# Charge/mass ratio of moving particles (SR)

1. Dec 9, 2013

### tartaneto

Hi guys. I have something stuck in my head since a few days and I'd like to have your opinion about that. I don't know if I am missing something in my assumptions so please feel free to enlighten me.
Here's the thing: considering two protons (or electrons), one fixed at the origin of the frame of reference and the second lying on the X axis fixed at a distance L. As we well know, the electric repulsive force is stronger than the gravitational pull between them and they will experience a repulsive resultant force in their frame of reference (they are at rest relative to each other but moving in relation to our frame of reference with a speed V).
In their frame of reference the charge/mass ratio is equal to e/mp where "e" is the charge of the proton and mp its mass. However in our frame of reference this ratio will be different because of the relativistic mass. The charge is invariable with the speed but the mass increases.
Consider what I will call "Planck's particle". This particle has the Planck mass and the Planck charge thus the square of the ratio charge/mass is equal to 4∏εG. If you calculate the interaction between two such particles you will find that the electric repulsive force equals the gravitational pull between them and they won't interact at all. If they are at rest they will keep at rest no matter how close or far they are from each other. So, let's go back to the protons. If their frame of reference is moving with such a speed V that the square of the charge/mass ratio will be equal to 4∏εG, these two protons will not interact with each other (in our frame of reference). So if we measure the resultant force of these two moving protons it will be zero. I've calculated that speed (it's quite high) and its value is:
v = c.[1-(mp^2/α.M^2)]^0,5 where "c" is the speed of light, mp is the mass of the proton (or electron), alpha is the fine-structure constant and M is the Planck mass. Also we would see them (if we'd be able to see them) closer than they really are since the distance L would be contracted from our point of view by the Lorentz factor.
Is that possible that we measure no resultant force at all while in their frame of reference they are experiencing a resultant force?
Thanks in advance for any help.

2. Dec 9, 2013

### Simon Bridge

No such thing as relativistic mass.

But what you are thinking is that the curvature of space depends on the energy density ... and a relativistic object has a lot of kinetic energy - which is energy, so it should give the proton more gravity.

Have a look at:
https://www.physicsforums.com/showthread.php?t=687833

You mean their separation would be less ... there would also be a magnetic field because they are a current.

3. Dec 9, 2013

### tartaneto

Thanks for the clarification on terminology, Simon. But how would that answer my question? The magnetic field experienced in our frame of reference is perpendicular to their motion and does not affect each other, or does it? I mean, the magnetic field won't generate any force between them (either repulsive or attractive) from our point of view and I am talking about the forces between them only, one pulling or pushing the other.

4. Dec 9, 2013

### dauto

You cannot use relativistic corrections to the (relativistic mass) and expect to be able to plug it in a non-relativistic approximation (Newtons gravitational theory) and get sensible results. You just proved that didn't you?

5. Dec 9, 2013

### Staff: Mentor

No, that is not possible. There are several problems with your argument.

First, the charge/mass ratio is invariant since the charge is invariant and the mass in the ratio is the invariant mass. Relativistic mass is not used for calculating the charge/mass ratio.

Second, gravity does not produce a force in GR, so two protons will always experience a net repulsion. Each will therefore always undergo proper acceleration away from the other. However, due to curvature of spacetime the distance between them could remain constant.

Third, you can use the weak field approximation to say that two protons should behave a certain way in their mutual rest frame, but when you boost that to relativistic velocities then the weak field approximation no longer holds and you must use the EFE, at least in a linearized form. In GR, gravity couples to momentum as well as energy, so it is not correct to simply say that increased energy leads to increased gravitation.

6. Dec 9, 2013

### Bill_K

Weak field means the same thing as linearized, Dale. You must be thinking of the slow motion approximation.

7. Dec 9, 2013

### Staff: Mentor

Oops, thanks for the correction. I thought they were different. I thought that the weak field approximation was for low masses AND slow motion. I thought that linearized was a small first-order perturbation from a background field that may or may not be weak.

8. Dec 9, 2013

### Histspec

In the early days of relativity, the charge/mass ratio was connected with the (electromagnetic) longitudinal/transverse mass, which was a predecessor of "relativistic mass". So by analyzing deflection curves, they indeed found a decrease of the charge-to-mass ratio defined that way.

https://en.wikipedia.org/wiki/Kaufmann-Bucherer-Neumann_experiments

Of course, from a modern perspective these experiments actually tested relativistic momentum.

9. Dec 9, 2013

### Simon Bridge

It helps you refine your question so you can get a better answer. Anyway, you also said you wanted:
... and I was sure that lots of people would jump in as well so wouldn't have to be complete.

I had hoped you used the points as the starting point of some more research.

Since there is no such thing as relativistic mass - central to your argument - then we cannot measure a different charge to mass ratio.

The moving pair of protons do not exert greater gravitation on each other because they are stationary with respect to each other anyway.

Your only actual question was:
There's a word missing there - I inserted a question mark. Is that word "while"?

It is entirely possible for us to think there is a force when there is none - that is where you get pseudo-forces. It's also possible for several forces to act to cancel each other out - so we'd measure zero force when there are many forces. We can also get a pseudoforce to cancel a real one - say we do a magnetic levitation experiment in an accelerating frame. (In GR, gravity is a pseudo-force.)

But I think you need to adjust the conceptual basis for your question so it is related to the stuff about the protons.

10. Dec 19, 2013

### tartaneto

Thank you all for the answers, it helped a lot

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