Transcript File

By: Victor Jeung
1.
2.
All laws of physics
are valid in all inertial
frame of references.
Light travels through
space at a speed of
c = 3.00 × 108 m/s,
relative to all inertial
frames of reference.


Simultaneity is not an
absolute concept
Two events that are
simultaneous in one
frame of reference are
not simultaneous in
another frame of
reference; that is in
motion with respect
to the first


Definition: Slowing down of time in a system
as seen by an observer in motion relative to
the system
∆𝑡𝑠
Equation: ∆𝑡𝑚 =
2
𝑣
1− 2
𝑐


∆𝑡𝑠 (proper time) is the time interval that
separates two events measured by an
observer to whom the events occur at the
same position
∆𝑡𝑚 is the time interval measured by an
observer external to the system

Objects cannot
have a speed
equal to or
greater than the
speed of light




Definition: Shortening of distances in a system
as seen by an observer in motion relative to the
system
𝑣2
− 2
𝑐
Equation: 𝐿𝑚 = 𝐿𝑠 1
𝐿𝑚 (proper length) is the length of the object
measured by an observer at rest relative to the
object
𝐿𝑠 is the length of the object measured by an
observer moving relative to the beginning and
end points of the length being measured
𝑚𝑣

Equation: 𝑝 =

𝑚 (rest mass) is the mass of an object when
the object is at rest
𝑣2
1− 2
𝑐


Conservation of mass-energy: rest mass and
energy are equal
Equations:
 𝐸𝑡𝑜𝑡𝑎𝑙 =
𝑚𝑐 2
𝑣2
1− 2
𝑐
 𝐸𝑟𝑒𝑠𝑡 = 𝑚𝑐 2
1.
a)
b)
c)
d)
The Star Wars spacecraft Tie Fighter flies at
a speed of 0.8c with respect to the Earth.
Luke Skywalker determines the time interval
between the two events on Earth is 20h.
What is the time interval for Tie Fighter?
16.7h
20h
30h
33.3h
1.
a)
b)
c)
d)
The Star Wars spacecraft Tie Fighter flies at
a speed of 0.8c with respect to the Earth.
Luke Skywalker determines the time interval
between the two events on Earth is 20h.
What is the time interval for Tie Fighter?
16.7h
20h
30h
33.3h
In this question, we are
determining ∆𝑡𝑚 .
Using the time dilation
equation, substitute the
given information for:
 ∆𝑡𝑠 = 20ℎ
 𝑣 = 0.8𝑐
and solve for ∆𝑡𝑚 .
∆𝑡𝑚 =
∆𝑡𝑚 =
∆𝑡𝑠
𝑣2
1− 2
𝑐
20ℎ
(0.8𝑐)2
1−
𝑐2
∆𝑡𝑚 = 33.3ℎ
2.
a)
b)
c)
d)
It is the year 3000. A rocket passes by at a
speed of 0.3c. The length of the rocket is
100m. What is the length at rest?
108.4m
104.8m
108m
104m
2.
a)
b)
c)
d)
It is the year 3000. A rocket passes by at a
speed of 0.3c. The length of the rocket is
100m. What is the length at rest?
108.4m
104.8m
108m
104m
In this question, we are
determining 𝐿𝑠 .
Using the length
contraction equation,
rearrange to solve for 𝐿𝑠
instead of 𝐿𝑚 by
substituting the given
information for:
 𝐿𝑚 = 100𝑚
 𝑣 = 0.3𝑐
and solve for 𝐿𝑠 .
𝑣2
𝐿𝑚 = 𝐿𝑠 1 − 2
𝑐
𝐿𝑚
𝐿𝑠 =
𝑣2
1− 2
𝑐
100𝑚
𝐿𝑠 =
(0.3𝑐)2
1−
𝑐2
𝐿𝑠 = 104.8𝑚
At what speed will the length of a 1.5m
hockey stick look 2/3 shorter?
a) 0.75c
b) 2.25𝑥108 m/s
c) a and b
d) 3𝑥108 m/s
3.
At what speed will the length of a 1.5m
hockey stick look 2/3 shorter?
a) 0.75c
b) 2.25𝑥108 m/s
c) a and b
d) 3𝑥108 m/s
3.
In this question, we are
determining 𝑣.
Using the length
contraction equation,
rearrange to solve for
𝑣 instead of 𝐿𝑚 by
substituting the given
information for:
 𝐿𝑚 = 1𝑚
 𝐿𝑠 = 1.5𝑚
and solve for 𝑣.
𝐿𝑚 = 𝐿𝑠
𝑣2
1− 2
𝑐
𝐿𝑚 2
𝑣 =𝑐 1−
𝐿𝑠
1𝑚 2
𝑣 =𝑐 1−
1.5𝑚
𝑣 = 0.75𝑐 𝑜𝑟 2.25𝑥108 m/s
What is the relativistic momentum of a
proton travelling at 0.95c through a particle
accelerator?
𝑘𝑔∙𝑚
a) 5 2
4.
𝑠
−27 𝑘𝑔∙𝑚
b) 5𝑥10
𝑠2
−27 𝑘𝑔∙𝑚
c) 5.1𝑥10
𝑠2
d)
None of the above
What is the relativistic momentum of a
proton travelling at 0.95c through a particle
accelerator?
𝑘𝑔∙𝑚
a) 5 2
4.
𝑠
−27 𝑘𝑔∙𝑚
b) 5𝑥10
𝑠2
−27 𝑘𝑔∙𝑚
c) 5.1𝑥10
𝑠2
d)
None of the above
In this question, we are
determining 𝑝.
Using the relativistic
momentum equation,
substitute the given
information for:
 𝑚 = 1.67𝑥10−27 𝑘𝑔
 𝑣 = 0.95𝑐
and solve for 𝑝.
𝑝=
𝑝=
𝑚𝑣
𝑣2
1− 2
𝑐
(1.67𝑥10−27 𝑘𝑔)(0.95𝑐)
(0.95𝑐)2
1−
𝑐2
𝑘𝑔 ∙ 𝑚
−27
𝑝 = 5.1𝑥10
𝑠2

Superstring theory is
the favoured version of
string theory
 Fundamental entities in
the universe are
microscopic, multidimensional strings

Strings are made of 10
or more dimensions,
but only 3 are seen



Einstein’s general relativity
theory incorporated in Kaluza’s
geometrical representation of
electromagnetic field results in
a five-dimensional universe
Strings are assumed to obey
Einstein’s equations in spacetime resulting in a possibility to
combine quantum mechanics
with general relativity
Gravity remains as the only
force that can be associated
with space-time
Einstein’s theory of special
relativity revises
conceptions of time, length
and energy allowing us to
analyze the effects of
motion at high speeds
 Einstein’s equation of total
relativistic energy makes it
possible to predict the
amount of energy available
from processes resulting in
a decrease of mass

 e.g. Energy available from the
complete conversion of coal

For further reading please refer to the pages found on the HowStuffWorks
website titled:
 How Special Relativity Works (http://science.howstuffworks.com/science-vs-
myth/everyday-myths/relativity.htm)
 What is string theory? (http://science.howstuffworks.com/science-vsmyth/everyday-myths/string-theory.htm)





EinsteinFest - Perimeter Institute for Theoretical Physics. (n.d.). Perimeter
Institute for Theoretical Physics. Retrieved January 4, 2012, from
http://www.perimeterinstitute.ca/en/Outreach/Online_Viewers/EinsteinFest/
Giancoli, D. C. (2005). Physics principles with applications (6th ed.). Upper
Saddle River, N.J.: Pearson/Prentice Hall.
Hirsch, A. J. (2003). Nelson physics 12. Toronto: Nelson Thomson Learning.
McFarland, E. L. (1991). Special relativity (2 ed.). Guelph: Department of
Physics, University of Guelph.
Wolfe, J. (n.d.). Relativity: Einstein's theory of relativity in animations and film
clips. Einstein Light. School of Physics at UNSW, Sydney, Australia. Retrieved
January 4, 2012, from http://www.phys.unsw.edu.au/einsteinlight/