Time dilation

Not doing your homework for you, but I'll let you know what you're looking for. Your textbook should explain this. It's a few very basic formulas.

Just change your speed from km/h into a fraction of c. Then use that fraction of c to calculate gamma (a value used to describe space and time contraction and dilation). Then there should be a formula for experienced time related to at-rest time, usually represented as tau and t.

>>8697151
It is not homework, I do not want to go past things before I feel confident I have grasped it. Thank you for your answer!

This was a "Think!" kind of question in the book

>>8697132
Your proper time (ie the time in your inertial frame) would be [math] \tau = t/ \gamma [/math] now just substitute in your values.

>>8697170
I may seem stupid but I can't get this done, would you mind walking me through the steps? I'm studying on distance and this is driving me crazy.

>>8697198
I would do it, just ask.
t. Physics student

>>8697198
This is what I get:

γ = 1/sqroot of (1-v^2/c^2) = 1
If i put t/sqroot of (1-v^2/c^2) = t

>>8697210
Can you do it please? It's hard do not have a teacher to ask sometimes

>>8697215
sqrt(1-(v/c)^2) is less than 1 whenever v is not zero, 1000km/h is just so slow that your calculator says it's 1

>>8697198
Okay, stop me when you no longer understand what I'm talking about. Take you plane to be an inertial frame, then the time within that frame (called the proper time) is related to the coordinate time (or the time in some frame at rest with respect to you) by the following: [eqn] \tau = \frac { 1 } { \gamma } t [/eqn] Where [math] \tau [/math] is the proper time, t is the coordinate time and [math] \gamma [/math] is the Lorentz factor. Which in turn is defined as [eqn] \gamma = \frac { 1 } { \sqrt { 1 - \frac { v^2 } { c^2 } [/eqn]. The first thing you should notice here is that, in your problem, [math] \gamma [/math] is going to be very small, since c is about [math] 10^6 [/math] times larger than v. So we expect that [math] \tau \approx t [/math] and indeed when we substitute our values into [math] \gamma [/math] we get that [math] v^2 / c^2 \approx 8.9 \times 10^{-13} [/math] so [math] \gamma \approx 1 [/math].

This shouldn't be that surprising since relativistic effects only become dominant about 20% of c.

>>8697230
Well let me try that again:
[eqn] \gamma = \frac { 1 } { \sqrt { 1 - \frac { v^2 } { c^2 } } } [/eqn] Also:
>In your problem [math] \gamma [/math] is going to be very small.

Should read:
>In your problem [math] \gamma [/math] is going to be on the order of 1.

>>8697230
>γ
>γ
> is going to be very small, since c is about 106
>10
>6
> times larger than v.

This makes much sense, I guess I expected something to be happening and was looking for an answer that was not there.

I very much appreciate you efforts here, thank you.

>>8697253
My mind is finally at peace

did anyone else get about -54µs per year?

>>8697216
Ok im sorry it's really late now where I live but I'll do it tomorrow morning. If this thread is dead by then just drop me an email at [email protected] :)

File: 20170224_081813.jpg (2MB, 4128x2322px) Image search: [Google]

2MB, 4128x2322px

:)

>>8699090
Oops I switched t and t'
Sorry it should be t=1/γ*t', so the time measured by you is shorter

>>8697393
d = sqrt(1-(v/c)^2)-1
v = 278 m/s
d = -4.299893774e-13 s/s
d * 3.1536e07 s = -1.356014501e-05 s
-13.56 µs per year