Gravity and charge question

1. Aug 28, 2008

duordi

This is how I understand a charge field changes with motion.

A charged particle, if it is accelerated to a velocity close to the speed of light, will flatten in the direction of motion and its charge field will also.
The field will strengthen perpendicular to the velocity vector and weaken parallel to the velocity vector.

Do gravitational fields do the same thing?
When in motion do they flatten and strengthen perpendicular to motion and weaken parallel to the direction of motion?

I am asking because I thought mass is easily accelerated perpendicular to a large velocity vector and difficult to accelerate in the direction of a large velocity vector.

These two statements seem opposite, which is correct, or does charge work different then gravity?

2. Aug 28, 2008

granpa

general relativity isnt something I know much about, what does the shape of the gravitational field have to do with the ability to accelerate the mass?

3. Aug 28, 2008

duordi

I can only tell you my limited understanding of it.

If you pushed a pea and it approached the speed of light it would increase in mass until it was an equivalent mass of the sun and the faster it went the more energy it would take to accelerate it.
You would never be able to accelerate the pea to the speed of light because it would have the mass of the entire universe and require the energy of the universe to get to the speed of light.

If however you push the pea on the side it would only resist the change in motion with the mass of a pea. So the mass gain is a directional thing.

Charge fields from a proton or an electron weaken in the direction of motion and amplify in a direction perpendicular to motion,

I was given this link in a previous answer which shows how a charge distribution changes with velocity.

http://physics.weber.edu/schroeder/mrr/MRRtalk.html

4. Aug 28, 2008

Mentz114

There is remarkably little interaction between gravity and electric and magnetic phenomena, aside from the determination of the paths of light rays.

For instance, the inhomogenous Maxwell equations when solved in curved space-time give the same solution in most space-times ( including Schwarzschild and FLRW) as in Minkowski space. There is also phenomenological evidence in gravitational time dilation.

I would therefore say that electric charges behave pretty much the same with or without gravity.

But that's an informed opinion on something I'm still studying.

M

5. Aug 28, 2008

JesseM

Take a look at pervect's post on this thread, he does mention that if an object is moving relative to you, gravitational tidal effects appear stronger in the direction transverse to its direction of motion. But he also mentions it's difficult to find anything equivalent to the notion of "force" from a moving object in GR.

Some links to other threads where pervect discussed this issue (unfortunately it seems that he no longer posts on this board) can be found here.

6. Aug 28, 2008

granpa

If that were true then particle accelerators wouldnt need to change their frequency as the particle becomes more massive. that is not the case.

7. Aug 28, 2008

JesseM

What do you mean by "change their frequency"? It is certainly true that for a moving object, it takes more energy to accelerate it in the direction of motion than perpendicular to the direction of motion.

8. Aug 28, 2008

granpa

A changed particle moving perpendicular to a uniform magnetic field will move in a circle at a frequency that for nonrelativistic speeds is independent of its velocity.

9. Aug 28, 2008

JesseM

What does that have to do with the issue of objects being harder to accelerate parallel to the direction of motion than perpendicular? Or was that not what you were referring to when you said "If that were true..."?

10. Aug 28, 2008

granpa

the op stated that the mass did not increase perpendicular to its motion. that is clearly not the case since the frequency does change as the particle increases in mass.

If it were, in fact, the case then you would be able to put a tremendous amount of momentum into a particle then easily curve it around till it was moving in the opposite direction with the result that momentum would not be conserved.

Last edited: Aug 28, 2008
11. Aug 28, 2008

MeJennifer

Huh?

12. Aug 28, 2008

JesseM

What I said in the statement you quoted was correct, it does take a greater force to accelerate an object a given amount (hence more work to increase its speed by a given increment) in the direction of motion than it does to accelerate it the same amount parallel to it. However I seem to have made a mistake in my response to granpa, somehow I thought that there was no increase in the force needed in the transverse direction compared to the force needed for the same object at rest, but looking here and here that seems to be wrong. Apparently for an object moving at speed v with relativistic gamma-factor $$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$, the force/acceleration relation in the direction parallel to the object's motion is f = m*(gamma^3)*a, while the force/acceleration relation in the direction transverse to the object's motion is f = m*gamma*a.

Last edited: Aug 28, 2008
13. Aug 28, 2008

duordi

Wow, that clears a lot up thanks for the reference!

14. Aug 28, 2008

MeJennifer

What you say runs counter to the principle of relativity.

Anyway there seems to be no point in arguing this with you since you think, or spin it such a way that, you are always right.

15. Aug 28, 2008

JesseM

I admitted I was wrong about something in that very post (the part where I said 'However I seem to have made a mistake in my response to granpa'). However, it is not wrong that one needs a greater force to accelerate a moving object parallel to its direction of motion than perpendicular to its direction of motion, and the sources I linked to (one of which was a physics textbook) confirmed that--did you look at them, and if so do you think they are incorrect? In no way does this run counter to the principle of relativity, since I am (and the sources are) referring to the force needed in a given frame to produce a given coordinate acceleration of the object in that same frame, and saying that if the object is moving relative to that frame then the force needed to accelerate it in the parallel direction is greater than the force needed to accelerate it in the transverse direction. This holds true in every inertial frame.

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16. Aug 28, 2008

MeJennifer

I see, so now you adopted the use of relativistic mass and coordinate acceleration to help people understand relativity.

17. Aug 28, 2008

JesseM

I didn't say anything about relativistic mass--in the equations I posted, f = m*(gamma^3)*a for a force parallel to the object's motion and f = m*gamma*a for a force transverse to its motion, "m" is the rest mass. And coordinate acceleration, which of course is just the derivative of coordinate velocity with respect to coordinate time, is used routinely in relativity textbooks, appearing for example in the force equation F = dp/dt.

18. Aug 28, 2008

MeJennifer

No kidding the headings of the article you referenced are:

The relativistic mass concept
Early developments: transverse and longitudinal mass

19. Aug 28, 2008

cosmik debris

JesseM is correct, that's why it is dangerous to think of relativistic mass as "mass". Mass is an invariant and is the "length" of the energy momentum vector (i.e. rest mass). relativistic mass is really an energy and is the time component of the energy momentum vector and is not invariant.

20. Aug 28, 2008

JesseM

It's not like when you cite a webpage to support a claim, every single thing on the page you cite is supposed to be relevant to what you're saying. From the context it's clear that I was just using that page to get the equations f = m*gamma^3*a and f = m*gamma*a, nothing else on the page was relevant to my post. The reason they mention these equations in an article about notions of "relativistic mass" is because you can define a "longitudinal mass" mL as mL = m*gamma^3, and a "transverse mass" mT as mT = m*gamma, and then the equations are f = mL*a and f=mT*a. But that's not what I was doing, I was just referring to the equations showing the force required to produce a given coordinate acceleration, which involved only gamma (a function of coordinate velocity) and the rest mass m.

Last edited: Aug 28, 2008