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## Time dilation in General Relativity

A student asked me last week if I could explain the difference between time dilation in Special Relativity (SR) and that in General Relativity (GR), so here is my attempt at doing so. Time dilation in SR comes about when something travels near the speed of light, and is due to the Lorentz transformations which ensure that experiments in any inertial frame are indistinguishable from each other.

I have already derived the Lorentz Transformations from first principles in this blog, and these equations are at the heart of SR, and show why time dilation occurs when one travels near the speed of light. In this blog here, I worked through some examples of time dilation in SR. But, what about time dilation in GR?

## How does time dilation come about in GR?

As I have already explained in this blog here, Einstein’s principle of equivalence tells us that whatever is true for acceleration is true for a gravitational field. So, to see how gravity affects time the easiest way is to consider how time would be affected in an accelerating rocket.

We will consider a rocket in empty space, away from any gravitational fields, which is accelerating with an acceleration $g$. We will have two people in the rocket, Alice and Bob. Alice is at the top end of the rocket, the nose end. Bob is at the bottom end of the rocket, where the tail is. Alice sends two pulses of light, one at time $t=0$, and the second one at a time $t= \Delta \tau_{A}$ later. They are received at the back of the rocket by Bob; the first pulse is received when the time is $t=t_{1}$, and the second one when the time is $t=t_{2} = t_{1} + \Delta \tau_{B}$, where $\Delta \tau_{B}$ is the time interval between flashes as measured by Bob.

This is illustrated in the figure below.

We can see how time dilation comes about in GR by considering a rocket accelerating in empty space with an acceleration $g$, and a light flashing from Alice at the front-end of the rocket and being received at the back-end by Bob.

We will set it up so that Bob’s position at time $t=0$ when the first flash is emitted by Alice is $z_{B}(0)=0$, and so his position at any other time is given by
$z_{B}(t) = \frac{1}{2}gt^{2} \text{ (Equ. 1) }$

(this just comes from Newton’s 2nd equation of motion $s=ut + \frac{1}{2}at^{2}$, see my blog here which derives those equations).

The position of Alice will just be Bob’s position plus the distance between them, which we will call $h$ (the height of the rocket), so
$z_{A}(t) = h + \frac{1}{2}gt^{2} \text{ (Equ. 2) }$

We will assume that the first pulse takes a time of $t=t_{1}$ to travel from Alice to Bob. The second pulse is emitted by Alice at a time $\Delta \tau_{A}$ after the first pulse, this is the time interval between each light pulse that Alice sends. This second pulse is received by Bob at a time of $t_{2} = t_{1} + \Delta \tau_{B}$, where $\Delta \tau_{B}$ is the time interval between pulses as measured by Bob using a clock next to him.

When the first pulse leaves Alice her position is $z_{A}(0)$, which from equation (2) is h, as she is at the top of the rocket. When Bob receives the pulse at time $t=t_{1}$ his position will be $z_{B}(t_{1})$ which, from equation (1) is $z_{B}(t_{1}) = \frac{1}{2} gt_{1}^{2}$. So, the distance travelled by the pulse is going to be
$z_{A}(0) - z_{B}(t_{1})= ct_{1} \text{ (Equ. 3) }$

as the speed of light is $c$ and it travels for $t_{1}$ seconds. Because the rocket is accelerating, the distance travelled by the second pulse will not be same (as it would be if the rocket were moving with a constant velocity). The distance travelled by the second pulse will be less, and is given by
$z_{A}(\Delta \tau_{A}) - z_{B}(t_{1} + \Delta \tau_{B}) = c(t_{1} + \Delta \tau_{B} - \Delta \tau_{A}) \text{ (Equ. 4) }$

We can use Equations (1) and (2), which give expressions for $z_{B} \text{ and } z_{A}$ as a function of $t$, to put in the values that $z_{A} \text{ and } z_{B}$ would have when $t = \Delta \tau_{A}$ for $z_{A}$ and $(t_{1} + \Delta \tau_{B})$ for $z_{B}$ respectively.

Substituting from Equations (1) and (2) into Equation (3) we have
$z_{A}(0) = h, \; \; z_{B}(t_{1}) = \frac{1}{2}gt_{1}^{2}$

which makes equation (3) become
$h - \frac{1}{2}gt_{1}^{2} = ct_{1} \text{ (Equ. 5) }$

Doing the same kind of substitution into equation (4) we have
$z_{A}(\Delta \tau_{A}) = h + \frac{1}{2}g \left( \Delta \tau_{A} \right)^{2} \rightarrow h$

$z_{B}(t_{1} + \Delta \tau_{B}) = \frac{1}{2}g(t_{1} + \Delta \tau_{B})^{2} \rightarrow \frac{1}{2}gt_{1}^2 +gt_{1} \Delta \tau_{B}$

assuming that we can ignore terms in $(\Delta \tau_{A})^{2} \text{ and } (\Delta \tau_{B})^{2}$

Substituting these expressions into equation (4) gives
$h - \frac{1}{2}gt_{1}^{2} - gt_{1} \Delta \tau_{B} = c(t_{1} + \Delta \tau_{B} - \Delta \tau_{A}) \text{ (Equ. 6) }$

We now subtract equation (6) from (5) to give
$gt_{1} \Delta \tau_{B} = c \Delta \tau_{A} - c \Delta \tau_{B} \text{ Equ. (7) }$

Re-arranging equation (5) as $\frac{1}{2}gt_{1}^{2} +ct^{1} -h$ and using the quadratic formula to find $t_{1}$ we can write that
$t_{1} = \frac{ -c \pm \sqrt{ c^{2} + 2gh } }{ g } \rightarrow \frac{ -c + \sqrt{ c^{2} + 2gh } }{ g }$

(we can ignore the negative solution because the time is always positive). We will next use the binomial expansion to write
$\sqrt{ c^{2} + 2gh } \approx c ( 1 + \frac{gh}{ c^{2} } )$

(where we have ignored terms in $\left( \frac{2gh}{c^{2}} \right)^{2}$ and higher in the Binomial expansion), and so we can write for $t_{1}$
$t_{1} \approx \left( \frac{ -c + c \left( 1 + \frac{gh}{ c^{2} } \right) }{ g } \right) \rightarrow gt_{1} = c \left( \frac{gh}{ c^{2} } \right)$

Substituting this expression for $gt_{1}$ into equation (7) we now have
$c \left( \frac{gh}{ c^{2} } \right) \Delta \tau_{B} = c \Delta \tau_{A} - c \Delta \tau_{B}$

We can cancel the $c$ in each term and bringing the terms in $\Delta \tau_{B}$ onto one side and the term in $\Delta \tau_{A}$ on the other side we have
$\Delta \tau_{B} \left( 1 + \frac{ gh }{ c^{2} } \right) = \Delta \tau_{A}$

and so
$\Delta \tau_{B} = \frac{ \Delta \tau_{A} }{ \left( 1 + \frac{ gh }{ c^{2} } \right) }$

and using the binomial expansion for $(1 + gh/c^{2})^{-1}$ (and ignoring terms in $\left( \frac{ gh }{ c^{2} } \right)^{2}$ and higher), we can finally write
$\boxed{ \Delta \tau_{B} = \Delta \tau_{A} \left( 1 - \frac{ gh }{ c^{2} } \right) \text{ (Equ. 8) } }$

Because $\frac{gh}{c^{2}}$ is always positive, this means that $\Delta \tau_{B}$ is always less than $\Delta \tau_{A}$, or to put it another way the time interval as measured by Bob at the back-end of the rocket will always be less than the time interval measured by Alice where the light pulses were sent. This means that Bob will measure time to be going at a slower rate than Alice, Bob’s time will be dilated compared to Alice.

From the principle of equivalence, whatever is true for acceleration is true for gravity, so if we now imagine the rocket stationary on the Earth’s surface, with the top end in a weaker gravitational field than the bottom end, we can see that a gravitational field will also lead to pulses arriving at Bob being measured closer together than where they were emitted by Alice. So, gravity slows clocks down!

A very important difference between time dilation in SR and time dilation in GR is that the time dilation in GR is not symmetrical. In SR, both observers in their respective inertial frames think it is the other person’s clock which is running slow. In GR, both Alice and Bob will agree that it is Bob’s clock which is running slower than Alice’s clock.

In a future blog I will do some calculations on this effect in different situations, but as you can see from Equation (8), the size of the dilation depends on the acceleration $g$ and the difference in height between $A \text{ and } B$. I will also discuss whether it is time dilation due to GR or time dilation due to SR which affect the satellites which give us GPS the more, as both effects have to be taken into account to get the accuracy we seek in the GPS position.

### 8 Responses

1. Good article, looking forward to your next post

The Science Geek
http://www.thesciencegeek.org

• Thanks!

2. on 03/03/2015 at 09:17 | Reply AbsolutelyRelative

I found it fun to derive all of the SR equations, thus including the Lorentz Transformation equations, by simply mentally analyzing “Motion”. In short, I had to do it this way since my parents pulled me out of school before I had a chance to receive any education in the field of physics.

(Have a look at http://goo.gl/fz4R0I to see how I did it if interested.)

3. […] we experience gravitational time dilation which is greater than that experienced by Tim Peake. In this blog here, I derived from the principle of equivalence the time dilation due to GR, and […]

4. […] means that the time dilation due to SR is negligible. The time dilation due to GR is given by (see my blog here on how to calculate this) , or 22 parts in . Compare this to the ISS, where it was about 1 part in […]

5. on 15/12/2018 at 00:19 | Reply hiroji kurihara

Time Dilation

Accuracy of atomic clock or optical lattice clock is written to be one second per 30 million years or per 30 billion years, etc. Effect of the gravity is not stated. On the other hand, on an atomic clock loaded on GPS, effect of gravity is said to be evident (with specific value : per a day). Are the two compatible ?

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6. Time dilation in SR

A light source is shinning (frequency is constant). Two observers are receding from the light source at the same speed (in the opposite direction). Two observers receive the same frequency. Where is the time dilation ?

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