A simple Proper Time Calculation

[unified]

Homework Statement

Suppose a spaceship starts from rest from a space station floating in deep space and accelerates at a rate of |a| relative to the space station for 1.0 Ms. It then decelerates for the same amount of time at the same constant rate |a| to arrive at rest at another space station and then repeats the trip back to its original space station with the same accelerations. How much less time has passed on the spaceship than on the space station?

Forgot to mention, $$|a| = 10\frac{m}{s^2}$$

Hints: We are supposed to calculate this by breaking the trip into 4 pieces and applying binomial approximation.

Homework Equations

$$\Delta \tau = \int_{t_A}^{t_B}([1 - v^2])^{(1/2)}$$

The Attempt at a Solution

I applied binomial theorem and for the first integral I get the following after integrating: $$\Delta \tau_1 = (t - \frac{1}{6}|a|^2t^3) |0Ms, 1Ms$$

The symmetry implies that $$\Delta \tau_1 = \Delta \tau_2 = \Delta \tau_3 = \Delta \tau_4$$ You can check the integrals to make sure this is true. I did, and got the same for each piece. So, $$\Delta \tau = 4 * \Delta \tau_1$$

Now, if you do this on your calculator you should get $$\Delta t - \Delta \tau = 740s$$

The book gets 84s. Either I'm making a mistake, or there is an error in the answers. I usually email the author if I can't see where I could be making a mistake, but I want to clarify with a second opinion before I question his answers.

 

[collinsmark]

Homework Statement

Suppose a spaceship starts from rest from a space station floating in deep space and accelerates at a rate of |a| relative to the space station for 1.0 Ms. It then decelerates for the same amount of time at the same constant rate |a| to arrive at rest at another space station and then repeats the trip back to its original space station with the same accelerations. How much less time has passed on the spaceship than on the space station?

Forgot to mention, $$|a| = 10\frac{m}{s^2}$$

Hints: We are supposed to calculate this by breaking the trip into 4 pieces and applying binomial approximation.

Homework Equations

$$\Delta \tau = \int_{t_A}^{t_B}([1 - v^2])^{(1/2)}$$

Can you elaborate on this equation? There seems to be something wrong with it. At the very least it needs some differential to integrate over (maybe $dt$ ?).

Also, this section would be a good place to list what the "binomial approximation" is. Doing a google search shows a wiki article, but that article is flagged with issues. Thus I think it's best to show us how it is defined in your textbook/coursework.

The Attempt at a Solution

I applied binomial theorem and for the first integral I get the following after integrating: $$\Delta \tau_1 = (t - \frac{1}{6}|a|^2t^3) |0Ms, 1Ms$$

Can you show us how you came up with that evaluation? It doesn't seem to be dimensionally correct*. How is the length dimension (meters) in $| a |$ getting canceled out?

*[Edit: Or is that only after you convert |a| = 10 m/s^2 to units of ls/s^2 ? Even so, please show us how you came up with that evaluation of the integral.]

The symmetry implies that $$\Delta \tau_1 = \Delta \tau_2 = \Delta \tau_3 = \Delta \tau_4$$ You can check the integrals to make sure this is true. I did, and got the same for each piece. So, $$\Delta \tau = 4 * \Delta \tau_1$$

Now, if you do this on your calculator you should get $$\Delta t - \Delta \tau = 740s$$

The book gets 84s. Either I'm making a mistake, or there is an error in the answers. I usually email the author if I can't see where I could be making a mistake, but I want to clarify with a second opinion before I question his answers.

 

[collinsmark]

Now, if you do this on your calculator you should get $$\Delta t - \Delta \tau = 740s$$

The book gets 84s. Either I'm making a mistake, or there is an error in the answers. I usually email the author if I can't see where I could be making a mistake, but I want to clarify with a second opinion before I question his answers.

All that said in my previous post,

for what it's worth, I worked out the problem using different methods and I too came up with an answer of ~740 seconds. So maybe there is an error in the book.

Edit:
Oh, wait! The way I solved the problem (as mentioned above), I assumed $|a| = 10 \frac{m}{s^2}$ is the proper acceleration. But the question states that the acceleration is with respect to the space station. That might change things a little. [But then again, without working out the math, I speculate that that would only make the differential greater than 740 sec, not less. So maybe there is a mistake in the book?]

 

[unified]

All that said in my previous post,

for what it's worth, I worked out the problem using different methods and I too came up with an answer of ~740 seconds. So maybe there is an error in the book.

Edit:
Oh, wait! The way I solved the problem (as mentioned above), I assumed $|a| = 10 \frac{m}{s^2}$ is the proper acceleration. But the question states that the acceleration is with respect to the space station. That might change things a little. [But then again, without working out the math, I speculate that that would only make the differential greater than 740 sec, not less. So maybe there is a mistake in the book?]

Thanks for taking a look at it. The dimensions are correct. I'm using relativistic units, so |a| has units of 1/s. I'll email the professor and see if he is in agreement with our answers.

 

[unified]

Homework Statement

Suppose a spaceship starts from rest from a space station floating in deep space and accelerates at a rate of |a| relative to the space station for 1.0 Ms. It then decelerates for the same amount of time at the same constant rate |a| to arrive at rest at another space station and then repeats the trip back to its original space station with the same accelerations. How much less time has passed on the spaceship than on the space station?

Forgot to mention, $$|a| = 10\frac{m}{s^2}$$

Hints: We are supposed to calculate this by breaking the trip into 4 pieces and applying binomial approximation.

Homework Equations

$$\Delta \tau = \int_{t_A}^{t_B}([1 - v^2])^{(1/2)}dt$$

The Attempt at a Solution

I applied binomial theorem and for the first integral I get the following after integrating: $$\Delta \tau_1 = (t - \frac{1}{6}|a|^2t^3) |0Ms, 1Ms$$

The symmetry implies that $$\Delta \tau_1 = \Delta \tau_2 = \Delta \tau_3 = \Delta \tau_4$$ You can check the integrals to make sure this is true. I did, and got the same for each piece. So, $$\Delta \tau = 4 * \Delta \tau_1$$

Now, if you do this on your calculator you should get $$\Delta t - \Delta \tau = 740s$$

The book gets 84s. Either I'm making a mistake, or there is an error in the answers. I usually email the author if I can't see where I could be making a mistake, but I want to clarify with a second opinion before I question his answers.

Integral updated to include a dt term.