Time Dilation, calculate the age of the returning astronaut

Yes, the last sentence is another example of the author having a misunderstanding of relativity. In order to properly understand relativity, it is important to state the reference frame explicitly. For example, you could say, "What do you think happens to the mass of an object if it is viewed from a reference frame that is moving at a constant velocity relative to the object's own reference frame?" This makes it clear that there are two reference frames involved and that the motion is relative. It is also important to specify the velocity of the reference frame, as the effect of relativity is dependent on the relative velocity between frames.
  • #1
apolloxiii

Homework Statement


"In 2010, a 20-year-old astronaut leaves her twin on Earth and goes on a rocket to explore the galaxy. The rocket moves at 2.7 x 10^8 m/s during the voyage. It returns to Earth in the year 2040. Using relativity, calculate the age of the returning astronaut."

Homework Equations


Δtm = Δts / √(1-((v^2)/(c^2)))

The Attempt at a Solution


It is a simple calculation, but I am having trouble knowing where to plug in the 30 years of time passed on earth. Subscript "s" is supposed to represent time passed with the observer at rest with respect to an event, and subscript "m" represents time passed with the observer in motion with respect to the event.
 
Physics news on Phys.org
  • #2
What do you think and why?

apolloxiii said:
time passed with the observer at rest with respect to an event

Note that this statement makes no sense. An event is a single instance in time at a single point in space.
 
  • #3
Orodruin said:
What do you think and why?

I think that the 30 years is inserted into the equation in the place of Δts, because 30 years has passed from the viewpoint of a stationary observer on earth. It seems to me that solving for Δtm would give you the change in time in the frame of reference of somebody inside the rocket, because that is the frame of reference that is in motion.

Orodruin said:
Note that this statement makes no sense. An event is a single instance in time at a single point in space.

What I meant to say, and as it says in my textbook, is that the subscript "s" means that the observer is at rest with respect to the event, and that "m" means that the observer is in motion with respect to the event.

 
  • #4
apolloxiii said:
What I meant to say, and as it says in my textbook, is that the subscript "s" means that the observer is at rest with respect to the event, and that "m" means that the observer is in motion with respect to the event.
Again, there is no such thing as being at rest "with respect to an event". An event in itself does not specify a state of motion.

Also note that a reference frame cannot simply "be in motion", motion is always relative to something.
 
  • #5
Orodruin said:
Again, there is no such thing as being at rest "with respect to an event". An event in itself does not specify a state of motion.

Also note that a reference frame cannot simply "be in motion", motion is always relative to something.

Thanks for the reply. So far studying physics I have not come across the term "event" in a technical sense, so I was not aware of that. I think that the book is using the word event to describe the rocket's trip. An observer on the Earth is stationary relative to the event of the rocket's voyage. I thought I should be placing the 30 elapsed years in the place of Δts, because 30 years on Earth have passed, and observers on Earth are stationary relative to the rocket.
 
  • #6
apolloxiii said:
I think that the book is using the word event to describe the rocket's trip.
If it does that you should immediately throw it away and go to get a better book.

apolloxiii said:
An observer on the Earth is stationary relative to the event of the rocket's voyage.
Again, this statement has no meaning. You can only be stationary in a reference frame or relative to another object. It is simply meaningless to call something stationary relative to an event so when you repeat this we have to guess what your intended meaning is and there is no way of really knowing.
apolloxiii said:
observers on Earth are stationary relative to the rocket.
This is not true. One thing we can definitely tell is that the Earth and rocket move relative to each other - their relative speed is given to be non zero in the problem.
 
  • #7
Orodruin said:
This is not true. One thing we can definitely tell is that the Earth and rocket move relative to each other - their relative speed is given to be non zero in the problem.

Ok, so what we are given in the problem is the rocket's speed...relative to the earth? The way I see it, the only information you need to do this problem is one object's speed relative to another, which we have been given. How do I know which variable to solve for? It makes sense to me that I should be solving for the variable of time which is associated with the object that is moving with respect to the other, Δtm, but this book is telling me that Δtm is 30 years, and I need to solve for the change in time experienced by the "stationary" observer.
 
  • #8
apolloxiii said:
the object that is moving with respect to the other
This represents a fundamental misunderstanding of relativity. Both objects are moving relative to the other.
I think it is very bad notation to call an object stationary and the other moving without explicit reference to a given inertial frame, but this might be on the author of your book.
 
  • #9
Orodruin said:
This represents a fundamental misunderstanding of relativity. Both objects are moving relative to the other.
I think it is very bad notation to call an object stationary and the other moving without explicit reference to a given inertial frame, but this might be on the author of your book.

How would you phrase it, giving explicit reference to a given inertial frame?
 
  • #10
Also, here is a direct quote from the book.
"Up to this point, you have learned that there is no such thing as absolute time or length. Just
by changing from one inertial reference frame to another, the amount of time and length
measured can change. For both of these concepts, you were able to derive equations that
govern how time and length vary from one at-rest reference frame to another. What do you
think happens to the mass of an object if it is viewed from a moving reference frame?"

Is that last sentence ("moving reference frame") another example of the author having a misunderstanding of relativity? Based on what you've said it seems that there is no such thing as a "moving reference frame", just inertial and accelerated.
 

What is time dilation?

Time dilation is a phenomenon in which time appears to pass differently for two observers, based on their relative velocities or gravitational fields.

How does time dilation affect an astronaut traveling at high speeds?

According to Einstein's theory of relativity, time dilation occurs for objects moving at high speeds. This means that for an astronaut traveling at high speeds, time will appear to pass slower for them compared to someone observing them from Earth.

What is the formula for calculating time dilation?

The formula for calculating time dilation is t = t0/√(1-v2/c2), where t is the time measured by the observer, t0 is the time measured by the moving object, v is the velocity of the object, and c is the speed of light.

How can time dilation be used to calculate the age of a returning astronaut?

If an astronaut travels at high speeds for a period of time and then returns to Earth, their time will have passed at a slower rate than the time on Earth. By using the formula for time dilation, the age of the astronaut can be calculated by comparing the difference in time between the two frames of reference.

What implications does time dilation have for space travel and exploration?

Time dilation plays a crucial role in space travel and exploration, as it allows for longer missions to be completed within a shorter amount of time. It also highlights the importance of precise timekeeping in space missions, as even small differences in time can have significant effects on the accuracy of calculations and navigation.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
3K
  • Special and General Relativity
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Special and General Relativity
2
Replies
46
Views
3K
  • Special and General Relativity
Replies
21
Views
508
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
917
Back
Top