Gravity and velocity

[[Tycho?]]

Ok, I can't keep this straight in my mind.

 

When an object accelerates close to c, does its gravity increase, or is it just inertial mass that increases? I dont think that gravity increases, but I want to make sure on this.

 

(you can explain your answer if you feel like it, but really just saying which is correct would be fine)

[Johnny5]

']Ok' date=' I can't keep this straight in my mind.

 

When an object accelerates close to c, does its gravity increase, or is it just inertial mass that increases? I dont think that gravity increases, but I want to make sure on this.

 

(you can explain your answer if you feel like it, but really just saying which is correct would be fine)[/quote']

 

Suppose that the relativistic formula for mass is correct. That means this:

 

[math] M = \frac{m_0}{\sqrt{1-v^2/c^2} } [/math]

 

In the formula above, m₀ is the rest inertial mass of some object, and M is the relativistic inertial mass of the object.

 

Now, if gravitation depends on relativistic inertial mass, then you have your answer.

 

Suppose that some huge object of mass M1 is just sitting in some inertial frame of reference, and nothing is exerting a gravitational force on it.

 

Then from far space, let something, say an asteroid, be zipping towards it. In fact, for the sake of argument, let it be a black hole.

 

In the rest frame of M1 (which is an inertial reference frame), let the speed of this asteroid be v. So we can say that the relative speed between the two objects is v.

 

That means that if we switch to the rest frame of the asteroid, and compute the speed of the black hole in the asteroids frame, we will also get v. So the asteroid isn't accelerating towards the black hole, via gravitational force yet, because it's so far away, at least initially.

 

Since both frames are inertial, the trajectory of each object in the frame of the other, is a straight line (or we could start off with the asteroid at rest in the black hole's reference frame, and then wait some time T, for the asteroid to begin being accelerating towards the black hole... in fact, this is easier).

 

In fact, for simplicity's sake, let the asteroid's center of inertia be headed straight for the Black hole's center of inertia.

 

At this point, you need to give me your formula for gravity.

[swansont]

']Ok' date=' I can't keep this straight in my mind.

 

When an object accelerates close to c, does its gravity increase, or is it just inertial mass that increases? I dont think that gravity increases, but I want to make sure on this.

 

(you can explain your answer if you feel like it, but really just saying which is correct would be fine)[/quote']

 

If the mass of the object increases, it must also increase if you are moving, because you can't say which one of you is actually at rest.

[ydoaPs]

what if one of them are standing on a planet? then they can tell who is accelerating.

[Johnny5]

what if one of them are standing on a planet? then they can tell who is accelerating.

 

Case where one of them is standing on the surface of a planet.

 

In such case, the centers of mass of each object are at rest with respect to one another. So how in the world could you specify which one is accelerating?

 

In other words, define a frame, whose origin is located where the center of mass of the two objects is, and let the center of mass of each object be at rest in this frame. This completely specifies what I will henceforth refer to as the CM frame.

 

In this frame, neither of the centers of mass of the two objects is moving, so that the speed v, of either of these two objects in this frame is equal to zero.

 

Acceleration in a frame is defined as:

 

[math] \vec a = \frac{d\vec v}{dt} [/math]

 

Where [math] v [/math] is velocity. The magnitude of the velocity vector is the speed of the thing in question, in whatever frame you have defined it's speed in.

 

Since the speed of the center of mass of either object in the CM frame is zero, neither is accelerating in that frame.

[ydoaPs]

Case where one of them is standing on the surface of a planet.

 

In such case' date=' the centers of mass of each object are at rest with respect to one another. So how in the world could you specify which one is accelerating?[/quote']

 

gravitational ACCELERATION. the guy standing on a planet is accelerating.

[Johnny5]

gravitational ACCELERATION. the guy standing on a planet is accelerating.

 

 

In what frame?

 

You can't just say something is accelerating, without explaining what frame it is accelerating in.

[swansont]

gravitational ACCELERATION. the guy standing on a planet is accelerating.

 

If that acceleration is independent of the center-of-mass velocity, it is immaterial. We're on earth, and we can tell we're rotating, but we can't tell with what speed the whole planet or solar system is moving, unless we reference it to something else.

[[Tycho?]]

If the mass of the object increases, it must also increase if you are moving, because you can't say which one of you is actually at rest.

 

Well whatever. Does the gravitation field increase as mass increases? In any case, can it happen, or is gravity not involved in this?

[swansont]

I refer you to the FAQ hosted by Baez here which says, in part:

 

Relativistic mass, which increases with the velocity and kinetic energy of an object, cannot be blindly substituted into formulae such as the one that gives the radius for a black hole in terms of its mass. One way to avoid this is to not speak about relativistic mass and think only in terms of invariant rest mass

 

There's also an included link about "relativistic mass." The cautionary tale here is that you need to define your terms and then be consistent in how you use them. Blind substitution into equations and mixing/marching gets you in trouble. GR is complicated, and trying to distill it down to a simple form causes problems. Which should not be particularly surprising.

[Johnny5]

gravitational ACCELERATION. the guy standing on a planet is accelerating.

 

Explain gravitational acceleration to me, when you get a chance.

 

Thank you

[[Tycho?]]

Explain gravitational acceleration to me' date=' when you get a chance.

 

Thank you[/quote']

 

 

Objects in a gravitational field experience an acceleration.

[Pangloss]

Which (as Einstein discovered) is identical in nature to any other kind of acceleration.

[Johnny5]

'']Objects in a gravitational field experience an acceleration.

 

Take an object, like a pebble in your hand. Hold it at a constant distance above the center of the earth.

 

Now, you have to choose a frame to define the acceleration of the object in.

 

In the rest frame of the sun, for sure that stone has a nonzero acceleration.

 

BUT, and here is the really important point. Frame switch to the rest frame of the earth.

 

Somewhere inside the earth, is the center of mass of the earth. And let the pebble be roughly spherical, so that we can ensure that it's center of mass is inside of it.

 

Ok so now there is a nonzero distance between the center of inertia of the earth, and the center of inertia of the pebble.

 

And since you aren't raising or lowering your hand in the earth frame, this center to center distance is constant in time (at least for now).

 

Now switch back to the rest frame of the sun.

 

In that frame, the earth is spinning, so that the radial vector from the center of the earth to the center of the pebble is rotating. And that means that the center of the pebble is accelerating in the sun frame.

 

Now switch to the earth frame where the radial vector is an axis of the frame.

 

Therefore, the location of the center of inertia of the pebble isn't moving in this frame. That means this now...

 

That means that the pebble is at rest in this frame.

 

Certainly the pebble could be spinning in your hand, but the center of mass of that pebble is at rest in this frame.

 

And when something is at rest in a frame, then its speed in the frame is zero. And this pebble is remaining at rest in this frame, hence it's speed isn't changing. Now here is the definition of acceleration:

 

[math] \vec a = \frac{d\vec v}{dt} [/math]

 

As you can see acceleration is a vector quantity. That means that it has magnitude, as well as direction.

 

Now, dt is strictly positive, and it certainly is non-infinite, hence the only way the LHS (left hand side) of the equation above can be zero, is if the numerator of the RHS is zero.

 

Now here is the numerator of the RHS:

 

[math] d\vec v [/math]

 

ok so

 

That quantity is a differential of a velocity vector.

 

Velocity is speed in a frame, times direction of motion in the frame.

 

Right now, we are in in a frame in which the center of mass of earth is at rest.

 

Now, there are many reference frames in which the center of mass of earth is at rest.

 

In some of them, a fixed point on the surface of earth may be orbiting the CM of earth. But other frames we can concieve of have fixed points on the surface being at rest, in addition to the CM of earth being at rest. And this is the type of frame that the pebble is in, because it is hovering at a height h over the the same spot on the ground.

 

So the point is, that the speed of the center of inertia of the pebble in this frame is zero, and hence the product of the speed of the pebble in this frame, and it's direction of motion in this frame must be zero, since its speed is zero, regardless of what it's direction of motion was previously, or later (since it's speed is zero over some interval of time in this frame, as measured by a clock at rest in this frame).

 

With all of this keenly in mind, it must be the case that the acceleration of the pebble in this frame of reference is zero.

 

It is only when you release the pebble, that it begins to fall, or using other words... it is only when you release the pebble that it begins to accelerate in this reference frame.

 

My thoughts on this matter are improving.

 

Now, once you release it, it has a measurable acceleration of g.

 

But whilst you are holding that pebble in your hand, it has NO acceleration in this frame.

 

On the other hand, you can feel the weight of the pebble at all moments in time for which the pebble has no acceleration.

 

Now, suppose you use Newton's second law of motion to describe the weight of this pebble, as being measured by your hand (or equivalently a scale which you are holding, with the pebble on top).

 

If you assume that the weight of the object is equivalent to its inertial mass, times its acceleration in this reference frame, then you will run into an immediate contradiction, and here's why:

 

Suppose you write that force is equal to inertial mass, times acceleration, like this:

 

[math] \vec F = M \vec a [/math]

 

We are already clear on the fact that the acceleration of the object in this frame is zero, and it's weight is certainly nonzero. So if we set W equal to F in the formula above we get this:

 

[math] \vec W = M \vec 0 [/math]

 

In an earlier discussion with Tom Mattson, Mr Mattson showed me a simple argument proving that a vector quantity is equal to the zero vector if and only if its magnitude is zero. So let's write this:

 

 

[math] W (dir \vec W) = (M)( 0 ) ( dir \vec 0 ) [/math]

 

In the formula above, from left to right, W is what is measured by the scale, (and is certainly nonzero), dir W points towards the center of the earth, M is the inertial mass of the pebble, 0 is the magnitude of the acceleration of the center of inertia of the pebble in this frame, and ( dir \vec 0 ) is the same vector as (dir \vec W), since the statement is supposedly true in this frame.

 

So dividing both sides by dir W, we obtain the following scalar equation:

 

[math] W = M 0 [/math]

 

It is an algebraic fact that zero times any finite quantity is equivalent to zero, and M is finite, hence M times zero is equal to zero, hence the RHS of the statement above is equivalent to zero. However, the LHS is the weight of the pebble as measured by the scale (or our hand), and is certainly not equivalent to zero, in fact it is positive. Therefore, we have reached a contradiction, W is, and isn't zero. Hence, it cannot be the case that the weight of the pebble is equivalent to its inertial mass times its acceleration in the frame.

 

But Newton's law is not F = M a.

 

Here is Newton's second law, expressed mathematically:

 

[math] \vec F = \frac{d\vec P}{dt} [/math]

 

In the formula above, P is the momentum of the pebble in this frame, note that the momentum of an object is a frame dependent quantity. In other words, it is not usually the case that the momentum of an object in one frame is the same as its momentum in another frame.

 

Now, the definition of momentum is as follows:

 

[math] \vec P = M \vec v [/math]

 

In the formula above, M is the inertial mass of an object (which is considered to be a property of the object, and not a frame dependent quantity), and vector v is the velocity of the object in the frame (which is intuitively a frame dependent quantity).

 

Now, using the differential calculus we can show that:

 

[math] \vec F = \frac{d\vec P}{dt} = \frac{d(M\vec v)}{dt} = M \frac{d\vec v}{dt} + \vec v \frac{dM}{dt} [/math]

 

We know that dv/dt =0, so let us continue our assumption that the weight of the pebble is given by Newton's second law. We then obtain:

 

[math] \vec W = \vec v \frac{dM}{dt} [/math]

 

Now, the LHS we know is nonzero, and on the RHS, there is multiplication by v, the speed of the pebble in this frame, which we know is zero, and remaining zero in this frame. Therefore, again we reach a contradiction. Therefore, it cannot be the case that the weight of the object satisfies Newton's second law in this reference frame. That is impossible.

[J.C.MacSwell]

Take an object' date=' like a pebble in your hand. Hold it at a constant distance above the center of the earth.

 

Now, you have to choose a frame to define the acceleration of the object in.

 

In the rest frame of the sun, for sure that stone has a nonzero acceleration.

 

BUT, and here is the really important point. Frame switch to the rest frame of the earth.

 

Somewhere inside the earth, is the center of mass of the earth. And let the pebble be roughly spherical, so that we can insure that it's center of mass is inside of it.

 

Ok so now there is a nonzero distance between the center of inertia of the earth, and the center of inertia of the pebble.

 

And since you aren't raising or lowering your hand in the earth frame, this center to center distance is constant in time (at least for now).

 

Now switch back to the rest frame of the sun.

 

In that frame, the earth is spinning, so that the radial vector from the center of the earth to the center of the pebble is rotating. And that means that the center of the pebble is accelerating in the sun frame.

 

Now switch to the earth frame where the radial vector is an axis of the frame.

 

Therefore, the location of the center of inertia of the pebble isn't moving in this frame. That means this now...

 

That means that the pebble is at rest in this frame.

 

Certainly the pebble could be spinning in your hand, but the center of mass of that pebble is at rest in this frame.

 

And when something is at rest in a frame, then its speed in the frame is zero. And this pebble is remaining at rest in this frame, hence it's speed isn't changing. Now here is the definition of acceleration:

 

[math'] \vec a = \frac{d\vec v}{dt} [/math]

 

As you can see acceleration is a vector quantity. That means that it has magnitude, as well as direction.

 

Now, dt is strictly positive, hence the only way the LHS (left hand side) of the equation above can be zero, is if the numerator of the RHS is zero.

 

Now here is the numerator of the RHS:

 

[math] d\vec v [/math]

 

ok so

 

That quantity is a differentital of a velocity vector.

 

Velocity is speed in a frame, times direction of motion in the frame.

 

Right now, we are in in a frame in which the center of mass of earth is at rest.

 

Now, there are many reference frames in which the center of mass of earth is at rest.

 

In some of them, a fixed point on the surface of earth may be orbiting the CM of earth. But other frames we can concieve of have fixed points on the surface being at rest, in addition to the CM of earth being at rest. And this is the type of frame that the pebble is in, because it is hovering at a height h over the the same spot on the ground.

 

So the point is, that the speed of the center of inertia of the pebble in this frame is zero, and hence the product of the speed of the pebble in this frame, and it's direction of motion in this frame must be zero, since its speed is zero, regardless of what it's direction of motion was previously, or later (since it's speed is zero over some interval of time in this frame, as measured by a clock at rest in this frame).

 

With all of this keenly in mind, it must be the case that the acceleration of the pebble in this frame of reference is zero.

 

It is only when you release the pebble, that it begins to fall, or using other words... it is only when you release the pebble that it begins to accelerate in this reference frame.

 

My thoughts on this matter are improving.

 

Now, once you release it, it has a measurable acceleration of g.

 

But whilst you are holding that pebble in your hand, it has NO acceleration in this frame.

 

On the other hand, you can feel the weight of the pebble at all moments in time for which the pebble has no acceleration.

 

This is a very commonly used frame, perhaps the most common. Be careful not to use this frame in extremes however. The speed of light is not constant in this frame.

[Johnny5]

This is a very commonly used frame, perhaps the most common. Be careful not to use this frame in extremes however. The speed of light is not constant in this frame.

 

ok.

[swansont]

Therefore, again we reach a contradiction. Therefore, it cannot be the case that the weight of the object satisfies Newton's second law in this reference frame. That is impossible.

 

We? You might want to consider that this physics has held up to scrutiny for a while, so the odds that you've found an inconsistency is quite small. When you reach a "contradiction" you might first want to check your own work.

 

 

Why is the pebble suspended above the earth, and not moving?

[Johnny5]

 

Why is the pebble suspended above the earth' date=' and not moving?[/quote']

 

The pebble is suspended above the earth, and not moving in this particular frame because... someone is holding it in their hand, at a constant height above the surface of the earth.

 

If we take the human out of the scenario, we might just as well have had the pebble sitting on the ground, or even better yet, have there be a scale resting at one location on the ground, and have the pebble placed on the scale.

 

Therefore, you can see that the pebble's weight it nonzero, by looking directly at the scale.

 

If we then equate the weight to dP/dt, since the v term is zero, the v dm/dt term is zero, and since the accleration a is zero, the ma term is zero as well. Hence you will have W>0 and W=0.

 

W is a real number, and we have trichotomy:

 

For any real number r: r=0 XOR r>0 XOR r<0

 

From which contradiction can be reached.

 

The assumption must be negated, therefore you cannot equate weight to dP/dt. Something else needs to be done.

 

 

For example, in quantum mechanics, it is impossible for anything to have a speed of absolute zero. I can prove this if you want. But if we make v extremely small, there could be a contribution from the v dm/dt term. But I assumed that the inertial mass m of the pebble was constant, which makes dm=0. I am not sure right now what to do next.

 

Actually, if we use quantum physics, then the acceleration term would be nonzero as well. Right now I just need to mull this over. Because we can sucessfully solve for the acceleration due to gravity by equating GMm/rr with ma.

 

you get a simple ode.

 

Something is bugging me.

 

Kind regards swansont

[swansont]

The pebble is suspended above the earth' date=' and not moving in this particular frame because... someone is holding it in their hand, at a constant height above the surface of the earth.

 

If we take the human out of the scenario, we might just as well have had the pebble sitting on the ground, or even better yet, have there be a scale resting at one location on the ground, and have the pebble placed on the scale.

[/quote']

 

Which means there's another force on it, since the net force is zero. That other force is absent in your analysis. Which is why W <> 0, contrary to one of your steps.

[Johnny5]

Which means there's another force on it, since the net force is zero. That other force is absent in your analysis. Which is why W <> 0, contrary to one of your steps.

 

Sum of forces equals zero... yes i know that

 

force down due to gravity...

 

force of hand on pebble due to electrodynamic repulsion.

 

Is this what you mean...

 

[math] \vec W + \vec F_e = \frac{d\vec P}{dt} = 0 [/math]

 

??

[J.C.MacSwell]

When it is not an inertial frame the forces do not have to net to zero to have the particle/body remain stationary in that frame. It often requires a pseudo force (coriollis etc.) to accomplish this.

[swansont]

Sum of forces equals zero... yes i know that

 

force down due to gravity...

 

force of hand on pebble due to electrodynamic repulsion.

 

Is this what you mean...

 

[math] \vec W + \vec F_e = \frac{d\vec P}{dt} = 0 [/math]

 

??

 

Yes.

[Johnny5]

When it is not an inertial frame the forces do not have to net to zero to have the particle/body remain stationary in that frame. It often requires a pseudo force (coriollis etc.) to accomplish this.

 

You better explain this one more.

 

Also, in a previous post in this thread, you stated the speed of light isn't constant in the frame in question. Could you say why please?

[Johnny5]

Originally Posted by Johnny5

Sum of forces equals zero... yes i know that

 

force down due to gravity...

 

force of hand on pebble due to electrodynamic repulsion.

 

Is this what you mean...

 

[math] \vev W + \vec F_e= \frac{d\vec P}{dt} = 0 [/math]

 

??

 

 

Yes.

 

I thought so. Yes, you are absolutely right. So what is the correct way to analyze this problem?

 

For sure the acceleration of the center of inertia of the pebble in this frame is zero, that is plain to see.

 

I don't want to write W=mg, I would rather write:

 

[math] \vec W = -\frac {GMm}{r^2} \hat r [/math]

 

Is this correct...

 

 

[math] -\frac {GMm}{r^2} \hat r + \vec F_e = 0 [/math]

 

Where F_e is the electric repulsion force

 

??

 

How would you write F_e

???

 

Which are the electrons which participate in the interaction?

[swansont]

How would you write F_e

???

 

Which are the electrons which participate in the interaction?

 

I wouldn't write it out as an electrostatic force. It's way too complicated and probably not relevant to the problem.