About the Equivalence Principle

  • I
  • Thread starter JohnnyGui
  • Start date
  • #1
756
49

Main Question or Discussion Point

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.

Rocket with A=0_1c.jpg


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?
 

Answers and Replies

  • #2
jbriggs444
Science Advisor
Homework Helper
2019 Award
8,776
3,534
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.
 
  • #3
756
49
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?
 
  • #4
130
47
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.
 
  • Like
Likes jbriggs444
  • #5
756
49
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?
 
  • #6
PeterDonis
Mentor
Insights Author
2019 Award
29,684
8,953
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.
 
  • Like
Likes vanhees71
  • #7
130
47
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.
 
  • #8
Janus
Staff Emeritus
Science Advisor
Insights Author
Gold Member
3,526
1,269
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
[tex] \frac{GMm}{r}- \frac{GMm}{r+h}[/tex]
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:
[tex] \frac{GMm}{r}[/tex]
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.
 
  • #9
PeterDonis
Mentor
Insights Author
2019 Award
29,684
8,953
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.
 
  • #10
756
49
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
[tex] \frac{GMm}{r}- \frac{GMm}{r+h}[/tex]
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:
[tex] \frac{GMm}{r}[/tex].
Thanks for the clear explanation. I concluded a similar formula which is:

Own Formula.jpg

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?
 
  • #11
jbriggs444
Science Advisor
Homework Helper
2019 Award
8,776
3,534
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.
 
  • #12
130
47
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.
 
  • #13
756
49
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:
Formula 1.jpg

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.
 
Last edited:
  • #14
PeterDonis
Mentor
Insights Author
2019 Award
29,684
8,953
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.
 
  • #15
756
49
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.
 
  • #16
PeterDonis
Mentor
Insights Author
2019 Award
29,684
8,953
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##.
 
  • #17
756
49
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) / Re

Again, sorry if I've missed your point.
 
  • #18
jbriggs444
Science Advisor
Homework Helper
2019 Award
8,776
3,534
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".
 
  • #19
756
49
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 Re? So there are other set of maximum potential differences depending on what radius (other than Re) you're calculating the height from?
 
  • #20
Nugatory
Mentor
12,785
5,390
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.
 
  • #21
jbriggs444
Science Advisor
Homework Helper
2019 Award
8,776
3,534
You mean, that this is the maximum height only when the height r is calculated from Re? So there are other set of maximum potential differences depending on what radius (other than Re) 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.
 
  • #22
PeterDonis
Mentor
Insights Author
2019 Award
29,684
8,953
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.
 
  • #23
PeterDonis
Mentor
Insights Author
2019 Award
29,684
8,953
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.
 
  • #24
756
49
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?
 

Related Threads on About the Equivalence Principle

Replies
1
Views
2K
Replies
100
Views
11K
Replies
14
Views
2K
  • Last Post
2
Replies
39
Views
4K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
14
Views
2K
Replies
14
Views
970
Replies
3
Views
864
  • Last Post
Replies
1
Views
1K
Top