General Relativity and Universal Gravitational Constant

[ktmsud]

TL;DR

Why is 'G' [Universal gravitational constant] in Einsteins field equation?

I don't know complex mathematics about general theory of relativity but i tried to understand it from internet. Somewhere i heard that in GTR gravity is explained as curvature of space-time not as force as in Newton's laws. If GTR doesn't explains gravity as force then why G which is constant of force equation is in Einstein's field equation? How and Why this constant appear in Field equations or GTR?

 

[Vanadium 50]

Somewhere i heard

Where? If you want us to clarify it, we need to see it.

How and Why this constant appear in Field equations or GTR?

I can show you "where". "Why" is "without it you get the wrong answer:
G_{\mu \nu} = 8\pi \frac{G}{c^4} T_{\mu \nu}

 

[Ibix]

If GTR doesn't explains gravity as force then why G which is constant of force equation is in Einstein's field equation?

You have it the wrong way around. $G$ appears in Newton's law of gravity because Newton's laws are an approximation to relativity, where $G$ can be seen as part of a unit conversion between stress-energy and curvature.

You can set it to 1 by a careful choice of units, if it bothers you.

 

[Dale]

TL;DR Summary: Why is 'G' [Universal gravitational constant] in Einsteins field equation?

If GTR doesn't explains gravity as force then why G which is constant of force equation is in Einstein's field equation? How and Why this constant appear in Field equations or GTR?

In any equation $G$ appears because you are using units that require it, such as SI units.

 

[ktmsud]

TL;DR Summary: Why is 'G' [Universal gravitational constant] in Einstein's field equation?

I heard that in GTR gravity is explained as curvature of space-time not as force as in Newton's laws

I forgot the source but I heard above quoted line more than once probably in YouTube videos. Is it common misconception about gravitation?
I just wanted to know, whether any reference of Newton's gravitational law is taken during derivation of that equation or Einstein found that it is the correct constant that should be inserted to get right formula independently? I know any particular value of G like $6.67*10^-11$ in SI unit is due to choice of system of units. But Why same G in two independent approach. Aren't their value same in same unit? Without knowing mathematics of gtr can it be understood?
ANSWERERS ARE ADVISED TO THINK ABOUT LETTER 'B' IN THE BEGINING OF THREAD BEFORE ANSWERING.

 

[berkeman]

ANSWERERS ARE ADVISED TO THINK ABOUT LETTER 'B' IN THE BEGINING OF THREAD BEFORE ANSWERING.

You have a peculiar sense of manners when asking others for their help. Just sayin'...

 

[ktmsud]

I am sorry for writing that line but how can a response like the first one from this thread help a high school student? I am here to learn. If it cannot be understood in simple terms. Okay! I'll quit. If question was not clearly stated let me clarify it. Everyone knows 'Everything is done in physics to correctly explain the behavior of nature'. Does stating it help a person who wants to know new thing?

 

[berkeman]

Somewhere i heard

The issue is that we don't allow that type of statement here at PF when starting a thread. We always require explicit references when asking questions in the technical forums. You are new here, so V50 also tried to help you after pointing that out.

Okay, back to the thread...

 

[Hill]

TL;DR Summary: Why is 'G' [Universal gravitational constant] in Einsteins field equation?

G which is constant of force equation

But G is not a "constant of force equation". It is rather a constant of gravitational acceleration: $$a=\frac {GM} {r^2}$$
It relates the source of gravity with its effects on movement of other bodies.
In GR, the source is more complicated than just a mass, and the effect is more complicated than just an acceleration, but G is still present in their relation.

 

[ktmsud]

Where?

We always require explicit references when asking questions

Here:

 

[berkeman]

LOL. YouTube videos are not an acceptable source. So many things to learn about how PF works!

(from the Rules under INFO at the top of the page):

Acceptable Sources:
Generally, discussion topics should be traceable to standard textbooks or to peer-reviewed scientific literature. Usually, we accept references from journals that are listed in the Thomson/Reuters list (now Clarivate):

https://mjl.clarivate.com/home

Use the search feature to search for journals by words in their titles.

 

[Dale]

I know any particular value of G like 6.67∗10−11 in SI unit is due to choice of system of units. But Why same G in two independent approach. Aren't their value same in same unit?

As I said above, it is simply a matter of the units. Nothing more. $G$ is simply a unit conversion factor.

In SI units, Newton’s gravitational law $G$ converts between $\mathrm{N}$ on the left and $\mathrm{kg^2 m^{-2}}$ on the right. In the Einstein field equations $G c^{-4}$ converts between $\mathrm{m^{-2}}$ on the left and $\mathrm{Pa}$ on the right.

In natural units the terms on both sides of both equations have the same units and no conversion is needed (or equivalently the conversion factor is 1)

The reason it shows up in both equations is simply because both equations use a combination of units that require it.

 

[Dale]

YouTube videos are not an acceptable source.

Although they are not acceptable sources for answers we usually permit them as sources for questions, as the OP did here. It is reasonable for someone to be confused from a pop sci presentation and come here for clarity.

 

[ktmsud]

As I said above, it is simply a matter of the units. Nothing more. $G$ is simply a unit conversion factor.

In SI units, Newton’s gravitational law $G$ converts between $\mathrm{N}$ on the left and $\mathrm{kg^2 m^{-2}}$ on the right. In the Einstein field equations $G c^{-4}$ converts between $\mathrm{m^{-2}}$ on the left and $\mathrm{Pa}$ on the right.

In natural units the terms on both sides of both equations have the same units and no conversion is needed (or equivalently the conversion factor is 1)

The reason it shows up in both equations is simply because both equations use a combination of units that require it.

I (Now) Understood function of G in those equations. We use Formula/Equations to derive other Formulae/Equations and which (to my knowledge) are either definitions, or laws or derived from other equations. I wanted to know if Einstein used any form of Newton's law of gravitation to derive his equation that G appeared on his equation? And if the derivation is not dependent on Newton's laws, how GTR defines 'G'? In Newton's Law it is easy to define as the gravitational force experienced by two unit masses separated by unit distance apart in vacuum. If there is no such things as Gravitational force in general relativity how is G defined?

With Dale's answers I almost understood what I originally wanted to understand.

 

[renormalize]

I wanted to know if Einstein used any form of Newton's law of gravitation to derive his equation that G appeared on his equation?

Yes he did. One piece (the time-time component) of the 10 Einstein field equations in general relativity matches, for weak gravitational fields, to exactly Newton's field equation for classical gravity. Since $G$ appears in the classical equations, this match shows that it also appears within the Einstein equations.

 

[Ibix]

If there is no such things as Gravitational force in general relativity how is G defined?

You can derive Einstein's Field Equations in various ways, but you end up with an unknown constant in the maths. So you do something like predict orbital periods of the planets, which depend on the unknown constant. Then you compare to real measurements and find the value of the constant. If you can't get a unique value out then your theory is incorrect - the thing it says is constant, isn't. If you do get a constant your theory isn't wrong (yet).

Einstein could take a shortcut, though. We already had a theory of gravity that made astoundingly good predictions as long as things were moving slowly and gravity was fairly weak. So he just had to work out the slow-speed weak-field form of his equations and compare to Newton's prediction, and deduce the value of the constant (the $8\pi G/c^4$ in post #2) from that.

So, historically, $G$ appears in the Einstein Field Equations because Newtonian gravity is a pretty good theory and Einstein's theory must make the same predictions in slow-speed weak-field cases. Conceptually, though, it's the other way around. The constant $G$ in Newton's approximation to GR is $c^4/8\pi$ times the constant that's in our better theory. We keep $G$, though, because why bother changing all the textbooks and making all of the easy maths messier?

 

[PeroK]

Whatever theory you use, the attraction between two masses is measurable. See, for example:

https://en.m.wikipedia.org/wiki/Cavendish_experiment

That establishes experimentally a (universal) constant $G$ that must be part of any theory of gravity. Existing experimental results aren't invalidated when a new theory is adopted.

Even more simply, if you drop an object it falls to Earth with a given acceleration that depends only on the mass and radius of the Earth and the gravitational constant. GR has to explain that as well, even if you don't want to call gravity a force.

 

[Ibix]

Existing experimental results aren't invalidated when a new theory is adopted.

This observation is sometimes called the Correspondence Principle, @ktmsud. New theories usually explain more phenomena than old ones but, as PeroK says, the old experimental results don't go away. The new theory must accurately predict them too. That means that the new theory must simplify to the old one in some circumstances (slow-speed weak-gravity, in the case of GR), otherwise it would make different predictions and be wrong. The new theory must correspond to the old in the cases we've already tested.

There's a whole formal mathematical process for working out what bits of a theory become too small to care about and under what circumstances. The process also tells you where your older theory will start making measurably wrong predictions, and that tells you where to look for confirmation or falsification of your new theory. In the case of GR it was already known that Newtonian gravity didn't quite predict the orbit of Mercury correctly, and Einstein was able to show that his theory did.

 

[A.T.]

If there is no such things as Gravitational force in general relativity how is G defined?

Even in Newtonian Mechanics you can define G without mentioning a "force", based on the acceleration of a falling test mass. Those accelerations must be predicted by GR as well, so the equivalent definitions of G apply there too.

 

[pervect]

I'd like to "upvote" A.T.'s observation, force / unit mass is the same thing as acceleration. This illustrates that forces don't really play a fundamental role in GR. As early as Gallileo, it was noted that all test masses fall at the same rate. This means we can just as easily talk about the acceleration of gravity on the Earth's surface (little g), as we could the "force" of gravity even in Newtonian theory.

In fact, in Newton's gravitational laws, there are two sorts of mass, gravitational mass and inertial mass. From the Newtonian point of view, it's just a coincidence that the two are always equal. GR basically argues that this is not a coincidence, that it is the result of a funamental law or principle, called "The equivalence principle".

There are actually several variant forms of the equivalence principle, the one I prefer (and the one I am talking about in this post) is the "weak" equivalence principle. This basically states the idea I presented earlier - the idea that all test masses (no matter what material they are composed of) fall at the same rate.

 

[DrGreg]

It's perhaps also pointing out that in relativity the concept of "acceleration" (specifically "proper acceleration") is a geometrical concept -- it's essentially the curvature of a curve (worldline) in spacetime, i.e. the distance-versus-time trajectory of an object that is accelerating relative to a freely falling object.

 

[Vanadium 50]

TL;DR Summary: Why is 'G' [Universal gravitational constant] in Einsteins field equation?

But do nut use the Einstein Field Equations, which I specifically asked about, in your answer.

but i tried to understand it [GE] from internet.

But don't treat mas if I understand what I said I understand OR I WILL YELL AT YOU.

We're really left with "explain this random YouTube video to me",. and most YouTube vidoes are junk. And take up a lot of time to watch. You've made it clear that our time is of no value to you, but it is of value to us.

It will serve you well to learn from this thrad - not just where the G's and c;s really go, but

1. Random YouTube videos are poor ways to learn and they may not even be correct.
2. When asking a question, don't use terms you don't understand.
3. Be the sort of person that people want to help.

If this message annoys you, I recommend printing it out and looking at it again in a couple of years.

 

[Dale]

Since there have been several posts mentioning the (correct) fact that general relativity reduces to Newtonian gravity in the usual limit, I think I should clarify my posts. The other posts explain why (if you use the same units) a constant that appears in Newtonian gravity must also appear in General Relativity. However, they do not explain why that constant appears in Newtonian gravity.

Again, that is due to the choice of units. Suppose that I made a unit system where length was measured in $\mathrm{m}$ and time in $\mathrm{s}$, but instead of measuring mass in its own new unit like SI suppose I made mass a derived unit with units of $\mathrm{m^3 s^{-2}}$. In these units, Newton’s gravitational law could be written as $$a=\frac{M}{r^2}$$ A unit mass of $1\mathrm{\ m^3 s^{-2}}$ would therefore be the mass that would cause a test mass to accelerate at $1 \mathrm{\ m s^{-2}}$ when separated by $1\mathrm{\ m}$. Thus in these units $G$ would not appear in the equations of Newtonian gravity nor in the equations of general relativity.

One $1\mathrm{\ m^3 s^{-2}}$ would be the same mass as $1.498 \ 10^{10}\mathrm{\ kg}$, but since they have different dimensions they would not be exactly interchangeable.

 

[pervect]

It's perhaps also pointing out that in relativity the concept of "acceleration" (specifically "proper acceleration") is a geometrical concept -- it's essentially the curvature of a curve (worldline) in spacetime, i.e. the distance-versus-time trajectory of an object that is accelerating relative to a freely falling object.

It's fairly easy to say that curvature can be related to relative acceleration between nearby geodesics per unit distance between said geodesics - which has units of (m / s^2) for the acceleration, and units of meters for the distance between them. When we divide the acceleration by the distance, we get units of (m/s^2) / meter, or units of 1/s^2. We can also note that mass and forces don't fundamentally enter into this geometric picture, because geodesics are a geometric concept, independent of forces - geodesics are curves that any test particle follows, we don't need to know the mass or composition of the test particle.

But I'm not sure how much sense that observation is going to make sense at the "B" level :(. I think it requires some background ideas of what a geodesic are, and that my foray here into what is basically the "geodesic deviation equation" is too brief to be comprehensible without already knowing the material.

Talking about space-time curvature at the "B" level is even harder. I tend to assume that most B-level posters are NOT all that familiar with special relativity, much less the geometric version of special relativity that is really needed here. Taylor & Wheeler's "Space-time Physics" is a good reference for this background material, but I'd put it at "I" level, and that's just the background for the "space-time" part of "space-time curvature", it doesn't talk about the "curvature" part at all.

Sector models (rather than the geodesic deviation equation) are an alternate approach to talk about curvature, which might also be useful, and there are even some papers about them that aren't too bad to follow. I talked a bit about this in https://www.physicsforums.com/threa...s-space-in-the-real-case.1050537/post-6972834.

Basically, as I see it, when we try to describe gravity as being due to space-time curvature, we need the reader to understand both space-time, and curvature, and if an understanding of either one (or most likely both) is lacking, putting it all together isn't going to work.

 

[PeterDonis]

curvature can be related to relative acceleration between nearby geodesics

That is spacetime curvature. @DrGreg was describing path curvature of a worldline (which is what proper acceleration is). Those are two different things.

 

[pervect]

That is spacetime curvature. @DrGreg was describing path curvature of a worldline (which is what proper acceleration is). Those are two different things.

I'm not sure I understand your point. What specific geometric entity (tensor, scalar) is "path curvature" supposed to represent in this context?

I was initially thinking you were making some point about space-time curvature being specific to a space-time manifold, (as opposed to a more general point). But I'm not clear if that's y our actual point. I will say that yes,, I was talking specifically about space-time in this context, as opposed to some other manifold that could have curvature.

 

[PeterDonis]

What specific geometric entity (tensor, scalar) is "path curvature" supposed to represent in this context?

The proper acceleration of the worldline. In other words, the 4-vector $A = d U / d\tau$, where $U$ is the 4-velocity, i.e., the unit tangent vector to the worldline, and $\tau$ is the proper time, i.e., the affine parameter along the worldline. In component notation it is $A^a = U^b \nabla_b U^a$.

I was initially thinking you were making some point about space-time curvature being specific to a space-time manifold, (as opposed to a more general point). But I'm not clear if that's y our actual point.

It's not. My actual point was just to clarify that @DrGreg, in what you quoted, was talking about path curvature, whereas you were talking about spacetime curvature, and that those are two different things.

I will say that yes,, I was talking specifically about space-time in this context, as opposed to some other manifold that could have curvature.

Path curvature is not the curvature of a manifold. It's the curvature of a worldline--basically the extent to which the worldline is not a geodesic (a geodesic has zero path curvature by definition).

 

[Dale]

a geodesic has zero path curvature by definition

Yes. A geodesic has zero path curvature even in a region where spacetime is curved. This highlights the fact that they are different