Twin Paradox - Revision history

Chaz at 14:25, 5 December 2024

2024-12-05T14:25:18Z

← Older revision		Revision as of 10:25, 5 December 2024
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	~~===~~Chaz Sporrer Fall 2024~~===~~		<strong>Chaz Sporrer Fall 2024</strong>

	==The Main Idea==		==The Main Idea==

Chaz at 14:24, 5 December 2024

2024-12-05T14:24:47Z

← Older revision		Revision as of 10:24, 5 December 2024
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	Chaz Sporrer Fall 2024		===Chaz Sporrer Fall 2024===

	==The Main Idea==		==The Main Idea==

Chaz at 14:03, 5 December 2024

2024-12-05T14:03:04Z

← Older revision		Revision as of 10:03, 5 December 2024
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	Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.		Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.

	Here is a video showing a twin traveling a distance D in spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0		Here is a video showing a twin traveling a distance D in a spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0

	Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0		Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0

Chaz at 13:57, 5 December 2024

2024-12-05T13:57:59Z

← Older revision		Revision as of 09:57, 5 December 2024
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	Here is a video showing a twin traveling a distance D in spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0		Here is a video showing a twin traveling a distance D in spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0

	Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0		Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0

Chaz at 13:57, 5 December 2024

2024-12-05T13:57:29Z

← Older revision		Revision as of 09:57, 5 December 2024
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	~~===~~Chaz Sporrer Fall 2024~~===~~		Chaz Sporrer Fall 2024
	~~Short Description of Topic~~

	==The Main Idea==		==The Main Idea==
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	The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math>		The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math>
	Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.		Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.

			Here is a video showing a twin traveling a distance D in spaceship from the perspective of the other twin on Earth: https://vimeo.com/1030116480?share=copy#t=0
			Here is a similar video showing the perspective of the twin in the spaceship (note the contraction of the distance the twin travels): https://vimeo.com/1030116491?share=copy#t=0

Chaz at 00:44, 3 December 2024

2024-12-03T00:44:30Z

← Older revision		Revision as of 20:44, 2 December 2024
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	==References==		==References==
	Krane, Kenneth S. Modern Physics. 4th ed., Wiley, 2019.		Krane, Kenneth S. <em>Modern Physics</em>. 4th ed., Wiley, 2019.

	[[Category:Which Category did you place this in?]]		[[Category:Which Category did you place this in?]]

Chaz at 00:42, 3 December 2024

2024-12-03T00:42:57Z

← Older revision		Revision as of 20:42, 2 December 2024
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	==References==		==References==
			Krane, Kenneth S. Modern Physics. 4th ed., Wiley, 2019.
	~~This section contains the the references you used while writing this page~~

	[[Category:Which Category did you place this in?]]		[[Category:Which Category did you place this in?]]

Chaz at 23:20, 2 December 2024

2024-12-02T23:20:56Z

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	Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?		Question 1: If there are two twins, and twin 1 travels a distance <math>d</math> away from twin 2 at a velocity <math>v_1</math> and then travels back to twin 2 at a velocity <math>v_2</math>, which twin will be older?

	Solution: ~~because~~ twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.		Solution: Because twin 1 changes inertial reference frames, he is actually moving relative to twin 2, and thus his clock will run slower. This means twin 2 will be older.

	===Middling===		===Middling===
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	[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]		[[File:Question_2_Twin_1_Reference_Frame.png]] [[File:Question_2_Twin_2_Reference_Frame.png]]

	Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years – 9.17\,years) = 0.83\,years</math>.		Solution: First, the distance of a light-year is equal to <math>(1\,year) \times c</math>. So time <math>t</math> to travel a round trip of <math>4</math> light-years = <math>4 \times c \times (1\,year)/0.4c = 10\,years</math>. Using the time dilation formula and rearranging for <math>t_0</math> (time according to twin 2), we get <math>t_0 = t \sqrt{1 - \frac{v^2}{c^2}} = (10\,years) \sqrt{1 - \frac{(0.4c)^2}{c^2}} = 9.17\,years</math>. This means we can subtract <math>t_0</math> from <math>t</math> to find how much older twin 1 is than twin 2. <math>(10\,years – 9.17\,years) = 0.83\,years</math>.

	Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.		Extra: We can further confirm our answer using the relativistic doppler effect. We know it takes twin 2 <math>(9.17 years/2) = 4.59\,years</math> to travel from Earth to planet B. If twin 1 were to send a signal to twin 2 each year, twin 1 would send a total of <math>10</math> signals to twin 2. The frequency that twin 2 will receive the signals depends on his velocity relative to twin 1. This frequency can be modeled by the relativistic doppler effect where <math>f_0</math> is the frequency observed by twin 2 and <math>f</math> is the frequency observed by twin 1.
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	Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 – u/c}{1 + u/c}}</math>		Relativistic Doppler Effect: <math>f_0 = f \sqrt{\frac{1 – u/c}{1 + u/c}}</math>

	While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 – 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on trip to planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.		While traveling to planet B, <math>f_0 = (1/year) \sqrt{\frac{1 – 0.4c/c}{1 + 0.4c/c}} = 0.655/year</math>. On the way back (towards Earth), <math>f_0 = (1/year) \sqrt{\frac{1 + 0.4c/c}{1 - 0.4c/c}} = 1.53/year</math>. So on the trip to the planet, he receives <math>(0.655/year)(4.59\,years) = 3.01</math> signals. On the way back, he receives <math>(1.53/year)(4.59\,years) = 7.02</math> signals. This means that over the entire journey, twin 2 receives <math>7.02 + 3.01 \approx 10</math> signals.

	===Difficult===		===Difficult===
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	The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.		The twin paradox gives a perspective on the relativity of time. Time does not just appear slower, but actually does pass slower for an object in motion. When you think about it, time dilation creates a form of time travel. If you could quickly accelerate on a spaceship to close to light speeds and then return to Earth, everyone on Earth would have aged far more than you. This would let you experience the distant future, but would be a risky endeavor as you could not go back in time.

	~~#How is this topic connected to something that you are interested in?~~
	~~#How is it connected to your major?~~
	~~#Is there an interesting industrial application?~~

	==History==		==History==
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	~~Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.~~

	== See also ==		== See also ==

	~~Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?~~

	===Further reading===		===Further reading===

	~~Books, Articles or other print media on this topic~~

	===External links===		===External links===

	~~Internet resources on this topic~~

	==References==		==References==

Chaz at 23:12, 2 December 2024

2024-12-02T23:12:51Z

← Older revision		Revision as of 19:12, 2 December 2024
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	==Examples==		==Examples==

	~~Be sure to show all steps in your solution and include diagrams whenever possible~~

	===Simple===		===Simple===

Chaz at 23:11, 2 December 2024

2024-12-02T23:11:57Z

← Older revision		Revision as of 19:11, 2 December 2024
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	To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.		To more easily see how this works, consider again the situation where one twin travels a distance <math>D</math> and back at a velocity <math>v</math>. The time <math>t</math> seen by the twin on Earth equals <math>\frac{2D}{v}</math>, where <math>2D</math> equals the length of the round trip. However, the twin on the spaceship will see the distance contract as a result of length contraction.
	The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math>		The formula for length contraction: <math> \Delta L = \Delta L_0\sqrt{1 - \frac{v^2}{c^2}} </math>
	Where <math>\Delta L_0</math> is the length measured by in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.		Where <math>\Delta L_0</math> is the length measured in the inertial frame, and <math>\Delta L</math> is the length measured by the observer moving at velocity <math>v</math> relative to the inertial frame. Thus, time <math>t_0</math> according to the twin on the spaceship is <math>\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}</math>. Thus, you can say <math>\frac{t}{t_0} = \frac{\frac{2D}{v}}{\frac{2D \sqrt{1 - \frac{v^2}{c^2}}}{v}}</math>. Rearranging, you get <math>t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}</math>, which is the equation for time dilation.