Activity

Video (and the notebook with the pictures I used) posted above.

I’ve uploaded the video, and I did the best I could to export the spacetime diagrams from that python notebook. There a link to them right after the link to the problems and answers (up above).

I made some corrections to the answer sheet and posted it on the September 7th meeting page.

You are correct! In my response to you, by t0 I just meant proper time (which is also t’ in this case). I completely see that the reduction works here, I feel I’m getting closer to having an intuition about when such reductions to proper time/length calculations are allowed (seems to be: whenever one event is set at the origin and is then compared…[Read more]

Mike, our answers match except for round-off error.

If we want to be picky:
1. For problem 1, our only difference is that I used 3 significant digits instead of your 2.
2. For problem 2, we took different approaches but I think either is valid, and our answers are off only due to the slightly different values we got for #1 (roundoff error).
3.…[Read more]

@john, @scott I put my first cut at the answers here. I’m not going to post them as the official answers until either Tina or one of you guys agrees with them. But have a look. I don’t think these all match either of what’s above, but I haven’t actually kept up on the conversation. Sorry!

I’m thinking that we’ll have a session to go over these…[Read more]

Scott, I agree with your first paragraph, but I don’t see why that means t₀ will have elements of both x and t in its calculation.

After looking at this yet again, I concluded that our approach to #1 is equivalent after all, but not our approach to #2. The key to understanding this is that when x and t are measured in the same frame,
x = vt.…[Read more]

I pasted a more readable version of what I’ve got below. I think the discrepancy here has to do with how v is conceptualized. v captures the relative velocity of two different frames. From the Earth frame, v captures the measured velocity of the muon, but it’s not going to be the velocity of the muon in it’s own frame of reference! For the muon,…[Read more]

Here’s my answers in a nicer format:

I worked problem 1 differently, but got the same answer (after accounting for significant digits).

1.) v = 0.99c

γ = 1 1 – v 2 c 2
= 7.09

t = 6.73 µs
t₀ = t / γ (t₀ is the proper time)
= 0.949 µs

————————-

I worked problem 2 differently, and got a very different answer.
2.) 
v = .99c = .99 * 3*10⁸ m/s

= 2.97*10⁸ m/s



t₀ =…[Read more]

Problems 3-5:

v = 4/5 LY/Y
d = 4 LY

————————-
3) time in Earth’s frame:
t = d/v
= 4 LY / (4/5 LY/Y)
= 5 Y
————————-
4) time in ship’s frame:
γ = 1 / √(1 – v²/c²)
= 5/3

t = t’/ γ (t, or proper time, is in frame where events happen at same place => ship’s frame)
= 5 Y / (5/3)
= 3 Y
—…[Read more]

Given:
v = .99c
x = 2000m
t = 6.73us

lambda = sqrt(1/[1-{v/c}^2]) = sqrt(1/[1-0.99^2]) ~= sqrt(50.25) ~= 7.1

1) t’ = lambda * (t – vx/(c^2)) = lambda * (t – .99x/c) ~= 7.1 * (6.73us – .99*(2000m)/(3*10^8 m/s)) = 7.1*(6.73us-6.6us) = 7.1*0.13us = 0.92us
2) x’ = lambda * (x – vt) ~= 7.1 * (2000m – .99(3*10^8 m/s)(6.73*10^-6 s)) = 7.1 * (2000m -…[Read more]

It’s 3:00pm West Cost time. I believe that’s 4:00pm Denver and 6:00pm Eastern.
The meetings are (almost) always Sunday at that time.
If we do a different day or time we’ll make sure to let everybody know.

Hi, Mike. What time is the meeting? I’m on Eastern Daylight Time.

–Joel