# Forces and reference frames

1. Jun 20, 2011

### lovetruth

Consider two point charges in space: one positive(+Q) and other negative(-Q), lying on the y-axis and separated by distance 'r'. In frame A, both charges are at rest so, only attractive electrostatic force (F_elec) acts on both the charges which is defined by coulombs formula. In another frame B, both charges are moving with velocity 'v' in positive x-direction so forces acting on the charges are attractive electrostatic force(F_elec) and repulsive magnetic force(F_mag).
It can be seen that the net attractive forces acting on the charges is greater in frame A than in frame B by an amount F_mag since, F_elec is same in both frame A & B.

Also,
F_mag = - F_elec ,when v=c [this result can be obtained by using coulomb law, ampere law, and the fact that epsilon*mew=1/(c^2)]

Conclusion : So an observer in frame B will see that net attractive force acting on charges is reduced than in frame A. Also, the net attractive force in frame B tends to zero as 'v' approaches 'c'.

Q: How is the above situation compatible with SR. SR says nothin about relativity of forces only about relativity of length and time.

Last edited: Jun 20, 2011
2. Jun 20, 2011

### Staff: Mentor

And what are the base units of force?

3. Jun 20, 2011

### lovetruth

mass*length*time^-2

But that doesnt answer the question.

4. Jun 20, 2011

### Bill_K

Of course it does, Lovetruth, special relativity says how force transforms from one reference frame to another, and also how electric and magnetic fields transform, which is what we need here. These may be found in any introductory relativity book:

E' = E
E' = γ(E + v/c x B)
B' = B
B' = γ(B - v/c x E)

where ∥ and ⊥ represent the components parallel and perpendicular to the relative motion. In your example, E' = γE, B' = γ(- v/c x E). In the moving frame the Lorentz force between the charges is

F' = q(E' + v/c x B') = q(γE + v/c x (γ(- v/c x E))) = qE γ (1-v2/c2) = qE

See? It's the same in the moving frame as it was in the original frame.

5. Jun 20, 2011

### lovetruth

Can u provide any internet links which can show these E & B field transformation because I have not read this anywhere. As far as i kno, only length, time, mass, and energy transforms. Please give me the list of all the entity that can transform.

I have spotted an error in ur calculation, in the last step, u have assumed :
gamma*(1-v^2/c^2)= 1 which is incorrect.

6. Jun 20, 2011

### Bill_K

Sorry you're right, the formula I quoted for the Lorentz force is actually the 3-force F = dp/dt. The force which is the same in both frames is the 4-force, dp/dτ where τ is the proper time. They differ by a factor of γ.
Where have you read? A much better idea is to get a book and read that. The book I quoted from was Jackson's Classical Electrodynamics chap 12. It's also in Misner, Thorne and Wheeler chap 1.

7. Jun 20, 2011

### lovetruth

I have googled electromagnetic lorentz transformation. U have made a slight mistake in the formula, the electric and magnetic field terms do not have same units so, a factor of velocity should be multiplied to the B in the equation(v/c is dimensionless).
But u have made a remarkable attempt to solve the problem. Could u please solve the problem with corrected equation and show that force is same in both frame A & B.

I hate the fact SR contains so many transformations!!??!!
I wonder whether the laws of physics and speed of light are also transformed(seriously).
I think the results of SR are quasi-true but there must be another simple theory with few equations which can predict the effects of SR without giving a obnoxious headache.

8. Jun 20, 2011

### lovetruth

Hey bill_k, can u explain why electromagnetic transformation is not used in the feynmann derivation of relation between Electric, magnetic field and relativity.

Here is the link of the problem i am referring: http://galileo.phys.virginia.edu/classes/252/rel_el_mag.html

In this case the EM transformation is ignored but the result of same force is obtained.

9. Jun 20, 2011

### Bill_K

Nonsense, the formulas are correct. They're in Gaussian units.

10. Jun 20, 2011

### lovetruth

I have attached image for the correct equation of transformation. Ur formula is wrong even in gaussian unit.

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11. Jun 20, 2011

### Staff: Mentor

Everything transforms! The question is only how it transforms.

There is nothing particularly unusual about SR with respect to transformations. Coordinate transformations are common to all branches of physics. The only difference is that in SR the coordinate transforms are 4D instead of 3D and that the metric is Minkowski instead of Euclidean. The Lorentz transform is certainly easier than doing a transformation from Euclidean to spherical coordinates.

12. Jun 20, 2011

### lovetruth

What about mass, energy, electric and magnetic field transformation?

13. Jun 20, 2011

### Staff: Mentor

Yes, they all transform. Mass transforms as a scalar, energy transforms as a component of a vector, and the electric and magnetic fields transform as the components of a tensor. Everything transforms one way or another.

14. Jun 20, 2011

### bcrowell

Staff Emeritus
It's all a lot simpler than you're imagining. Pretty much everything of physical interest in relativity transforms as some kind of tensor.

Rest mass is a rank-0 tensor.

Energy is part of the energy-momentum vector, which is a rank-1 tensor.

The electric and magnetic fields are the ingredients of the Maxwell tensor, which is a rank-2 tensor.

-Ben

15. Jun 20, 2011

### lovetruth

Does the sum of electric and magnetic force changes on charges when reference frame are changed (like in the question i have posted)?

16. Jun 20, 2011

### pervect

Staff Emeritus
The 3-force, dp/dt, depends on your reference frame. So the short but possibly misleading answer is yes.

However, the 4-force, dp/dtau, does NOT depend on the reference frame. To be more precise, the 4-force transforms via the Lorentz transform.

There are two differences - the second formula uses proper time tau,rather than coordinate time t, and it uses the energy momentum 4-vector, rather than the momentum 3-vector.

The 4-force is actually d(energy-momentum-4-vector)/dtau, where the energy-momentum 4-vector is

E, px, py, pz

E being the energy, px, py, and pz, being the x, y, and z components of momentum.

Like the 4-force, or any 4-vector in SR, the energy momentum 4-vector transforms via the Lorentz transform.

I'd suggest picking up some textbook like Griffiths.

https://www.amazon.com/Introduction-Electrodynamics-3rd-David-Griffiths/dp/013805326X

You should be able to get a copy from your local library, possibly via inter-library loan. If you like it, you might even want to buy it, if you have the book budget money.

Last edited by a moderator: Apr 26, 2017
17. Jun 20, 2011

### bcrowell

Staff Emeritus
I know some people like to think about it this way, and I'm sure it's a self-consistent logical system, but it's not the way I think about it myself. I would just point out to lovetruth that if you write a 4-force in terms of its (t,x,y,z) components in a particular frame, then act on it with a Lorentz transformation, its components in the new frame are different numbers. So in that sense, it *does* change, and the Lorentz transformation says how it changes.

In lovetruth's OP, I believe the conclusion is correct. Let three-force F act between the two charges in the original frame. The four-force in that frame is (0,0,F,0) (based on the rule that the spacelike part of the force four-vector equals the force three-vector multiplied by gamma, and the charges have gamma=1 in the original frame). Lorentz-transforming to the new frame gives (0,0,F,0) again (because a boost in the x direction doesn't change a y-component). Again applying the rule that the spacelike part of the force four-vector equals the force three-vector multiplied by gamma, the corresponding three-force is $F/\gamma$.

My calculation agrees with the statement in Lovetruth's #1 that the three-force is smaller in the new frame.

My calculation agrees with the calculation in Bill K's #4 that F' = qE γ (1-v2/c2), since this quantity equals qE/γ=F/γ. (As lovetruth and Bill_K agreed in #5,6, there was a trivial algebra mistake when Bill equated this to qE.)

-Ben

Last edited: Jun 20, 2011
18. Jun 20, 2011

### pervect

Staff Emeritus
I might not have been too clear. I do think of 4-vectors as geometric objects, and when I'm thinking of them as geometric objects I think of them as "unchanging" in some larger philosophical sense.

But this is potentially confusing because, as Ben points out, the components of these geometric objects do change as you change the reference frames. It's just that they all change in a standardized manner.

I tried to address that in my post,but I might not have spelt it out too clearly.

What really doesn't change is the norm of the four-vector.

19. Jun 20, 2011

### bcrowell

Staff Emeritus
Hi, pervect,

I think we're basically in agreement. We just have different words that we use to express the same math. Just wanted to head off any possible confusion.

-Ben

20. Jun 21, 2011

### lovetruth

It will be great if someone can name an experiment in which reduced EM forces on charged particles are observed, like: Interaction of alpha and beta particles as a function of their velocities in a bubble chamber.

If the EM force is reduced by changing frame then, all 4 fundamental forces(EM,gravity,strong,and weak forces) should have 'magnetic counterparts force' which can reduce the net force by 1/gamma [like Gravito-magnetic(made-up term)]. These 'magnetic counterparts' must exist otherwise reality will not be same in every frame.

I also have a great doubt about the problem in my OP: In the new frame B, should I
1. Apply newtons law of motion with reduced EM force and neglect time dilation
2. Use electrostatic force as in frame A and apply time dilation
3. Use reduced EM force and also apply time dilation
Some how i feel reduced EM force is tantamount to time dilation and therefore, applied individually(options 1 & 2).

21. Jun 21, 2011

### lovetruth

I have read on the internet that net forces on charged bodies does not change when frames are changed. This result is also shown by Feynmann in his books.
galileo.phys.virginia.edu/classes/252/rel_el_mag.html

So the question still remains: Whether net force on charged particles changes when frames are changed?

22. Jun 21, 2011

### bcrowell

Staff Emeritus
Cathode ray tubes are a good example.

You need to distinguish between forces and fields here. All four-forces transform the same way. That doesn't mean that all fields have the same behavior.

The way to approach this without getting confused is to stop using three-vectors and just use four-vectors, which transform in a consistent way. For example, you can express the equivalent of Newton's second law in terms of four-vectors.

-Ben

23. Jun 21, 2011

### bcrowell

Staff Emeritus
I've gone through both pages carefully, and I don't see any statement that "net forces on charged bodies does not change when frames are changed."

Fowler's page at virginia.edu presents a simplified version of a derivation by Purcell, referring the reader to Purcell at the end for more details. One of the simplifications is that Fowler makes an approximation at "[...]so relativistic contraction will increase their density to[...]" Purcell does not make the approximation. This means that all of Fowler's results are only valid to leading order in v. At the end, he says, "This purely electrical force is identical in magnitude to the purely magnetic force in the other frame! So observers in the two frames will agree on the rate at which the particle accelerates away from the wire, but one will call the accelerating force magnetic, the other electric." He does not state this as a general law that forces and accelerations are frame-invariant (which would be false). He is also only saying this in the context of an approximation to leading order in v. His statement can't be true to all orders in v, because the 3-force f and the 3-acceleration a do not satisfy F=ma in all frames, if m is kept constant. What is the same in all frames is F=mA, where F is the four-force and A is the four-acceleration:
http://en.wikipedia.org/wiki/Four-force
http://en.wikipedia.org/wiki/Four-acceleration
In the present case, where the velocity is perpendicular to the force, F=mA reduces to $f=m\gamma a$. This is the well known result for what some older books call the "transverse mass:"
http://en.wikipedia.org/wiki/Mass_in_special_relativity#Transverse_and_longitudinal_mass
Fowler is only doing a calculation of f and a to leading order in v. To leading order in v, $\gamma=1$, and this is the justification for his claim that, in this context, f=ma holds in both frames.

If you want to see an exact treatment, check Purcell out of the library. It's a wonderful book.

-Ben

24. Jun 21, 2011

### lovetruth

Cathode ray are acted by electric and magnetic source which are stationary. I wanted an experiment in which electric and magnetic field is produced by a moving charged particle.
The best experiment would be observing the trajectories of alpha and beta particles which have same velocity and are at some distance to each other. The trajectories could be seen in cloud or bubble chamber. If anyone here is nuclear physicist by any chance then, this will be a great experiment.

25. Jun 21, 2011

### lovetruth

Is there a way of using only 3-force with some changes and avoiding 4-force concept.