About the Equivalence Principle

[JohnnyGui]

Hey all,

I was trying to compare the similarity between time dilation due to gravity and the scenario of an accelerating rocket through the equivalence principle and there's something tha blocks my understanding about their similarity.

I found a good distance-time graph of the accelerating rocket scenario a post of ghostpy which I found in this thread: https://www.physicsforums.com/threa...-time-dilation-effects-inside-of-them.741028/ . So credits to him.

The y-axis represents the distance and the x-axis the time. Two observers are standing at y=1 and y=0. The upper one is the emitter who is sending light pulses to the receiver below while they're both accelerating.

Here's are some conclusions that I deducted from one another that confuses me about how time dilation due to gravity should be similar to an accelerating rocket scenario.
1. If the accelerating rocket scenario is used to explain time dilation due to gravity, then shouldn't the emitter (upper observer) experience a lower acceleration than the receiver since he's further away from the gravitational field?
2. If 1. is true, then this means that the receiver below should measure even shorter intervals than the time intervals shown in the graph above since the receiver is having a larger acceleration than the emitter.
3. If 2. is true, then that means that if an emitter is sitting in a zero gravity field, the receiver should measure an even shorter time interval between light pulses than in point 2. Much to the extent that in this scenario, acceleration isn't even needed anymore for time dilation. A mere constant velocity of the receiver would let the receiver still measure shorter time intervals of light pulses because the emitter is standing still.

If these 3 points are true, then what is it that makes acceleration needed to use it as an equivalence principle for the time dilation due to gravity if point 3 shows that just velocity is enough for time dilation? And why does the equivalence principle show in the rocket scenario that they're both undergoing the same acceleration while in reality they should have a different force of g due to height differences?

 

[jbriggs444]

1. If the accelerating rocket scenario is used to explain time dilation due to gravity, then shouldn't the emitter (upper observer) experience a lower acceleration than the receiver since he's further away from the gravitational field?

Both are in the gravitational field. You are not going to retrieve a sensible intuition by trying to apply an inverse square force law in a regime where the gravity is not a force and has no source.

 

[JohnnyGui]

Both are in the gravitational field. You are not going to retrieve a sensible intuition by trying to apply an inverse square force law in a regime where the gravity is not a force and has no source.

Then how can the equivalence principle compare it to time dilation due to gravity if gravity IS a force and has a source?

From what I understand, time dilation due to gravity is happening because of the difference in gravity force at different distances from the source. How can an accelerating rocket scenario be an equivalent to this when in the rocket scenario both the emitter and the receiver are experiencing the same gravity force?

 

[Vitro]

Then how can the equivalence principle compare it to time dilation due to gravity if gravity IS a force and has a source?

From what I understand, time dilation due to gravity is happening because of the difference in gravity force at different distances from the source. How can an accelerating rocket scenario be an equivalent to this when in the rocket scenario both the emitter and the receiver are experiencing the same gravity force?

Because it is not due to difference in gravitational "force" like you think, it's due to difference in potential, the amount of work per unit mass needed to move a test object from one place to the other. An object "higher up" has more potential than a "lower" one even in uniform gravitational field or acceleration.

 

[JohnnyGui]

Because it is not due to difference in gravitational "force" like you think, it's due to difference in potential, the amount of work per unit mass needed to move a test object from one place to the other. An object "higher up" has more potential than a "lower" one even in uniform gravitational field or acceleration.

Ah, that explains the difference in a uniform gravitational field. This raised some questions for me though:
1. Are you referring to the gravitational potential as in m x g x h? If so, that means obviously that the higher one gets, the higher the potential difference and thus the higher the factor time gets dilated with. However, eventually g decreases with height to the extent that g gets to zero when one is in space, thus giving a gravitational potential of 0. How can a negative potential difference in this case translate to a larger time dilation factor?

2. If it's about the difference in gravitational potential that causes time dilation, why is there acceleration involved in the first place? Is the factor by which the gravitational potentials differ the same as the Doppler shift factor by which the time intervals of emitted and received light pulses differ due to acceleration?

 

[PeterDonis]

why does the equivalence principle show in the rocket scenario that they're both undergoing the same acceleration while in reality they should have a different force of g due to height differences?

Because if you're talking about a region of spacetime that is large enough to see the difference in acceleration with height, then that region is too large for the EP to apply. (It must be, because the variation of acceleration with height is different in an accelerating rocket than it is on a planet.) The EP only applies over a region of spacetime small enough that no difference in acceleration can be observed, to the accuracy of observation you are using.

If it's about the difference in gravitational potential that causes time dilation, why is there acceleration involved in the first place?

Because you put it into the initial scenario. The EP does not just cover the accelerating case. It also covers the case of objects in free fall; in a small enough region of spacetime, free-falling objects behave as if they are in flat spacetime, even if globally the spacetime they are in is curved.

 

[Vitro]

Ah, that explains the difference in a uniform gravitational field.

And also in an acceleration field in the absence of any source of gravity, they are equivalent. You also have a difference in potential between the front and the back of an accelerating rocket in empty space.

This raised some questions for me though:
1. Are you referring to the gravitational potential as in m x g x h?

Without the m, yes, but that's only valid in a uniform field where g is constant, otherwise you need to integrate g(h) between the two heights. Also notice that g is just an acceleration, you can substitute it with a.

 

[Janus]

Ah, that explains the difference in a uniform gravitational field. This raised some questions for me though:
1. Are you referring to the gravitational potential as in m x g x h? If so, that means obviously that the higher one gets, the higher the potential difference and thus the higher the factor time gets dilated with. However, eventually g decreases with height to the extent that g gets to zero when one is in space, thus giving a gravitational potential of 0. How can a negative potential difference in this case translate to a larger time dilation factor?

mgh only work for gravitational potential if g is considered constant over the entire range of h.
If you want to take the decrease of g with distance, then you have to use
$$\frac{GMm}{r}- \frac{GMm}{r+h}$$
Where M is the mass of the gravity source
r is the distance from the center of the gravity source you are measuring from.
h is the difference in height.
If you are measuring with respect to the surface of the Earth, then r is the radius of the Earth. As r+h goes to infinity, the second term goes to zero, and the difference in potential increases to a maximum of:
$$\frac{GMm}{r}$$

2. If it's about the difference in gravitational potential that causes time dilation, why is there acceleration involved in the first place? Is the factor by which the gravitational potentials differ the same as the Doppler shift factor by which the time intervals of emitted and received light pulses differ due to acceleration?

With acceleration, you have a Doppler shift due to the difference in velocities between emission and reception. This works out to be the same as the resultant shift in frequency light would undergo as it lost energy climbing against a uniform gravity field between the two ( or gain in energy falling in the field)

In a planetary gravity field, light still loses energy climbing against it and gains energy when falling with the field, and thus you get a similar frequency shift.

 

[PeterDonis]

With acceleration, you have a Doppler shift due to the difference in velocities between emission and reception.

Note that this is true even if the acceleration does not vary with height--i.e., in the limit where we are considering a region of spacetime small enough that no variation with height is not observable (which is the limit in which the EP applies). What causes the difference in velocities is acceleration itself, not any change in acceleration with height.

 

[JohnnyGui]

mgh only work for gravitational potential if g is considered constant over the entire range of h.
If you want to take the decrease of g with distance, then you have to use
$$\frac{GMm}{r}- \frac{GMm}{r+h}$$
Where M is the mass of the gravity source
r is the distance from the center of the gravity source you are measuring from.
h is the difference in height.
If you are measuring with respect to the surface of the Earth, then r is the radius of the Earth. As r+h goes to infinity, the second term goes to zero, and the difference in potential increases to a maximum of:
$$\frac{GMm}{r}$$.

Thanks for the clear explanation. I concluded a similar formula which is:

I've got 2 questions regarding this
1. I noticed that the difference here is that you've left out the m in the formula m x g x h while I didn't. Why did you?

2. From this above formula you can conclude the following about m x g x h: That as someone increases in height, the decrease in g overcompensates the increase in h which leads to the fact that the gravitational potential decreases when someone increases in height (talking about a planetary gravity field, not a uniform one). Is this conclusion correct? This means that the lower a gravitational potential is the faster time goes since the higher one gets away from a gravity field, the faster time goes according to GR.

With acceleration, you have a Doppler shift due to the difference in velocities between emission and reception. This works out to be the same as the resultant shift in frequency light would undergo as it lost energy climbing against a uniform gravity field between the two ( or gain in energy falling in the field)

In a planetary gravity field, light still loses energy climbing against it and gains energy when falling with the field, and thus you get a similar frequency shift.

So can I say that the factor by which time gets dilated can be concluded from either 1) the difference in velocities between emission and reception or 2) the amount of energy that is lost between the two? They both would give the same factor of time dilation?

Note that this is true even if the acceleration does not vary with height--i.e., in the limit where we are considering a region of spacetime small enough that no variation with height is not observable (which is the limit in which the EP applies). What causes the difference in velocities is acceleration itself, not any change in acceleration with height.

This is what I indeed stated in point 3 in my OP. That acceleration is merely causing a difference in velocities between reception and emission which is causing the time dilation. This is in case of a uniform gravity field. I think that in a planetary gravity field, time gets dilated because of a difference in gravitational potential instead of a difference in velocities (or maybe both).
If this is correct, is there a way to explain how a difference in gravitational potential causes time dilation just like it is shown in the graph in my OP why a difference in velocities causes time dilation?

 

[jbriggs444]

Thanks for the clear explanation. I concluded a similar formula which is:
[re-rendering in TeX]
$\frac{G\ \cdot\ m^2\ \cdot\ M}{r} - \frac{G\ \cdot\ m^2\ \cdot\ M}{r-h}$
I've got 2 questions regarding this
1. I noticed that the difference here is that you've left out the m in the formula m x g x h while I didn't. Why did you?

Because if you are looking at the gravitational field strength, the mass of the test mass is irrelevant. The field is what it is. The fact that you ate too much last night does not mean that you get to blame the increase in the bathroom scale reading on an increase in the Earth's gravitational field.

There is certainly no point in putting in the square of the test mass.

2. From this above formula you can conclude the following about m x g x h: That as someone increases in height, the decrease in g overcompensates the increase in h which leads to the fact that the gravitational potential decreases when someone increases in height (talking about a planetary gravity field, not a uniform one). Is this conclusion correct?

No. You'll have to explain why you think this. As you go higher, your potential increases always. There is no question of compensation or overcompensation.

 

[Vitro]

2. From this above formula you can conclude the following about m x g x h: That as someone increases in height, the decrease in g overcompensates the increase in h which leads to the fact that the gravitational potential decreases when someone increases in height (talking about a planetary gravity field, not a uniform one). Is this conclusion correct?

No, both @Janus and I already told you m x g x h only works for constant g, and he gave you the right formula for a spherical (planetary) field. In that formula the potential increases with h, not the other way around.

 

[JohnnyGui]

No. You'll have to explain why you think this. As you go higher, your potential increases always. There is no question of compensation or overcompensation.

No, both @Janus and I already told you m x g x h only works for constant g, and he gave you the right formula for a spherical (planetary) field. In that formula the potential increases with h, not the other way around.

Yes, I'm aware that m x g x h is for constant g. But @Janus said that if you want to take the decrease of g with distance, then you have to use the following formula:

This formula gives a potential difference between different observers for which I agree that it increases with height. But if you look at the 2 fractions individually, which gives the individual gravitational potential for an object at a certain height with g being a function of h, you can see that as height increases (r or r + h), the denominator increases and thus the potential decreases. I might have misunderstood this, please elaborate if I do.

 

[PeterDonis]

the denominator increases and thus the potential decreases

You're missing a minus sign. The potential at radius $r$ is $- G M m / r$. So as the denominator increases from $r$ to $r + h$, the potential gets less negative, i.e., higher.

Janus's formula should really be written:

$$
- \frac{GMm}{r} + \frac{GMm}{r + h}
$$

That makes explicit the fact that the "difference" in this case is negative, i.e., the potential increases (gets less negative) as height increases.

 

[JohnnyGui]

You're missing a minus sign. The potential at radius $r$ is $- G M m / r$. So as the denominator increases from $r$ to $r + h$, the potential gets less negative, i.e., higher.

Janus's formula should really be written:

$$
- \frac{GMm}{r} + \frac{GMm}{r + h}
$$

That makes explicit the fact that the "difference" in this case is negative, i.e., the potential increases (gets less negative) as height increases.

Thanks. I think my confusion lied in the fact that I was considering potential being something else than potential difference. For example, we just use m x g x h to calculate a gravity potential because we don't have to substract it from m x g x 0. That's why I thought I just have to use one of the fractions from Janu's formula to calculate the potential at (r + h) while in reality you'd have to substract it from the potential at r first to get the gravity potential. With this in mind, potential indeed increases with height to a maximum of (G x m x M) / r like Janu said.

Sorry if I'm nagging but is it possible to answer my question about your quote in my post #10? I'm curious about it.

 

[PeterDonis]

we just use m x g x h to calculate a gravity potential because we don't have to substract it from m x g x 0.

Yes, you do; when you use $mgh$ as your gravitational potential you are implicitly assuming that the point of "zero" potential is at $h = 0$. If $h$ becomes negative, i.e., if you dig a hole and go down to the bottom of it, the potential will be negative, using that convention. The "zero" of potential is an arbitrary choice. When I wrote that the potential was $- GMm / r$, I was defining the "zero" of potential to be at $r \rightarrow \infty$. This works out nicely mathematically because the limit of $- GMm/r$ as $r \rightarrow \infty$ is zero anyway, so we don't have to put any extra constant term in the formula. But I could just as easily have defined the potential at radius $r$ as

$$
- \frac{GMm}{r} + \frac{GMm}{R_e}
$$

where $R_e$ is the radius of the Earth. That would be the same convention as you were using, that the "zero" of potential is at the Earth's surface--but note that as $r$ increases above $R_e$, i.e., as height increases, the potential still increases. In fact, if you calculate the difference between the potential at any two values of $r$, the constant term $GMm/R_e$ drops out of the calculation. So the choice of "zero" for the potential does not affect the sign of potential differences, i.e., how potential changes with height.

potential indeed increases with height to a maximum of (G x m x M) / r

There is no "maximum" at any finite height. The potential increases with height forever, all the way out to $r \rightarrow \infty$.

 

[JohnnyGui]

Yes, you do; when you use $mgh$ as your gravitational potential you are implicitly assuming that the point of "zero" potential is at $h = 0$. If $h$ becomes negative, i.e., if you dig a hole and go down to the bottom of it, the potential will be negative, using that convention. The "zero" of potential is an arbitrary choice. When I wrote that the potential was $- GMm / r$, I was defining the "zero" of potential to be at $r \rightarrow \infty$. This works out nicely mathematically because the limit of $- GMm/r$ as $r \rightarrow \infty$ is zero anyway, so we don't have to put any extra constant term in the formula. But I could just as easily have defined the potential at radius $r$ as

$$
- \frac{GMm}{r} + \frac{GMm}{R_e}
$$

where $R_e$ is the radius of the Earth. That would be the same convention as you were using, that the "zero" of potential is at the Earth's surface--but note that as $r$ increases above $R_e$, i.e., as height increases, the potential still increases. In fact, if you calculate the difference between the potential at any two values of $r$, the constant term $GMm/R_e$ drops out of the calculation. So the choice of "zero" for the potential does not affect the sign of potential differences, i.e., how potential changes with height.

Sorry if I missed it, but what confused me is that when calculating potential difference in a uniform gravity field, one would have to substract the potential with the lowest height from the potential with the highest height. For example: m x g x h1 - m x g x h2 where h1 > h2. This means that the more positive the potential difference is, the larger the time difference factor where time goes faster at the potential with the heighest height.

However, when calculating the potential difference while taking into account that the g is changing with height according to (GmM) / r² substracting the potential with the lowest height from the potential with the heighest height would give a potential difference with a negative value and all of a sudden the formula has to be rearranged to give it a positive one. I don't understand why this rearrangement has to happen because it didn't have to when using m x g x h

There is no "maximum" at any finite height. The potential increases with height forever, all the way out to $r \rightarrow \infty$.

Your formula suggests that if r approaches infinity, the potential difference would approach (GMm) / R_e

Again, sorry if I've missed your point.

 

[jbriggs444]

However, when calculating the potential difference while taking into account that the g is changing with height according to (GmM) / r2 substracting the potential with the lowest height from the potential with the heighest height would give a potential difference with a negative value and all of a sudden the formula has to be rearranged to give it a positive one. I don't understand why this rearrangement has to happen because it didn't have to when using m x g x h

No rearrangement is required. You are still subtracting the potential at the lowest height from the potential at the highest height. However, the formula for the potential is $-\frac{GM}{r}$. Note the minus sign. [Note also the absence of m in the formula -- we measure gravitational potential per unit mass, not for a specific mass m]

If you subtract the potential at the surface of the Earth ($-\frac{GM_e}{r_e}$) from the limiting potential of 0, you get $+\frac{GM_e}{r_e}$.

Your formula suggests that if r approaches infinity, the potential difference would approach (GMm) / Re

Yes. However, Peter's point was that this is not a "maximum" in the strict mathematical sense, since it is not attained at any finite point. Instead, it is the "least upper bound" on the set of potential differences. Or the "limit of the potential difference as the distance increases without bound".

 

[JohnnyGui]

No rearrangement is required. You are still subtracting the potential at the lowest height from the potential at the highest height. However, the formula for the potential is $-\frac{GM}{r}$. Note the minus sign. [Note also the absence of m in the formula -- we measure gravitational potential per unit mass, not for a specific mass m]

If you subtract the potential at the surface of the Earth ($-\frac{GM_e}{r_e}$) from the limiting potential of 0, you get $+\frac{GM_e}{r_e}$.

Sorry, I wasn't clear when calling it rearrangement. What I meant was, why does there have to be a minus sign for a potential with a higher distance while this wasn't necessary for m x g x h? In both cases you want to calculate the potential difference the same way.

Yes. However, Peter's point was that this is not a "maximum" in the strict mathematical sense, since it is not attained at any finite point. Instead, it is the "least upper bound" on the set of potential differences. Or the "limit of the potential difference as the distance increases without bound".

You mean, that this is the maximum height only when the height r is calculated from R_e? So there are other set of maximum potential differences depending on what radius (other than R_e) you're calculating the height from?

 

[Nugatory]

Sorry, I wasn't clear when calling it rearrangement. What I meant was, why does there have to be a minus sign for a potential with a higher distance while this wasn't necessary for m x g x h? In both cases you want to calculate the potential difference the same way.

The potential increases with increasing height, in both cases.
The magnitude of $mgh$ increases with increasing height, while the magnitude of $GM_Em/r$ decreases with increasing height. The minus sign ensures that the potential at larger $r$ values is higher (that is, less negative) than at lower $r$ value. It will appear naturally when you calculate the potential from first principles, from the integral of $W=Fd$.

Do not be distracted by the quirk that $mgh$ is always positive while $-GmM_E/r$ is always negative - that's just a result of choosing the point of zero potential to be the surface of the Earth ($h=0$) in one case and at an infinite distance from the Earth in the other.

 

[jbriggs444]

You mean, that this is the maximum height only when the height r is calculated from R_e? So there are other set of maximum potential differences depending on what radius (other than R_e) you're calculating the height from?

No, that is not what I mean.

I mean that there is no maximum distance from Earth and, accordingly, no maximum for the potential. There are no points "at infinity". Only points that are arbitrarily far away.

 

[PeterDonis]

when calculating potential difference in a uniform gravity field, one would have to substract the potential with the lowest height from the potential with the highest height.

Yes.

when calculating the potential difference while taking into account that the g is changing with height according to (GmM) / r2 substracting the potential with the lowest height from the potential with the heighest height would give a potential difference with a negative value

No, it wouldn't. The potential at the lower height will be

$$
- \frac{GMm}{r}
$$

The potential at the higher height will be

$$
- \frac{GMm}{r + h}
$$

When you subtract the first from the second, you get a positive number. Try it and see.

Your formula suggests that if r approaches infinity, the potential difference would approach (GMm) / Re

But $r$ can only approach infinity; it can never reach it. There is no finite value of $r$ at which the potential attains a maximum value, as others have pointed out.

 

[PeterDonis]

What I meant was, why does there have to be a minus sign for a potential with a higher distance

When you're using the formula $- GMm/r$ for the potential, there is a minus sign at all distances, not just at the higher distance. See my previous post.

 

[JohnnyGui]

When you're using the formula $- GMm/r$ for the potential, there is a minus sign at all distances, not just at the higher distance. See my previous post.

So if I understand this correctly:

- Suppose you want to lift up an object from r1 to r2 (away from the Earth). Since the force changes with height, one would have to sum up the gravity potentials of small Δr's between r1 and r2. Thus, one would have to do the integral of gravity potential from r1 to r2.

The integral then turns out to be (GMm / r1) - (GMm / r2)

- If we choose r2 to be infinity then the energy to bring up an object from r1 to r2 would be equal to GMm / r1 since (GMm / r2) would be 0.
- If we make r2 a finite number again that is > r1, then from the perspective of infinity ∞, lifting up an object from r1 to r2 would have to be:

(Integral of gravity potential from r1 to ∞) - (Integral of gravity potential from r2 to ∞) = (GMm / r1) - (GMm / r2)

- This means that from the perspective of ∞, bringing an object down from r2 to r1 would be the negative of that, thus: - ((GMm / r1) - (GMm / r2)) which is equal to - (GMm / r1) + (GMm / r2)

This might be a different way of explaining it but it helps me understand it better. Regardless of that, is this correct?

 

[PeterDonis]

is this correct?

Yes.

 

[JohnnyGui]

Yes.

Great. I see that most of the time the formula for gravitational potential is noted as -GmM / r . Based on what I said, this formula shows the energy that is needed to bring an object from a radius of infinity down to a radius of r (for example the radius of Earth). Correct?

That being said, now that I understand that just a difference in potential energy causes time dilation even if g is constant, does this mean that in case of a varying g with height, time dilation should change even stronger over a Δh compared to the same Δh in a constant g? I'm thinking this because the difference between the velocities between emitter and receiver would be larger in case of a varying g with height.

 

[jbriggs444]

difference between the velocities between emitter and receiver would be larger in case of a varying g with height.

It is the difference in potential, not the difference in velocity. Do you think that a varying g means an increasing g?

 

[JohnnyGui]

It is the difference in potential, not the difference in velocity. Do you think that a varying g means an increasing g?

No, decreasing with distance. The thing is, that according to the Equivalence Principle, the difference in velocities between emitting and receiving is the cause for time dilation, even if the acceleration of the emitter and receiver are the same. If it's the difference in gravity potential that causes this instead, then explaining gravitational time dilation by using the Equivalence Principle would be obsolete since it's explaining it by the difference in velocities.

 

[PeterDonis]

Based on what I said, this formula shows the energy that is needed to bring an object from a radius of infinity down to a radius of r (for example the radius of Earth). Correct?

Yes. With the convention for the "zero point" of potential energy that that formula implies, an object at infinity has zero potential energy, and an object at a finite radius has negative potential energy, which means the process of moving the object from infinity to a finite radius can be made to produce energy.

does this mean that in case of a varying g with height, time dilation should change even stronger over a Δh compared to the same Δh in a constant g? I'm thinking this because the difference between the velocities between emitter and receiver would be larger in case of a varying g with height.

You're mixing up two scenarios. The scenario with constant g is a local approximation, in a small patch of spacetime; that is the scenario in which the equivalence principle applies. The scenario with g varying with height is not local--if you can see any variation of g with height, then you are dealing with a region of spacetime that is too large to apply the equivalence principle. So you can't make the comparison you are trying to make here; there is no such thing as a constant g with height over a large range of height.

according to the Equivalence Principle, the difference in velocities between emitting and receiving is the cause for time dilation

Only locally. You can't use the equivalence principle to understand time dilation over a large range of heights. See above.

If it's the difference in gravity potential that causes this instead, then explaining gravitational time dilation by using the Equivalence Principle would be obsolete since it's explaining it by the difference in velocities.

No, this is not correct. The equivalence principle and potential energy are not two different competing explanations for time dilation. They are two different aspects of the same explanation.

 

[JohnnyGui]

No, this is not correct. The equivalence principle and potential energy are not two different competing explanations for time dilation. They are two different aspects of the same explanation.

Ah, I see. The problem is, I can understand time dilation if it's explained mathematically by the difference in velocity, as shown in de graph in my OP. But how can time dilation be explained by the difference in gravity potential? Especially when the equivalence principle can't be used for time dilation over a large range of heights?

 

[vanhees71]

In the non-relativistic limit the pseudo-metric reads
$$g_{00}=1-\frac{2MG}{r}, \quad g_{0j}=g_{j0}=0, \quad g_{ij}=-\delta_{ij}.$$
Here, latin indices run from 1 to 3, and I set $c=1$. From this you get
$$\mathrm{d} \tau^2 = \left (1-\frac{2MG}{r} \right) \mathrm{d} t^2-\mathrm{d} \vec{x}^2, \quad r=|\vec{x}|$$
A clock's proper time is thus
$$\mathrm{d} \tau = \mathrm{d} t \sqrt{1-\frac{2MG}{r}} \simeq \mathrm{d} t \left (1-\frac{MG}{r} \right).$$
The coordinate time $t$ is the time a clock very far from the mass $M$ (which is located around the origin of the spatial coordinate system).

 

[JohnnyGui]

In the non-relativistic limit the pseudo-metric reads
$$g_{00}=1-\frac{2MG}{r}, \quad g_{0j}=g_{j0}=0, \quad g_{ij}=-\delta_{ij}.$$
Here, latin indices run from 1 to 3, and I set $c=1$. From this you get
$$\mathrm{d} \tau^2 = \left (1-\frac{2MG}{r} \right) \mathrm{d} t^2-\mathrm{d} \vec{x}^2, \quad r=|\vec{x}|$$
A clock's proper time is thus
$$\mathrm{d} \tau = \mathrm{d} t \sqrt{1-\frac{2MG}{r}} \simeq \mathrm{d} t \left (1-\frac{MG}{r} \right).$$
The coordinate time $t$ is the time a clock very far from the mass $M$ (which is located around the origin of the spatial coordinate system).

Thanks. I've still yet to delve into pseudo-metrics and its formula derivations.

I've got 2 questions regarding your formula:

1. Is the Equivalence Principle explanation some kind of alternative substitute for the gravitational potential explanation in case of a local scenario?
2. If so, does this mean that if you apply the gravitational potential formula in your post for a local scenario, it would dilate the time by the same factor the Equivalence Principle would? Thus, time dilated by a infinitesimal small change in g (local) would be approximately equivalent to the time dilated by the difference in velocities?

 

[jbriggs444]

2. If so, does this mean that if you apply the gravitational potential formula in your post for a local scenario, it would dilate the time by the same factor the Equivalence Principle would? Thus, time dilated by a infinitesimal small change in g (local) would be approximately equivalent to the time dilated by the difference in velocities?

It's not the change in g over distance. It's the integral of g over distance.

 

[JohnnyGui]

It's not the change in g over distance. It's the integral of g over distance.

Sorry, that's what I meant in the question.

 

[PeterDonis]

I can understand time dilation if it's explained mathematically by the difference in velocity

Which only works locally, in a small enough patch of spacetime such that there is no difference in g within the patch. So this explanation doesn't help you over a large enough range of height that differences in g are observable.

how can time dilation be explained by the difference in gravity potential?

Because that's how the math works. It's different math from the math you're using in a local inertial frame, but it still works.

To put it another way, the time dilation between observers at different heights in a gravity well is simply part of the geometry of spacetime in this particular scenario. That is the explanation.

Especially when the equivalence principle can't be used for time dilation over a large range of heights?

The equivalence principle can't tell you anything about the geometry of spacetime anyway, because by definition it only works in a small enough patch of spacetime that tidal gravity (which is the physical observation referred to by "the geometry of spacetime") is unobservable.

Basically, you're looking at things backwards. The geometry of spacetime is the fundamental entity in GR; the equivalence principle is just one of many possible approximations. You're trying to make the EP the fundamental entity and derive everything else from it. That won't work.