Exploring Newton's "G" in Relativity

[Hebob80]

The Big "G"

I've been studying relativity for a bit over 2 years now, in this time I've come across Newton's G in various equations. For example: http://en.wikipedia.org/wiki/Gravitational_time_dilation#Outside_a_non-rotating_sphere

t_o = t_f sqrt(1 - 2GM/rc²)

It was my understanding that Einstein disproved Newton's concept of gravity, so I'm really wondering - what am I not understanding that is making it okay for us to use Newton's G?

As far as I know, G was derived by:

F= G ( m₁ m₂ / r²)
or
G = Fr² / m₁ m₂

Which, I thought was an inaccurate statement...

Someone help

 

[Dale]

Hi Hebob80, welcome to PF!

Newton's theory of gravity is known to work very well in many situations. One of the prerequisites of any new theory is that it must reduce to the old theory in all situations where the old theory is know to work. This means that GR must reduce to Newtonian gravity in the classical limit. So it is not surprising to see G in the equations of GR.

 

[Hebob80]

My 8th grade algebra teacher told me that pi was 7/22, and I don't mean to be rude, but by your argument, that works very well in many situations as well; however, it does nothing towards proving it is a valid variable to use. If you were to say that the equations that use G were only approximations, I would find that acceptable, is this the case?

 

[Dale]

My 8th grade algebra teacher told me that pi was 7/22, and I don't mean to be rude, but by your argument, that works very well in many situations as well; however, it does nothing towards proving it is a valid variable to use. If you were to say that the equations that use G were only approximations, I would find that acceptable, is this the case?

No, the equations that use G are not only approximations to Newtonian physics.

I don't understand your argument. Pi is a mathematical constant which can be calculated to any arbitrary degree of precision using only math. What does it have to do with G which is a physical constant whose value can only be determined by experiment?

If G were not in the Einstein Field Equations then how would you be able to get Poisson's equation in the classical limit?

 

[DrGreg]

My 8th grade algebra teacher told me that pi was 7/22

I hope he or she said 22/7, not 7/22! Still wrong, of course.

If you were to say that the equations that use G were only approximations, I would find that acceptable, is this the case?

Yes, Newton's equations are only approximations to the GR equations, but it's the same G in all of them.

For example, according to Newton the acceleration due to gravity at a distance r from the centre of a planet of mass M is

$$\frac{GM}{r^2}$$​

but in GR the equivalent formula is

$$\frac{GM}{r^2 \cdot \sqrt{1 - \frac{2GM}{rc^2}}}$$​

although in this case r is not "ruler distance" but equals $C/2\pi$ where C is the circumference of a circle drawn round the planet at that height. (This is for the idealised case of a non-rotating spherically symmetric planet; it gets more complicated for real planets that are ellipsoidal and rotating.)

The two formulas are approximately the same provided r is very large compared with 2GM/c² (which is tiny for planets and many stars).

 

[Hebob80]

Then, perhaps, I was not clear enough in my original question, let me make this as simple as possible:

If:

F = G ( m₁ m₂ / r² ) <--- this is wrong
G = Fr² / m₁ m₂ <--- this is wrong
G <--- this is wrong

Then:

Every equation that uses G is wrong.

No?

 

[Dale]

Those equations are not wrong. They have been tested for centuries and have been confirmed many times with a wealth of empirical data.

You seem to think that new theories disprove existing theories in domains where they have been experimentally verified. They do not. New theories extend or generalize existing theories to new domains where they have not been verified or where they have been contradicted. But it is absolutely essential the new theories agree with previous theories wherever the previous theories have been verified.

 

[Hebob80]

Ah HA! Thank you for helping me with this... although I don't like it in the least

I can assure you, I'll be back with more questions!

 

[TurtleMeister]

I have wondered about this myself. It appears from DrGreg's equations that the greater the mass and density, the greater the discrepancy between Newton and GR. And since the value of G is usually obtained by torsion balance experiments, would it be true that the measured value would only be inaccurate (within limits of the equipment and method of course) if the test masses were extremely dense and massive?

 

[Hebob80]

$$\frac{GM}{r^2 \cdot \sqrt{1 - \frac{2GM}{rc^2}}}$$​

although in this case r is not "ruler distance" but equals $C/2\pi$ where C is the circumference of a circle drawn round the planet at that height.

For this equation, do the measurements of distance or mass need to be in any particular unit?

 

[HallsofIvy]

No, but the value of G will change depending upon the units of distance and mass.

 

[DrGreg]

For this equation, do the measurements of distance or mass need to be in any particular unit?

No, but the value of G will change depending upon the units of distance and mass.

Exactly. If you measure distance in metres, acceleration in m/s² and mass in kg, then G has the value G = 6.67428 × 10⁻¹¹ m³ kg⁻¹ s⁻², but in other units it would have different value e.g. 0.8650 cm³ g⁻¹ hr⁻².

Relativists often choose units so that G = c = 1.

 

[Hebob80]

I have wondered about this myself. It appears from DrGreg's equations that the greater the mass and density, the greater the discrepancy between Newton and GR. And since the value of G is usually obtained by torsion balance experiments, would it be true that the measured value would only be inaccurate (within limits of the equipment and method of course) if the test masses were extremely dense and massive?

TurtleMeister brings up a good point. If Newton's equations and Einstein's equations disagree at some point, any point, doesn't that by default make one of them incorrect?

 

[Dale]

No, it makes Newton's equations a specific limit of the EFE. Solutions to Newton's equations are also solutions to Einstein's. They agree.

Should we avoid discussing plane waves simply because they are a specific case of Maxwell's equations rather than the full general Maxwell equations?

 

[Buckleymanor]

I have wondered about this myself. It appears from DrGreg's equations that the greater the mass and density, the greater the discrepancy between Newton and GR. And since the value of G is usually obtained by torsion balance experiments, would it be true that the measured value would only be inaccurate (within limits of the equipment and method of course) if the test masses were extremely dense and massive?

What if the test masses were small and less dense, would they not still be influenced by other surrounding objects more dense and massive like the earth.

 

[TurtleMeister]

Maybe I should rephrase my question. In post #5 DrGreg gave a GR equivalent to Newton's equation for acceleration. And as he stated, the two are approximately equal while r is very large compared with 2GM/c². However, if r is not very large compared with 2GM/c², then the GR equation gives the more accurate result. My question is, does the same also apply to the Newtonian equation used for the determination of the value of G (G = Fr² / m₁ m₂)? For example, if I want to do a torsion balance experiment using very massive test masses of very great density, what GR equivalent equation must I use in place of the Newtonian equation in order to obtain the most accurate result?

What if the test masses were small and less dense, would they not still be influenced by other surrounding objects more dense and massive like the earth.

The earth? No. the whole point of using a torsion balance is to cancel the influence of the Earth's gravitational field. Other surrounding objects? Yes. But the distances between the test masses and the attractor mass is usually very short, making the influence of surrounding objects insignificant. But this really has nothing to do with what we're talking about.

 

[Buckleymanor]

Maybe I should rephrase my question. In post #5 DrGreg gave a GR equivalent to Newton's equation for acceleration. And as he stated, the two are approximately equal while r is very large compared with 2GM/c². However, if r is not very large compared with 2GM/c², then the GR equation gives the more accurate result. My question is, does the same also apply to the Newtonian equation used for the determination of the value of G (G = Fr² / m₁ m₂)? For example, if I want to do a torsion balance experiment using very massive test masses of very great density, what GR equivalent equation must I use in place of the Newtonian equation in order to obtain the most accurate result?


The earth? No. the whole point of using a torsion balance is to cancel the influence of the Earth's gravitational field. Other surrounding objects? Yes. But the distances between the test masses and the attractor mass is usually very short, making the influence of surrounding objects insignificant. But this really has nothing to do with what we're talking about.

Sorry for enquireing you might like to have a look at wikis big G

The accuracy of the measured value of G has increased only modestly since the original Cavendish experiment. G is quite difficult to measure, as gravity is much weaker than other fundamental forces, and an experimental apparatus cannot be separated from the gravitational influence of other bodies.

I just took for granted that other bodies included the Earth but it's obviously wrong.

 

[Hebob80]

No, it makes Newton's equations a specific limit of the EFE.

Specific limit? Can you clarify what, exactly this means?

 

[edguy99]

TurtleMeister brings up a good point. If Newton's equations and Einstein's equations disagree at some point, any point, doesn't that by default make one of them incorrect?

They are based on different assumptions:
In Einstein's second paper on relativity in 1905, he explicitly concludes
"Radiation carries inertia between emitting and absorbing bodies". It is important that not only does something receive a "kick" from the momentum of the energy, but the internal inertia (i.e., the inertial mass) of the body is actually increased. (from mathpages.com)

A note from Newton's Principles definition #4 in 1687 conflict with Einstein:
Impressed Force - This force conflicts in the action only; and remains no longer in the body when the action is over.

 

[Dale]

Specific limit? Can you clarify what, exactly this means?

If you take the EFE and solve them for the situation where you have a central point mass and a test mass which is far away from the central mass and moving slowly then you get Newton's equations.

This is similar to how Coulomb's law is a limit of Maxwell's equations for a single point charge which is stationary. Maxwell's equations hold in general. Coulomb's law is not wrong, it is simply a solution of Maxwell's equations which only holds in certain circumstances.