Calculate Gravitational Time Dilation: Step-by-Step Guide"

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In summary, the conversation revolved around a user's request for help with calculating gravitational time dilation. Other users provided resources and explanations, as well as an example calculation. Eventually, the original user attempted their own calculation with a different example and asked for verification. The final calculation was .999997877, which is very close to the actual value of 1.
  • #1
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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!:biggrin:
 
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  • #3
hover said:
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!:biggrin:

I suggest this link in wiki.
 
  • #5
I'll look at this stuff tomarrow and will ask questions then. Thanks for the replies by the way!:biggrin:
 
  • #6
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
hover said:
Can some one give me a good, detailed example of using these equations please?

The formula is [tex]\Delta t_r=\Delta t_\infty \sqrt{1-\frac{2GM}{rc^2}}[/tex]
Where:
[itex]\Delta t_\infty[/itex] is a time duration that an observer at infinity (where there is no gravity) measures at his own clock and
[itex]\Delta t_r[/itex] is the time duration that this observer measures (or better: calculates) on a clock of an object at radius [itex]r[/itex] from a massive object with mass [itex]M[/itex]. [itex]G[/itex] is the gravitational constant.
You can see that [itex]\Delta t_r[/itex] gets smaller when [itex]r[/itex] gets smaller, indicating the time dilation for an object near a massive object. Once [itex]r[/itex] nears the value [itex]2GM/c^2[/itex], [itex]\Delta t_r[/itex] 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.
 
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  • #8
So let me punch in numbers for this equation and someone tell me if i am right.

[tex]\Delta t_r=\Delta t_\infty \sqrt{1-\frac{2GM}{rc^2}}[/tex]

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.

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

If i did my calculations right, for every 1 second out in space .9981 seconds goes by on earth. Is this correct??
 
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  • #9
hover said:
So let me punch in numbers for this equation and someone tell me if i am right.

[tex]\Delta t_r=\Delta t_\infty \sqrt{1-\frac{2GM}{rc^2}}[/tex]

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.

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

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

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

http://www.google.com/search?hl=en&...h)+/+(+(radius+of+earth)+*+c^2)+=&btnG=Search
 
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  • #10
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
hover said:
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.

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
nakurusil said:
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.

Just making sure that was the only thing.
 
  • #13
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
pervect said:
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.

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 let's 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??
 
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  • #15
hover said:
So let's 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))

Minor typo in the above

(2*G*1.98892×10^30kilograms)/(695500000meters*c^2)=4.24648794×10-6
Yep - or at least I get the same thing

1*Square root(1-4.24648794 × 10-6)=.999997877

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.
 

1. Can someone help me with this experiment?

Yes, I am here to assist you with any questions or concerns you may have about the experiment. What specifically do you need help with?

2. Can you explain this concept to me?

Of course, I would be happy to explain the concept to you. Can you tell me what specifically you are having trouble understanding?

3. Can you check my data and see if I did it correctly?

Yes, I can review your data and provide feedback. Can you send me your data and explain what you are trying to achieve?

4. Can you help me interpret the results of my experiment?

Absolutely, I am here to help you understand the results. Can you provide me with more information about your experiment and what you are trying to determine?

5. Can you recommend any further experiments to test my hypothesis?

I would be happy to suggest additional experiments for you to conduct. Can you share your hypothesis and what you have already tested so far?

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