- #1

gtguhoij

- 33

- 2

- Homework Statement
- A train with length 15cs moves at speed 3c/5. (1cs is one “light-second.” It equals (1)(3 x 10^8 m/s)(1 s) = 3 10^8 m.) How much time does it take to pass a person standing on the ground, as measured by that person? Solve this by working in the frame of the person, and then again by working in the frame of the train.

- Relevant Equations
- ##gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}##

Because the book doesn't have an answer. Can someone confirm if I am correct?

The question can be found on page 49 of the link below.

I added arrows to the picture just to separate the equations not to say greater then.

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

first step solve for gamma

##v = \frac {3c} { 5 } = .6c##

## gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}##

## gamma = \frac 1 {\sqrt {1 - \frac {(.6c)^2 } {c^2}}} = ##

## \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}} ##

## \frac 1 {\sqrt {1 - \frac{.36c^2 } {c^2}}} = ##

## \frac 1 {\sqrt {1 - .36 }} = ##

## \frac 1 {\sqrt .64} = ##

## \frac 1 {.8} = ##

## 1.25##

Next calculate in my frame

## v = .6c ##

## d = 3 * 10 ^8 m/s ##

## t = ?##

I want time

##t = 1.67##

I plug the numbers into the Lorentz transform for time to get the moving frame

I get t' = 0.

So the answer are t' = 0 and t = 1.67.

The question can be found on page 49 of the link below.

I added arrows to the picture just to separate the equations not to say greater then.

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

first step solve for gamma

##v = \frac {3c} { 5 } = .6c##

## gamma = \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}}##

## gamma = \frac 1 {\sqrt {1 - \frac {(.6c)^2 } {c^2}}} = ##

## \frac 1 {\sqrt {1 - \frac {v^2} {c^2}}} ##

## \frac 1 {\sqrt {1 - \frac{.36c^2 } {c^2}}} = ##

## \frac 1 {\sqrt {1 - .36 }} = ##

## \frac 1 {\sqrt .64} = ##

## \frac 1 {.8} = ##

## 1.25##

Next calculate in my frame

## v = .6c ##

## d = 3 * 10 ^8 m/s ##

## t = ?##

I want time

##t = 1.67##

I plug the numbers into the Lorentz transform for time to get the moving frame

I get t' = 0.

So the answer are t' = 0 and t = 1.67.