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In Example 2.5, we noted that Anna could go wherever she wished in as little time as desired by going fast enough to length-contract the distance to an arbitrarily small value. This overlooks a physiological limitation. Accelerations greater than about 30g are fatal, and there are serious concerns about the effects of prolonged accelerations greater than 1g. Here we see how far a person could go under a constant acceleration of 1g, producing a comfortable artificial gravity.

(a) Though traveller Anna accelerates, Bob, being on near-inertial Earth, is a reliable observer and will see less time go by on Anna's clock (dt') than on his own (dt). Thus,, whereuis Anna's instantaneous speed relative to Bob. Using the result of Exercise 117(c), withgreplacingF/m, substitute for u, then integrate to show that

(b) How much time goes by for observers on Earth as they “see” Anna age 20 years?

(c) Using the result of Exercise 119, show that when Anna has aged a timet’, she is a distance from Earth (according to Earth observers) of

(d) If Anna accelerates away from Earth while aging 20 years and then slows to a stop while aging another 20. How far away from Earth will she end up and how much time will have passed on Earth?

Short Answer

Expert verified

The expression for the time in Bob’s frame in terms of Anna’s frame is determined by integrating the equation for.430 million yearswould go by as observers on earth as they see Anna age 20 years. And during Anna’s journey, the observers on Earth would have agedand Anna would have covered a distance of .

Step by step solution

01

Determine the expression for  and then integrate

Anna is accelerating at with respect to Bob who is stationary on earth. The time on Anna’s clock as observed by Bob with respect to his clock will be given by

dt'=1-u2/c2dt

The relativistic velocity is given by the following.

role="math" localid="1657611855240" u=11+Ftmc2Ftm=gt1+gtc2

Putting the expression for in,

role="math" localid="1657612015181" width="198" height="174">dt'=1-(gt)2c21+(gt/c)2dt=1-(gt/c)21+(gt/c)2dt=11+(gt/c)2dt

Integrating the above expression,

As the clocks of Anna and Bob were synchronized at, the integration constant. Therefore,

or

02

Use the above expression to determine the time for which the observers on earth would have aged

Suppose Anna was traveling at a constant velocity comparable to the speed of light, the time required for Anna to cover some distance with respect to Bob’s frame is as follows.

In this situation, Anna is accelerating at under constant force with respect to stationary Bob. The expression for time passed on Earth’s in terms of time passed on Anna’s frame as derived in the previous part is as follows.

Hence, the time in years would be

.

So, 430 million years would go by as observers on earth as they see Anna age 20 years. For Anna’s clock to run so slowly, the velocity must have been very close to the speed of light. As you might recall from Exercise 118, it takes 6.8 years with respect to an observer on Earth for Anna to reach the velocity of 0.99c from stationary.

03

Derive the expression for the relativistic position of Anna

In Exercise 119, we derived an expression for the relativistic position of an object under constant force accelerating at is given below.

Replacing with and using the expression of in terms of which we derived in the previous part, i.e., , we get,

Using the identity of Hyperbolic functions, that is, , yields as

Step 4: Determine the position of Anna when she has aged 20 yrs

Let’s consider the first half of the question and determine the distance traveled by Anna as she ages 20 years as measured by Bob.

Hence, the distance in light-years is

.

This value is the same as the value in part(b) because Bob sees Anna traveling at almost the speed of light. From Anna’s frame, it takes Anna 20 years to stop. Bob will see Anna move the same distance as before. Therefore, Bob will see Anna travel a total distance of. As a result, we can say that during Anna’s journey the observers on Earth would have aged.

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Most popular questions from this chapter

Equation (2-30) is an approximation correct only if the gravitational time-dilation effect is small. In this exercise, it is also assumed to be small. but we still allow for a nonuniform gravitational field. We start with (2-29), based on the Doppler effect in the accelerating frame. Consider two elevations, the lower at r1 and the upper at r1+dr. Equation (2·29) becomes

f(r1+dr)f(r1)=(1-gr1drc2)

Similarly, if we consider elevationsdata-custom-editor="chemistry" r1+dr and data-custom-editor="chemistry" r1+2dr, we have

f(r1+2dr)f(r1+dr)=(1-gr1+drdrc2)

We continue the process, incrementing r by dr, until we reach r2.

f(r2)f(r2-dr)=(1-gr2-drdrc2)

Now imagine multiplying the left sides of all the equations and setting the product equal to the product of all the right sides. (a) Argue that the left side of the product is simply f(r2)/f(r1). (b) Assuming that the termgdr/c2 in each individual equation is very small. so that productsof such termscan be ignored, argue that the right side of the product is

1-1c2g(r)dr

(c) Deduceg(r) from Newton’s universal law of gravitation, then argue that equation (2-31) follows from the result, just as (2-30) does from (2-29).

How fast must be a plane 50 m long travel to be found by observer on the ground to be 0.10 nm shorter than 50 m?

From the Lorentz transformation equations, show that if time intervals between two events,t andt' , in two frames are of opposite sign, then the events are too far apart in either frame for light to travel from one to the other. Argue that therefore they cannot be casually related.

Refer to Figure 2.18. (a) How long is a spaceship? (b) At what speed do the ships move relative to one another? (c) Show that Anna’s times are in accord with the Lorentz transformation equations. (d) Sketch a set of diagrams showing Anna’s complementary view of the passing of the ships. Include times in both frames.

Question: Here we verify the conditions under which in equation (2-33) will be negative. (a) Show that is equivalent to the following:

vc>u0c21+u02/c2

(b) By construction v, cannot exceed u0, for if it did, the information could not catch up with Amy at event 2. Use this to argue that if u0<c, then must be positive for whatever value is allowed to have(x-1)20. (c) Using the fact that , show that the right side of the expression in part (a) never exceeds . This confirms that when u0>cv need not exceed to produce a negativet'3 .

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