# I am attempting a condition where Newton and Einstein agree

1. Jun 7, 2007

### duordi

On this site

http://www.geocities.com/physics_world/mass_paper.pdf

It states that the 3 force can be written as....

F = M1 x A1 + M2 x A2

Where 1 is the longitudinal component of acceleration and 2 is the transverse component of acceleration.

It appears that if there is no transverse acceleration then the 3 force resolves to a form of Newtons equation.

If I wanted to determine the gravitational effect from a distant mass which had no transverse acceleration or velocity but very large longitudinal acceleration or velocities....
Can I use Newtons gravitational equation, substitute relativistic mass and get a correct calculated force.

F = G x ( Mass of observer ) x (Relativistic mass of distant object)
-----------------------------------------------------------
( Distance as measured by the observer)^2

Last edited: Jun 7, 2007
2. Jun 7, 2007

### cristo

Staff Emeritus
No, I'm afraid you can't.

3. Jun 7, 2007

### duordi

How can I limit the motion or other constraints so it works?

Duane

4. Jun 7, 2007

### rbj

duordi, you should learn how to use LaTeX here:

$$F = m_1 a_1 + m_2 a_2$$

$$F = G \frac{m_{obs} m_{obj(rel)}}{r^2}$$

looking at real math equations is nice and is something you don't get on USENET where you have to use "ASCII math".

5. Jun 8, 2007

6. Jun 8, 2007

### pervect

Staff Emeritus
No, you can't, not if you want to get a sensible answer. Interpreting gravity as a "force" works only at low velocities. At high velocities, the effects of spatial curvature become very important (in the usual Schwarzschild coordinates) but cannot be modeled as a "force". (The effect of spatial curvature near a large mass shows up in the deflection of light via a massive body, for instance.)

To calculate the motion of the body, it's easiest not to deal in forces at all, but to simply calculate the geodesic path.

If you really want to calculate forces (rather than the motion of the bodies), what you can calculate is the tidal force from the distant mass, which is well defined. It's also what you can actually measure. If you are in free-fall around a massive body, your acceleration will always be zero because you are in free fall, you won't be able to measure your acceleration due to gravity. There is no "gravitationally neutral" test particle that you can compare your motion to, so you can't determine the "force" of gravity by any sort of local experiment. All you can do is look at your accelerometer to confirm that you are in free-fall, which gives you no information about the "force" of gravity.

But while you can't measure the "force" of gravity, you can measure the tidal acceleration, the force per meter, by measuring the change in your accelerometer reading per unit distance.

You might ask - "If we can measure force/ meter, why can't we measure the total force"?

The answer is that to convert the force/meter into a total force, you have to integrate along a particular path. In Newtonian physics, this integral is independent of the path. In GR, the integral is NOT in general independent of the path, so there isn't any single value for force.

As far as the answer to your question goes in terms of tidal forces, the easiest way to phrase it is this:

If you have a test particle stationary near a massive body, the tidal force will be equal to the Newtonian tidal force if the Schwarzschild coordinate r is substituted for the "distance", i.e. the stretching components of the tidal force in the radial direction will be 2GM/r^2, and the compressive tidal forces will be GM/r^2, where r is the schwarzschild coordinate.

If you have a moving particle at the same location as the stationary particle, the tidal force readings won't change if the body moves directly towards or away from the central mass.

If you have a moving particle at the same location as the stationary particle but it is moving in such a direction as to orbit the massive body (i.e it is moving transversely), the tidal forces will increase. This has been worked out in detail in some old posts, there are some particular examples that were just mentioned in a recent thread on the speed of gravity https://www.physicsforums.com/showpost.php?p=1347429&postcount=10.

Last edited: Jun 8, 2007
7. Jun 9, 2007

### duordi

Three questions

How do I use LaTeX?

One of the posts said I should, but I can not find a button for it anywhere.

If GR does not have force then how does it handle inertia (force) or is this another hole in my understanding?

After doing some reading on GR it seams that one of the reasons GR and Newtons laws cannot be merged is that GR assigns gravity to energy not mass. Because of this even gravity fields ( which contains energy ) can cause gravity.

There was one example where Newtons gravitational force equation with relativistic mass substituted for stationary mass was the same as GR.
I could not determine what restrictive conditions cause this.

If you have time an explanation for the feeble minded would be helpful.

BTW I did order an intro to GR By Wheeler and someone else.
It was recommended on this site.
It is not my intension to remain in the dark forever.

Duane

8. Jun 9, 2007

### MeJennifer

Looks to me that you are confused here Pervect.

Do you have a specific reference for your claim that "spatial curvature becomes very important at high velocities"?

9. Jun 9, 2007

### pervect

Staff Emeritus
Oh, you're the one that's confused (as usual). (If this seems perhaps a bit too direct to be entirely polite, so are your remarks).

Perhaps, though, I've contributed to your confusion by being somewhat terse.

As far as references go - the deflection of light (which moves at 'c') is twice what one would expect. For a picture that shows how the space curvature in Schwarzschild coordinates causes light to deflect by the extra amount, see MTW pg 32 (box 1.6). It's not convenient to reproduce the diagram, but I'll type in the text:

A more mathematical approach would go like this:

Write down the geodesic equation for the Schwarzschild metric, parameterized in terms of proper time as we are considering the deflection of a particle. To consider light deflection, one would have the same equation in terms of an affine parameter rather than proper time.

$$\frac{d^2x^\lambda }{d\tau^2} + \Gamma^{\lambda}_{~\mu \nu }\frac{dx^\mu }{d\tau}\frac{dx^\nu }{d\tau} = 0\ ,$$

Note that the Christoffel symbol, $$\Gamma^0{}_{00}$$, represents a term which acts just like a Newtonian force as far as the equations of motion go in the Newtonian limit - it gives the second time (proper time) derivative of a position coordinate $d^2 x^i / d \ \tau^2$ a constant value.

In the Newtonian limit, for small velocities $dt / d\tau$ is nearly equal to one, and all the other partial derivatives in the geodesic equation are very small. So one can represent gravity as a force, represented given by the Christoffel symbol $\Gamma^0{}_{00}$

But in the case of high velocites, one cannot neglect the other partial derivatives.

One of the reasons I didn't go through all of this before, besides the fact that it takes time, was that I didn't think many readers had the background to understand it. (I'm not even sure if you do, Mejennifer, but it's worth a shot).

[add]Another thing that may be confusing you is that in this case I am explaining things in a very coordinate-dependent way. Note that to talk about space-curvature at all, one must pick out a specific coordinate system - in this case, I've picked out the Schwarzschild coordinate system.

Last edited: Jun 9, 2007
10. Jun 11, 2007

### MeJennifer

Using statements like "At high velocities, the effects of spatial curvature become very important (in the usual Schwarzschild coordinates)" do not help people increase their understanding, it is likely only to mislead them.

First of all curvature is a coordinate free property of spacetime.
Second, there is not such thing as a separate space and time curvature. What is space and what is time can be different for different classes of observers.
In a Schwarzschild coordinate chart the r and t coordinates do not represent a physical space and time. Pretending it does is completely wrong and is only going to confuse people.

For instructional purposes I recommend the Gullstrand-PainlevĂ© coordinate chart over the Schwarzschild coordinate chart.

11. Jun 11, 2007

### rbj

that was me... didn't want to nag, it's just that you can express yourself better mathematically if you use tex.

first of all, in the Math & Science Tutorials, there is a sticky thread near the top. this thread refers to this quick little cheat sheet.

https://www.physicsforums.com/misc/howtolatex.pdf

also, when you quote me (or anyone else using TeX for math) you can see, in the quoted text, what we are doing to make the equations we desire in our posts.

in addition, even though the keywords to kick it into TeX mode and out are different, the LaTeX language and syntax inside of Wikipedia is the same. so here is another good reference:

http://en.wikipedia.org/wiki/WP:MATH

but use "[ tex ]" and "[ /tex ]" instead of "< math >" and "< /math >" as you would editing Wikipedia.

i'll let someone more, bigger, better versed in GR than me take this.