# Relativity and Circular Magnetic Field

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1. Mar 24, 2015

### userunknown

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference, why is the magnetic field around the wire is circular?

2. Mar 24, 2015

### Orodruin

Staff Emeritus
This is a misleading statement at best. What is true is that electric and magnetic fields do transform into each other under Lorentz transformations. This does not mean that you can always find a frame where the magnetic field is equal to zero.

3. Mar 24, 2015

### harrylin

Hi userunknown, welcome to Physicsforums.

As Orodruin's clarified, that often-heard statement is quite wrong as it is not in general possible to totally transform a magnetic field away.
You can read some old discussions by means of the Search function (top right) as well as with the "similar discussions" at the bottom of this page.

One interesting old discussion is here (and it also shows how tricky it can get):

4. Mar 24, 2015

### userunknown

OK.
I was trying to ask why magnetic field is $$\vec{v} \times \vec{r}$$ so it has a different direction from the magnetic force. Electric field and force have same direction, but this not true for magnetic field and force.

Thanks...

5. Mar 24, 2015

### jartsa

Those red arrows show how an object, that has charges flowing inside of it, will tend become oriented.

In other words when a compass needle is not aligned with the red arrows, it feels a torque that tends to make it aligned with the red arrows. If we want, we can say that the torque is caused by electric repulsion and attraction.

6. Mar 24, 2015

### userunknown

Relativity is the reason behind the cross product operation in magnetic field and force calculation. Right?

7. Mar 24, 2015

### jartsa

The reason that magnetic field around a wire is circular is that we like it that way, for similar reasons as in the case of angular momentum: https://www.physicsforums.com/threads/direction-of-angular-momentum.771452/

An alternative convention might be magnetic field pointing to the direction of the Lorentz-force.

The reason that there is a cross product operation in magnetic field and force calculation is that we have chosen the first alternative not the second one.

8. Mar 24, 2015

### stevendaryl

Staff Emeritus
What's physically meaningful is not the direction of the magnetic field, but the direction of the magnetic force. With a circular magnetic field, that means that for a positively charged particle moving in the same direction as the current (or a negatively charged particle moving in the opposite direction), the magnetic force is toward the wire.

As harrylin's link discusses, this is sort of understandable using Special Relativity, although it's a little complicated to figure out all the details. What is true is that if we assume that
1. In the frame of the wire, the positive charges (nuclei of atoms) are stationary. They have some average linear charge density along the wire, $\lambda_{+}$.
2. In the frame of the wire, the negative charges (electrons) have some average charge density $\lambda_{-}$ and they are moving at some drift velocity $v_d$. The current would just be $\lambda_{-} \cdot v_d$, which will be in the opposite direction of $v_d$, since $\lambda_{-}$ is negative.
3. In the frame of the wire, the wire is electrically neutral: $\lambda_{+} + \lambda_{-} = 0$
Now consider the frame of a particle moving in the same direction as the drift electrons. In this frame, the space between atoms is lorentz-contracted, so the space between atoms is decreased, so the charge density for positive charges is increased. The electrons are moving in the same direction as the particle, so they are not lorentz-contracted as much. So in this frame, the wire has a net positive charge. So a negatively charged particle moving in the same direction as the drift electrons, parallel to the wire will be attracted to the wire.

The difficult part of the argument is explaining exactly why the wire is electrically neutral in the frame of the wire, and not in some other frame.

9. Mar 24, 2015

### userunknown

Really? I have been looking for an answer like 5 days. Was that the answer? Ahhh, man.

So let me show whether I got it right.

Magnetic field could be defined as pointing to the direction of Lorentz force as well. But scientist wanted it to be like it that way. So there is no deeper meaning, no deeper scientific, relativistic reason. Right?

10. Mar 24, 2015

### stevendaryl

Staff Emeritus
Umm, I wouldn't say that, exactly. What's important about the magnetic field, as I said, is not the direction it points in, but the direction the magnetic force points in. To say: "The magnetic field points in the x-direction" is just a convention for summarizing the information:
1. A positive charge moving in the positive y-direction experiences a magnetic force in the negative-z direction.
2. A positive charge moving in the positive z-direction experiences a magnetic force in the positive-y direction.
3. A positive charge moving in the negative-y direction experiences a magnetic force in the positive-z direction.
4. A positive charge moving in the negative-z direction experiences a magnetic force in the negative-y direction.
5. (And the opposite for a negative charge)
These facts are neatly summarized by saying:
$\vec{F} = q \vec{v} \times (B \hat{x})$

It's just a convention to say which "direction" the B field points in. It's also just a convention to define what the cross product $\times$ does. But the two conventions have to work together so that $\vec{v} \times \vec{B}$ is in the correct direction.

11. Mar 24, 2015

### Orodruin

Staff Emeritus
No, this is wrong. The field would then depend on the velocity of the particle it is acting on and you would lose the independence of the field from whatever is moving in the field.

12. Mar 24, 2015

### stevendaryl

Staff Emeritus
The more modern way of doing electrodynamics is to treat $B$ as a tensor, rather than a vector.

13. Mar 24, 2015

### userunknown

I don't understand what you are saying, sir. The magnetic field does depend on the velocity of the particle. It looks like you are talking against biot-savart law.

Also, thanks stevendaryl

14. Mar 24, 2015

### Orodruin

Staff Emeritus
If the field was pointing in the same way as the force, then the direction of the force could not be velocity dependent, unless the field itself depended on the velocity of the moving object. This is not the way we do physics. The field is independent of the velocity of the test particle.

A second order antisymmetric tensor to be more specific. I would also say the even more modern way would be to treat B and E as a second order antisymmetric tensor in space-time, but this probably is not at the level the OP is comfortable with.

15. Mar 24, 2015

### Staff: Mentor

The magnetic field does not depend on the velocity of the particle that it is acting on (aka the test charge). The velocity of the test charge is not even in the Biot Savart law. You are thinking of the source of the field, not the test charge.

16. Mar 25, 2015

### userunknown

Then it looks like there is a source for magnetism. Right? We need another way so, there is only one way left to define it. This is what you mean with "then the direction of the force could not be velocity dependent". Right?

Magnetic field is not a physical thing. Right?

Thank you all...

17. Mar 25, 2015

### Orodruin

Staff Emeritus
Not right. The magnetic field is every bit as physical as the electric or gravitational field.

18. Mar 25, 2015

### pervect

Staff Emeritus
I'm not sure what you mean by "a physical thing". If by "a physical thing" you mean a tensor (a common but not universal interpretation), I do not believe the magnetic field in isolation is a tensor - I'm not sure why steve appears to be saying otherwise :(.

A combination of the electric and magnetic fields in a 4 x 4 matrix known as the Faraday tensor is a tensor. See for instance http://en.wikipedia.org/w/index.php?title=Electromagnetic_tensor&oldid=652946718

I apologize if tensors are too advanced, the point about tensors that I think is relevant is that a tensor quantity when known in one frame of reference can be converted to any other desired frame of reference by the tensor transformation laws. This property allows tensors to be regarded as an observer-independent geometrical object. While the components of the tensor depend on the specific reference frame you adopt, specifying the value of a tensor's components in one frame of reference defines uniquely the value of the tensors components in any other frame of reference. It is this property that allows tensors to be regarded as "physical things" independent of the frame of reference.

In the case of electromagnetic field, when you know both E and B in any frame, you can transform the values to any other frame you choose, finding the values of E and B in the new frame. But if you only know the electric field in one frame (and don't know the magnetic field in that frame), you do not have enough information to compute the value of E in a different frame. This makes knowing the value of just E, or just B, an incomplete description of the electromagnetic field. For a complete description, you need both. When you have both, then you can represent the state of the electromagnetic field via the Faraday tensor. The lack of completeness makes E and B part of something bigger, the Faraday tensor. The completeness of the description of the electromagnetic field by the Faraday tensor makes it the best candidate for a "physical thing", i.e. a complete representation of the electromagnetic field that can be converted to any desired frame of reference by standardized transformation laws.

In tensor notation, the electromagnetic force law becomes $f^a = F^a{}_b u^b$, where $f^a$ is the four-force and $F^{ab}$ is the Farady tensor. Note that there are no cross products or need for cross products in this notation, which is what I suspect steve's point was (and that he tried to avoid mentioning tensors explicitly for simplicity).

19. Mar 25, 2015

### Orodruin

Staff Emeritus
Any pseudovector in three dimensional space has a 1-to-1 correspondance to a rank two anitsymmetric tensor. It is also the space-space part of the electromagnetic field tensor in relativity, which means it is a tensor (rank 2 antisymmetric) under coordinate changes on space. In some sense, it is every bit as much a tensor as the stress tensor (i.e., the space-space part of the energy-momentum tensor, which is very commonly used in solid mechanics). Of course, vectors are also tensors ... The point here is that the magnetic field is essentially the tensor $F^{ij}$, where the $i,j$ are spatial indices, which has a 1-to-1 correspondance to the pseudovector $B_i = \varepsilon_{ijk} F^{jk}/2$.

20. Mar 25, 2015

### stevendaryl

Staff Emeritus
Sheesh! Your'e getting way beyond the level of the original poster.

21. Mar 25, 2015

### stevendaryl

Staff Emeritus
Why do you think that? You can measure the magnetic field, so it's definitely physical. What I was saying was that the DIRECTION of the magnetic field is not particularly meaningful. It's just a convention to say that the magnetic field "points" in this direction or that direction.

It's a coincidence of 3-dmensional space that the magnetic field can be thought of as a vector.

In general, the magnetic field has components $F_{ij}$, the meaning of which is that $q F_{xy} V_x$ is the force in the y-direction on a particle of charge $q$ moving in the x-direction at speed $V_x$. In general, there could be 9 components to such a tensor*, but in the special case of the magetic field tensor, we have:

$F_{xx} = F_{yy} = F_{zz} = 0$
$F_{xy} = - F_{yx}$
$F_{yz} = - F_{zy}$
$F_{zx} = - F_{xz}$

So there are only 3 independent components: $F_{xy}, F_{yz}, F_{zx}$. It's only the coincidence of the number 3 that allows us to define:

$B_x = F_{yz}$
$B_y = F_{zx}$
$B_z = F_{xy}$

That association is purely a convention, and there is nothing physically meaningful about it. But the field itself is certainly physically meaningful.

*: As pervect points out, the relativistic way of talking about the magnetic field and the electric field is to combine them into a single tensor $\mathcal{F}_{\mu \nu}$, where $\mathcal{F}_{tx} = E_x$, $\mathcal{F}_{ty} = E_y$, $\mathcal{F}_{tz} = E_z$.

22. Mar 25, 2015

### pervect

Staff Emeritus
I assume you are refering to the same tensor I linked to in my original post, the Faraday tensor, which is indeed antisymmetric. To quote the wiki reference, the Faraday tensor can be written as:

$F^{\mu\nu} = \left | \begin{matrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{matrix} \right |$

However, if you are claiming there is a one-one correspondence between $\vec{B}$ and this tensor, I have to disagree. The Faraday tensor has more degrees of freedom, thus for any specified value of $\vec{B}$ there are an infinite number of Faraday tensors that each have a different value for $\vec{E}$, this is not a one-one mapping but a one-many mapping.

Another argument that $\vec{B}$ isn't a tensor - if a tensor is zero in one coordinate system, it's zero in all. But this is not true of the B field, you can have a zero B-field in one frame where a charge is stationary, and a non-zero B-field in another frame moving with respect to the first where the B-field is not zero.

[quote = "stevendaryl"] Sheesh! Your'e getting way beyond the level of the original poster.[/quote]

Sorry about that. The non-technical version of the point I'm trying to get across is that the B-field and the E-field together form a complete description of the electromagnetic field, while neither the E-field nor the B-field in isolation can make that claim. I was a bit surprised there was any argument about the claim, or the rather more techincal issue of noting that considering in isolation, the B-field is not a tensor.

23. Mar 25, 2015

### stevendaryl

Staff Emeritus
No, not the 4x4 tensor, but the 3x3 inner matrix which is a tensor relative to spatial transformations (but not spacetime transformations).

24. Mar 25, 2015

### Orodruin

Staff Emeritus
I am claiming nothing of the sort. I am claiming there is a 1-to-1 correspondence of the space-space part of this tensor and the magnetic field. The space-space part of any rank 2 tensor in Minkowski space is a rank 2 3-tensor.

When we are saying that E and B are vectors/tensors, we are obviously not referring to boosts, but only to 3-vectors/tensors. The electric field is also not a 4-vector.

25. Mar 25, 2015

### Orodruin

Staff Emeritus
It does have some meaning in that the component $F_{xy}$ is left invariant under rotations around the z-axis. It therefore makes sense to map it to the z-component of the vector. The vector only transforms as a (pseudo)vector if the relationship is taken to be $F_{ij} = \varepsilon_{ijk} B_k$. It is worth noting that you cannot map the components of the tensor to the components of a (pseudo)vector any way you like and get something that has the correct transformation properties.