I see, I wasn't quite sure of how the time dilation equation worked when I started this thread, but I understand it more clearly now.
Thank you! The graphs and explanation made this concept much easier to understand, and I didn't expect to learn the relativistic doppler when I started this...
I was trying to say that an event happens at coordinate time 1.1547 in the rest inertial frame of the planet, but this event happens at the spaceship's proper time of 1.
So the time dilation equation, t' = γt. Where in the rest inertial frame of the planet, t' is the coordinate time, t is...
Okay the frame I was dealing with would be the frame of the planet in the diagram on post 17. At t = 1 and x = 0 in the frame of the planet, t' =1.1547. 1.1547 is the coordinate time, but the proper time on the spaceship is 1. The observer does not see the spaceship clock at 1 when the...
Isn't \nu_{source} the observer time and \nu_{observed} the spaceship time? When the spaceship time is -1 the observer time is -0.577.
I think I understand now:
The equations t' = γ(t-xβ) and x' = γ(x-tβ) replicate an event from 1 frame to another.
When the clock on the planet strikes...
So in the rest frame of the spaceship, the coordinate time and proper time are only equal at x = = 0. If an event happens anywhere else, the time that the event happens is the coordinate time, but there is no proper time because the spaceship does not pass through the event.
\nu_{observed} =...
Could you please explain what you mean by a coordinate effect, the way I see it the coordinate time is the proper time in the reference frame that the calculations are based on.
The equation I found was:
\nu_{observed} = \nu_{source} \sqrt{\frac{1+\beta}{1-\beta}}
\nu_{observed} comes...
Thank you for all your help! =)
So to summarize:
When the observer's clock ticks 1 second, the spaceship's clock ticks 0.866s from time dilation.
At 1 second, the observer sees an event and sees the spaceship's clock at 0.577s due the relativistic doppler effect (I couldn't find the...
So the 0.866s represents a fraction of distance between the points for the black line and not the total distance, but the black dots are a distance 1.1547 away. Does this mean that when the planet's clock tick 1 second, the observer sees an event, but when the clock ticks 1.1547 seconds, the...
So t' = γ(t - vx/c2) and t' = γ(t - xβ) are the same except x in the second equation is in the unit of light seconds. I assume this also applies to x' = γ(x-tβ), where x' = γ(x-vt), v = β/c and c = 1.
As for the 0.75s and 0.25s, I realized I forgot the square root when I solved for γ, so...
The equation t = γ(t' + x'v/c2) creates t = γ(t' + t'β) with β=v/c and t' = x'/c, or t' = γ(t - xv/c2) into t' = γ(t - tβ).
I am not sure where t' = γ(t - xβ) comes from.
Are you talking about t = γ(t' + x'v/c2) or t' = γ(t - xβ)?
I think I am starting to understand this now, the second equation doesn't describe an interval of time, but a moment in time.
This equation t' = γ(t-xβ) (should this be t' = γ(t-tβ) instead?) calculates the time an event happens in the frame of s' given the time and position of the event...
So x is the distance between frames, then this is my problem.
Suppose a spaceship travels past at 0.5c, the spaceship passes an observer on a planet.
The observer measures 1 second on his clock and a clock on the spaceship will spin for 0.75s using the time dilation formula.
However, after 1...
When I first learned time dilation I was given this equation proven using the light clock example.
t = γt'
However, when I looked into the topic in more detail, I was given this equation.
t = γ(t' + xv/c2)
If x represents the distance between 2 frames, then does x have to change...
So that means I would need to compare the series to \frac{3}{n^{2}} instead.
I can prove \frac{3}{n^{2}} through the integral test. Though speaking of the integral test, it doesn't seem like the test works for the original series since I am getting tan-1(∞). So would the integral test be...
Homework Statement
Show that:
\sum \frac{3}{n^{2} + 1}
converges from n = 1 to ∞
Homework Equations
If Ʃbn converges, and Ʃan < Ʃbn.
Ʃan also converges.
The Attempt at a Solution
\sum \frac{1}{n^{2}} converges
\sum \frac{3}{n^{2} + 1} = 3 * \sum \frac{1}{n^{2} + 1}...