Solving Two People Clapping Problem on Road

[TristanRE]

Homework Statement

"Two people stand a distance L apart along an east–west road. They both clap their hands at
precisely noon in the ground frame. You are driving eastward down this road at speed 4c/5. You
notice that you are next to the western person (W) at the same instant (as measured in your frame)
that the eastern person (E) claps. Later on, you notice that you are next to a tree at the same
instant (as measured in your frame) that the western person claps. Where is the tree along the
road? (Describe its location in the ground frame by computing how far to the east of W it is.)"

Homework Equations

"rear clock ahead": t = Lv/c^2
gamma = 1/sqrt(1-v^2/c^2)
d=vt
t_A = gamma*t_B

The Attempt at a Solution

So I have obtained 2 answers, and I'm unsure which is correct.

By looking at it from my frame, and noting that "Rear clock ahead" tells me that E will clap t = Lv/c^2 =4L/5c before W, I know that (in my frame) the tree will travel towards me at v = 4c/5, and will travel for time t. Therefore: d = v*t = 4c/5*4L/5c=16L/25.

I'm confused though, for the questions asks me to give the distance from the ground frame. Is the above correct? I believe that its not, and that I have to adjust the distance obtained by the gamma factor (3/5) because of length contraction.

Therefore L(ground) = L(for me)/gamma = 16L/25*5/3 = 16L/15.

What gives me confidence in this is that I get the same answer if I instead adjust the time it takes the tree to move (in the ground frame, the time that I move for) using the equality t(ground) = gamma*t(me) (for the ground frame).

I'm having a hard time justifying either of these in my head. Which is correct?

Thanks for the help.

 

[Nugatory]

You will get better results if you use the Lorentz transforms instead of thinking in terms of length contraction and time dilation. You have three relevant events (West claps, East claps, next to the tree) and you have enough of the x, x', t, and t' coordinates for each to work out the ones you don't have.

 

[TristanRE]

I gave it a go with the LTs, thought I'm not sure I'm going through with it correctly.

Here's what I did:
Between claps
Ground frame: x = L, t = 0
x' = Y(x-vt) = 5L/3
t' = Y(t-vt/c^2) = -5L/3c

But now, for between the West clap and meeting the tree (what I need to know):
x' = 0, t' = -5L/3c

I get x = Y(x'+vt') = 20L/9 which definitely doesn't seem right. Did I miss something?

 

[TSny]

Between claps
Ground frame: x = L, t = 0
x' = Y(x-vt) = 5L/3
t' = Y(t-vt/c^2) = -5L/3c

But now, for between the West clap and meeting the tree (what I need to know):
x' = 0, t' = -5L/3c

I get x = Y(x'+vt') = 20L/9 which definitely doesn't seem right. Did I miss something?

I had to stare a while at what you did here, but I think your procedure is OK. (I had to keep in mind that your symbols x, t, x', t' are intervals between events.)

However, I believe you made a mistake in the equation t' = Y(t-vt/c^2) = -5L/3c. Should the second t in the middle expression be x? That's probably just a typo. Nevertheless, I don't get -5L/(3c).

 

[TristanRE]

My apologies! I get so used to shortening it (no deltas) that I often forget they were ever there.

It was a typo. I ran it again though, and I obtained 4L/3c for t'. Not sure how I did that. =/

I thought about it again though, and I believe I got the values for x' and t' incorrect for the second set. I meet the tree when W claps, therefore, in my frame (the primes): x' = ?. t' = 0.

In this case, I'm not sure where to go since I am lacking a value for x', and it is needed to calculate either value x or t. And the previous value for distance shouldn't apply, as it corresponds to the distance between W clapping and E clapping, yes?

I'm really not sure where to go with this, it seems like an endless circle that doesn't connect all three. But let me have a go at it again tomorrow, and I'll get back to it here! (long day)

 

[TSny]

It was a typo. I ran it again though, and I obtained 4L/3c for t'.

OK. If you give the correct interpretation to t' here, then you are getting close to the answer for the original question.

I thought about it again though, and I believe I got the values for x' and t' incorrect for the second set. I meet the tree when W claps, therefore, in my frame (the primes): x' = ?. t' = 0.

Yes, Δt' is zero for the two events: (1) W claps (2) Car meets tree

Instead, now think about the two events: (1) Car meets W person (2) Car meets tree

(I think it's best to use the Δ symbol when referring to differences in x or differences in t for two specified events. It keeps things clearer for me, anyway.)

 

[PeroK]

Another idea is to leave ##v## and ##\gamma## in general and derive a formula for the distance to the tree in terms of these variables and ##L##. Then you could check that formula makes sense in the limits of ##v## small and close to ##c## before plugging in the specific ##v## you have.