- #1

- 81

- 4

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 81

- 4

- #2

You don't need special relativistic calculation to see that. Maxwell equations unambiguously give the magnetic field as a result of a time-varying electric field.Using special relativistic calculation it can be showed that magnetism is a result of electric charges in motion

- #3

- 81

- 4

I know. But that's not my point. My point is that this suggests that there should be an analog of magnetism in gravitation as well?You don't need special relativistic calculation to see that. Maxwell equations unambiguously give the magnetic field as a result of a time-varying electric field.

- #4

Science Advisor

Gold Member

- 28,019

- 6,407

There are a number of significant differences between Gravity and Electromagnetism so why would one expect the 'analog' you suggest? The speed of travel of disturbances in both is the same, so they do have at least one common feature.I know. But that's not my point. My point is that this suggests that there should be an analog of magnetism in gravitation as well?

- #5

- 10,569

- 11,471

There is - gravitoelectromagnetism. Frame dragging, as detected by Gravity Probe B is one such effect.I know. But that's not my point. My point is that this suggests that there should be an analog of magnetism in gravitation as well?

I don't know a lot about it; I suspect you can't push the analogy too hard.

- #6

- 10,569

- 11,471

Arguable. Maxwell's equations are intrinsically relativistic - in fact it was their incompatibility with Newtonian physics that led to the discovery of relativity. So one could make the case that you need special relativity (whether you or the entire 19th century scientific community realize it or not!) to address anything to do with electromagnetism.You don't need special relativistic calculation to see that.

- #7

- 1,565

- 134

Gravity is ridiculously weak, that is why we don't observe those magnetic forces.

Let's say the gravitational force between a moving train and the Earth is million Newtons in trains frame. In platforms frame the force is million Newtons divided by gamma.

At speed 100 m/s gamma is 1.0000000000000557, so there is a 0.0000000557 Newtons magnetic force between the Earth and the train in platform frame. (It's a repulsive force)

Let's see what changes occur when that train leaves the platform and accelerates to 100 m/s.

In train frame: Earth gains energy and contracts, the gravitational force changes by a factor of gamma squared.

In platform frame: Train gains energy, the gravitational force changes by a factor of gamma.

So between the frames there is a disagreement of gamma, just like relativity's force transformation predicts.

Let's say the gravitational force between a moving train and the Earth is million Newtons in trains frame. In platforms frame the force is million Newtons divided by gamma.

At speed 100 m/s gamma is 1.0000000000000557, so there is a 0.0000000557 Newtons magnetic force between the Earth and the train in platform frame. (It's a repulsive force)

Let's see what changes occur when that train leaves the platform and accelerates to 100 m/s.

In train frame: Earth gains energy and contracts, the gravitational force changes by a factor of gamma squared.

In platform frame: Train gains energy, the gravitational force changes by a factor of gamma.

So between the frames there is a disagreement of gamma, just like relativity's force transformation predicts.

Last edited:

- #8

- 1,565

- 134

There are a number of significant differences between Gravity and Electromagnetism so why would one expect the 'analog' you suggest? The speed of travel of disturbances in both is the same, so they do have at least one common feature.

We don't want a balloon sticking to a cat by electrostatic force to fall off when we go to another frame, do we?

So the force of gravity must change the same way as the electro-static force, when frame is changed.

- #9

- 10,569

- 11,471

The analogy is only exact for an approximation to gravity, so you need to be a little careful with this line of reasoning. Essentially, you can't talk quite so casually about "changing frames" in GR as you can with EM in SR.So the force of gravity must change the same way as the electro-static force, when frame is changed.

- #10

- 1,565

- 134

The analogy is only exact for an approximation to gravity, so you need to be a little careful with this line of reasoning. Essentially, you can't talk quite so casually about "changing frames" in GR as you can with EM in SR.

1: Let's consider a cat and a balloon standing side by side in vacuum. A force meter attached to the cat and the balloon measures electric force X. Then the cat and the balloon accelerate to high speed, the force meter still measures force X. The force between the cat and the balloon is X/gamma, says relativity.

2: Let's consider two spaceships standing side by side in vacuum. A force meter attached to spaceships measures gravitational force X. Then the two spaceships accelerate to high speed, the force meter still measures force X. The force between the two spaceships is X/gamma, says relativity.

Do you see something wrong with the above?

- #11

- 10,569

- 11,471

The force due to gravity is zero, for starters.Do you see something wrong with the above?

If you are weighing something on Earth, you are measuring the electromagnetic repulsion between the object and the surface of the balance.

- #12

- 1,565

- 134

The force due to gravity is zero, for starters.

If you are weighing something on Earth, you are measuring the electromagnetic repulsion between the object and the surface of the balance.

The force due to gravity is zero, for starters.

If you are weighing something on Earth, you are measuring the electromagnetic repulsion between the object and the surface of the balance.

Well, I don't understand what you are saying.

But let's say there are two co-moving planets in vacuum, a long rod keeps the planets apart. The force compressing the rod is X, the force is measured at the middle of the rod where gravity is not too strong.

As you perhaps say, X is same as electromagnetic repulsion between rod-molecules.

And electromagnetic repulsion between objects obeys special relativity. So X/gamma is the force of the electromagnetic repulsion in different frames, in the rod frame gamma is 1. The force of gravity is zero? Or is it X/gamma? Or something else?

- #13

- 10,569

- 11,471

Since you are talking about the interaction of fields you need to worry about momentum and energy carried by the field, and force is a really clumsy tool for this.

Note also that the Coulomb field is an exact flat space solution of Maxwell's equations. Newton's formula for gravity is not an exact solution of Einstein's equations. If the "electrostatic" components of the fields are different in form, why would you expect their "magnetic" behaviour to be the same?

- #14

- 1,565

- 134

If the "electrostatic" components of the fields are different in form, why would you expect their "magnetic" behaviour to be the same?

First postulate of relativity is the reason.

Like with pendulum clocks. We just transform an hour of pendulum clock time to another frame without thinking too much about all that stuff you mentioned.

- #15

- 10,569

- 11,471

But there isn't a gravitational force to transform.

- #16

- 1,565

- 134

But there isn't a gravitational force to transform.

Well then let's transform the rate of exchange of momentum. Which is same thing as force. Note that I did not say gravitational force.

So we agree with a person standing on a planet about how much momentum is exchanged between the planet and the person during the time that the person's clock proceeds one second, but we disagree about how much time the second takes, if we observe the planet moving. So the rate of exhcange of momentum transforms as 1/gamma.

The above is correct only when the direction of the force is perpendicular to the motion. And note again that I did not say gravitational force.

- #17

- 10,569

- 11,471

As you say, this is the same thing as the force. Why do you think the answer will change?Well then let's transform the rate of exchange of momentum.

But it's easy to see the problem here - both of the planets are in free-fall. Neither is accelerating; accelerometers on the ships will prove that. Nevertheless the distance between them starts to decrease, faster and faster. This situation is not covered by special relativity.

If want to go into momentum transfer you need to define some kind of coordinate system large enough to cover both rockets, since a local analysis has two problems - neither planet is accelerating, and there is no local reason for them to be getting closer. So your coordinates must cover a region which is not (even approximately) flat if the planets are attracted to each other through gravity. Then you can analyse any point using Lorentz transforms, but you can't analyse the whole problem with them, since they relate coordinate systems in flat spacetime.

You can do it approximately, treating gravity as a field in flat spacetime. And then the analogy with electromagnetism is exact, as I understand it. But if you account for curvature correctly you have a problem you cannot analyse in the special relativistic manner you are trying.

- #18

I actually wrote a paper on this force. It does exist but it is absurdly weak, and you need GR to calculate it exactly. It is in the same direction of gravity, it is essentially a gravitational imbalance that exists because length contraction creates more density in one body (the one moving according to you) and there is no length contraction in the other body (the one at rest according to you) so it's gravity becomes weaker in your frame. But in a frame where you move in the opposite direction with velocity -v, the effect reverses and the gravitational imbalance moves the other way. I can give more exact figures, diagrams, and logic trees if you want.

Last edited by a moderator:

- #19

At equator the force is about 10^{-7} times the gravity ,if I remember correctly.

It is kindof similar to the GR at low fields. ( No spacetime curvature though)

- #20

- 1,565

- 134

What are the sizes of your objects, I mean relative sizes? If a large object contracts, I can see that small objects at many locations near the large object see large parts of the large object to get closer, which intuitively means an increase of force of gravity at those locations.

- #21

If you are moving towards a large body, the length of the body would contract according to you (if its spherical it will become and ellipsoid with the eccentricity in the direction of your motion), but you would still feel the same gravitational force as the center of mass has not changed. The grav force will be the same if we use spec rel and Newtonian gravity. But in GR more density mean more pressure and more gravity, so there may be a change in gravity.What are the sizes of your objects, I mean relative sizes? If a large object contracts, I can see that small objects at many locations near the large object see large parts of the large object to get closer, which intuitively means an increase of force of gravity at those locations.

The reason I analogized this force to magnetism is because it arises from gravity in analogy to how magnetism arises from an electrostatic force. Both objects have to be of a gravitational order of magnitude (around 10^12 kg - gives you around one Newton of force). My original derivation has two parallel rods of length L and mass around 10^12. If you have a test particle at the Lagrange point of the two rods (where gravity cancels out, and the test particle is at rest with respect to both rods, you feel no force. However, if one rod is moving with velocity v relative to the other rod and the test particle it undergoes length contraction from the perspective of the test particle. Meanwhile the other rod is stationary with respect to the test particle and remains at length L from the perspective of the test particle. So the density of the rod moving at v increases and its length decreases by sqrt[1-(v/c)^2]. Now the test particle will feel a force because the center of mass of the moving rod has shifted. The gravitational force is radially coming out of the rod and the rod has contracted to the point that there are points over the region L-sqrt[1-(v/c)^2]*L where only one rod is exerting force. The test point now feels a force. This force is the gravitational analog to magnetism and points in the same direction as gravity with a magnitude smaller than gravity.

Now let's remember the idealizations I've made here. I'm using special relativity but discussing gravity. Special relativity is not accurate in reference frames that experience a gravitational force. I'm also using Newtonian gravity, not GR gravity. So we need GR to see what really happens here. But these assumptions are fairly accurate as the gravitational force is low.

Share:

- Replies
- 32

- Views
- 437

- Replies
- 11

- Views
- 709

- Replies
- 2

- Views
- 401

- Replies
- 5

- Views
- 571

- Replies
- 7

- Views
- 947

- Replies
- 15

- Views
- 1K

- Replies
- 3

- Views
- 433

- Replies
- 16

- Views
- 298

- Replies
- 32

- Views
- 2K

- Replies
- 2

- Views
- 8K