billllib
- 76
- 2
- Homework Statement
- https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf
- Relevant Equations
- L' = L/y
t' = ty
V = u-v / 1 - uv / (c^2)
https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf
The questions start at page 44
Whenever I refer to y, y = gamma.
1.1
This question is primarily deriving LV/C^2?
How does 2LV / c^2-v^2 becomes
2Lv / c^2(1-v^2/c^2)1.4
On the solution page it shows fig 1.61 and fig 1.62. It talks about the light having to reach my eyes. Where am I standing in 1.61 and 1.62? In fig 1.61 starting from A going vertical, why is it equal to L and not LB? Also why LB/y = L squareroot 1-B^2?
Just to confirm 1.62 represent the image rotated more but has more width due to length contraction not applying?
Also I slightly confused what the answer is for 1.4 it just shows a bunch of angles and numbers.
1.5
How is (a^2cL / (c-v )-L) )/ v derivived?
I get L / c-v = (Lv / c^2) + c1.7
skip
(L-L/y)/v part confused
1.8
How does the stick start as a "/" shape then bend in a "V" shape then switch direction to a "\" shape
1.9
where are the marks at the end of the board?
1.10
why "y" in cookie cutter opposite dough
1.11
picture B why is the distance L/y between the balls?
1.12
For 1.12
L/y and in y I plug in the values of 3c/5 and c/2?ignoreIs picture 1.12 correct?
Why is 3c/5 - c/2? Shouldn't c/2 - 3c/5 be first because c/2 is located before 3c/5?
I get 4c/10 . Is that correct? I can explain.
putting all the numbers together I get
V = v-u / 1 - vu / (c^2)
(3c/5) - (c/2) = c/10
1 - (3c/5)(c/2) = (10/ 10) - 6c^2/10
gives
c / 10 / (10 / 10 - 6c^2 / 10 ) /c^2
first I go (10) (c^2)/1 = 10c^2
# c^2 cancel out
(c / 10) / (10 / 10 - 6c^2 / 10c^2 )
=
(c / 10 ) / (10 / 10 - 6 / 10)
=
(c / 10) / (4 / 10)
=
# change (4 / 10) to 40
=
(c / 10) / 40
=
(c/10) (40/1) = 40c/10 = 4c/10 = 2c/5
1.54 part don't understand
1.13
1.15
I get
(c/n) + v / ( 1 + (cv/ n) / (c^2) ) How do I get 1.68?
Why multiply by ( 1-v/nc) (1-v/nc ) in 1.69 and can someone show the steps how I get the answer in 1.69
1.17 (probably don't need)
How do i get the 2u? I get 2V. Does it make difference? 1.19
I am incredibly confused first it says don't use time dilation or length contracation I assume I can't use gamma. I tried reading the answer and just got more confused. Also why can I use gamma for einstein velocity addition formula but not for the example below.
A train is standing still of length L. A person is running in the train from left to right . A person is running with velocity v to the right on the ground.
What is the velocity of the person running on the train? v1 = ball = V_ball
v2 = train = V_train
C = photonGround frame
First:
V_Ball_train = V_ball + V_train
V_C_train = V_C + V_train
then
apply gamma:
V_Ball_train = V_ball + V_train / 1 + V_ball V_train / c^2
V_C_train = V_C + V_train / 1 + V_C V_train / c^2
or
V_Ball_train / y
V_C_train / y
plus both velocity together in ground frame
V_ground_frame = 1 + V_Ball_train V_C_train / c^2
I want speed of train to be 0
train frame
First:
V_Ball_train = V_ball + V_train
V_C_train = V_C + V_train
then
apply gamma:
V_Ball_train = -V_ball - V_train / 1 - V_ball (-V_train) / c^2
V_C_train = -V_C - V_train / 1 - V_C (-V_train) / c^2
or
V_Ball_train / y
V_C_train / y
plus both velocity together in train frame
V_train_frame = 1 + V_Ball_train V_C_train / c^2
plus velocities from both frames combined
But it is completely different the answer. Can someone help?
ground frame
velocity_person_running_left _on_train_frame
u = velocity of train
v = velocity of person
velocity_person_running_left_ground_frame = u + v / 1 + uv/c^2
velocity_person_running_right_ground_frame = u + v / 1 + uv/c^2V_ground_frame = velocity_person_running_right _on_train_frame -
velocity_person_running_left _on_train_frame
Train frame
u = velocity of person
v = velocity of train
The train is moving to the right. In order to get velocity_person_running_left_on_train_frame do I go <------(v_person_running_left) + (v_train)-----> or
(v_train)--------> + <------(v_person_running_left)(v_train) when getting new velocity
I assume it is the first options.
velocity_person_running_right _on_train_frame = u - v / 1 - uv/c^2
velocity_person_running_left _on_train_frame = -u -v /1 - (-u)(-v)/c^2
V_train frame = velocity_person_running_right _on_train_frame - velocity_person_running_left _on_train_frame
combining frames
u = V_train_frame
v = V_ground_frame
V = u - v / 1 - uv/c^2 = u/c^2
The questions start at page 44
Whenever I refer to y, y = gamma.
1.1
This question is primarily deriving LV/C^2?
How does 2LV / c^2-v^2 becomes
2Lv / c^2(1-v^2/c^2)1.4
On the solution page it shows fig 1.61 and fig 1.62. It talks about the light having to reach my eyes. Where am I standing in 1.61 and 1.62? In fig 1.61 starting from A going vertical, why is it equal to L and not LB? Also why LB/y = L squareroot 1-B^2?
Just to confirm 1.62 represent the image rotated more but has more width due to length contraction not applying?
Also I slightly confused what the answer is for 1.4 it just shows a bunch of angles and numbers.
1.5
How is (a^2cL / (c-v )-L) )/ v derivived?
I get L / c-v = (Lv / c^2) + c1.7
skip
(L-L/y)/v part confused
1.8
How does the stick start as a "/" shape then bend in a "V" shape then switch direction to a "\" shape
1.9
where are the marks at the end of the board?
1.10
why "y" in cookie cutter opposite dough
1.11
picture B why is the distance L/y between the balls?
1.12
For 1.12
L/y and in y I plug in the values of 3c/5 and c/2?ignoreIs picture 1.12 correct?
Why is 3c/5 - c/2? Shouldn't c/2 - 3c/5 be first because c/2 is located before 3c/5?
I get 4c/10 . Is that correct? I can explain.
putting all the numbers together I get
V = v-u / 1 - vu / (c^2)
(3c/5) - (c/2) = c/10
1 - (3c/5)(c/2) = (10/ 10) - 6c^2/10
gives
c / 10 / (10 / 10 - 6c^2 / 10 ) /c^2
first I go (10) (c^2)/1 = 10c^2
# c^2 cancel out
(c / 10) / (10 / 10 - 6c^2 / 10c^2 )
=
(c / 10 ) / (10 / 10 - 6 / 10)
=
(c / 10) / (4 / 10)
=
# change (4 / 10) to 40
=
(c / 10) / 40
=
(c/10) (40/1) = 40c/10 = 4c/10 = 2c/5
1.54 part don't understand
1.13
1.15
I get
(c/n) + v / ( 1 + (cv/ n) / (c^2) ) How do I get 1.68?
Why multiply by ( 1-v/nc) (1-v/nc ) in 1.69 and can someone show the steps how I get the answer in 1.69
1.17 (probably don't need)
How do i get the 2u? I get 2V. Does it make difference? 1.19
I am incredibly confused first it says don't use time dilation or length contracation I assume I can't use gamma. I tried reading the answer and just got more confused. Also why can I use gamma for einstein velocity addition formula but not for the example below.
A train is standing still of length L. A person is running in the train from left to right . A person is running with velocity v to the right on the ground.
What is the velocity of the person running on the train? v1 = ball = V_ball
v2 = train = V_train
C = photonGround frame
First:
V_Ball_train = V_ball + V_train
V_C_train = V_C + V_train
then
apply gamma:
V_Ball_train = V_ball + V_train / 1 + V_ball V_train / c^2
V_C_train = V_C + V_train / 1 + V_C V_train / c^2
or
V_Ball_train / y
V_C_train / y
plus both velocity together in ground frame
V_ground_frame = 1 + V_Ball_train V_C_train / c^2
I want speed of train to be 0
train frame
First:
V_Ball_train = V_ball + V_train
V_C_train = V_C + V_train
then
apply gamma:
V_Ball_train = -V_ball - V_train / 1 - V_ball (-V_train) / c^2
V_C_train = -V_C - V_train / 1 - V_C (-V_train) / c^2
or
V_Ball_train / y
V_C_train / y
plus both velocity together in train frame
V_train_frame = 1 + V_Ball_train V_C_train / c^2
plus velocities from both frames combined
But it is completely different the answer. Can someone help?
ground frame
velocity_person_running_left _on_train_frame
u = velocity of train
v = velocity of person
velocity_person_running_left_ground_frame = u + v / 1 + uv/c^2
velocity_person_running_right_ground_frame = u + v / 1 + uv/c^2V_ground_frame = velocity_person_running_right _on_train_frame -
velocity_person_running_left _on_train_frame
Train frame
u = velocity of person
v = velocity of train
The train is moving to the right. In order to get velocity_person_running_left_on_train_frame do I go <------(v_person_running_left) + (v_train)-----> or
(v_train)--------> + <------(v_person_running_left)(v_train) when getting new velocity
I assume it is the first options.
velocity_person_running_right _on_train_frame = u - v / 1 - uv/c^2
velocity_person_running_left _on_train_frame = -u -v /1 - (-u)(-v)/c^2
V_train frame = velocity_person_running_right _on_train_frame - velocity_person_running_left _on_train_frame
combining frames
u = V_train_frame
v = V_ground_frame
V = u - v / 1 - uv/c^2 = u/c^2
