Hey I have a question, right now I am in a modern physics class and i cannot figure out this questions becuase i keep contradicting myself.
Two firecrackers explode simultaneously 125 m apart along a railroad track, which we take to define the x axis of an inertial reference frame ( the Home Frame). A train ( the Other Frame) moves at a constant 25 m/s in the +x direction relative to the track frame. How far apart are the explosions as measured in the train's frame of reference?
1.) How far apart are the explosions as meaured in the train frame?
2.) Assume instead that the explosions are not simultaneous, the firecracker farther ahead in the +x direction explodes 3.0 s before the other. Now how far apart would the explosions be in the trains reference?
so is the answer for #2, 50 m or am i understanding something wrong?
// Edit -- whoops -- I just saw the subtitle and am removing the Lorentzian transforms for pure Galilean transforms:
t' = t
x' = x - vt
t' = t
x' = x - vt
QUOTE (mhouck+Sep 7 2009, 06:42 PM)
Hey I have a question, right now I am in a modern physics class and i cannot figure out this questions becuase i keep contradicting myself.
One way out of your difficulties it to label each event (a point in space and time) and then assign coordinates in each of the different frames.
QUOTE (mhouck+Sep 7 2009, 06:42 PM)
Two firecrackers explode simultaneously 125 m apart along a railroad track, which we take to define the x axis of an inertial reference frame ( the Home Frame).
Label events as F1 (t=0,x=0) and F2 (t=0,x=125m). In this frame the events are simultaneous Δt(F1,F2)=0 and happen | Δx(F1,F2) | = 125 meters apart.
QUOTE (mhouck+Sep 7 2009, 06:42 PM)
A train ( the Other Frame) moves at a constant 25 m/s in the +x direction relative to the track frame. How far apart are the explosions as measured in the train's frame of reference?
1.) How far apart are the explosions as measured in the train frame?
1.) How far apart are the explosions as measured in the train frame?
We are going to use t' and x' to represent things in the train's frame of reference. This frame needs and origin event, and to make things simple we can use F1 as the origin.
So F1 = (t=0,x=0) = (t'=0,x'=0) So since we need to figure out what | Δx'(F1,F2) | is, we need to be able to know the x' coordinate of F2.
t'(F2) = t(F2) = 0s
x'(F2) = x(F2) - v t(F2) = 125m - 0v = 125m
So F1 = (t=0,x=0) = (t'=0,x'=0) So since we need to figure out what | Δx'(F1,F2) | is, we need to be able to know the x' coordinate of F2.
t'(F2) = t(F2) = 0s
x'(F2) = x(F2) - v t(F2) = 125m - 0v = 125m
QUOTE (mhouck+Sep 7 2009, 06:42 PM)
2.) Assume instead that the explosions are not simultaneous, the firecracker farther ahead in the +x direction explodes 3.0 s before the other. Now how far apart would the explosions be in the trains reference?
Lets think about this geometrically for a bit.
Say the train, which according to the Home frame moves 25 meters each second would see the firecracker further along explode 5 seconds later than the first firecracker. During that time, the train would have moved 5s × 25(m/s) = 125 m in that same direction, so the distance between any point on the train would be the same distance from the first firecracker as from the second. So Δx'(F3,F4) = 0 = 125m - 5s × 25(m/s). But for this problem, the firecracker further down the track explodes not 5 seconds later, but 3 seconds earlier, so we need an extra minus sign.
Now back to the algebra.
F3 = (t=0,x=0) = (t'=0,x'=0)
F4 = (t=-3s,x=125m)
t'(F4) = t(F4) = -3s
x'(F4) = x(F4) - v t(F4) = 125m - (25 m/s)(-3 s) = 200m
If Since this is the very first course in relativity, then instead of the above Lorentz transformations, they may want you to use the simpler Galilean transformations where γ(v) = 1 and 1/c² = 0 and the answers are just 125 meters and 200 meters. The key is again labeling your events and thinking geometrically.
// Edit: Appendix with Lorentz transforms which are nearly identical at this speed to Galilean transforms
t'(F2) = γ(v)(t(F2) - vx(F2)/c²) = γ(v)(0 - (125 m)v/c²) = ... we really don't care ... ≈ -34.77 fs
x'(F2) = γ(v)(x(F2) - vt(F2)) = γ(v)(125m - 0v) = γ(v)(125m) = (1/√(1 - (v/c)²))(125m) = (1/√(1 - ((25 m/s)/(299792458 m/s))²))(125m) ≈ 125.000000000000435m = 125m + 435 fm
t'(F4) = γ(v)(t(F4) - vx(F4)/c²) = γ(v)(-3s - (25 m/s)(125 m)/c²) ≈ -3.000000000000045 s = -3 s - 45 fs
x'(F4) = γ(v)(x(F4) - vt(F4)) = γ(v)(125m - (25 m/s)(-3 s)) = γ(v)(200m) = (1/√(1 - (v/c)²))(200m) = (1/√(1 - ((25 m/s)/(299792458 m/s))²))(200m) ≈ 200.000000000000695m = 200m + 695 fm
Lets think about this geometrically for a bit.
Say the train, which according to the Home frame moves 25 meters each second would see the firecracker further along explode 5 seconds later than the first firecracker. During that time, the train would have moved 5s × 25(m/s) = 125 m in that same direction, so the distance between any point on the train would be the same distance from the first firecracker as from the second. So Δx'(F3,F4) = 0 = 125m - 5s × 25(m/s). But for this problem, the firecracker further down the track explodes not 5 seconds later, but 3 seconds earlier, so we need an extra minus sign.
Now back to the algebra.
F3 = (t=0,x=0) = (t'=0,x'=0)
F4 = (t=-3s,x=125m)
t'(F4) = t(F4) = -3s
x'(F4) = x(F4) - v t(F4) = 125m - (25 m/s)(-3 s) = 200m
// Edit: Appendix with Lorentz transforms which are nearly identical at this speed to Galilean transforms
t'(F2) = γ(v)(t(F2) - vx(F2)/c²) = γ(v)(0 - (125 m)v/c²) = ... we really don't care ... ≈ -34.77 fs
x'(F2) = γ(v)(x(F2) - vt(F2)) = γ(v)(125m - 0v) = γ(v)(125m) = (1/√(1 - (v/c)²))(125m) = (1/√(1 - ((25 m/s)/(299792458 m/s))²))(125m) ≈ 125.000000000000435m = 125m + 435 fm
t'(F4) = γ(v)(t(F4) - vx(F4)/c²) = γ(v)(-3s - (25 m/s)(125 m)/c²) ≈ -3.000000000000045 s = -3 s - 45 fs
x'(F4) = γ(v)(x(F4) - vt(F4)) = γ(v)(125m - (25 m/s)(-3 s)) = γ(v)(200m) = (1/√(1 - (v/c)²))(200m) = (1/√(1 - ((25 m/s)/(299792458 m/s))²))(200m) ≈ 200.000000000000695m = 200m + 695 fm
QUOTE (mhouck+Sep 7 2009, 06:42 PM)
Hey I have a question, right now I am in a modern physics class and i cannot figure out this questions becuase i keep contradicting myself.
Two firecrackers explode simultaneously 125 m apart along a railroad track, which we take to define the x axis of an inertial reference frame ( the Home Frame). A train ( the Other Frame) moves at a constant 25 m/s in the +x direction relative to the track frame. How far apart are the explosions as measured in the train's frame of reference?
1.) How far apart are the explosions as meaured in the train frame?
2.) Assume instead that the explosions are not simultaneous, the firecracker farther ahead in the +x direction explodes 3.0 s before the other. Now how far apart would the explosions be in the trains reference?
so is the answer for #2, 50 m or am i understanding something wrong?
alright thank you very much, it all makes sense now that I have an example, the spacetime diagrams are hard to interpret the first time you see them.
Thanks again
Two firecrackers explode simultaneously 125 m apart along a railroad track, which we take to define the x axis of an inertial reference frame ( the Home Frame). A train ( the Other Frame) moves at a constant 25 m/s in the +x direction relative to the track frame. How far apart are the explosions as measured in the train's frame of reference?
1.) How far apart are the explosions as meaured in the train frame?
2.) Assume instead that the explosions are not simultaneous, the firecracker farther ahead in the +x direction explodes 3.0 s before the other. Now how far apart would the explosions be in the trains reference?
so is the answer for #2, 50 m or am i understanding something wrong?
alright thank you very much, it all makes sense now that I have an example, the spacetime diagrams are hard to interpret the first time you see them.
Thanks again
What seems to make this different than the standard Einsteinian gedankinexperiment is that question is about distance, and not time.
Imagine that construction crews have set up a pair of blinking lights 60 m apart to mark a construction zone along a road. A car moves along the road with the speed of 30 m/s toward the lights. Let us take the road to be the Home Frame and the car to be the Other Frame.
(a) In the Home Frame, the light farther from the car (light 2) blinks .66s before the other light (light 1) blinks as the car approaches both lights. What is the time interval t'(2)-t'(1) between the blink events in the car's frame?
(
What is the x displacement x'(2)-x'(1) between events in the car frame?
Using Galilean Transformations
I got that the time between the two events was the same in both frames and that the displacement was 79 m for the car frame, and I wanted to make sure that was right.
(a) In the Home Frame, the light farther from the car (light 2) blinks .66s before the other light (light 1) blinks as the car approaches both lights. What is the time interval t'(2)-t'(1) between the blink events in the car's frame?
(
What is the x displacement x'(2)-x'(1) between events in the car frame?Using Galilean Transformations
I got that the time between the two events was the same in both frames and that the displacement was 79 m for the car frame, and I wanted to make sure that was right.
QUOTE (mhouck+Sep 8 2009, 01:01 AM)
What is the time interval t'(2)-t'(1) between the blink events in the car's frame?
You need to learn how to do your homework all by yourself. If one of us does it for you, you don't get to learn anything.
You need to learn how to do your homework all by yourself. If one of us does it for you, you don't get to learn anything.
I did, is my answer right or not?
QUOTE (mhouck+Sep 8 2009, 03:13 AM)
I did, is my answer right or not?
In Galilean relativity, yes. (attention, the distance between events is 79.8 not 79)
In special relativity, no.
In Galilean relativity, yes. (attention, the distance between events is 79.8 not 79)
In special relativity, no.
haha alright awesome, i figured out what i was doing wrong earlier, thanks
AlexG - I have no idea what you mean -- this is a exercise about Galilean Relativity, which is perfectly valid if you have no evidence that c is a constant to all observers. And since it is likely an introductory course, they surely aren't going to develop that evidence from the success of Maxwell's equations and the homopolar generator. 
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