# Understanding General Relativity

Hi all,

I'm having a discussion with someone who believes they understand GR, here is their understanding of how gravity works...

"The time and space at your feet are different from the time and space at your head.

This has been experimentally verified, especially in terms of the time component. We have the technology to measure time accurately enough to actually make these kinds of experiments, and they have been made. A clock placed at your feet will run slower than a clock sitting on your shoulder.

It is the dilation in spacetime (both space and time) between your head and your feet that cause your body to accelerate downward toward the massive earth.

And because this is indeed the cause of gravity on the Earth, this means that you body is actually exerting a downward force onto the Earth's surface."

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I'm having a discussion with someone who believes they understand GR, here is their understanding of how gravity works...

"The time and space at your feet are different from the time and space at your head.
Distance form a gravitational source will have an effect on time, so your head ages at a different rate to your feet!

It is the dilation in spacetime (both space and time) between your head and your feet that cause your body to accelerate downward toward the massive earth. And because this is indeed the cause of gravity on the Earth, this means that you body is actually exerting a downward force onto the Earth's surface."
Hmm... The gravitational force can be thought more of as the bending of spacetime by an object. If you imagine placing a spherical object on a simply supported large piece of paper it would create a similar "distortion" and things would effectively "fall into" the hole towards the object... its a similar concept... if you travelled fast enough around the "side" of this slope, you would effectively orbit the object. It's hard to visualise in 4 dimensional spacetime though.

We do exert a gravitational force on the earth, if you look up the equation you can establish how much the earth is "pulled" up towards you if you were to jump... it's obviously negligible, but it does happen.

Dale
Mentor
I'm having a discussion with someone who believes they understand GR, here is their understanding of how gravity works...

"The time and space at your feet are different from the time and space at your head.

This has been experimentally verified, especially in terms of the time component. We have the technology to measure time accurately enough to actually make these kinds of experiments, and they have been made. A clock placed at your feet will run slower than a clock sitting on your shoulder.

It is the dilation in spacetime (both space and time) between your head and your feet that cause your body to accelerate downward toward the massive earth.

And because this is indeed the cause of gravity on the Earth, this means that you body is actually exerting a downward force onto the Earth's surface."

How is it that gravity is the result of the difference in time dilation between ones' hand and head?

How can a body in free fall be exerting a force? Isn't the surface of the earth an accelerated frame, hence it is the surface accelerating upwards towards the body in free fall?

PAllen
2019 Award
How is it that gravity is the result of the difference in time dilation between ones' hand and head?

How can a body in free fall be exerting a force? Isn't the surface of the earth an accelerated frame, hence it is the surface accelerating upwards towards the body in free fall?
Forces balance. Earth's surface exerts force on you, holding you stationary. You exert force on earth, resisting being prevented from free falling. The two forces are equal.

Ken G
Gold Member
How is it that gravity is the result of the difference in time dilation between ones' hand and head?
How can a body in free fall be exerting a force? Isn't the surface of the earth an accelerated frame, hence it is the surface accelerating upwards towards the body in free fall?
Your friend is not talking about a body in free fall, but rather the situation of when you are standing on the ground. The surface of the Earth is indeed an accelerated frame, and so is the person standing on it (they can feel the force on their feet). That force is needed to make you follow a noninertial path-- the inertial path is the one where you are in free fall, and if you are in free fall, you exert no forces on the Earth at all (though you do slightly curve the spacetime the Earth is moving through).

Dale
Mentor
How is it that gravity is the result of the difference in time dilation between ones' hand and head?
Here is a rough analogy. Suppose that you have an axle with two wheels rolling across a surface. If both wheels are rotating at the same speed then you are going straight, but if one wheel is rotating slower than the other then you are going to be turning.

There is a precise geometric sense in which we are "travelling" through time (or rather spacetime), and the fact that the clock at your feet is "rotating" slower means that your path through spacetime is "turning". The only way to stop "turning" is to go into free-fall where you wouldn't detect any difference in time between your head and your feet, but the force of the ground on your feet makes you keep "turning".

How can a body in free fall be exerting a force? Isn't the surface of the earth an accelerated frame, hence it is the surface accelerating upwards towards the body in free fall?
A body in free-fall is not exerting a force. However, a body on the surface of the earth is not in free-fall, it is being accelerated upwards. By Newton's 3rd law it doesn't matter which object you say is exerting the force. The upwards force on your feet is equal and opposite the downwards force on the earth.

Bill_K
How is it that gravity is the result of the difference in time dilation between ones' hand and head?
The gradient of time dilation is the gradient of g00. But for a static field in linearized GR, g00 = 1 - 2Φ, where Φ is the Newtonian potential, and so the gradient of g00 is also the Newtonian gravitational force.

Forces balance. Earth's surface exerts force on you, holding you stationary. You exert force on earth, resisting being prevented from free falling. The two forces are equal.
He said that we exert a force on the Earth while in free fall.

...if time is going faster at your head than at your feet, it forces your motion in x,t space (x is the vertical coordinate of your height above the ground) to curve (so you change x coordinate, not just t coordinate like you might normally imagine for an object that starts out stationary and has no forces on it). That's because the direction that is "perpendicular to time" is curving downward, if time is going faster (in a relative way) at your head than at your feet.
Would a test particle follow the same path considering it doesn't have that same difference in time dilation?

I would object to your friend's language that this is what gravity is "really doing"
That is exactly what I was objecting.

Your friend is not talking about a body in free fall, but rather the situation of when you are standing on the ground. The surface of the Earth is indeed an accelerated frame, and so is the person standing on it (they can feel the force on their feet). That force is needed to make you follow a noninertial path-- the inertial path is the one where you are in free fall, and if you are in free fall, you exert no forces on the Earth at all (though you do slightly curve the spacetime the Earth is moving through).
That's what I was trying to explain but he told me I was out to lunch claiming the surface couldn't possibly be an accelerated frame because that would mean the surface of the earth would be expanding outwards,the earth would grow in size continuously.

Here is a rough analogy. Suppose that you have an axle with two wheels rolling across a surface. If both wheels are rotating at the same speed then you are going straight, but if one wheel is rotating slower than the other then you are going to be turning.

There is a precise geometric sense in which we are "travelling" through time (or rather spacetime), and the fact that the clock at your feet is "rotating" slower means that your path through spacetime is "turning". The only way to stop "turning" is to go into free-fall where you wouldn't detect any difference in time between your head and your feet, but the force of the ground on your feet makes you keep "turning".
Again, as asked above, would a test particle follow the same path? As well, considering that the difference in time dilation between one's head and feet is infinitesimally small, how can it account for such a major role in the path followed?

A body in free-fall is not exerting a force. However, a body on the surface of the earth is not in free-fall, it is being accelerated upwards. By Newton's 3rd law it doesn't matter which object you say is exerting the force. The upwards force on your feet is equal and opposite the downwards force on the earth.
But, in relativity, it is the surface that is accelerated upwards and not the object exerting a force on the surface.

Bill_K
considering that the difference in time dilation between one's head and feet is infinitesimally small, how can it account for such a major role in the path followed?
Because, as I pointed out in #8 above, it is not the difference in time dilation, it is the gradient of the time dilation that represents the Newtonian gravitational force.

A.T.
...claiming the surface couldn't possibly be an accelerated frame because that would mean the surface of the earth would be expanding outwards...
It is a common misconception, that acceleration towards/away from a point implies a change in distance to that point. It simply doesn't. Not even in flat space-time where you can have circular motion with a centripetal acceleration towards a center, without ever coming closer to the center. In curved-spacetime the opposite case is also possible: The Earth surface can accelerate away from the the Earth's center, without getting further away from it.

A.T.
As well, considering that the difference in time dilation between one's head and feet is infinitesimally small, how can it account for such a major role in the path followed?
The gradient in the time-dilatation dominates only for objects that move much slower than light through space. Such objects move mostly through time, so the distortion of the time-distances has the most effect.

But, in relativity, it is the surface that is accelerated upwards and not the object exerting a force on the surface.
If the object is in freefall it doesn't exert any forces on the surface in GR. If the object rests on the surface, they exert equal but opposite contact forces on each other

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It is a common misconception, that acceleration towards/away from a point implies a change in distance to that point.
That's always confused me as well.

So I am sitting here perfectly still but accelerating towards the Earth at 9.8m/s^2. How does classical mechanics or GR explain that? Newton certainly knew about it.

WannabeNewton
That's always confused me as well.

So I am sitting here perfectly still but accelerating towards the Earth at 9.8m/s^2. How does classical mechanics or GR explain that? Newton certainly knew about it.
Well it is quite easy to explain with a classical analysis because the reaction force (normal force) from the ground on you tends to cancel out all normal components of force acting on you if you treat your motion as being at rest. You don't even need to invoke GR for that.

A.T.
So I am sitting here perfectly still but accelerating towards the Earth at 9.8m/s^2.
Away from the Earth's center, not towards. That is what an accelerometer resting on the surface measures.
How does classical mechanics or GR explain that?
Newton would say it doesn't really accelerate, but is affected by the force of gravity. Einstein would say the accelerometer works fine, and is indeed accelerated upwards.

It is a common misconception, that acceleration towards/away from a point implies a change in distance to that point. It simply doesn't. Not even in flat space-time where you can have circular motion with a centripetal acceleration towards a center, without ever coming closer to the center. In curved-spacetime the opposite case is also possible: The Earth surface can accelerate away from the the Earth's center, without getting further away from it.
That's what I was trying to explain, but was told that was wrong even when it is a clear representation of the effects of an accelerated surface.

The gradient in the time-dilatation dominates only for objects that move much slower than light through space. Such objects move mostly through time, so the distortion of the time-distances has the most effect.

If the object is in freefall it doesn't exert any forces on the surface in GR. If the object rests on the surface, they exert equal but opposite contact forces on each other
That is my understanding as well. Thanks.

Because, as I pointed out in #8 above, it is not the difference in time dilation, it is the gradient of the time dilation that represents the Newtonian gravitational force.
Only the Newtonian force?

Only the Newtonian force?
Yes. In a weak gravitational field, the time-time component of the metric reproduces the Newtonian force. The lowest order deviations from this (so-called "post-Newtonian corrections") are due to the remaining components of the metric, i.e. due to curvature in space. These components are much smaller, which is why we can get away with just using Newtonian gravity in many circumstances.

Dale
Mentor
Again, as asked above, would a test particle follow the same path? As well, considering that the difference in time dilation between one's head and feet is infinitesimally small, how can it account for such a major role in the path followed?
Yes, a test particle would follow the same path, but the analogy starts to break down. At the level of a single test particle I don't know an analogy any more and all I can do is explain the actual math.

So the actual math describes spacetime around the earth as curved. In a curved space any coordinate system you draw is going to have to have curved lines. The way that you physically see if your timelike coordinate is curved is to place accelerometers at rest in your coordinates. If the accelerometer is at rest in your coordinate and reads something other than 0 then it means that your time coordinate is curving. So, in the case of the usual (Schwarzschild) coordinates outside the earth you find that the time coordinate is curving outward at a rate of g.

A free-falling test particle travels in a straight line known as a geodesic. We know that it is a straight line because an accelerometer attached to a free-falling particle reads 0. In the frame of the curved coordinates the geodesics appear to curve.

But, in relativity, it is the surface that is accelerated upwards and not the object exerting a force on the surface.
In relativity both happen. The surface is accelerated upwards, it runs into the object and exerts a contact force on the object to make it accelerate upwards also, and by Newton's 3rd law the object also exerts a force on the object.

Dalespam:
If the accelerometer is at rest in your coordinate and reads something other than 0 then it means that your time coordinate is curving. So, in the case of the usual (Schwarzschild) coordinates outside the earth you find that the time coordinate is curving outward at a rate of g.
That is a weak field approximation??..... right...so you are neglecting the minor curvature of space??

Dale
Mentor
That is a weak field approximation??..... right...so you are neglecting the minor curvature of space??
It is always a little messy to partition curvature into time curvature and space curvature, so I am not sure if my comments are restricted to a weak field approximation. They very well could be, but I am not certain.