# Can someone help me with this?

1. Feb 7, 2007

### hover

Can someone help me with this??

I've always wanted to learn about calculating gravitational time dilation but i don't know where to start. Can someone show me how to calculate gravitational time dilation please??

Thank you!

2. Feb 7, 2007

3. Feb 7, 2007

### nakurusil

I suggest this link in wiki.

4. Feb 7, 2007

### pervect

Staff Emeritus
5. Feb 7, 2007

### hover

I'll look at this stuff tomarrow and will ask questions then. Thanks for the replies by the way!!!

6. Feb 9, 2007

### hover

I just looked at the links you guys sent me. They have some info but i'm having a hard time trying to understand the stuff. Can some one give me a good, detailed example of using these equations please? I'm hoping to pick up what is in the example, and maybe if i really understand it i can do another example and show you guys how to do it.

Thanks!!!!

7. Feb 10, 2007

### Mortimer

The formula is $$\Delta t_r=\Delta t_\infty \sqrt{1-\frac{2GM}{rc^2}}$$
Where:
$\Delta t_\infty$ is a time duration that an observer at infinity (where there is no gravity) measures at his own clock and
$\Delta t_r$ is the time duration that this observer measures (or better: calculates) on a clock of an object at radius $r$ from a massive object with mass $M$. $G$ is the gravitational constant.
You can see that $\Delta t_r$ gets smaller when $r$ gets smaller, indicating the time dilation for an object near a massive object. Once $r$ nears the value $2GM/c^2$, $\Delta t_r$ nears zero. This is the Schwarzschild radius for the massive object, which is usually unreachable because it is inside the object, unless the object is a black hole.

Last edited: Feb 10, 2007
8. Feb 10, 2007

### hover

So let me punch in numbers for this equation and someone tell me if i am right.

$$\Delta t_r=\Delta t_\infty \sqrt{1-\frac{2GM}{rc^2}}$$

So say that someone far away in space measures 1 sec. That person compares his measured time to time on the surface of the earth. So let me punch that in.

$$\Delta t_r=\Delta t_\infty \sqrt{1-\frac{2*G*5.9742 × 10^24 kilograms}{6378100 meters*186000^2}}$$
(2*G*5.9742*10^24)/(6378100*186000)
7.9695828*10^14/2.206567476*10^17=.003611756
1*Square root(1-.003611756)
$\Delta t_r=.9981$

If i did my calculations right, for every 1 second out in space .9981 seconds goes by on earth. Is this correct??

Last edited: Feb 10, 2007
9. Feb 10, 2007

### pervect

Staff Emeritus
Nope. The actual number is a lot closer to 1 - the difference is only .7 parts per billion.

See http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html

Looking over your calculation, one error appears to be using 186000 mi/hr for the speed of light rather than 3*10^8 meters/second.

Google calculator can help with this sort of calculation, see for instance

Last edited: Feb 10, 2007
10. Feb 10, 2007

### hover

So i did it wrong.... Can someone please show me the correct steps on how to do this using that equation because i'm starting to get lost now.

11. Feb 10, 2007

### nakurusil

You did it almost right. You need to replace 186000 with 3x10^8. The speed of light must be expressed in SI units (i.e. m/s). pervect just told you that.

12. Feb 10, 2007

### hover

Just making sure that was the only thing.

13. Feb 10, 2007

### pervect

Staff Emeritus
While I like to help people out with physics problems, I think getting people to be as self-reliant as is reasonable is the best way to help them. "Teach a man to fish" vs giving him fish, etc.

I think that if you actually look at the calculations on the hyperphysics webpage you will see that they get an actual number.

If you can't find the number at first glance, read on towards the end of the page. It's there.

If you redo your calculation and get a different number, then it's time to post again.

If you redo your calculation and get the same number, then you can crow about getting it right :-).

14. Feb 11, 2007

### hover

Thats what i'm trying to do. I'm trying to get a couple examples out of you guys so i can figure out how to do them myself.

So lets use a new example for me to figure out. The sun has a mass of 1.98892×10^30kilograms and a radius 695500000meters. So plugging that in

1* Square root((2*G*1.98892×10^30kilograms)/(695500000meters*c^2))
(2*G*1.98892×10^30kilograms)/(695500000meters*c^2)=4.24648794×10-6
1*Square root(1-4.24648794 × 10-6)=.999997877

Can someone tell me if this is correct?? Did i do it right??

Last edited: Feb 11, 2007
15. Feb 11, 2007

### pervect

Staff Emeritus
Minor typo in the above

Yep - or at least I get the same thing

Yep, I get the same number. The only remaining thing to do is to insure that you're interpreting this number correctly. When compared via (for instance) light signals, the clock on the surface of the sun will run slowly compared by the clock at infinity. Another way of saying this - radiation falling from infinity to the sun's surface will be blue-shifted, making the interval between pulses shorter. So that a 1 second period signal "at infinity" will be blueshifted to a .999997877 second period signal on the surface of the sun.