Time Dilation

  • Thread starter cd27
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  • #1
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Hey, for some odd reason i can't post a new thread, so i'll jsut post it here, seeing as this is the most related topic here. i found this formula for time dilation, and used it to see how much time dilation would occur if you went 90% the speed of light for 20 minutes and back. i came up with 51 minutes. my teacher came up with 51 years...who is right?

T=T'/sqrt1-(velocity sq./speed of light sq)

T=1200 seconds/sqrt1-(167400 m/secs/186000 m/secs)

T=1200 seconds/sqrt 1-0.81

T=1200 seconds/sqrt 0.19

T=1200 seconds/ 0.4358898944

T=2952.988806

Subtract T from T' and you get time dilation=51.76629355 minutes (converted from seconds).

Key:

T=time measured by person who is stationary
T'=time measured by person in motion
V=velocity/speed
C=speed of light (186000 m/secs)

my teacher says the formula is flawed and it's really 51 years, not 51 minutes. who's right here?
 

Answers and Replies

  • #2
551
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cd27 said:
C=speed of light (186000 m/secs)

Where did you get this value of c from? c = 3 x 10^8 ms^-1.

Also, what did you mean when you said you couldn't post a new thread? You posted one fine..
 
  • #3
Doc Al
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[tex]\Delta T = \Delta T' / \sqrt{1 - v^2/c^2}[/tex]
For v = .9c and [itex]\Delta T'[/itex] = 20 minutes, you get:
[tex]\Delta T = 20 / \sqrt{1 - (.9)^2}} = 20 / \sqrt{0.19} = 45.9 [\rmtex{ minutes}][/tex]

FYI: c = 186,000 miles/sec

[Note: I split this thread from another thread in TD.]
 
  • #4
DaveC426913
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Intuitively, 90% of the speed of light is on the low end of the scale of time dilatory effects. So is 20 minutes.

To get years of dilation, especially after only 20 minutes, you'd need to be travelling at better than 99.9999999% of C.

Your prof's wrong.
 
  • #5
551
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Doc Al said:
FYI: c = 186,000 miles/sec

Ahh, whenever I see m I think metres :redface:.
 
  • #6
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thank you guys really appreciated!
 
  • #7
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Most baffling Time Dilation Paradox

Please Point out the ridicules of the below senario

There 3 points A, B and D equidistant from each other. From point A, Rocket A1 is traveling towards point C ie. center of line BD; At speed such that clocks in A1 have slowed down to half the rate of clocks in C1 Rocket at point C.

Form inside A1 with most powerful telescope in the universe it can be seen so. The funny this is everything that is seen happening in Rocket C1 at point C is at twice the speed of normal motion.

Then suddenly an alien ship D1 is spotted traveling from point D to B. When the alien ship passed by point C the scientist at C1 measured and made a note of its speed as 99% speed of light. Scientist at A1 read this written note at C1 through their Telescope and then measured the speed of D1 with their own devices and note it as 198% speed of light, all the things at point C were fast forwarded at twice the speed in A1 as rate of time there had doubled compared to that on A1.
 
  • #8
HallsofIvy
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The first ridiculous thing I see is that you are "hijacking" someone else's thread for a completely different topic- please start your own thread.

The second ridiculous thing is the poor grammar that makes this very difficult to read.

The third ridiculous thing is the phrase "note it as 198% speed of light" which is impossible.
 
  • #9
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well, i'd answer that if i had time, but i don't, sorry! now, Doc, your calculation was correct, but i thought you had to SUBTRACT T from T'? (seeing as the equation appears to be for getting the times seen by the stationary mover [T]) This would account for 1/2 a round trip, meaning to multiply the result of T-T' to get the amount of time dilation for a "there and back" trip. the time dilation is then a total of 51.8 minutes, correct?

Robo, i will TRY to see if i can't work with your question and get an answer by tomorrow. i can't make any promises though...my luck someone else will already have it answered by then, lol.
 
  • #10
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on the contrary to what einstein believed (that once one reaches speed of light time ceased to move [as shown by this formula i am using, time dilation is reduced to zero] now if you go faster, well it becomes a negative number and you can't square root a negative number. therefore einstein said that you "can't reach the speed of light, and if you did, it would have to be changed"...my belief is that there is a "time barrier" and this formula can not describe it once it is broken...but we won't get into that (i don't want this thread blocked).

robo, as you question is a very good one, lots of turns and twists, mfeans you used alot of thinking with it. however, as i said above, this formula can not describe it and you would need to create a new one, which would probably get you the nobel prize.

Ivy, i couldn't say that it is impossible, b/c i haven't exactly gone the speed of light myself, but knowing how Ivy is such a great person, i suppose he has, haven't you Ivy? now how rediculus does it sound for you to say that something is impossible if you've never tried to do it!

however robo, this is not a good thread to post that in...it's way off topic.
 
  • #11
Janus
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cd27 said:
well, i'd answer that if i had time, but i don't, sorry! now, Doc, your calculation was correct, but i thought you had to SUBTRACT T from T'? (seeing as the equation appears to be for getting the times seen by the stationary mover [T]) This would account for 1/2 a round trip, meaning to multiply the result of T-T' to get the amount of time dilation for a "there and back" trip. the time dilation is then a total of 51.8 minutes, correct?

Okay, generally we don't call the difference between T' and T the time dilation. The time dilation is the factor by which they differ. In this case, T is 2.18 times larger than T'.
 
  • #12
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i don't exactly understand what you're saying...
 
  • #13
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cd27 said:
i don't exactly understand what you're saying...
Not to sound patronizing, but it's all relative. If you say difference, you imply subtracting one thing from another. And which would be subtracted from which? Instead, provide ratios that show how things change between your current frame of reference and the dialated frame of reference.
 
  • #14
Doc Al
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cd27 said:
now, Doc, your calculation was correct, but i thought you had to SUBTRACT T from T'? (seeing as the equation appears to be for getting the times seen by the stationary mover [T]) This would account for 1/2 a round trip, meaning to multiply the result of T-T' to get the amount of time dilation for a "there and back" trip. the time dilation is then a total of 51.8 minutes, correct?
Now I see where you get your answer from. (I missed it before.) But, as Janus explained, what we call the "time dilation" is a factor, not a difference in clock readings. If you want the difference between the two clock readings (assuming they are synchronized at the start of the trip, and that the trip takes a total of 40 minutes as recorded by the "moving" clock) then you are correct. The difference in clock readings will be about 51.8 minutes. (Certainly not years!)
 
  • #15
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well, then tell me EXACTLY how i would get the factor? all i know is to subtract, and i guessed that b/c it seemed to be the most logical thing to do.
 
  • #16
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also, i seem to not be able to post new threads in this site, i don't know why, i'll post it somewhere else in another thread and it ends up in a new thread in another place. i don't get it, i imagine someone else is doing it for me, but i dunno. how do i go about getting this screen name re-signed up?
 
  • #17
Doc Al
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cd27 said:
well, then tell me EXACTLY how i would get the factor? all i know is to subtract, and i guessed that b/c it seemed to be the most logical thing to do.
You probably don't realize it, but you obviously must know how to get the time dilation factor... else you wouldn't have been able to do the calculation.

Refer to post #3 in this thread, in which I wrote:

[tex]\Delta T = \Delta T' / \sqrt{1 - v^2/c^2}[/tex]

That factor of [itex]1/ \sqrt{1 - v^2/c^2}[/itex], often called gamma ([itex]\gamma[/itex]), is what we're talking about.
 
  • #18
Doc Al
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cd27 said:
also, i seem to not be able to post new threads in this site, i don't know why, i'll post it somewhere else in another thread and it ends up in a new thread in another place. i don't get it, i imagine someone else is doing it for me, but i dunno. how do i go about getting this screen name re-signed up?
Where are you trying to create a new thread? This thread got moved (by me) into its current forum because you tagged it onto an existing thread in TD (where it didn't belong). Note: New threads are not allowed in TD, so if that's where you are trying to create one, it won't work.
 
  • #19
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Doc Al said:
You probably don't realize it, but you obviously must know how to get the time dilation factor... else you wouldn't have been able to do the calculation.


Doc...thanks man..lol....also, i'm only in algebra one...nice touch huh? i'm a big science nerd, but you don't know things you haven't heard of...well, sometimes. i jsut haven't been through all the classes yet, i just seem to understand very quickly. in fact, that's what my "theory" or "hypothesis" is all from, just me being able to understand certain things better or quicker than others. so, can you explain this stuff to a person who is in algebra one :)
 
  • #21
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thank you!!!
 
  • #22
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cd27 said:
however robo, this is not a good thread to post that in...it's way off topic.

I thought this was a Time Dialation topic, sorry I goofed up.

Hijacked to where ?

I just wanted to see what PF scientist r made up of, will they takeup the challange or choose to ignore. Got the point ....
 
  • #23
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lol...well that was a good way to do it, usually if you want to ask a question about something that's thought to be "impossible" by the majority of the scientific community, this ntire site would be a bad choice to do it at, even though it is a large site. your question was justified, but many scientists like to aggree with only mathematical proof tta because you can't square a negative or zero number, you can't reach the speed of light.
 

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