Series of questions about Special Relativity

In summary, there is a discussion about using LaTeX and posting one question per thread with an attempt at a solution. The thread is now closed.
  • #1
billllib
77
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
 
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  • #2
billllib said:
Whenever I refer to y, y = gamma.
Time to learn LaTeX... :smile:

https://www.physicsforums.com/help/latexhelp/

1585088761638.png
 
  • #3
Should I delete the thread and use LaTeX ?
 
  • #4
You should probably post one problem per thread.
 
  • #5
billllib said:
Should I delete the thread and use LaTeX ?

That, and limit yourself to one question per thread, and show your attempt at a solution for each question. Remember that we cannot just give you answers. All we can do is help you to figure them out for yourself.

This thread is closed.
 

1. What is Special Relativity?

Special Relativity is a theory proposed by Albert Einstein in 1905 that explains the relationship between space and time in the absence of gravity. It states that the laws of physics are the same for all observers in uniform motion, and the speed of light is constant regardless of the observer's frame of reference.

2. How is Special Relativity different from Classical Mechanics?

Classical Mechanics, developed by Isaac Newton, is based on the concept of absolute time and space. In contrast, Special Relativity introduces the idea of relative time and space, where measurements of time and space are dependent on the observer's frame of reference. Additionally, Special Relativity takes into account the constant speed of light, which is not a factor in Classical Mechanics.

3. What is the significance of the speed of light in Special Relativity?

The speed of light, denoted by the letter "c", is a fundamental constant in Special Relativity. It is the maximum speed at which all matter and information can travel in the universe. It is also the same for all observers, regardless of their frame of reference, which is a key principle in Special Relativity.

4. Can Special Relativity be applied to everyday situations?

Yes, Special Relativity has been extensively tested and verified through experiments and observations. It is used in many modern technologies, such as GPS systems, particle accelerators, and nuclear power plants. Although the effects of Special Relativity are not noticeable in everyday life, they are crucial for understanding the behavior of matter and energy in extreme conditions.

5. Are there any limitations to Special Relativity?

Special Relativity has been very successful in explaining and predicting the behavior of matter and energy in the universe. However, it is limited to situations involving objects moving at constant speeds in a straight line. It does not apply to situations involving acceleration, gravity, or objects moving at speeds close to the speed of light. In these cases, General Relativity, another theory proposed by Einstein, is needed.

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