Confirm if I'm Correct? "Page 49 Q: Time in Moving Frame?

[gtguhoij]

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.

 

[kuruman]

Let's start with the cs unit. It is the distance by which light travels in one second. If light travels 3×10⁸ meters in one second, it will travel the length of the train in 15 seconds. How many meters is that?

To find how long it takes the train to pass the person on the ground, I suggest that you first figure out how long the train is from that person's point of view. Then divide that length by the speed of the train.

Do you really believe that in the train's frame of reference, a train that is 15cs long passes the person on the platform in zero time? If you assume that the front of the train passes the person at t' = 0, you must find the time t'' at which the end of the train passes the person. Then t'' is the time you are looking for.

BTW, I am curious why you use the symbol "d" and not "c" for the speed of light? Or did I miss what $d = 3 * 10 ^8 m/s$ stands for?

 

[gtguhoij]

Whoops careless mistake on the d = 3 x10^8 m/s it should be d = 3 x10^8 m

What is really confusing me is what is the initial position and time of both ends of the train and the initial position and time of the person standing still. Can you help with this question? Well I guess I can just assume t = 0 x = 0 for stationary person.

 

[kuruman]

OK, d = 3 x10^8 m is the distance that light travels in one second. How is that relevant to this problem?

What is really confusing me is what is the initial position of both ends of the train and the initial position of the person standing still. Can you help with this question?

OK, say the person at t = 0 is at x = 0 in the rest frame of the platform and that the front of the train at rest is also at x = 0. Where do you think the end of train is if the train is 15cs long? You are free to choose the origins at t = 0 and t' = 0 however you please, but the usual formulation of the Lorentz equations assumes that at t = 0 and t' = 0 the origins coincide, i.e. x = 0 and x' = 0.

 

[gtguhoij]

I drew a spacetime diagram I got t = 1 sec and t' = 1.25. Is this correct now? I also forgot the units in the graph for time. It is seconds

 

[kuruman]

It is not correct. How can a train that is 15cs long pass by the person on the platform in 1 s? To do that, it must travel faster than the speed of light. Not possible. You don't need spacetime diagrams to do this. I offered you suggestions which you ignored so far. I will try one more time and then I'll quit.

Imagine being a person on the platform. When you see the front of the train pass by you, you start a clock running. When you see the back of the train pass by you, you stop the clock. The train is moving at 0.6c. What does the stopwatch read? If you cannot answer this relativistically, consider first a non-relativistic warm up problem:

Non-relativistic problem
A freight train is 120 m long and travels at 10 m/s. How long does it take to go by a person standing near the tracks? The strategy for answering this is the same as the strategy for the relativistic case and should get you started.

 

[gtguhoij]

Here is my attempt. I first did your question The answer I first got was t = 12 seconds.
Then I did the problem above. I got t = 25 seconds. I am not sure if I need to use relativity of simultaniety for the time in the stationary person frame. Do I ? Then I calculated the time in the train frame. The answer I got was t' = 18.75 seconds. I think this reduces to t' = t gamma because x = 0. Did I make any mistakes? What would a spacetime diagram look like?

 

[kuruman]

The non relativistic answer is correct. The equivalent relativistic calculation in the stationary person frame is not. Do you understand length contraction? The 15cs is the length of the train as seen by an observer that moves with the train. Will the stationary person see the same length? If not (the answer is not) what length does the person on the platform see?

 

[gtguhoij]

I think so. I assumed the value 15LS is already the contracted length. I guess I was wrong. I assumed L' = 15 LS was L = 15LS. The definition is moving objects are length contracted.

 

[kuruman]

Unless otherwise stated, the length of an object in special relativity problems is the proper length. Thus, the 15cs is the length according to the train engineer. So what is the length of the train as measured by the person on the platform?

 

[gtguhoij]

Here is my attempt.

I have velocity = .6c and d' = 15LS.
t' = d'/v = 25 Seconds
x = gamma(x'+vt')
x = 1.25 (15LS + (.6c)(25 Seconds) ) =
x = 1.25 (15LS + 15LS) =
x = 1.25 (30LS) =

x = 37.5 LS
Is this correct? Or should x' = 0?
After thinking about I think x ' = 0 let me correct the answer.
x = gamma(x'+vt')
x = 1.25 (0 + (.6c)(25 Seconds) ) =
x = 1.25 (0 + 15LS) =
x = 1.25 (15LS) = 18.75 LS

 

[kuruman]

x' denotes the coordinate where an event occurs as seen in the primed (moving) frame. It could be zero depending on the event that occurs. For example, if the event is the front of the train being at the origin at time $t'=0$, then $x'_{\mathrm{front}}=0$. Since the back of the train in the moving frame is $L_0 = 15~\mathrm{cs}$ behind the front, then at $t'=0$ you have $x'_{\mathrm{back}}=-L_0.$

It seems you are unsure about the Lorentz contraction. Here is the bottom line. The observer moving with the train sees the length of the train to be 15 cs. The observer at rest on the platform sees the train as having a length

1. Greater that 15 cs
2. Equal to 15 cs
3. Less than 15 cs

Please pick an answer and justify it as best as you can.
Hint: https://en.wikipedia.org/wiki/Length_contraction#Basis_in_relativity

 

[kuruman]

Can I just confirm something that was already stated before I answer.
In the train frame x'_rear = 0 and x'_front = 15 when the train passes?

What was stated before was that at t' = 0, x'_front = 0 which would make x'_rear= -15cs. We assumed that the origin in the primed frame is at the front. Don't confuse x with x'. The primed origin is at the front of the train and moves with it. The unprimed origin is fixed at the person on the platform. In the primed frame the rear of the train is always at -15 cs.

 

[gtguhoij]

Here is my thought process.

How do I figure out the distance for an person on the ground frame? I use the fact that moving clocks move slow. So I try the equation x′=gamma(x+vt).
But I don't have time of the rear clock.
Then I try to calculate trear=dv. trear=.6c. I know this is wrong because t isn't ′t=0 . Also the same logic can be applied to x_rear.

So I decide to switch frames
I discover xrear=−15LS I have no idea why xrear=−15LS. I use the fact that moving clocks move slow. So I try the equation
t=gamma(t′−x′vc2)=. But since I don't have time I calculate time' using $t'_{rear} = \frac {d′] {v} =$

−15LS−.6c=2.5seconds

So now I go back to the
t=gamma(t′−x′vc2)=
1.25(2.5Seconds−(−15LS)(−.6cc2))=
1.25(2.5Seconds)−(+9seconds))=
1.25(2.5seconds−9seconds) = 1.25(-7.5 seconds) = -9.4. But -9.4 doesn't make sense. I am sorry the latex keeps disappearing.

In the prime frame person coordinates are unknown and x_front = 0

Just noticed I screwed up the - and + sign on the lorentz transform will edit later,

 

[kuruman]

Please stay on track and answer the multiple choice question I asked you in post #12. Then we will talk some more.

 

[gtguhoij]

greater then 15LS

Here is my reasoning. It should be -.6c. Also x'_rear = -15. I forgot the minus sign in my drawing. I am little confused why in the equation x'_rear needs to be positive.

 

[kuruman]

greater then 15LS

Incorrect. Please read about length contraction and time dilation then try answering again. It looks like you do not understand these two ideas which are important in special relativity. If you understood them, you would be able to answer the question without calculating anything.

It is not clear to me what is "it" that should be -.6c. That is a velocity in the negative x-direction.

Your calculation of x = 18.75 is half correct. It lacks units which should be cs and it should have a negative sign up front. That's because $x'_{\text{rear}}=-15\text{cs} = -L.$ Let's use symbols instead of numbers - we will substitute at the very end. There are two events that happen here:
Event 1 - the front of the train passes the stationary observer.
Event 2 - the rear of the train passes the stationary observer.
So far we know the following for Event 1:
$x_{\text{front,1}}=0~;~~t_{\text{front,1}}=0~;~~x'_{\text{rear,1}}=-L~;~~ t'_{\text{rear,1}} =0.$
You can use the Lorentz transformation (as you tried to do) to find the the position of the rear in the stationary frame. The correct calculation is

$x_{\text{rear,1}}=\gamma(x'_{\text{rear,1}}+vt'_{\text{rear,1}})=\gamma(-L+0)=-\gamma L$.

Since $\gamma > 1$, you might jump to the hasty conclusion (as you seem to have done) that this means that in the stationary framer the train is longer than its proper length L. That would be true if the rear of the train were at that distance at time $t_{\text{rear,1}}=0.$ But $t_{\text{rear,1}}\neq 0$ and you need to find it.

The problem is asking you to find $t_{\text{rear,2}}$ which is the time in the stationary frame at which the rear passes by the observer. I will give you a generous hint. In the stationary frame $\Delta x_{\text{rear}}=v\Delta t_{\text{rear}}~~\Rightarrow~x_{\text{rear,2}} -x_{\text{rear,1}}=v(t_{\text{rear,2} }-t_{\text{rear,1}})$. Figure out which quantities in the last equation you already know and which you can find out and then solve for $t_{\text{rear,2}}.$

 

[gtguhoij]

Also you mention 2 events isn't event 1 happen at t = 0 and t' = 0?

Are there other ways to solve the answer?

Instead of t1 and t2 I am going to use t0 and t1.
Why can't I use ${t'rear1} =$ $\frac {d'rear1} {v}$
Then once I have $t'rear1$ I use the Lorentz transform for time to get t_rear_1.
$trear0 = 0$ I think

Then take $deltatrear$ and get $trear1- trear0 =$ ?
Would that work and why not?

 

[kuruman]

Are there other ways to solve the answer?

There is another way but you you were not able to follow it, so I tried to start you along a path that uses the Lorentz transformation equations. This "other" way is a shortcut to the other path and requires you to (a) find the length of the train as observed by the stationary observer and (b) divide that length by the speed of the train. In post #16 you found that this length is 18.75cs. That is incorrect. If you want to take the shortcut, you have to find the correct length. To do that I suggested,

Please read about length contraction and time dilation then try answering again. It looks like you do not understand these two ideas which are important in special relativity. If you understood them, you would be able to answer the question without calculating anything.

Do it and tell me what the length of the train as measured by the stationary observer is. If you cannot do it, then you have to derive that length using the Lorentz transformations which is where I tried to guide you in the second half of post #17.

Instead of t1 and t2 I am going to use t0 and t1.

The subscripts refer to Event 1 and Event 2 as defined in post #17. Why do you have the urge to redefine them? OK by me, you may rename them if it's important to you but t0 and t1 are meaningless. In relativity the clock time is local in the sense that it is associated with a position. That is why you need another subscript to identify whether the clock time is associated with the rear or the front.

Why can't I use ${t'rear1} =$ $\frac {d'rear1} {v}$

Why do you think this expression is correct? You cannot guess what it is but you have to get it using the Lorentz transformation. Also, what does ${d'rear1}$ stand for? If you use a new symbol, you have to define it.

Then once I have $t'rear1$ I use the Lorentz transform for time to get t_rear_1.

Yes.

$trear0 = 0$ I think

No. You must use the Lorentz transformation to find what it is.

Then take $deltatrear$ and get $trear1- trear0 =$ ?
Would that work and why not?

Yes, $\Delta t_{\text{rear}}$ is by definition $t_{\text{rear,1}}-t_{\text{rear,0}}$ and what your are looking for.

 

[gtguhoij]

$d'rear1$ is just the second event for x_rear in the prime frame.

I will post doing it your way and my way and then ask a few question and speculate why I am wrong using the example in post 18 my way and post 17 your way.

Thanks for the help.

 

[kuruman]

$d'rear1$ is just the second event for x_rear in the prime frame.

So you are saying that $d'rear1$ is what would be more appropriately called $x'_{\text{rear,1}}$, namely the position, in the primed frame, of the rear of the train when the rear of the train passes by the stationary observer? If that's the case, then $x'_{\text{rear,1}}=-L$, no? In the primed frame, the rear of the train moves simultaneously with the front and the proper distance $L$ between them is maintained.

I will post doing it your way and my way and then ask a few question and speculate why I am wrong using the example in post 18 my way and post 17 your way.

Thanks for the help.

OK.

 

[gtguhoij]

Sorry for the delay. I just calculated t_rear_1. I didn't calculate t'_rear_1 I will do that after assuming I got the correct answer.Here is my thought process.

event 0

t = 0s

x = gamma(x' + vt')

x_rear = 1.25( -15LS + (-.6c)(0) )

x_rear = -18.75LS

x_front = 1.25(0 + -.6c(0))

x_front = 0

t = gamma (t' + x'v/c^2)

t_rear = 1.25(0 +(-15LS)(-.6c/c^2) =

t_rear = 11.25 seconds

t_front = 1.25( 0 +(0)(-.6c/c^2)

t_front = 0

why is t_rear not t = 0 both are stationary?

event 1t = gamma (t' + x'v/c^2)t_front =1.25( 9 sec + (0)( .6c /c^2) = 11.25 sect_rear = 1.25( 9 sec + (-15LS)( .6c /c^2) = 0

x = gamma(x' + vt')

x_front = 1.25(0 + (9sec)(.6c) ) = 6.75LS

x_rear = 1.25(-15 LS + (9sec)(.6c) ) = 12 LS

The definition of a time interval is 2 events at the same place at different times.I think I use the person on the ground because it is located at the front of the box for event 0 and event 1 it is located at the rear of the box. The box also fits the definition of time interval . So I go time interval = t1 - t0 and I get 12 seconds.

 

[kuruman]

why is t_rear not t = 0 both are stationary?

Because if it were zero, it would contradict the axioms of special relativity. The thinking that says that it is zero is only an approximation that works only when at velocities much much smaller than the speed of light,

t = gamma (t' + x'v/c^2)
t_rear = 1.25(0 +(-15LS)(-.6c/c^2)

The substitution is incorrect. v = 0.6 c, not -0.6 c. Time is a coordinate. A negative sign in either frame means a time earlier than when the clocks are synchronized.

Up to this point I can follow what you have done and it is correct except for the fudged sign that pointed ot above. After this point I get confused because a whole lot of stuff is jumbled together in the text and in the figure you posted. I think you are also confused about how to go about answering the question. Look at post #19, the last sentence at the bottom. You don't care about the front of the train, where it is or what it does after it passes the stationary observer.. The events are:
Event 0 = the front of the train is at the observer
All times and positions associated with event 0 must involve the front of the train. That you did.
Event 1 = the rear of the train is at the observer
All times and positions associated with event 1 must involve the rear of the train. That you must do.

Remember your goal. The stationary observer's clock starts at event 0. The time when the train passes him is event 1. You are looking for:
Passing time for stationary observer = $t_{\text{rear,1}}-t_{\text{rear,0}}.$
Passing time for train rider = $t'_{\text{rear,1}}-t'_{\text{rear,0}}.$

You already know the event 0 times. Focus your efforts on calculating the event 1 primed and unprimed times and positions of the rear.

 

[pai535]

I'd like to add:

I think that it is easier to figure out what is "proper" - time or length (of "something") can be called "proper" when it is measured by an observer who is at rest w.r.t. (that "something"). Proper time is shortest time interval, while proper length is longest length. Given that $\gamma > 1$, it is easy to know what need to multiply $\gamma$ or what need to divide $\gamma$.

Also, to measure length in this question, what you need to do is to measure the time of two events: the front of train passes observer and the rear of train passes observer. Then multiply the time interval by velocity. I don't recommend measuring the x-coordinates since I found it confusing.

 

[gtguhoij]

Thanks for the help and the extreme patience.

Let us start with the spaceship frame.

Event 0 $x ~ rear = -15$ and $t = 0$.

Event 1 $x ~ rear = -15$ and $t = -25 sec$

So there are 2 ways to solve this. $t ~ rear = \frac { ~delta ~ x ~ rear} {v}$ or just $delta ~ t ~ rear$

To get $delta ~ t_rear$ I go $t ~ rear ~ 1 - t ~ rear ~ 0$. To get $delta ~ x ~ rear$ I go $x ~ rear ~ 1 - x ~ rear ~ 0$.

I currently have delta $t~ rear$ in the spaceship frame. I need $delta t'~ rear$ for the person on the ground frame. I can get this by using the Lorentz transform. Then I repeat the process above except in the non prime frame.

I think the exact numbers are $delta ~ t~ rear = -25 sec$ This is because v = .6c and x'~ rear = -``15 in the equation.

Can I use a delta Lorentz transform or only event 1 at a time?
$delta~t~rear~$
Assuming I can't I will just do it the long way
$t' = gamma (t - \frac {x v} {c^2})$
$t'~ rear ~ 0 = 1.25 (0 - \frac {-15LS .6c } {c^2}) = 1.25 (9) = +11.25 sec$
$t_rear_1' = 1.25 ( -25 sec - \frac {-15LS .6c}{c^2} ) = 1.25 (-16 secs) = -20 sec$
$delta t = t1 - t0$
$delta t = (-20 sec) - (+11.25 sec) = -31.25$
Also I could use the invariant interval.

Are there any other ways to solve for time?

In the picture I use event 1 and event 2 but in the equation above I use event 0 and event 1.

Also you mentioned I assume if given a length or time in a question it is always proper time?