Ch 26 HW Space-Time Diagrams and Relativity of Simultaneity « previous 6 of 26 n
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Ch 26 HW
Space-Time Diagrams and Relativity of Simultaneity
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Space-Time Diagrams and Relativity of Simultaneity
Learning Goal:
To introduce space-time diagrams to show that the speed of light postulate necessarily makes simultaneity a relative concept.
In this problem, we will introduce space-time diagrams, also called Minkowski diagrams, to provide a graphical way to think about this and related problems of simultaneity. In a space-time diagram, quantity c×time rather than time is plotted vertically and the x position horizontally.
An object (or a light pulse) is represented by a line, called a world line. On the space-time diagram; a world line is straight if the object is moving in a straight line in a region of space not seriously perturbed by nearby massive objects (e.g., black holes).
We will first consider the world line of light, or a particle of light, traveling at speed c. A particle of light is called a photon. If a photon starts at x0 and travels along the x-axis, then its position vs. time is x(t)=x0±ct. Therefore, the slope of a photon's world line is ±1. The orange lines represent the world line of light/photons (traveling in the positive x direction).
Part A
Imagine now that another reference frame is moving toward positive x at velocity v, which is a significant fraction of the speed of light. What is the world line of the origin of this coordinate system if this origin coincides with that of the primary reference frame at t=0?
Indicate the letter of the graph in the introductory figure and the color of the line (orange, green, black, or blue) that could be the world line of this particle, separated by a comma. For example, if you select the second graph and the color black you would enter b,black.
Hints
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Part B
Now find the world line of another point, also moving with velocity v in the positive x direction, that has position x0 at t=0, where x00.
Indicate the letter of the graph in the introductory figure and the color of the line (orange, green, black, or blue) that could be the world line of this particle, separated by a comma. For example, if you select the second graph and the color black you would enter b,black.
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Space-Time Diagrams and Relativity of Simultaneity
Learning Goal:
To introduce space-time diagrams to show that the speed of light postulate necessarily makes simultaneity a relative concept.
In this problem, we will introduce space-time diagrams, also called Minkowski diagrams, to provide a graphical way to think about this and related problems of simultaneity. In a space-time diagram, quantity c×time rather than time is plotted vertically and the x position horizontally.
An object (or a light pulse) is represented by a line, called a world line. On the space-time diagram; a world line is straight if the object is moving in a straight line in a region of space not seriously perturbed by nearby massive objects (e.g., black holes).
We will first consider the world line of light, or a particle of light, traveling at speed c. A particle of light is called a photon. If a photon starts at x0 and travels along the x-axis, then its position vs. time is x(t)=x0±ct. Therefore, the slope of a photon's world line is ±1. The orange lines represent the world line of light/photons (traveling in the positive x direction).
Part A
Imagine now that another reference frame is moving toward positive x at velocity v, which is a significant fraction of the speed of light. What is the world line of the origin of this coordinate system if this origin coincides with that of the primary reference frame at t=0?
Indicate the letter of the graph in the introductory figure and the color of the line (orange, green, black, or blue) that could be the world line of this particle, separated by a comma. For example, if you select the second graph and the color black you would enter b,black.
Hints
ct ct ct ct ct ctExplanation / Answer
SEE EDIT BELOW.
In (A) the location of the origin of the moving frame, such that xo =0 at to=0 is
x = vt
to use the Minkowski coordinates rewrite the right side as;
x = (v/c)ct
In the Minkowski diagram this is the eq. of a straight line thru the origin with positive slope "v/c". Since c>v this slope is less than "1" and therefore not as steep as the light line (orange).
So the answer is Graph C, green line.
B) The equation of this moving particle such that at to=0, xo not= 0 is;
x = xo + (v/c)ct
This has same slope as (A) , but does not go through the origin. The only possible answer is then Graph C, blue line. This graph would make xo negative at to = 0, but that's fine.
BTW, your setup discussion is a little fuzzy. The orange light line shown on the graphs goes thru the origin. So the equation would only be x = xo + ct , if xo was actually zero. So the eq. of the light line is, x = ct.
EDIT__________________________________...
I forgot that ct is on the vertical axis. That definitely changes things.
In (A) if you wrote this equation in typical fashion so the slope was measured from the horizontal it would be;
ct = (c/v)x
This is thru the origin with positive slope greater than 1, so the correct answer is Graph D, green line.
In (B) the eq. should be;
ct = - (c/v)xo + (c/v)x
Again, same slope as (A) but not thru the origin. This would have to be Graph D, blue line. And at t=0 we see from the graph that x=xo and is positive.