# Invariance of Lorentz Force

1. Jul 27, 2010

### mysearch

I am trying to establish whether the force defined by the Lorentz equation below is invariant under the Lorentz transforms:

[1] $$F = F_E + F_B = qE + qvB$$

In the context of this equation, [q] is moving with velocity [v] such that it is acted on by both an electric E-force and magnetic B-force. If this frame of reference were transformed into a stationary (*) frame such that v=0, then $$F_B*$$ would be zero, i.e.

[2] $$F* = (F_E*) + 0$$

So is the Lorentz force [F] invariant?

If so, I presume you can equate [1] and [2] as follows

[3] $$F = F_E + F_B = (F_E*) + 0 = F*$$

[4] $$F_E* = F_E + F_B$$

However, I believe the magnetic force can be related to the electric force as follows:

[5] $$F_B=F_E \frac {v^2}{c^2}$$

Again, if this is correct, it would appear that [4] can be transposed as follows:

[6] $$F_E* = F_E (1+ \frac {v^2}{c^2})$$

As such, the implication appears to be that $$F_E* > F_E$$ and the invariance of the Lorentz force [F=F*] leads to the conclusion that the electrostatic force $$F_E$$ in the stationary frame is transformed into $$F_E+F_B$$ in the moving frame. Would welcome any clarification of these points. Thanks

2. Jul 27, 2010

### starthaus

The general Lorentz force :

$$F=q(vXB+E)$$

is not Lorentz invariant . The reason is that the "magnetic" component
$$F_b=qvXB$$

is not Lorentz invariant since it depends on particle speed.

On the other hand, the Lorentz force can be recast in a "covariant" form, the proofs are non-trivial, i.e. they require quite a lot of of math. For example:

$$f_b=\gamma(v)(qvXB, (qvXB).v/c)=\gamma(v)(qvXB,0)$$

is covariant
Likewise:

$$f_e=\gamma(v)(qE, qE.v/c)$$

is covariant. I can write up the proof for you.

Last edited: Jul 27, 2010
3. Jul 27, 2010

### bcrowell

Staff Emeritus
A force can't be Lorentz-invariant, because it's a vector, not a scalar.

4. Jul 27, 2010

### Staff: Mentor

As mentioned by both starthaus and bcrowell, the Lorentz force, being a vector, cannot possibly be invariant. It is, however, covariant. Einstein went into quite some detail on the Lorentz force in a moving frame in part 2 (Electrodynamical Part) of his seminal paper:
http://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf

This includes everything that you need to know to work out the Lorentz force in special relativity. In general relativity the Lorentz force can be cast in a manifestly covariant tensor-based form:

$$f^{\mu}=q U^{\lambda}F_{\lambda} ^{\mu}$$

5. Jul 27, 2010

### Altabeh

No. The scenario must turn out to be amiss; as it has to be portrayed in such a way that the result stands for the non-invariance of the Lorentz force under Lorentz transformations.

You can consider a point charge $$q$$ being inert in a laboratory frame on Earth (your stationary frame.) A stationary observer only observes an electric field while an inertial observer measures both electric and magnetic fields. In the laboratory frame, the charge would experience a Lorentz force $$F=F_E=qE$$; whereas in any inertial frame the same charge experiences a Lorentz force $$F'=qE'+qv\times B'$$ and thus it can be seen that $$F$$ and $$F'$$ are not identical; leading to the non-invariance of the Lorentz force as already have been pointed out here by many other.

AB

6. Jul 28, 2010

### mysearch

Thanks for all the knowledgeable insights. I am not sure that I fully understand the idea of covariance and while I have quickly looked at the following 2 wikipedia references, I am not sure that I am really any the wiser, at this point. However, it is obviously a gap in my understanding – so thanks.
http://en.wikipedia.org/wiki/Lorentz_covariance
http://en.wikipedia.org/wiki/General_covariance

However, the following quote seems to capture an intuitive insight, which makes sense to me because in the example I have been trying resolve, the force always changes direction when the frame of reference is switched. However, I had always thought of velocity being invariant, but possibly the inference relates only to its magnitude, not its direction.
I raised the example, cited above, in the diagram attached in another post (#3)/thread:
https://www.physicsforums.com/showpost.php?p=2811129&postcount=3

While I had looked at a number of solutions under the general heading of ‘Relativistic Electrodynamics’ there were certain assumptions that I couldn’t transpose into a discrete model involving just 2 charges as outlined in post #3 above, which related to the propagation delay of the E and B fields, when the respective charges are moving with relativistic velocities. While accepting the statement about the non-invariance of the Lorentz force in this thread, I was looking for a way to balance the net force on charge [q] in the diagram of post #3.

Apologises for taking the following quotes out of context, but in some ways, they gave me the impression that force may be somehow invariant, when possibly they only meant to make a comment about the net effect. The thread above also outlines how some of my confusion over force arose. However, the net force is problematic in the example cited, because I couldn’t get the electric and magnetic force vectors to align, when accounting for the propagation delay and so I was looking to understand where this model was wrong. Anyway thanks for the correction and would wlecome any further in-sights to the other problem.

Last edited: Jul 28, 2010
7. Jul 28, 2010

### starthaus

Check out the third attachment https://www.physicsforums.com/blog.php?b=1887 [Broken]. I just wrote it specifically for you.

Last edited by a moderator: May 4, 2017
8. Jul 28, 2010

### bcrowell

Staff Emeritus
In Newtonian mechanics, velocity is not invariant. For something to be invariant means that it's the same in all frames of reference. For example, mass and charge are invariant in Newtonian mechanics. In a frame of reference defined by my desk, a tree outside has a velocity of zero. In a frame of reference defined by a car driving by, the tree has a nonzero velocity.

Typically, there are more things in relativity that are frame-dependent (not invariant) than in Newtonian mechanics. Since velocity is frame-dependent in Newtonian mechanics, we would definitely expect it to be frame-dependent in relativity. Although force is frame-independent in Newtonian mechanics, it is frame-dependent in relativity.

A mathematical way of making the whole picture very simple is that in relativity, we only care about quantities that transform as tensors. A rank-0 tensor, such as charge, is a scalar; it's frame-independent. A rank-1 tensor is a four-vector, i.e., a vector with four components, three spacelike and one timelike. All four-vectors transform in the same way, as defined by the Lorentz transformation. Since force is a vector in Newtonian mechanics, the only way we could possibly describe it meaningfully in relativity is as part of a four-vector, in which case it will transform as a four-vector, and will be frame-dependent.

9. Jul 28, 2010

### starthaus

To continue from where bcrowell left off, the four-vector constructed by using the 3-vector Lorentz force is "covariant". This means (among other things) that it transforms as the "prototype" 4-vector (x,y,z,ct) and that its norm is frame-invariant, exactly like the one for the "prototype".

10. Jul 29, 2010

### mysearch

Thanks again for the insight into some of the maths. While I have a basic understanding of 4-vectors and tensors, it been quite sometime since I had any reason to revisit these ideas in any depth, so Starthaus’ papers in #7 will be a useful place to start.

My comment about velocity did not relate to either a desk or a tree in which you orientate your frame of reference to either the stationary or moving frame, but rather a more general case. For example, let the infamous hypothetical rocket travel between A and B, which we might align along the x-axis for convenience. Let a stationary observer measure the distance between A-B as [x] and the time taken as [t]. I am assuming an observer on board would measure this time as [t’] and the distance to be [x’]. However, it was my understanding that both would determine the magnitude of the velocity as [v=x/t=x’/t’] such that that [v] remains ‘constant’ in both frame of reference.

Anyway, to some extent, my main focus at this point was not the subject of relativity itself but its implications on electrodynamics as cited in the problem here:
https://www.physicsforums.com/showpost.php?p=2811129&postcount=3

May be you could applied the maths to explain why the propagation delay of the E-field 'seems' to cause a problem. Thanks