Where can I find a good textbook for special relativity?

[Taylor_1989]

Homework Statement

Hi all I am having trouble working part b of a question that I am currently doing. I have attached the question below in a pdf file. I am really not sure where to start, I have looked in several book and can only think of relating to the light clock thought experiment. But the problem is I can't see where the $$(ct)^2$$ comes from.

Homework Equations

$$t'=ϒ(t-(ux/c^2))$$

$$x'=ϒ(x-ut)$$

The Attempt at a Solution

I did attempt to set $$t'=0$$ but this got me no where.

 

[stevendaryl]

Well, using your two equations for $x'$ and $t'$, see if you can figure out what the combination

$(x')^2 - c^2 (t')^2$

is, in terms of $x$ and $t$. Then look at the special case where $t'=0$.

 

[Taylor_1989]

Before I attempt what you have told me. can I ask how u got that equation

 

[PeroK]

Homework Statement

Hi all I am having trouble working part b of a question that I am currently doing. I have attached the question below in a pdf file. I am really not sure where to start, I have looked in several book and can only think of relating to the light clock thought experiment. But the problem is I can't see where the $$(ct)^2$$ comes from.

You need to give some thought to question (a) (i). This does not require any equations or calculations.

 

[Taylor_1989]

You need to give some thought to question (a) (i). This does not require any equations or calculations.

It is because it is the rest frame, so it will be the proper length of the rod.

 

[PeroK]

It is because it is the rest frame, so it will be the proper length of the rod.

Pretend you've never heard the words "rest frame" or "proper length".

 

[PeroK]

@Taylor_1989 In fact, let me ask you a practical question. Forget relativity for a moment.

1) There's a train standing at a station platform and you are asked to measure its length. How do you do it? Hint: this one is easy.

2) A fast train comes through the station without stopping. How do you measure its length?

Remember: this is a practical, non-relativistic question.

 

[Taylor_1989]

If at rest I would use a tape measure, whilest moving I would use a stop watch

 

[PeroK]

If at rest I would use a tape measure, whilest moving I would use a stop watch

That's not an answer! You have to be prepared to think if you want to learn physics.

 

[Taylor_1989]

I am really not sure. How else would I measure a train standing still? I really don't have a clue. Also I am trying to think but I am struggling quite bad area of physics , so no need for that comment.

 

[PeroK]

I am really not sure. How else would I measure a train standing still? I really don't have a clue. Also I am trying to think but I am struggling quite bad at this pasifc area , so no need for that comment.

Imagine I was going to carry out your measurements for you. You said: "use a tape measure". So, I get a tape measure. Now what? You've said nothing about what to do with the tape measure. You've given me no procedure or even any hint. You haven't even distinguished between measuring the height of the train and the length of the train.

That said, here is a better idea. I would mark the platform at regular intervals (let's say every metre is accurate enough). Then, I'd measure where the two ends of the train are and subtract one reading from the other. For example, if the front of the train is at position $106m$ and the back of the train is at position $3m$, then I'd have an approximate measurement of $103m$ for the length of the train.

Now, what about a moving train? Let's assume the plaform is still marked every metre. And you have a stopwatch. Give me a procedure for measuring (approx) the length of the train as it speeds through the station. Hint: this is not so easy to do on your own. You may need a friend with another stopwatch.. Or, as many friends as you can muster!

 

[Taylor_1989]

I keep thinking that if I start my stop watch at the same time as my friend then when the train passes him then we both stop our stop watch. But that make no sense, it would not give me the length of the train. Erm, I feel like I should have each friend standing at each marker then they start and stop there stop watch when the front and back of the train passes them. But even then it make nonsense to me I still can't figure how to measure the length of the train.

 

[PeroK]

I keep thinking that if I start my stop watch at the same time as my friend then when the train passes him then we both stop our stop watch. But that make no sense, it would not give me the length of the train. Erm, I feel like I should have each friend standing at each marker then they start and stop there stop watch when the front and back of the train passes them. But even then it make nonsense to me I still can't figure how to measure the length of the train.

You're on the right track with one person at each metre interval. Each stops their watch when the front of the train passes them. If you stand at the end of the platform (where the train comes in), when do you stop your watch?

It goes without saying that you all synchronise your watches in advance!

 

[Taylor_1989]

There is something fundamental I am missing here. My lecture note don't really explain it, the book I have describes a light clock and and video on you tube I have been watching uses a graph with ct being on the y axis.

 

[Taylor_1989]

I would stop my stop watch as soon as it come in right, then I would compare the difference in time intervals with my freinds, correct?

 

[Taylor_1989]

Am I allso on the right lines look at the
Minkowski diagram in special relativity

 

[PeroK]

There is something fundamental I am missing here. My lecture note don't really explain it, the book I have describes a light clock and and video on you tube I have been watching uses a graph with ct being on the y axis.

The light clock is something different. The question you posted was set for a good reason: to get you to analyse the notion of length for a moving object. Let me help you a bit more with the train measurement.

You stop your watch when the rear of the train passes. Let's say you are standing at point $0m$ at the end of the platform. This establishes the time that the rear of the train was at this point. Let's say your stopwatch reads $57s$. Then you go to each of your friends and find which one also stopped at $57s$. This establishes where the front of the train was at $57s$.

So, you have simultaneous measurements of the position of both ends of the train. And, for a moving object,that is as good a definition of its length as you are likely to get!

For the stationary train you did not need simultaneous measurements. You could measure the position of the rear of the train, then stroll up the platform to the front of the train, and measure its position at your leisure.

If you read Einstein's 1905 paper, that is the sort of analysis he undertook because, as he put it, "Insufficient consideration of this circumstance lies at the root of the difficulties which the electrodynamics of moving bodies at present encounters.".

 

[PeroK]

Am I allso on the right lines look at the
Minkowski diagram in special relativity

Our discussion so far was about the meaning of the length of a moving body. We have not yet progressed to SR and spacetime diagrams. This discussion was fundamental, because if you haven't thought carefully about what "length" actually means, then it is going to be impossible to understand length contraction.

You should be in a position to answer (a) (i) now. Then, you might try (a) (ii).

 

[Taylor_1989]

Okay I am with you there. So why is t'=0 when really t'1=t'2 or any time for that fact. is t'=0 the difference in time if so is that not delta t. I am trying to get use to the notation as well. t is the none moving frame and t' is the moving frame, correct? So in ur example t' would be me and my friends standing on the platform?

 

[Taylor_1989]

Yes I understand the question u posted me and thank you for your patients.

I would like to add that my lecture never went through any of this neither any graph stuff.

 

[PeroK]

Okay I am with you there. So why is t'=0 when really t'1=t'2 or any time for that fact. is t'=0 the difference in time if so is that not delta t. I am trying to get use to the notation as well. t is the none moving frame and t' is the moving frame, correct? So in ur example t' would be me and my friends standing on the platform?

In our example, we were dealing with only one reference frame: the platform. We were not comparing measurements with those of a second frame.

It is usual to have the second frame $S'$ moving to the right relative to the unprimed frame $S$. Neither frame, however, is moving and neither frame is stationary. Measurements in the two frames are related by the Lorentz Transformation.

For question (a) (ii) you will have to think carefully about how to get a measurement of length $L'$ for the rod in the frame ($S'$) in which it is moving. You will have to think carefully about how to apply the Lorentz Transformation.

 

[Taylor_1989]

from the length transformation I was given in the work sheet is this correct.
$$x'=y(x-ut)$$
$$L'=x_2'-x_1'$$
$$L'=y(x_2-ut-x_1+ut)$$
$$L'=yL$$
where $$L=x_2-x_1$$

Yis my gamma

 

[PeroK]

from the length transformation I was given in the work sheet is this correct.
$$x'=y(x-ut)$$
$$L'=x_2'-x_1'$$
$$L'=y(x_2-ut-x_1+ut)$$
$$L'=yL$$
where $$L=x_2-x_1$$

Yis my gamma

That means that $L' = \gamma L$ and $L' > L$. That would be length dilation, not contraction. So, unfortunately, that is not right.

What you have done is transformed simultaneous measurements in $S$ into non-simultaneous measurements in $S'$. But, as we discussed, you need simultaneous measurements in $S'$.

How I would do it is to think about each end of the rod and its path through spacetime in the $S'$ frame. The ends are at at $(0, t)$ and $(L, t)$ in $S$. Can you transform these spacetime paths into the $S'$ frame?

 

[Taylor_1989]

okay so I had another go is this correct?

$$(0,t) & (L,t)$$
$$x=y(x'+ut)$$

$$x_2-x_1=y(x_2'-x_1'+ut-ut)$$

$$L-0=y(L')$$

 

[PeroK]

okay so I had another go is this correct?

$$(0,t) & (L,t)$$
$$x=y(x'+ut)$$

$$x_2-x_1=y(x_2'-x_1'+ut-ut)$$

$$L-0=y(L')$$

That's almost doing it using the inverse Lorentz Transformation, which is another approach. Can you explain those steps?

I'll be offline now.

 

[PeroK]

@Taylor_1989 You basically have two approaches.

Calculate where both ends of the rod are simultaneously in the $S'$ frame. You know that one end is at $x_1' = 0$ at $t' = 0$, so you try to find out where the other end is at $t' = 0$. It's easiest to use $t' = 0$, but it's not a bad exercise to check that you get the same result for $L'$ for any time $t'$.

Or, you can assume you have found $x_2'$ at $t' = 0$ and transform that back to the $S$ frame. You know that must transform to $x_2= L$, so this gives you a solvable equation for $x_2'$.

 

[Taylor_1989]

@Taylor_1989 You basically have two approaches.

Calculate where both ends of the rod are simultaneously in the $S'$ frame. You know that one end is at $x_1' = 0$ at $t' = 0$, so you try to find out where the other end is at $t' = 0$. It's easiest to use $t' = 0$, but it's not a bad exercise to check that you get the same result for $L'$ for any time $t'$.

Or, you can assume you have found $x_2'$ at $t' = 0$ and transform that back to the $S$ frame. You know that must transform to $x_2= L$, so this gives you a solvable equation for $x_2'$.

I have not forgot about this question, I just have had a lot of assingments in that past week. I am going to retry the question. I am slightly confused that how can t'=t is there is not such this as absolute time. I have been watching videos and reading up a lot, so I might have got my info muddled

 

[Taylor_1989]

@PeroK, Ok I have another attempt at the question, but for part a)i) I am having trouble in putting it into words. Here What I have written so far. The reason is that in frame s the rod is stationary, so if I were to measure the rod at t=t_1 x=L_1 with a tape measure and the take the tape measure to the other end of the rod which would be t_2 and x=L_2 this would give me the length of the rod as its not moving in my ref thus dose not matter what time I start and fin measuring the rod it will give the actual length. Now in regards to L' this frame is moving relativite to me, so what I percive as proper length will not me the same in the L' frame and vice versa if I was in the L' frame and I measure the rod this would be my stationary frame and L would be moving w/r to me in L' therefore as there is no simultaneity between the ref frame t in the must me t'=0 so that the the proper length can be obtained in both ref frames.

 

[PeroK]

@PeroK, Ok I have another attempt at the question, but for part a)i) I am having trouble in putting it into words. Here What I have written so far. The reason is that in frame s the rod is stationary, so if I were to measure the rod at t=t_1 x=L_1 with a tape measure and the take the tape measure to the other end of the rod which would be t_2 and x=L_2 this would give me the length of the rod as its not moving in my ref thus dose not matter what time I start and fin measuring the rod it will give the actual length. Now in regards to L' this frame is moving relativite to me, so what I percive as proper length will not me the same in the L' frame and vice versa if I was in the L' frame and I measure the rod this would be my stationary frame and L would be moving w/r to me in L' therefore as there is no simultaneity between the ref frame t in the must me t'=0 so that the the proper length can be obtained in both ref frames.

I wish you could see this without the clutter of SR terminology! As I tried to explain before, the difficulty of measuring the length of a moving object is not related to SR, as such, but is a consequence of the measuring process. Note that:

(Classical) If something is moving you need to establish where the two ends are at the same time (in order to determine its length). This is all you need to say. That's true, as I tried to explain, for objects at non-relativistic speeds, like a normal train passing through a station.

But:

(SR) Simultaneity is relative. And this leads to different measurements of length in different reference frames. (Even once you have the measuring process sorted out.)

The biggest error you will get if you measure the ends at different times has nothing to do with SR, but simply with the fact that the object has moved. Consider, for example, a train:

You observe the rear of the train to be in London at 12 noon. Your friend observes the front of the train in Edinburgh, 400 miles away, at 5pm. If you allow this as a measurement of the length of the train, then you have an answer of 400 miles for its length.

 

[Taylor_1989]

Sorry, I have difficulty with put what I am thinking into words. Let me try again. So if I measure and rod in the rest frame it dose not matter what times I measure it at, as it is at rest. So in the S' frame I would have to measure the rod at t'1=t'2 so I don't measure the motion of the object instead?

 

[Taylor_1989]

@PeroK so now for part b using that I can use the Lorentz transformations like so:

$$L=x_2-x_1$$ $$L'=x'_2-x_1'$$ $$t'_1=t'_2$$

$$x_2-x_1=y((x'_2-x'_1)-(ut'_2-ut'_1))$$

$$L=yL'$$

This is because

$$t'_2-t'_1=0$$

 

[PeroK]

@PeroK so now for part b using that I can use the Lorentz transformations like so:

$$L=x_2-x_1$$ $$L'=x'_2-x_1'$$ $$t'_1=t'_2$$

$$x_2-x_1=y((x'_2-x'_1)-(ut'_2-ut'_1))$$

$$L=yL'$$

This is because

$$t'_2-t'_1=0$$

Those are the correct algebraic steps in the right order. So, perhaps that's good enough! Let me give you the explanation (and tidy up the notation a bit):

Suppose the two ends of the rod in $S'$ are simultaneously at $x'_1$ and $x'_2$ at some time $t'$.

The length of the rod in $S'$ is therefore given by $L' = x'_2 - x'_1$.

These two events transform to positions $x_1$ and $x_2$ in $S$ given by:

$x_1 = \gamma (x'_1 + Vt'), \ x_2 = \gamma (x'_2 + Vt')$

(Note the sign of $V$ in the Lorentz Transformation, as we are going from $S'$ to $S$.)

Note: as the rod is stationary in $S$ the times of those two events are not important for a length measurement in $S$, which is:

$L = x_2 - x_1 = .\gamma (x'_2 - x'_1) = \gamma L'$

Do you understand the logic of this?

 

[Taylor_1989]

Those are the correct algebraic steps in the right order. So, perhaps that's good enough! Let me give you the explanation (and tidy up the notation a bit):

Suppose the two ends of the rod in $S'$ are simultaneously at $x'_1$ and $x'_2$ at some time $t'$.

The length of the rod in $S'$ is therefore given by $L' = x'_2 - x'_1$.

These two events transform to positions $x_1$ and $x_2$ in $S$ given by:

$x_1 = \gamma (x'_1 + Vt'), \ x_2 = \gamma (x'_2 + Vt')$

(Note the sign of $V$ in the Lorentz Transformation, as we are going from $S'$ to $S$.)

Note: as the rod is stationary in $S$ the times of those two events are not important for a length measurement in $S$, which is:

$L = x_2 - x_1 = .\gamma (x'_2 - x'_1) = \gamma L'$

Do you understand the logic of this?

Yes I do and many thanks. Could you recommend a textbook? I am a first year physics undergrad. I taught myself A-level math, and was not allowed to A-level physics as a private candidate. Which I am regretting now. As I feel if I actually I missed out on a lot of theory and now trying to play catch. Would it be worth me getting a a-level book that contains special relativity or just find a first year?

 

[PeroK]

Yes I do and many thanks. Could you recommend a textbook? I am a first year physics undergrad. I taught myself A-level math, and was not allowed to A-level physics as a private candidate. Which I am regretting now. As I feel if I actually I missed out on a lot of theory and now trying to play catch. Would it be worth me getting a a-level book that contains special relativity or just find a first year?

I like T M Helliwell's book on SR, and it's available at a reasonable price in the UK.