# Electromagnetic 4-vectors

1. May 21, 2005

### Mortimer

(See thread "Relativistic velocity in the time dimension", post 11)

I delved into this some time ago. An interesting source I found in Feynman's Lectures on Physics, volume II, in particular because of his non-standard way of approaching these things. Do you have suggestions for additional sources?

While studying this matter I wondered if the scalar component $$\phi$$ for the electric potential in the 4-vector for electromagnetic potential $$(\phi, \vec{A})$$ could not simply be regarded a vector potential from the perspective of 4D as a result of the temporal component $$\gamma c$$ in the 4-velocity vector $$\gamma(c,\vec{v})$$ for the charge? After all, this temporal velocity component would present the charge as a "current" in the time dimension (a "temporal current" if you wish), which would logically allow the association of a vector potential with it.

Last edited: May 21, 2005
2. May 21, 2005

### dextercioby

Make up your mind.Is it a scalar,or a vector...?To get the answer,take the 4-vector $A^{\mu}$ and perform a finite spatial rotation,along one of the 3 axis (nevermind which).How does $\phi$ transform...?As a vector,or as a scalar...?

The rest is pure speculation.

Daniel.

EDIT:There's something wrong with the LaTex code.Not my fault.

Last edited: May 21, 2005
3. May 21, 2005

### Mortimer

As a scalar.
But in a Lorentz transformation it seems to transform into a vector (the vector potential $$A$$) in coordinates outside the axis of motion, doesn't it? (this is a question, not a statement)

I'm just trying to find a logical and physical meaning for the components in this 4-vector. It is called a 4-vector so why not try to find an explanation where $$\phi$$ is a vector component?

I'm not pretending to be an allround expert on this but am just curious.

4. May 21, 2005

### dextercioby

What Lorentz transformation...?Do you mean Lorentzian boost...?Well,that's not what gives the vector/scalar/tensor character of a classical field,but behavior under the rotation group SO(3),a subgroup of the proper Lorentz group SO(1,3).(I'm refering to proper rotations).

The logics comes from group theory...It is not only called,but also proven to be a 4 vector.Actually,once you use the geometry of the Minkowski space,it's not a vector,but a covector/one-form/covariant vector,but,because the space-time is metrizable,it doesn't matter,as one could easily build the vector/contravariant vector using the metric.

Daniel.

5. May 21, 2005

### Mortimer

I only know of one Lorentz transformation in SR that transforms 4-vectors between reference frames. Are there more?

I don't see how switching from covariant to contravariant is relevant in this discussion about the 4-vector for electromagnetic potential but maybe I am missing the point here.

I know that the 4-vector for potential is proven to be a 4-vector. That's simply because it transforms like a 4-vector. The odd thing is that a scalar entity like the electrostatic potential can be a component of a 4-vector. Actually, the same exists in the 4-vector for energy-momentum, where energy is the scalar. Take any vector with 4 components in Euclidean geometry and you see that the individual components each represent a vector in the basis. Not so in these Minkowski 4-vectors.

6. May 21, 2005

### dextercioby

There are 6 generators to SO(1,3):3 for spatial rotations and 3 for Lorentzian boosts.

There's nothing odd to it.ANY zero-th component of a 4-vector is a scalar under SO(3)not only energy,scalar potential,but even time...

Daniel.

7. May 21, 2005

### Mortimer

Aha. So the zeroth component in the 4-velocity is supposed to be a scalar too in SO(3). Still odd... Also in SO(1,3) or SO(4)?

8. May 23, 2005

### cefarix

This is quite interesting...I've come to the same conclusion within my unified field theory.

The magnetic 4-vector potential I've labelled as the Q-field. According to this then, E, B, and A are defined as follows:
A = Q - Q_w = Q - c*Q_t
B = cross A
E = Q_w = c*Q_t

-- Please don't confuse Q with charge here. Q is a 4-vector field and not a scalar.

Charge then can be defined as being the momentum along the time (w-) axis divided by mass. I believe the crucial difference within my theory, though, is that a compression factor tau_g is associated with the Q-field of energy, and this leads to gravitation which behaves in accordance with general relativity's predictions in the weak electromagnetic field limit.

The compression factor is used to rescale the magnitude of the Q-field to be equal to c. Larger energies have lower values of tau_g, leading to a higher distortion of spacetime, to say it in geometric terms, or to a higher density of spacetime, to look at it in fluid mechanics/optical terms. Gravitation is then caused by energies following geodesics in Riemannian geometry, as in GR, or alternatively, the refraction of the paths of energies as they travel through a spacetime with continuously changing densities.

You might note that with this definition, charge comes out to be a constant, since w-momentum divided by mass is always c. The sign of the charge is determined by whether the Q-field is parallel or antiparallel with the time axis of the frame of reference of the observer.

The interesting thing to note is that photons have no charge...their 4-velocity is oriented completely in the spatial dimensions and hence their Q-field is completely orthogonal to the time axis.

Hmm...sorry for rambling off here. The topic really caught my attention, I hope I didn't piss anyone off

9. May 23, 2005

### dextercioby

No,not SO(3),that's for euclidean space (3D),SO(1,3) is the group of proper Lorentz transformations.

Daniel.

10. Jun 1, 2005

### pmb_phy

That is not usually referred to as a scalar. It is best to call it a covariant component. When the term "covariant" is applied to a number it means that this number changes with a change in coordinates. E.g. proper-time is invariant while coordinate time is covariant.
Its usually referred to as the 4-potential. Is that what you were getting at?

Pete

11. Jun 2, 2005

### Mortimer

Hi Pete,
Essentially yes. But I was trying to get all components of the 4-vector on "equal footing". If the vector potential$\vec A$ is a property of a moving charge that is related to the magnetic fieldvector $\vec B$ according to $\vec B=\nabla \times \vec A$ then why could not a similar basis exist for the electrostatic potential of that charge by just interpreting the temporal component of its 4-velocity as its motion in time. That motion should logically lead to a "magnetic field" in the timedimension and we could say $\vec B_{time}=\nabla \times \vec A_{time}=\nabla \times \phi$. In particular if we regard time as a true 4th (spatial) dimension this should be allowed.
Am I talking complete nonsense now or just a little bit

Last edited: Jun 3, 2005
12. Jun 2, 2005

### dextercioby

Yes,that $\nabla\times\phi$ is complete nonsense.

Daniel.

13. Jun 2, 2005

### Mortimer

I know what you are saying. This is no correct math. True. Change $$\phi$$ into $$\vec \phi_{BeingAVectorIfViewedFromAFrameThatIsRotatedToTheTimedimension}$$. I must be sounding quite cryptic now but don't now how to say it more clear.

14. Jun 2, 2005

### pmb_phy

The fact that they are compomponents of a 4-vector implies they are on equal footing mathematically. One must be cautious here. The fact that they are on equal footing mathematically does not mean that they must be on the same footing physically.

Pete

15. Jun 3, 2005

### Mortimer

Thanks Pete. You have given me some things to think about. I'll probably come back on it sometime but will first study EM 4-vectors in more depth.

16. Jun 5, 2005

### Mortimer

Question to PMB_PHY, but if someone else can help, you're welcome.
I'm missing something in the 4-vectors for charge/current density and potential. Is there no Lorentz invariant quantity? Looking at e.g. $p^{\mu}$ the invariant is $m_0c$. $J^{\mu}=(\rho,J_x,J_y,J_z)$ and $A^{\mu}=(\phi,A_x,A_y,A_z)$ are never associated with such invariants. Is my perception wrong? I checked various textbooks. None give a clue. Thanks for any help.

17. Jun 5, 2005

### pmb_phy

The magniture is, of course, invariant. But the invariant number is a function of the 4-vector. Give it a whirl. Try a few special cases and see what you get.

Pete

18. Jun 6, 2005

### Mortimer

I"ve worked out the current density 4-vector in the meantime and have come to some interesting similarities with the energy-momentum 4-vector.
Apparently, the invariant in $J^{\mu}$ is $\rho_0c$, which was the kind of quantity that I was missing and which is similar to $m_0c$ in the 4-momentum $p^{\mu}$.

19. Jun 6, 2005

### dextercioby

Well,$\rho_{el}^{0}c$ is the first component of the 4 vector and is a scalar/invariant wrt to $SO(3)\leq SO(1,3)$.

Daniel.

20. Jun 6, 2005

### pmb_phy

That is a special case which applies only to charge distributions where there is a frame in which all charges are at rest. It wouldn't work for a gas of charged particles.

Pete