Superficial discussion of Flat and Curved Space Time

This post is a response to a post by exchemist on another thread.

Quoting from Richard A Mould's book on Basic Relativity, Chapter 8, page 221:
"To anticipate things a bit, general relativity makes a statement about how the metrical properties of space-time are affected by the presence of matter in space-time. Until the relationship between matter and metric is explicitly stated, we cannot be said to have left the domain of special relativity, even when working with non-inertial frames of reference. The move from special relativity to general relativity is like the move from a two-dimensional flat plane to a two-dimensional curved surface such as the surface of a sphere.

Every time I read that statement I have the same reaction. "Come on man, speak English !"

The best way I can think of this is,
If you got mass in the problem, you are doing general relativity. Else, you are doing special relativity.

Space-time was invented for special relativity. Its default structure is "flat", meaning, if you plot x vs t for some object, the x and t axes are straight (flat) lines. They define a plane. A point object moving at constant relative velocity takes a straight line path in that x/t plane. If the point object has no relative velocity, it takes a straight vertical path. On the other hand, a point object with a uniform acceleration takes a curved path in the x/t plane. That can only happen if there is a force acting on it.

General relativity says that energy (including mass) curves space-time. In particular, if you plot x vs t, the t axis is curved. When I first read about general relativity, I went out and stared up at the sky looking for some indication of the curvature of space-time created by the earth. It's not out there. The curvature in on the graph.

Here's a trick. You can take the curved path of a uniformly accelerating point object in a flat x/t plane and exchange the t axis with it. Now the point object follows a straight path and the t axis is curved. You have converted from flat space-time to curved space-time. The straight object path means you took away the force. The curved t axis means you added gravity. Equivalence Principle.

You should see an image file here.
SpaceTimeDiagrams.png



So you can describe the motion of an accelerating point object using either special or general relativity. But be forewarned. If you want to describe the motion using general relativity you have to know tensor calculus. Gulp.
 
mikelizzi said:
You should see an image file here.
SpaceTimeDiagrams.png
Click to expand...
No idea why that image won't display. I can display it directly in my browser, but SciFo doesn't like it. And it's not because of the uppercase characters in the URL either.


Anyway, I've re-uploaded it (and even waive my usual 'deputy moderator' fee):


SpaceTimeDiagrams.png
 
DaveC426913 said:
No idea why that image won't display. I can display it directly in my browser, but SciFo doesn't like it. And it's not because of the uppercase characters in the URL either.


Anyway, I've re-uploaded it (and even waive my usual 'deputy moderator' fee):


View attachment 5492
Click to expand...
The image displays only when I am logged in. I thought that was normal. It's a *.png file. Would a different file format behave differently?
 
Huh
mikelizzi said:
The image displays only when I am logged in. I thought that was normal. It's a *.png file. Would a different file format behave differently?
Click to expand...
I see it on my phone now, but it showed as broken on my lappie.

.png is a supported format; I double-checked.

It's a mystery...
 
mikelizzi said:
This post is a response to a post by exchemist on another thread.

Quoting from Richard A Mould's book on Basic Relativity, Chapter 8, page 221:
"To anticipate things a bit, general relativity makes a statement about how the metrical properties of space-time are affected by the presence of matter in space-time. Until the relationship between matter and metric is explicitly stated, we cannot be said to have left the domain of special relativity, even when working with non-inertial frames of reference. The move from special relativity to general relativity is like the move from a two-dimensional flat plane to a two-dimensional curved surface such as the surface of a sphere.

Every time I read that statement I have the same reaction. "Come on man, speak English !"

The best way I can think of this is,
If you got mass in the problem, you are doing general relativity. Else, you are doing special relativity.

Space-time was invented for special relativity. Its default structure is "flat", meaning, if you plot x vs t for some object, the x and t axes are straight (flat) lines. They define a plane. A point object moving at constant relative velocity takes a straight line path in that x/t plane. If the point object has no relative velocity, it takes a straight vertical path. On the other hand, a point object with a uniform acceleration takes a curved path in the x/t plane. That can only happen if there is a force acting on it.

General relativity says that energy (including mass) curves space-time. In particular, if you plot x vs t, the t axis is curved. When I first read about general relativity, I went out and stared up at the sky looking for some indication of the curvature of space-time created by the earth. It's not out there. The curvature in on the graph.

Here's a trick. You can take the curved path of a uniformly accelerating point object in a flat x/t plane and exchange the t axis with it. Now the point object follows a straight path and the t axis is curved. You have converted from flat space-time to curved space-time. The straight object path means you took away the force. The curved t axis means you added gravity. Equivalence Principle.

You should see an image file here.
SpaceTimeDiagrams.png



So you can describe the motion of an accelerating point object using either special or general relativity. But be forewarned. If you want to describe the motion using general relativity you have to know tensor calculus. Gulp.
Click to expand...
This much is fine. And indeed I never learnt tensors at university, as one doesen’t need them for quantum chemistry. SR I have some understanding of, for simple scenarios like the example of atmospheric muon lifetime.

My problem has been that I always thought GR started from treating acceleration and gravitation as in some way indistinguishable, which is how I picked up the - evidently incorrect - notion that one needs GR in order to handle either.
 
exchemist:

It is true that in GR acceleration due to gravity and the "gravitational force" of Newtonian physics are indistinguishable. Or, more accurately, in GR we can dispense with the idea of a "gravitational force" completely, since all observable effects of so-called "gravitational forces" can be reproduced by using an appropriately curved spacetime, in which all objects in gravitational "free fall" simply follow shortest-path trajectories through the curved spacetime.

This idea that the effects of gravity are indistinguishable from the effects of a curved spacetime is one version of the "equivalence principle" of GR, which more broadly states that all the laws of physics take the same forms in all inertial frames of reference, where "inertial frames" explicitly include frames that are in gravitational free-fall.

To give you an idea of why GR is possible at all, we can look to experimental evidence.

Newton's 2nd law of motion is F=ma, which says that the (net) force determines the acceleration of an object. For example, suppose there's an electrical force on an object, due to an electrical field E. Then Newton's 2nd law tells us the acceleration that the object will experience:

a = qE/m

Notice that, in this case, the acceleration depends on the mass of the object and its electrical charge.

This sort of thing happens with other kinds of forces as well. Typically, acceleration depends on the mass of the object and some other property that it has. But when it comes to gravity, something rather odd happens. Newton tells us that the force of gravity on an object is F=mg, where g is the "gravitational field strength" and m is the mass. Putting F=ma, as usual, we find:

a=g

But wait! This is independent of the mass. This is familiar - experimentally, in the absence of other forces, all objects are observed to accelerate at the same rate under gravity, regardless of mass. This is not true for electrical and other forces (compare above, for electric forces).

It is because of this property of gravity - that all objects experience the same acceleration, regardless of mass, in a "gravitational field", that it is possible to formulate GR. The "curvature of spacetime" can be the same for all objects, because they all react to gravity in the same way. Of course, having said that, to make sure that GR is consistent with observation we need to make sure that spacetime is curved in just the right way. Also, observationally, we know that mass somehow "causes" gravity, so that idea needs to be built into the "new" model as well; the curvature of spacetime has to be appropriate for the amount of mass/energy that is causing the gravitational effect.

There are still "forces" in GR - just no gravitational forces any more. The main point is that the "special" frames in which the laws of physics "work" in GR include frames that are in gravitational free-fall, because gravity is not a force. This idea of inertial frames being special is consistent with the Newtonian/Galilean definition of an "inertial frame" as being one in which no net force acts on the observer; but in the Newtonian picture gravity is a force and so can be a "net force" - something that is impossible in GR.
 
James R said:
exchemist:

It is true that in GR acceleration due to gravity and the "gravitational force" of Newtonian physics are indistinguishable. Or, more accurately, in GR we can dispense with the idea of a "gravitational force" completely, since all observable effects of so-called "gravitational forces" can be reproduced by using an appropriately curved spacetime, in which all objects in gravitational "free fall" simply follow shortest-path trajectories through the curved spacetime.

This idea that the effects of gravity are indistinguishable from the effects of a curved spacetime is one version of the "equivalence principle" of GR, which more broadly states that all the laws of physics take the same forms in all inertial frames of reference, where "inertial frames" explicitly include frames that are in gravitational free-fall.

To give you an idea of why GR is possible at all, we can look to experimental evidence.

Newton's 2nd law of motion is F=ma, which says that the (net) force determines the acceleration of an object. For example, suppose there's an electrical force on an object, due to an electrical field E. Then Newton's 2nd law tells us the acceleration that the object will experience:

a = qE/m

Notice that, in this case, the acceleration depends on the mass of the object and its electrical charge.

This sort of thing happens with other kinds of forces as well. Typically, acceleration depends on the mass of the object and some other property that it has. But when it comes to gravity, something rather odd happens. Newton tells us that the force of gravity on an object is F=mg, where g is the "gravitational field strength" and m is the mass. Putting F=ma, as usual, we find:

a=g

But wait! This is independent of the mass. This is familiar - experimentally, in the absence of other forces, all objects are observed to accelerate at the same rate under gravity, regardless of mass. This is not true for electrical and other forces (compare above, for electric forces).

It is because of this property of gravity - that all objects experience the same acceleration, regardless of mass, in a "gravitational field", that it is possible to formulate GR. The "curvature of spacetime" can be the same for all objects, because they all react to gravity in the same way. Of course, having said that, to make sure that GR is consistent with observation we need to make sure that spacetime is curved in just the right way. Also, observationally, we know that mass somehow "causes" gravity, so that idea needs to be built into the "new" model as well; the curvature of spacetime has to be appropriate for the amount of mass/energy that is causing the gravitational effect.

There are still "forces" in GR - just no gravitational forces any more. The main point is that the "special" frames in which the laws of physics "work" in GR include frames that are in gravitational free-fall, because gravity is not a force. This idea of inertial frames being special is consistent with the Newtonian/Galilean definition of an "inertial frame" as being one in which no net force acts on the observer; but in the Newtonian picture gravity is a force and so can be a "net force" - something that is impossible in GR.
Click to expand...
Yes that’s a good explanation. I had forgotten that it all comes back to the rather “suspicious” equivalence of inertial and gravitational mass.
 
James R said:
exchemist:

It is true that in GR acceleration due to gravity and the "gravitational force" of Newtonian physics are indistinguishable. Or, more accurately, in GR we can dispense with the idea of a "gravitational force" completely, since all observable effects of so-called "gravitational forces" can be reproduced by using an appropriately curved spacetime, in which all objects in gravitational "free fall" simply follow shortest-path trajectories through the curved spacetime.

This idea that the effects of gravity are indistinguishable from the effects of a curved spacetime is one version of the "equivalence principle" of GR, which more broadly states that all the laws of physics take the same forms in all inertial frames of reference, where "inertial frames" explicitly include frames that are in gravitational free-fall.

To give you an idea of why GR is possible at all, we can look to experimental evidence.

Newton's 2nd law of motion is F=ma, which says that the (net) force determines the acceleration of an object. For example, suppose there's an electrical force on an object, due to an electrical field E. Then Newton's 2nd law tells us the acceleration that the object will experience:

a = qE/m

Notice that, in this case, the acceleration depends on the mass of the object and its electrical charge.

This sort of thing happens with other kinds of forces as well. Typically, acceleration depends on the mass of the object and some other property that it has. But when it comes to gravity, something rather odd happens. Newton tells us that the force of gravity on an object is F=mg, where g is the "gravitational field strength" and m is the mass. Putting F=ma, as usual, we find:

a=g

But wait! This is independent of the mass. This is familiar - experimentally, in the absence of other forces, all objects are observed to accelerate at the same rate under gravity, regardless of mass. This is not true for electrical and other forces (compare above, for electric forces).

It is because of this property of gravity - that all objects experience the same acceleration, regardless of mass, in a "gravitational field", that it is possible to formulate GR. The "curvature of spacetime" can be the same for all objects, because they all react to gravity in the same way. Of course, having said that, to make sure that GR is consistent with observation we need to make sure that spacetime is curved in just the right way. Also, observationally, we know that mass somehow "causes" gravity, so that idea needs to be built into the "new" model as well; the curvature of spacetime has to be appropriate for the amount of mass/energy that is causing the gravitational effect.

There are still "forces" in GR - just no gravitational forces any more. The main point is that the "special" frames in which the laws of physics "work" in GR include frames that are in gravitational free-fall, because gravity is not a force. This idea of inertial frames being special is consistent with the Newtonian/Galilean definition of an "inertial frame" as being one in which no net force acts on the observer; but in the Newtonian picture gravity is a force and so can be a "net force" - something that is impossible in GR.
Click to expand...
What about the equation that says F=m(1).m(2)/r^2 ?

Does that change anything?
Does m=g pop out or come of that also?

It seems to me that g is not constant in time in that equation.

Is the first equation an approximation where the first mass is hugely greater than the second mass so that the size of the second mass can be ignored?

What if the two masses are the same size or one is double?

Do we still get m=g from the maths?
 
geordief:
geordief said:
What about the equation that says F=m(1).m(2)/r^2 ?

Does that change anything?
Does m=g pop out or come of that also?
Click to expand...
That equation gives the force on m(2) due to m(1) (or vice-versa because of Newton's 3rd law). Let's take it as the force on m(2) due to m(1) for now.

The acceleration of m(2) that results from this force is found using Newton's 2nd law F=ma. That is:

m(2) a(2) = G m(1) m(2)/r^2

or

a(2) = G m(1)/r^2.

Notice that the acceleration of mass 2 doesn't depend on m(2).

It seems to me that g is not constant in time in that equation.
Click to expand...
The value a(2), above, could equally be given the symbol g(1), because it is the acceleration (of any other mass) due to the gravity of m(1). Notice that g(1) only depends on the mass m(1) that "causes" the gravitational force on some other object (m(2)) placed at a distance r from m(1). So g(1) can be considered to be a property of the "gravitational field" caused by m(1). At a given distance, all objects will accelerate at the same rate towards m(1).
Is the first equation an approximation where the first mass is hugely greater than the second mass so that the size of the second mass can be ignored?
Click to expand...
In the derivation above, you will see that m(2) cancels out, no matter how big m(2) is compared to m(1).

The assumption that, say, m(1) is much greater than m(2) is only important if we want to consider how m(1) moves, along with m(2). Under their mutual gravitational attraction, the acceleration of m(1) due to the gravity of m(2) will be

a(1) = G m (2)/r^2.

But if m(1) is much larger than m(2), then a(1) might well be negligible compared to a(2). This is the usual situation for small objects (like people) being gravationally attracted to large ones (like the Earth). If you fall of a building, you accelerate towards the ground at about 9.8 metres per second per second, and the Earth's acceleration towards you is negligibly small in comparison - to the extent that we mostly just pretend the Earth doesn't accelerate up to meet you as you fall, which it does (just a bit).
What if the two masses are the same size or one is double?
Click to expand...
If the two masses are the same, then the two objects will accelerate towards one another at equal rates, as you can see from the a(1) and a(2) values.

Usually, the r value is changing with time as objects fall towards one another under gravity. Therefore, the closer together they get, the greater their accelerations. For a situation like a person falling off a building, though, the value of r doesn't change very much, since the height of the building is very small in comparison to the radius of the Earth, so again we can approximate that a(2) = g(1) is approximately 9.8 metres per second per second close to Earth's surface (where "close") means within a few hundred kilometres.
 
James R said:
geordief:

That equation gives the force on m(2) due to m(1) (or vice-versa because of Newton's 3rd law). Let's take it as the force on m(2) due to m(1) for now.

The acceleration of m(2) that results from this force is found using Newton's 2nd law F=ma. That is:

m(2) a(2) = G m(1) m(2)/r^2

or

a(2) = G m(1)/r^2.

Notice that the acceleration of mass 2 doesn't depend on m(2).


The value a(2), above, could equally be given the symbol g(1), because it is the acceleration (of any other mass) due to the gravity of m(1). Notice that g(1) only depends on the mass m(1) that "causes" the gravitational force on some other object (m(2)) placed at a distance r from m(1). So g(1) can be considered to be a property of the "gravitational field" caused by m(1). At a given distance, all objects will accelerate at the same rate towards m(1).

In the derivation above, you will see that m(2) cancels out, no matter how big m(2) is compared to m(1).

The assumption that, say, m(1) is much greater than m(2) is only important if we want to consider how m(1) moves, along with m(2). Under their mutual gravitational attraction, the acceleration of m(1) due to the gravity of m(2) will be

a(1) = G m (2)/r^2.

But if m(1) is much larger than m(2), then a(1) might well be negligible compared to a(2). This is the usual situation for small objects (like people) being gravationally attracted to large ones (like the Earth). If you fall of a building, you accelerate towards the ground at about 9.8 metres per second per second, and the Earth's acceleration towards you is negligibly small in comparison - to the extent that we mostly just pretend the Earth doesn't accelerate up to meet you as you fall, which it does (just a bit).

If the two masses are the same, then the two objects will accelerate towards one another at equal rates, as you can see from the a(1) and a(2) values.

Usually, the r value is changing with time as objects fall towards one another under gravity. Therefore, the closer together they get, the greater their accelerations. For a situation like a person falling off a building, though, the value of r doesn't change very much, since the height of the building is very small in comparison to the radius of the Earth, so again we can approximate that a(2) = g(1) is approximately 9.8 metres per second per second close to Earth's surface (where "close") means within a few hundred kilometres.
Click to expand...
Thanks for that .

By the way ,on another forum I happened to see DaveC426913 saying he cannot login to sciforums over the past week.

He has contacted the owner without reply so far.

Can the mods log him back in ?
 
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