# How Do Lorentz Transformations Affect Clock Readings in a Moving Rocket?

• Colin Martin
In summary, the problem involves finding the reading of the clock in the nose of a rocket after receiving a return signal from the ground. The key equations used are the Lorentz transformations and the time dilation formula. By transforming the event of the ground signal being sent to the rocket's frame, the time and location in the rocket frame can be determined. Then, using the time taken for the signal to reach the nose of the rocket, the reading of the clock in the nose can be calculated.
Colin Martin

## Homework Statement

Hello I need some advice on how to figure out Part F of the following problem. I was able to find the correct answer but in a very illogical way. I was wondering if I could get some help understanding the lorentz transformations necessary. Here is the full problem, I have parts a-e solved and put the answer for f.
[/B]
A rocket that has a proper length of 746 m is moving to the right at a speed of 0.882c. It has two clocks—one in the nose and one in the tail—that have been synchronized in the frame of the rocket. A clock on the ground and the clock in the nose of the rocket both read zero as they pass by each other.

(a) At the instant the clock on the ground reads zero, what does the clock in the tail of the rocket read according to observers on the ground? A: 2.19 µs

(b) When the clock in the tail of the rocket passes the clock on the ground, what does the clock in the tail read according to observers on the ground? A: 2.82 µs

(c) When the clock in the tail of the rocket passes the clock on the ground, what does the clock in the nose read according to observers on the ground? A: .63 µs

(d) When the clock in the tail of the rocket passes the clock on the ground, what does the clock in the nose read according to observers on the rocket? A: 2.82 µs

(e) At the instant the clock in the nose of the rocket reads 1.00h, a light signal is sent from the nose of the rocket to an observer standing by the clock on the ground. What does the clock on the ground read when the observer on the ground receives the signal? A: 3.99 h

(f) When the observer on the ground receives the signal, he immediately sends a return signal to the nose of the rocket. What is the reading of the clock in the nose of the rocket when that signal is received at the nose of the rocket? A: 18.3 h

## Homework Equations

x = ɣ(x' + vt')
t = ɣ(t' + (vx')/c^2)
ɣ = 1/sqrt(1-(v^2/c^2))

## The Attempt at a Solution

I have the Earth as farm S and the ship as frame S'.[/B]
I know that the signal is received at Earth at t = 3.99h from part e. I then know due to the speed that the ship is a distance of 3.52 c * h away so light would take another 3.52h to reach that point. I am unable to figure out what to transform or how to calculate the last section where the light catches the ship. Any help would be very appreciated.

But during those 3.52 h, the ship moves further away from the Earth and so the signal does not take 3.52 h to reach the ship in the Earth frame.

I suggest you transform the event of the Earth signal being sent to the rocket's frame. This will make the problem much easier to solve.

In the rocket frame of reference, what is the time t' and location x' when, in the Earth frame of reference, x = 0 and t = 3.99 hr? How long does it take for a signal to travel from this x' location to x' = 0 (the nose of the rocket)?

Chestermiller said:
In the rocket frame of reference, what is the time t' and location x' when, in the Earth frame of reference, x = 0 and t = 3.99 hr? How long does it take for a signal to travel from this x' location to x' = 0 (the nose of the rocket)?
t' = gamma (t - vx/c^2)
t' = 2.12 (3.99 - 0)
t' = 8.46 hrs

x' = gamma (x - vt)
x' = 2.12 (0 - .882c * 3.99)
x' = -7.46 c*hr

So time to travel from x' = -7.46ch to x' = 0 at a signal speed of c = 7.46 hrs.
So now I have the time it takes the signal to reach the nose of the rocket in the frame of the rocket which converted to the frame of the Earth is:
t = gamma (t' + vx'/c^2)
I'm slightly confused on which values to use at this point.
t = 2.12 (7.46 + (c)(-7.46ch)/c^2)
Which is getting me 0.
Do I use t' = 8.46 found earlier or am I setting this up incorrectly.

Colin Martin said:
t' = gamma (t - vx/c^2)
t' = 2.12 (3.99 - 0)
t' = 8.46 hrs

x' = gamma (x - vt)
x' = 2.12 (0 - .882c * 3.99)
x' = -7.46 c*hr

So time to travel from x' = -7.46ch to x' = 0 at a signal speed of c = 7.46 hrs.
So now I have the time it takes the signal to reach the nose of the rocket in the frame of the rocket which converted to the frame of the Earth is:
t = gamma (t' + vx'/c^2)
I'm slightly confused on which values to use at this point.
t = 2.12 (7.46 + (c)(-7.46ch)/c^2)
Which is getting me 0.
Do I use t' = 8.46 found earlier or am I setting this up incorrectly.
Once you have determined the location and time of the signal in the rocket frame of reference, you can forget about the Earth frame of reference and focus entirely on the rocket frame. As you said, in the rocket frame, the signal was sent at t'=8.46 hr, and it then takes Δt'=7.46 hr for the signal to reach the nose of the rocket. So, in the rocket frame of reference, at what time does the signal reach the nose?

Chestermiller said:
Once you have determined the location and time of the signal in the rocket frame of reference, you can forget about the Earth frame of reference and focus entirely on the rocket frame. As you said, in the rocket frame, the signal was sent at t'=8.46 hr, and it then takes Δt'=7.46 hr for the signal to reach the nose of the rocket. So, in the rocket frame of reference, at what time does the signal reach the nose?

So if it's sent at t' = 8.46 and takes 7.46 hours to reach the nose of the rocket. In the rocket frame it will reach the rocket at t'=15.92 hrs. Which isn't giving me the correct answer of 18.3 hrs which is why I felt the need to do some form of transformation. Is there something I should do with the 15.92 beyond that answer?

I suggest you keep all of your digits when doing computations, or at least round the answer to a number you can trust given the number of digits you used throughout your computation.

By symmetry, the ratio between your answer in (e) and 1 h should be the same as the ratio between your answer in (f) and that in (e). It is therefore impossible that both given answers are correct.

I get 15.9 hr also. I am unable to duplicate the 18.3

Chet

The answers 3.99 for (e) and 15.92 for (f) satisfy Orodruin's criterion in post #7. In particular, 3.992=15.92. So they must be right.

Chet

## What are Lorentz Transformations?

Lorentz Transformations are a set of equations used in the theory of special relativity to describe how measurements of space and time change for an observer moving at a constant velocity relative to another observer.

## Why are Lorentz Transformations important?

Lorentz Transformations are important because they allow us to understand the effects of relative motion on the measurements of space and time. They are a fundamental component of the theory of special relativity, which has been proven accurate in countless experiments and has greatly advanced our understanding of the universe.

## How do Lorentz Transformations affect time and space measurements?

Lorentz Transformations show that time and space are not absolute, but are relative to the observer's frame of reference. This means that measurements of time and space will appear different to observers moving at different velocities.

## What is the equation for Lorentz Transformations?

The equation for a Lorentz Transformation is t' = γ(t - vx/c^2), where t' is the time measured by the moving observer, t is the time measured by the stationary observer, v is the relative velocity between the two observers, c is the speed of light, and γ is the Lorentz factor, which is equal to 1/√(1 - v^2/c^2).

## Can Lorentz Transformations be applied to objects with mass?

Yes, Lorentz Transformations can be applied to objects with mass. In fact, it was through the application of Lorentz Transformations that Albert Einstein developed his famous equation E=mc^2, which shows the relationship between mass and energy in special relativity.

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