How to calculate time dilation

[Bjarne]

Which equation do I have to use to be able to calculate the total time dilation in an accelerating framework?

Let say a space craft is accelerating away from us
I need to calculate the time dilation for a certian period

According to special relativity time dilation quadruples every time velocity doubles

Edited January 29, 2016 by Bjarne

[ajb]

If you are talking about just special relativity, then you can use Rindler coordinates which describe uniformly accelerating observers. From there I think you can define a notion of time dilation. I would have to check carefully.

[swansont]

In SR you have to know v(t) and integrate the dilation over the time interval see http://www.scienceforums.net/topic/74683-acceleration-is-not-important-in-the-twin-paradox/page-20#entry779626That will tell you how much the proper time has changed

 

Since accelerations are indistinguishable from gravity, you should be able to calculate it using an equivalent to the gravitational redshift if the acceleration is constant. That's what they did in the following paper

 

MEASUREMENT OF THE RED SHIFT IN AN ACCELERATED SYSTEM USING THE MOSSBAUER EFFECT IN Fe57

H. J. Hay, J. P. Schiffer, T. E. Cranshaw, and P. A. Egelstaff

 

The expected shift can be calculated in two

ways. One can treat the acceleration as an effec-

tive gravitational field and calculate the difference

in potential between the source and absorber, or

one can obtain the same answer using the time

dilatation of special relativity

 

Here v is constant, because it's a rotating system

[Sensei]

According to special relativity time dilation quadruples every time velocity doubles

Where did you get such idea?

 

[math]t'=\frac{t}{\gamma}[/math]

or

[math]t'=t*\gamma[/math]

 

Where

[math]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

Gamma for v=0.25c is 1.0327955589886444

Gamma for v=0.5c is 1.1547005383792517

 

1.1547005383792517 / 1.0327955589886445027144707732753 = 1.118033988749894848204586834366

Doesn't look like quadruple to me..

 

That's classical physics kinetic energy that is quadrupled when velocity is doubled.

E.K.=1/2*m*v^2

for v=1,m=1,E.K.=1/2*1*1^2=0.5 J

for v=2,m=1,E.K.=1/2*1*2^2=2 J (4 times as for v=1)

for v=4,m=1,E.K.=1/2*1*4^2=8 J (4 times as for v=2)

[Bjarne]

Where did you get such idea?

 

[math]t'=\frac{t}{\gamma}[/math]

or

[math]t'=t*\gamma[/math]

 

Where

[math]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

Gamma for v=0.25c is 1.0327955589886444

Gamma for v=0.5c is 1.1547005383792517

 

1.1547005383792517 / 1.0327955589886445027144707732753 = 1.118033988749894848204586834366

Doesn't look like quadruple to me..

 

That's classical physics kinetic energy that is quadrupled when velocity is doubled.

E.K.=1/2*m*v^2

for v=1,m=1,E.K.=1/2*1*1^2=0.5 J

for v=2,m=1,E.K.=1/2*1*2^2=2 J (4 times as for v=1)

for v=4,m=1,E.K.=1/2*1*4^2=8 J (4 times as for v=2)

 

Right. I mean by low speed (almost).

For example:

10 000 m/s = 5,6 E-8

20 000 m/s = 2,2 E-7

[imatfaal]

As Sensei pointed out you really need to consider gamma rather than gamma minus 1 which is what you are doing - but yes gamma minus 1 will quadruple when v doubles for values where v is lots lower than c. This ratio will increase as the values get closer to c and is never actually as low as 4