too, because the operators live in the boosted frame.

Thanks, jost, you've been a help.

Last edited: Nov 13, 2007
10. Nov 13, 2007

### meopemuk

Note that in quantum electrodynamics the potential does not form a 4-vector. Boost transformations are accompanied by gauge transformations

$$A'^{\alpha}(x) = \Lambda^{\alpha}_{\mu} A^{\mu} (\Lambda x)+ \partial_{\mu} \Omega (x, \Lambda)$$

See eq. (5.9.31) in Weinberg's "The quantum theory of fields", vol. 1

Eugene.

11. Nov 13, 2007

### Mentz114

Hi Eugene,

that computes. It clearly needs more than a boost to preserve invariance.

so transforming the differentials is adding a term to the derivatives to satisfy a gauge condition ?

M

12. Nov 13, 2007

### jostpuur

You're welcome.

I just got one QFT exercise done finally, so I could relax now and put couple of minutes to PF again

Here's the thing as concretely as it can be:

If phi transforms into phi' at each fixed point in space time, then we have

$$\phi(t,x) = \phi'(t',x')$$

when (t,x) and (t',x') are describing the same space time point. (This is 1 time dimension plus 1 spatial dimension now)

$$t = \frac{t'+ux'/c^2}{\sqrt{1-u^2/c^2}}$$
$$x = \frac{x'+ut'}{\sqrt{1-u^2/c^2}}$$

So

$$\partial_{t'}\phi'(t',x') = D_{t'}\phi(t,x) = D_{t'}\phi\big(\frac{t'+ux/c^2}{\sqrt{1-u^2/c^2}}, \;\frac{x'+ut'}{\sqrt{1-u^2/c^2}}\big) = \frac{\partial_t + u\partial_x}{\sqrt{1-u^2/c^2}} \phi(t,x)$$

by the chain rule. In other words,

$$\partial_{t'} = \frac{\partial_t + u\partial_x}{\sqrt{1-u^2/c^2}}$$

and similarly also

$$\partial_{x'} = \frac{(u/c^2)\partial_t + \partial_x}{\sqrt{1-u^2/c^2}}$$

It is possible to see that the vector

$$(\frac{1}{c}\partial_t,\;-\partial_x)$$

transforms as a four vector.

Where from does the need for the gauge transformation arise? The interaction term of the EM-potential and the Dirac's field at least is invariant with the usual transformations.

Last edited: Nov 14, 2007
13. Nov 13, 2007

### meopemuk

The non-covariant boost transformation law for the electromagnetic potential is originated from specific boost transformations of photon creation/annihilation operators. Weinberg's book describes all this in detail. For example, on page 250 he writes:

We have thus come to the conclusion that no four-vector field can be constructed from the annihilation and creation operators for a particle of mass zero and helicity $\pm 1$.

The fact that electromagnetic fields do not transform covariantly plays a big role in Weinberg's justification for the "minimal coupling" interaction in QED. On page 251 he writes:

As we will see in more detail in Chapter 8, we will be able to use a field like $a^{\mu}(x)$ as an ingredient in Lorentz-invariant physical theories if the couplings of $a^{\mu}(x)$ are not only formally Lorentz-invariant (that is, invariant under formal Lorentz transformations under which $a^{\mu} \to \Lambda^{\mu}_{\nu} a^{\nu}$), but are also invariant under the 'gauge' transformations $a^{\mu} \to a^{\mu} + \partial_{\mu} \Omega$. This is accomplished by taking the couplings of $a^{\mu}$ to be of the form $a_{\mu}j^{\mu}$, where $j^{\mu}$ is a four-vector current with $\partial _{\mu}j^{\mu} = 0$

In other words, the non-covariant nature of the photon field $a^{\mu}(x)$ seemingly prevents us from using this field in constructing Lorentz-invariant interactions. The only (known) way around this obstacle is to couple this field with the conserved current. To me this is the most sensible explanation of why we need the gauge invariance in particle physics.

Eugene.

14. Nov 13, 2007

### jostpuur

15. Nov 13, 2007

### meopemuk

Hi jostpuur,

Regarding gauge symmetry, I strongly recommend you to read Weinberg's volume 1. In Weinberg's logic, this symmetry is not a fundamental law of nature (as many other authors present), but a rather technical condition which follows from three truly fundamental laws - Lorentz invariance, cluster separability, and charge conservation.

Eugene.

16. Nov 13, 2007

### Mentz114

Jost, the derivation is very interesting thank you. I hope you got the gauge symmetry/charge conservation worked out. That's another good thread.

I've got the Lorentz transformation programmed now and having fun transforming anything that stands still long enough.

M