Attempting To Understand Relativity And Time Dilation - Physics Forums

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Attempting to understand relativity and time dilation
  • Thread starter Thread starter Prierin
  • Start date Start date Dec 2, 2015
  • Tags Tags Dilation Relativity Time Time dilation
Click For Summary The discussion revolves around the correct application of the time dilation formula in relativity, specifically t = t0/(1-v^2/c^2)^(1/2). A user initially calculated the time dilation incorrectly, consistently arriving at 51.28 instead of the expected 32. After clarification on the formula's exponent, they realized that using 0.5 instead of 1/2 resolved their issue, yielding the correct answer. The confusion stemmed from how the exponent was input into the calculator, leading to misinterpretation of the calculations. Ultimately, the user confirmed that understanding the exponent's role was key to achieving the correct result. Prierin Messages 5 Reaction score 0

Homework Statement

t = t0/(1-v2/c2)1/2

Homework Equations

t = 10/(1-.95c^2/c^2)1/2

The Attempt at a Solution

[/B] The provided solutiojn to the example given above is: t = 10/(1- (.95c)2/c2)1/2 t = 10/(1- .952)1/2 t = 10/ .312 t = 32 Unfortunately, no matter what I do the answer I continuously come up with is 51.28. I've tried four different approaches and come up with the same EXACT answer each time without fail. I am not so bold as to say the example above is incorrect as this level of math isn't my forte, so I am stuck. Can anyone explain to me WHY the answer would be 32 rather than 51.28? I am not familiar with the 1/2 at the end of the formula and I suspect that may be what is throwing my numbers off. I am nto clever enough, it seems, to get my calculator to understand that portion correctly. Thanks in advance. (On a side note, I have also seen this forumal written as t=t0/(1-v^2/c^2)^1/2. Not sure which is correct.) Physics news on Phys.org
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andrewkirk Science Advisor Homework Helper Insights Author Gold Member Messages 4,140 Reaction score 1,741
Prierin said: (On a side note, I have also seen this forumal written as t=t0/(1-v^2/c^2)^1/2. Not sure which is correct.)
That's what's correct, and your use of the other one is probably what's causing you problems. Prierin Messages 5 Reaction score 0 OK, that's good to know, thanks. However. even with that I am still coming up with the same result of 51.28. if c= 299792458 and .95c = 284802835.1 10/(1-284802835.1^2/299792458^2)^1/2 = 51.28205128205128205128 The example answer, however, is 32, so I am still confused as how they reached that conclusion. EDIT: I just attempted to edit the formula somewhat and FINALLY gor 32 as an answer by using .5 rather than 1/2. 10/(1-284802835.1^2 / 299792458^2)^.5 = 32.02 I had a feeing the 1/2 was being interpreted incorrectly by the calculator as a division rather than a fraction... still, I'll need confirmation from someone who knows the math better than I before I celebrate Last edited: Dec 2, 2015 Ray Vickson Science Advisor Homework Helper Dearly Missed Messages 10,704 Reaction score 1,723
Prierin said:

Homework Statement

t = t0/(1-v2/c2)1/2

Homework Equations

t = 10/(1-.95c^2/c^2)1/2

The Attempt at a Solution

[/B] The provided solutiojn to the example given above is: t = 10/(1- (.95c)2/c2)1/2 t = 10/(1- .952)1/2 t = 10/ .312 t = 32 Unfortunately, no matter what I do the answer I continuously come up with is 51.28. I've tried four different approaches and come up with the same EXACT answer each time without fail. I am not so bold as to say the example above is incorrect as this level of math isn't my forte, so I am stuck. Can anyone explain to me WHY the answer would be 32 rather than 51.28? I am not familiar with the 1/2 at the end of the formula and I suspect that may be what is throwing my numbers off. I am nto clever enough, it seems, to get my calculator to understand that portion correctly. Thanks in advance. (On a side note, I have also seen this forumal written as t=t0/(1-v^2/c^2)^1/2. Not sure which is correct.)
The correct formula is t = \frac{t_0}{\sqrt{ 1 - v^2/c^2} }= \frac{t_0}{(1 - v^2/c^2)^{1/2} } If you use that you will get 32.03. The "1/2" is not a fraction or anything like it; it is an exponent (= the "1/2"th power = square root). Prierin Messages 5 Reaction score 0 BEAUTIFUL! I knew I was doing something wrong. Now, I have calculated the formula both ways: 10/(sqrt(1-284802835.1^2/299792458^2)) & 10/(1-284802835.1^2/299792458^2)^.5 and have come up with the same EXACT answer of 32.02563076101742669665 (or 32.03) so now I am a happy chappie! Thanks and kudos! Now if I can just fogure out how to save values in the calculator LOL andrewkirk Science Advisor Homework Helper Insights Author Gold Member Messages 4,140 Reaction score 1,741
Prierin said: 10/(1-284802835.1^2/299792458^2)^1/2 = 51.28205128205128205128 The example answer, however, is 32, so I am still confused as how they reached that conclusion. EDIT: I just attempted to edit the formula somewhat and FINALLY gor 32 as an answer by using .5 rather than 1/2. 10/(1-284802835.1^2 / 299792458^2)^.5 = 32.02 I had a feeing the 1/2 was being interpreted incorrectly by the calculator as a division rather than a fraction... still, I'll need confirmation from someone who knows the math better than I before I celebrate
It wasn't being interpreted incorrectly, it was being punched in incorrectly. The standard rule for precedence of operators (ie which arithmetic operations get done before others) is that exponentiation is done before multiplication and division, which are done before addition and subtraction. So if you punched in 10/(1-284802835.1^2/299792458^2)^1/2 then that means 10/((1-284802835.1^2/299792458^2)^1)/2 which is $$\bigg[\frac{10}{(1-\frac{(284802835.1)^2}{(299792458)^2})^1}\bigg]\div 2$$ which is not what you want. Prierin Messages 5 Reaction score 0 I meant to say, the calculator wasn't misinterpreting - I was. LOL It makes sense now. As I pointed out, if I use ^.5 the answer comes out correctly so now I can breathe a little easier. I didn't *kneed* to know any of this math, but I have a curious mind and like to understand how things work...

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